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<channel>
	<title>Lysario - by Panagiotis Chatzichrisafis</title>
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	<link>https://lysario.de</link>
	<description>&#34;ούτω γάρ ειδέναι το σύνθετον υπολαμβάνομεν, όταν ειδώμεν εκ τίνων και πόσων εστίν ...&#34;</description>
	<lastBuildDate>Thu, 15 Apr 2021 17:28:24 +0000</lastBuildDate>
	<language>de-DE</language>
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			<item>
		<title>Export data from Scilab for PSTricks plots</title>
		<link>https://lysario.de/scilab/export-data-from-scilab-for-pstricks-plots-in-latex</link>
		<comments>https://lysario.de/scilab/export-data-from-scilab-for-pstricks-plots-in-latex#comments</comments>
		<pubDate>Sun, 25 Jan 2015 13:36:28 +0000</pubDate>
		<dc:creator>Panagiotis</dc:creator>
				<category><![CDATA[Programming]]></category>
		<category><![CDATA[Scilab]]></category>

		<guid isPermaLink="false">https://lysario.de/?p=1080</guid>
		<description><![CDATA[It is possible to load data from a file for plots with PSTricks in TeX. According to the PSTricks manual all that is needed is a file with a list of coordinate pairs delimited by curly braces , parentheses, commas. One easy solution is to use the write command from scilab. Define the following function: [...]]]></description>
				<content:encoded><![CDATA[It is possible to load data from a file for plots with PSTricks in TeX. According to the <a href="https://mirror.ctan.org/graphics/pstricks/base/doc/pstricks-doc.pdf" target="_blank">PSTricks manual </a>all that is needed is a file with a list of coordinate pairs delimited by curly braces , parentheses, commas.<span id="more-1080"></span>
<br />
One easy solution is to use the write command from scilab. Define the following function:
<pre><code>
function pstrickswrite2d(filename,data_matrix)
//Author: Panagiotis Chatzichrisafis
//Date: 25.01.2015
//Description: 
//The function pstrickswrite2d(filename,data_matrix) has two inputs: 
//<filename>: The path and the filename which shall be used to write 
//            formated output for PSTricks 
//<data_matrix>: The matrix containing the data to be written into the file provided by
//               <filename> argument.
//The output written to the <filename> file has a format which is suitable for 
//the TeX PSTricks package fileplot or dataplot commands.  
    _num_decimal_places_=5; 
    file_desc=file('open',filename,'unknown','formatted');
    _max_num_ = max(abs(data_matrix));
    _num_integer_digits_ = ceil(log10(_max_num_)); 
    _num_format_(1) ='F';
    // add number of integer digits and decimal digits plus 2 for sign information
    _num_format_(2) = string(_num_integer_digits_+_num_decimal_places_+2); 
    _num_format_(3) ='.'; 
    _num_format_(4) = string(_num_decimal_places_);
    _num_format_ =strcat(_num_format_);
    _format_(1)   ='( ''{'' '   ;
    _format_(2)   = _num_format_; 
    _format_(3)   = ', '','' '  ;
    _format_(4)   = _num_format_; 
    _format_(5)   = ' ''}'' )'  ;
    _format_=strcat(_format_);
    write(file_desc,data_matrix,_format_); 
    file('close',file_desc);
endfunction
</blockquote>
</code></pre> 

and load it into the scilab workspace.

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		</item>
		<item>
		<title>Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 94 exercise 4.2</title>
		<link>https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2</link>
		<comments>https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2#comments</comments>
		<pubDate>Tue, 24 Jun 2014 18:54:07 +0000</pubDate>
		<dc:creator>Panagiotis</dc:creator>
				<category><![CDATA[Kay: Modern Spectral Estimation, Theory and Application]]></category>
		<category><![CDATA[Solved Problems]]></category>

		<guid isPermaLink="false">https://lysario.de/?p=1062</guid>
		<description><![CDATA[In [1, p. 94 exercise 4.2] we are asked to consider the estimator (1) where (2) for the process of Problem 4.1. We are informed that this estimator may be viewed as an averaged periodogram. In this point of view the data record is sectioned into blocks (in this case, of length 1) and the [...]]]></description>
				<content:encoded><![CDATA[In <a href="https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2#lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2bib1">[1, p. 94 exercise 4.2]</a> we are asked to consider the estimator 
<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_baf985c94e5d3d36eec3ca5399353886.png" align="bottom" class="tex" alt="\hat{P}_{AVPER}(0)=\frac{1}{N}\sum\limits_{m=0}^{N-1}\hat{P}_{PER}^{m}(0)" /></td><td><a name="lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2eq1"> (1)</a></td></tr>
</table><br/> where 
<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_111f972a11d36418820c12baa2503f8a.png" align="bottom" class="tex" alt="\hat{P}_{PER}^{m}(0)= x^{2}[m]" /></td><td><a name="lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2eq2"> (2)</a></td></tr>
</table><br/> for the process of Problem 4.1. We are informed that this estimator may be viewed as an averaged periodogram. In this point of view the data record is sectioned into blocks (in this case, of length 1) and the periodograms for each block are averaged. We are asked to find the mean and variance of <img src="https://lysario.de/wp-content/cache/tex_c5e2f02b864c4893bd8e1b3fe1ea1e7e.png" align="bottom" class="tex" alt=" \hat{P}_{AVPER}(0) " /> and compare the result to that obtained in <a href="https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2#lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2bib2">[2]</a>.  
<span id="more-1062"></span>
<br /> <strong>Solution:</strong>
We note that according to the boundaries of problem 4.1 ( for which the process is white Gaussian with zero mean) we obtain the fact that the sum <img src="https://lysario.de/wp-content/cache/tex_17170bea01e2e71414bbc3c1720d276b.png" align="bottom" class="tex" alt=" U = \sum_{m=0}^{N-1}\frac{x^{2}[n]}{\sigma_{x}^{2}} " /> is distributed according to a <img src="https://lysario.de/wp-content/cache/tex_854d28882806eb796d2bead7e5ff06a1.png" align="bottom" class="tex" alt="\chi^{2}(N)" /> distribution <a href="https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2#lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2bib3">[3, p. 682]</a> with mean <img src="https://lysario.de/wp-content/cache/tex_95e55ec89cb98743dc5ece7304a057b4.png" align="bottom" class="tex" alt="E\left\{U\right\}=N" /> and variance <img src="https://lysario.de/wp-content/cache/tex_9174bd2c274f7814f499aebc7b2494d9.png" align="bottom" class="tex" alt=" Var\left\{U\right\}=2N" />. 
The average periodogram is obtained by <img src="https://lysario.de/wp-content/cache/tex_419a83a4fe3c7d2e5a0a8cae5dc75c5d.png" align="bottom" class="tex" alt="\hat{P}_{PER}(0) =\frac{1}{N} U \cdot \sigma_{x}^{2}" /> and thus the mean and the average is obtained by


<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_09d98e20833e8834e8847b11a79be1d6.png" align="bottom" class="tex" alt="E\left\{\hat{P}_{PER}(0)\right\} = \sigma_{x}^{2} E\left\{U\right\} = \sigma_{x}^{2}" /></td><td></td></tr>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_f3e3f832281ecddc8cf22c3105a4f683.png" align="bottom" class="tex" alt="Var\left\{\hat{P}_{PER}(0)\right\}  = \sigma_{x}^{2} Var\left\{U\right\}  = 2 \sigma_{x}^{2}." /></td><td></td></tr>
</table><br/>

Thus taking the average periodogram of one sample section has no statistical advantage for the result of the estimator. 

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		</item>
		<item>
		<title>Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 94 exercise 4.1</title>
		<link>https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-1</link>
		<comments>https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-1#comments</comments>
		<pubDate>Sat, 01 Feb 2014 16:40:19 +0000</pubDate>
		<dc:creator>Panagiotis</dc:creator>
				<category><![CDATA[Kay: Modern Spectral Estimation, Theory and Application]]></category>
		<category><![CDATA[Solved Problems]]></category>

		<guid isPermaLink="false">https://lysario.de/?p=1050</guid>
		<description><![CDATA[In problem [1, p. 94 exercise 4.1] we will show that the periodogram is an inconsistent estimator by examining the estimator at , or (1) If is real white Gaussian noise process with PSD (2) we are asked to find the mean an variance of . We are asked if the variance converge to zero [...]]]></description>
				<content:encoded><![CDATA[In problem <a href="https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-1#lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-1bib1">[1, p. 94 exercise 4.1]</a> we will show that the periodogram is an inconsistent estimator by examining the estimator at <img src="https://lysario.de/wp-content/cache/tex_9030c9fdd1390d3ed9f1a55b1611aa74.png" align="bottom" class="tex" alt="f=0" />, or 

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_938765f09480f77020c3aad219444ca0.png" align="bottom" class="tex" alt="\hat{P}_{PER}(0)=\frac{1}{N}\left(\sum\limits_{n=0}^{N-1}x[n]\right)^{2}." /></td><td><a name="lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-1eq1"> (1)</a></td></tr>
</table><br/>
If <img src="https://lysario.de/wp-content/cache/tex_d3baaa3204e2a03ef9528a7d631a4806.png" align="bottom" class="tex" alt="x[n]" /> is real white Gaussian noise process with PSD 
<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_e9bddf9fead41d9cae3623239ebbf84f.png" align="bottom" class="tex" alt="P_{xx}(f)=\sigma_{x}^{2}" /></td><td><a name="lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-1eq2"> (2)</a></td></tr>
</table><br/> we are asked to find the mean an variance of <img src="https://lysario.de/wp-content/cache/tex_63930453a899931c436c344a06617115.png" align="bottom" class="tex" alt="hat{P}_{PER}(0)" />. We are asked if the variance converge to zero as <img src="https://lysario.de/wp-content/cache/tex_524ce044c402a61b400bc14273b8cdda.png" align="bottom" class="tex" alt="N \rightarrow \infty" />. The hint provided within the exercise is to note that 
<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_df83a5d6a6944d47bfd3d103cb81d6c4.png" align="bottom" class="tex" alt="\hat{P}_{PER}(0)= \sigma_{x}^{2} \left(\sum\limits_{n=0}^{N-1}\frac{x[n]}{\sigma_{x}\sqrt{N}}\right)^{2}" /></td><td><a name="lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-1eq3"> (3)</a></td></tr>
</table><br/> where the quantiiy inside the parenthesis is <img src="https://lysario.de/wp-content/cache/tex_694c07565dcea1b4e3f4cee2097e2dac.png" align="bottom" class="tex" alt=" \sim N(0,1)" />
<span id="more-1050"></span> 
<strong>Solution:</strong>
Because <img src="https://lysario.de/wp-content/cache/tex_3dd5e12134e10cee884cdc9f326c4244.png" align="bottom" class="tex" alt="Y = \sum_{n=0}^{N-1}\frac{x[n]}{\sigma_{x}\sqrt{N}} " /> is distributed according to a normal distribution <img src="https://lysario.de/wp-content/cache/tex_a164405f332c9acc6e0f494b39ae24dd.png" align="bottom" class="tex" alt="\sim N(0,1)" /> the squared variable <img src="https://lysario.de/wp-content/cache/tex_ef052608e86be972ab329503345a5f31.png" align="bottom" class="tex" alt="Y^{2}=\frac{\hat{P}_{PER}(0)}{\sigma_{x}^{2}}" /> is distributed according to a <img src="https://lysario.de/wp-content/cache/tex_d0731fa4038a5b554d12c42d72b3bb5e.png" align="bottom" class="tex" alt="\chi^{2}(1)" /> with mean <img src="https://lysario.de/wp-content/cache/tex_66ae9e0797c52dd08b5c192a4816e4cc.png" align="bottom" class="tex" alt="E\left\{Y\right\}=1" /> and variance <img src="https://lysario.de/wp-content/cache/tex_17f2390c947cfdaaefd8c45986e6ea83.png" align="bottom" class="tex" alt="Var\left\{Y\right\}= 2" />. From the previous relations we obtain the mean and the variance of the periodogram <img src="https://lysario.de/wp-content/cache/tex_eee635b7562e359753032f7b0f4bdf08.png" align="bottom" class="tex" alt="\hat{P}_{PER}(0)" /> as 

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_7163e5d588ce560d1abc71d9b5253e50.png" align="bottom" class="tex" alt="E\left\{\hat{P}_{PER}(0)\right\} = \sigma_{x}^{2} E\left\{Y\right\} = \sigma_{x}^{2}" /></td><td></td></tr>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_99b7abaf23de25bf2e21d1c6a262e846.png" align="bottom" class="tex" alt="Var\left\{\hat{P}_{PER}(0)\right\}  = \sigma_{x}^{2} Var\left\{Y\right\}  = 2 \sigma_{x}^{2}." /></td><td></td></tr>
</table><br/>
We see that while the mean converges to the true power spectral density, the variance does not converge to zero. Thus the estimator is inconsistent. QED.

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		<item>
		<title>Installing OpenCV 2.4.7 on MacOS Snow Leopard and get it work with macam drivers</title>
		<link>https://lysario.de/programming/intalling-and-working-with-opencv-on-macos-snow-leopard</link>
		<comments>https://lysario.de/programming/intalling-and-working-with-opencv-on-macos-snow-leopard#comments</comments>
		<pubDate>Fri, 31 Jan 2014 18:57:32 +0000</pubDate>
		<dc:creator>Panagiotis</dc:creator>
				<category><![CDATA[Programming]]></category>

		<guid isPermaLink="false">https://lysario.de/?p=996</guid>
		<description><![CDATA[I recently wanted to update my OpenCV installation and try to find out why my usb camera &#8211; using the macam drivers &#8211; was not working with the libraries installed on my system. So this time i had the intention to dig into the code of OpenCV release. That meant using the easy way &#8211; [...]]]></description>
				<content:encoded><![CDATA[I recently wanted to update my OpenCV installation and try to find out why my usb camera &#8211; using the macam drivers &#8211; was not working with the libraries installed on my system. So this time i had the intention to dig into the code of OpenCV release. That meant using the easy way &#8211; mac ports &#8211; to install the OpenCV libraries was not an option.
<span id="more-996"></span>

The latest OpenCV software can be downloaded at the <a href="https://opencv.org/downloads.html">OpenCV homepage</a>. I downloaded OpenCV version 2.4.7 and extracted it within a local folder i created solely for OpenCV. The package comes with a cmake file (CMakeLists.txt) for creating different packages and setting up the build environment for OpenCV. cmake is open source software and is it&#8217;s purpose can be described as cross plattform build generator.That means it is used in order to provide environments for different plattforms that can be used to create a build for some project. Several <a href="https://www.cmake.org/cmake/help/documentation.html">tutorials </a>and <a href="https://www.cmake.org/cmake/resources/webinars.html">webinars </a>may be found on the homepage of the developing company <a href="https://www.kitware.com/">kitware</a>. I strongly recommend the webinar &#8220;Introduction to CMake Course&#8221; in order to get an understanding of the cmake process. cmake uses special generator commands that allow you to deploy the project for different purposes. You may like to create make files for Unix for example, or prepare the project to be used with XCode. The actual available generators may be displayed (as of cmake version 2.8.3) by typing into the commandline the command cmake &#8211;help.

Here&#8217;s the excerpt of my terminal output to the command:
<blockquote>
<pre>[...]</pre>
<pre>Generators</pre>
<pre>The following generators are available on this platform:
 Unix Makefiles                 = Generates standard UNIX makefiles.
 Xcode                          = Generate Xcode project files.
 CodeBlocks - Unix Makefiles    = Generates CodeBlocks project files.
 Eclipse CDT4 - Unix Makefiles  = Generates Eclipse CDT 4.0 project files.
 KDevelop3                      = Generates KDevelop 3 project files.
 KDevelop3 - Unix Makefiles     = Generates KDevelop 3 project files.</pre>
<pre>[...]</pre>
</blockquote>
<pre></pre>
In my case i wanted to create Xcode projects out of the source directory. For this purpose i created a new path into the directory where i downloaded and extracted the opencv software (let&#8217;s call this place
<pre>mypath_to_opencv</pre>
where
<pre>mypath_to_opencv/opencv-2.4.7</pre>
is the path to the opencv-2.4.7 sources). Start  your terminal and type:
<blockquote>
<pre>mkdir xcode
cd xcode
cmake -G Xcode mypath_to_opencv/opencv-2.4.7</pre>
</blockquote>
With the last command cmake is told to generate Xcode projects out of the files at the location were opencv-2.4.7 was extracted. The Xcode projects will be generated into the path where we called cmake. In the above mentioned case this path is mypath_to_opencv/xcode.

If you only want to use the libraries for using the OpenCV API you may like to read <a href="https://tilomitra.com/opencv-on-mac-osx/">Tilo Mitras article</a> about the procedure, while i still suggest that the easiest approach is using <a href="https://trac.macports.org/browser/trunk/dports/graphics/opencv/Portfile">macports</a>.

Now let&#8217;s go back to our Xcode generated build. If you look through your build tree you &#8216;ll find several Xcode Proejects. The top level of your build tree (xcode) includes for example  mypath_to_opencv/xcode/OpenCV.xcodeproj . The samples directory, in case the sample build has been enabled in ccmake, has the Xcode project file mypath_to_opencv/xcode/samples/c/c_samples.xcodeproj

In my case the standard cmake configuration had activated the <a href="https://docs.opencv.org/modules/ocl/doc/introduction.html">OCL dependency </a>which caused during the build in my systems problem because i have not installed any OCL driver on my system.

I had to reconfigure the system and deactivate OCL by the command (yes the command below is <strong>c</strong>cmake (not cmake) a commandline configuration tool for cmake)
<blockquote>
<pre>ccmake -G Xcode <mypath_to_opencv>/opencv-2.4.7</pre>
</blockquote>
and switching OCL off cmake variable WITH_OPENCL.

<a href="https://lysario.de/wp-content/uploads//2013/12/Bildschirmfoto-2013-12-14-um-19.32.12.png"><img class="alignnone size-medium wp-image-1009" alt="Bildschirmfoto 2013-12-14 um 19.32.12" src="https://lysario.de/wp-content/uploads//2013/12/Bildschirmfoto-2013-12-14-um-19.32.12-300x180.png" width="300" height="180" /></a>

So now i was ready for experimentation. The error message i got when building and running OpenCV XCode examples that were accessing a camera was &#8220;QTKit didn&#8217;t find any attached Video Input Devices !&#8221;. The same error was reported in a post by karlphillip at <a href="https://stackoverflow.com/questions/14451246/qtkit-didnt-find-any-attached-video-input-devices">stackoverflow.</a> Unfortunately i couldn&#8217;t fix my problem with the solution provided there. Debugging didn&#8217;t provide any clues why this was happening and somehow i came across a post by Andrew Janke on <a href="https://stackoverflow.com/questions/15861156/homebrew-opencv-32-bit-on-mac-10-7-5">stackoverflow</a> which gave me the right momentum.

The Macam version that i am using (macam-cvs-build-2009-09-25) could be  build without any changes for 32 bit system while the 64 bit build failed. OpenCV (2.4.7) on the other hand is generated with 32/64 bit architecture support when running the previous mentioned cmake commands. It is possible to reconfigure the Xcode project by changing the architecture option in ccmake and regenerate the xcode build directory. Trying to built OpenCV with 32 bit support failed on my system due to incompatible libraries that were installed on my machine. The following libraries  were only for 64 bit architecture available:
<ol>
	<li><a href="https://sourceforge.net/projects/libdc1394/">libdc1394.dylib</a></li>
	<li>libavcodec.dylib (part of ffmpeg)</li>
	<li>libavformat.dylib (part of ffmpeg)</li>
	<li>libavutil.dylib (part of ffmpeg)</li>
	<li>libswscale.dylib (part of ffmpeg)</li>
	<li>libbz2.dylib</li>
</ol>
My problem now was, when i build examples  i.e. example_video_homography that the 64 bit libraries were used with the 32 bit macam driver. So i tried to compile everything with a 32 bit architecture.

Using macports i adapted the <a href="https://superuser.com/questions/63198/install-32-bits-ports-on-snow-leopard">opt/local/etc/macports/macports.config file to make builds for 32-bit architecture</a>

build_arch            i386

I reinstalled the above mentioned libraries. Voila ! Happy end, my good old USB camera is now working with the OpenCV libraries !
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		<title>Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.19</title>
		<link>https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19</link>
		<comments>https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19#comments</comments>
		<pubDate>Fri, 17 May 2013 21:09:39 +0000</pubDate>
		<dc:creator>Panagiotis</dc:creator>
				<category><![CDATA[Kay: Modern Spectral Estimation, Theory and Application]]></category>
		<category><![CDATA[Solved Problems]]></category>

		<guid isPermaLink="false">https://lysario.de/?p=925</guid>
		<description><![CDATA[In [1, p. 62 exercise 3.19] we are asked to find for the multiple sinusoidal process the ensemble ACF and the temporal ACF as , where the &#8216;s are all uniformly distributed random variables on and independent of each other. We are also asked to determine if this random process is autocorrelation ergodic. Solution: We [...]]]></description>
				<content:encoded><![CDATA[In <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19bib1">[1, p. 62 exercise 3.19]</a> we are asked to find for the multiple sinusoidal process

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_18c06f6747166204af88405a37dafba0.png" align="bottom" class="tex" alt="x[n]=\sum\limits_{i=1}^{P}A_{i}\cos(2\pi f_{i}n+\phi_{i})" /></td><td></td></tr>
</table><br/>
the ensemble ACF and the temporal ACF as <img src="https://lysario.de/wp-content/cache/tex_b123b761a8f85f26145f5995ed64e3ed.png" align="bottom" class="tex" alt="M\rightarrow \infty " />, where the <img src="https://lysario.de/wp-content/cache/tex_d86907b87ab7cce2415bf86f6898a1a0.png" align="bottom" class="tex" alt="\phi_{i}" />&#8216;s are all uniformly distributed random variables on <img src="https://lysario.de/wp-content/cache/tex_809eac9fb66b81e416c504d099b0e44d.png" align="bottom" class="tex" alt="[0, 2 \pi)" /> and independent of each other. We are also asked to determine if this random process is autocorrelation ergodic.
<span id="more-925"></span>
<br /> <strong>Solution:</strong>
We note that the joint p.d.f of the uniformly random variables <img src="https://lysario.de/wp-content/cache/tex_d86907b87ab7cce2415bf86f6898a1a0.png" align="bottom" class="tex" alt="\phi_{i}" /> is given by <img src="https://lysario.de/wp-content/cache/tex_a1ebe26f6cf0afe603d7c0cd7164344a.png" align="bottom" class="tex" alt="f(\boldsymbol{\phi})=\prod_{i=1}^{P}f(\phi_{i})=\frac{1}{(2\pi)^{P}}" />, with domain <img src="https://lysario.de/wp-content/cache/tex_3c43f5bb5c9839c582da65d4081633ee.png" align="bottom" class="tex" alt="\mathbb{D}= \mathbb{A}^{P}, \; \mathbb{A} =[0, 2 \pi) " /> we can proceed to calculate the ensemble autocorrelation function which is defined as:

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_86d1714abfafbac31f80e60727d3add6.png" align="bottom" class="tex" alt="r_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_ed33a5bfeea3e43bd2105939d30e1f9e.png" align="bottom" class="tex" alt="E \left\{ x^{\star}[n] x[n+k] \right\} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_50afb8c1bf2b06eed664c222b2d2df15.png" align="bottom" class="tex" alt="\int\limits_{\mathbb{D}}\sum\limits_{i=1}^{P}A_{i}\cos(2\pi f_{i}n+\phi_{i})\sum\limits_{\ell=1}^{P}A_{\ell}\cos(2\pi f_{\ell}(n+k)+\phi_{\ell}) f(\boldsymbol{\phi})d\boldsymbol{\phi} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_a63a9a0c41a61fa99e52aae396cc86bc.png" align="bottom" class="tex" alt="\sum\limits_{i=1}^{P}\sum\limits_{\ell=1, \ell\neq i}^{P} \int_{0}^{2\pi} \int_{0}^{2\pi}\frac{A_{i}A_{\ell}}{(2\pi)^{2}}\cos(2\pi f_{i}n+\phi_{i})\cos(2\pi f_{\ell}(n+k)+\phi_{\ell}) d\phi_{i}d\phi_{\ell} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_7cbe209a69b54bef20eb9302e9acb520.png" align="bottom" class="tex" alt="+ \sum\limits_{i=1}^{P}\int_{0}^{2\pi} \frac{A_{i}^{2}}{(2\pi)}\cos(2\pi f_{i}n+\phi_{i})\cos(2\pi f_{i}(n+k)+\phi_{i})d\phi_{i} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_f59990cd0ecc01009a055fee0e76d657.png" align="bottom" class="tex" alt="\sum\limits_{i=1}^{P} \sum\limits_{\ell=1, \ell\neq i}^{P} \int_{0}^{2\pi} \frac{A_{i}A_{\ell}}{(2\pi)^{2}} \underbrace{\left[ \sin(2\pi f_{i}n+\phi_{i})\right]_{0}^{2\pi}}_{=0}\cos(2\pi f_{\ell}(n+k)+\phi_{\ell}) d\phi_{\ell} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_7cbe209a69b54bef20eb9302e9acb520.png" align="bottom" class="tex" alt="+ \sum\limits_{i=1}^{P}\int_{0}^{2\pi} \frac{A_{i}^{2}}{(2\pi)}\cos(2\pi f_{i}n+\phi_{i})\cos(2\pi f_{i}(n+k)+\phi_{i})d\phi_{i} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_23735c6a6acc79fe24328afd5a59fbe2.png" align="bottom" class="tex" alt="\sum\limits_{i=1}^{P}\int_{0}^{2\pi} \frac{A_{i}^{2}}{(2\pi)}\cos(2\pi f_{i}n+\phi_{i})\cos(2\pi f_{i}(n+k)+\phi_{i})d\phi_{i} " /></td><td></td></tr>
</table><br/>
As in exercise <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19bib2">[2]</a> we can use the trigonometric relation <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19bib3">[3, p. 810]</a> :

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_5e6ac7ca6f1e20ba2b21d2b8675d6004.png" align="bottom" class="tex" alt="2\cos\frak{A}\cos\frak{B} = \cos(\frak{A-B}) + \cos(\frak{A+B}) " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eqeq:Korn2000:cosmul"> (1)</a></td></tr>
</table><br/>

(<img src="https://lysario.de/wp-content/cache/tex_132c521320500dae60013966a39e421b.png" align="bottom" class="tex" alt="\frak{A}=2\pi f_{i}(n+k)+\phi_{i}" />, <img src="https://lysario.de/wp-content/cache/tex_df383ec5289b2c09305a73440bfc0caf.png" align="bottom" class="tex" alt="\frak{B}=2\pi f_{i}n+\phi_{i}" />) to further simplify the expression for the ensemble autocorrelation function:


<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_86d1714abfafbac31f80e60727d3add6.png" align="bottom" class="tex" alt="r_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_e1c33d821ba68f224f565b7a13f1bbf4.png" align="bottom" class="tex" alt="\sum\limits_{i=1}^{P}\int_{0}^{2\pi} \frac{A_{i}^{2}}{(4\pi)}\cos(2\pi f_{i}k)d\phi_{i} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_a377d641520edffae085476dfc9c6900.png" align="bottom" class="tex" alt="+\sum\limits_{i=1}^{P}\int_{0}^{2\pi}\frac{A_{i}^{2}}{(4\pi)}\cos(2\pi f_{i}(2n+k)+2 \phi_{i})d\phi_{i} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_e1c33d821ba68f224f565b7a13f1bbf4.png" align="bottom" class="tex" alt="\sum\limits_{i=1}^{P}\int_{0}^{2\pi} \frac{A_{i}^{2}}{(4\pi)}\cos(2\pi f_{i}k)d\phi_{i} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_e0f28f57d09e6a6c6e1d976a8fbfe1c0.png" align="bottom" class="tex" alt="+\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{(8\pi)}\underbrace{\left[\sin(2\pi f_{i}(2n+k)+2 \phi_{i})\right]_{\phi_{i}=0}^{2\pi}}_{=0} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_f54e1abbfc3ee19e95daf8167a092683.png" align="bottom" class="tex" alt="\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2}\cos(2\pi f_{i}k) " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eqeq:319:relensemblautocorr"> (2)</a></td></tr>
</table><br/>
Having obtained the ensemble autocorrelation function we can proceed to obtain the temporal autocorrelation function, which we hope will be a good approximation <img src="https://lysario.de/wp-content/cache/tex_ca6151c7d846302aeb171d16bcaf24ce.png" align="bottom" class="tex" alt="\hat{r}_{xx}[k]" /> to the ensemble autocorrelation function <img src="https://lysario.de/wp-content/cache/tex_86d1714abfafbac31f80e60727d3add6.png" align="bottom" class="tex" alt="r_{xx}[k]" />. By definition the temporal autocorrelation function is given by

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_ca6151c7d846302aeb171d16bcaf24ce.png" align="bottom" class="tex" alt="\hat{r}_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_7a001667ad6d1295e121ad3d94ee3ca1.png" align="bottom" class="tex" alt="\frac{1}{2M+1} \sum\limits_{n=-M}^{M} x[n+k]x^{\star}[n] " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_460d6f71098d84498b22dfd5173da282.png" align="bottom" class="tex" alt="\frac{1}{2M+1} \sum\limits_{n=-M}^{M}\Bigg[ \left(\sum\limits_{i=1}^{P}A_{i}\cos(2\pi f_{i}(n+k)+\phi_{i})\right) \cdot " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_cd7b099c92313c3d503a8da21822744b.png" align="bottom" class="tex" alt="\cdot \left(\sum\limits_{\ell=1}^{P}A_{\ell}\cos(2\pi f_{\ell}n+\phi_{\ell}) \right)\Bigg]" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_387929a935b3b542b609cff1c3e6a4f5.png" align="bottom" class="tex" alt="\sum\limits_{i=1}^{P}\sum\limits_{\ell=1,\ell \neq i}^{P}\sum\limits_{n=-M}^{M} \frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}(n+k)+\phi_{i})
\cos(2\pi f_{\ell}n+\phi_{\ell})" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_21853504aba169aad754235a887b7f21.png" align="bottom" class="tex" alt="+ \sum\limits_{i=1}^{P}\sum\limits_{n=-M}^{M}\frac{A_{i}^{2}}{2M+1}\cos(2\pi f_{i}(n+k)+\phi_{i})
\cos(2\pi f_{i}n+\phi_{i}) " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eqeq:319:directtempautocorr"> (3)</a></td></tr>
</table><br/>

The second sum of the previous relation can be simplified by one of the derived formulas of <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19bib2">[2]</a>, for <img src="https://lysario.de/wp-content/cache/tex_1faca3014cb54c80d2ce267f0ebbf162.png" align="bottom" class="tex" alt="e^{j2\pi f_{0}}\neq 1" /> :

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_f6c5134b6588fc5d1212e915f6303ed0.png" align="bottom" class="tex" alt="\frac{1}{2M+1}\sum\limits_{n=-M}^{M}A^{2}\cos(2\pi f_{0}n+\phi)\cos(2\pi f_{0}(n+k)+\phi) =" /></td><td></td></tr>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_782e22a1374307a9c5d5186839e377f9.png" align="bottom" class="tex" alt="\frac{A^{2}}{2} \cos(2\pi f_{0}k) + \frac{A^{2}}{2(2M+1)}\cos(2\pi f_{0}k+2\phi)\frac{\sin(2 \pi f_{0}(2M+1))}{\sin(2\pi f_{0})} " /></td><td></td></tr>
</table><br/>
Thus the temporal autocorrelation function may be expressed, as long as <img src="https://lysario.de/wp-content/cache/tex_126ca901524609363174a599613c4ccf.png" align="bottom" class="tex" alt="e^{j2\pi f_{i}}\neq 1, \; i=1,...,P" /> as:

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_ca6151c7d846302aeb171d16bcaf24ce.png" align="bottom" class="tex" alt="\hat{r}_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_387929a935b3b542b609cff1c3e6a4f5.png" align="bottom" class="tex" alt="\sum\limits_{i=1}^{P}\sum\limits_{\ell=1,\ell \neq i}^{P}\sum\limits_{n=-M}^{M} \frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}(n+k)+\phi_{i})
\cos(2\pi f_{\ell}n+\phi_{\ell})" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_9c65a2045cce144deae94222d8276882.png" align="bottom" class="tex" alt="+ \sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2} \cos(2\pi f_{i}k) " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_00ef85d820c183a7349b4c57baf51344.png" align="bottom" class="tex" alt="+\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2(2M+1)}\cos(2\pi f_{i}k+2\phi_{i})\frac{\sin(2 \pi f_{i}(2M+1))}{\sin(2\pi f_{i})}" /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eqeq319:temporalautocorrelsum"> (4)</a></td></tr>
</table><br/>

In order to simplify the relation further, especially the first sum, we will again use (<a href="#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eqeq:Korn2000:cosmul">1</a>), with <img src="https://lysario.de/wp-content/cache/tex_402f04540f32abc416b801ae7579fff5.png" align="bottom" class="tex" alt="\frak{A} =2\pi f_{i}(n+k)+\phi_{i} " /> and <img src="https://lysario.de/wp-content/cache/tex_c25ed8fa37badd79424b47d4bad84f24.png" align="bottom" class="tex" alt="\frak{B} = 2\pi f_{\ell}n+\phi_{\ell}" />, and thus :


<br/><table>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_b16bbb465f1bc587003747f2d5c93551.png" align="bottom" class="tex" alt="\sum\limits_{n=-M}^{M}\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}(n+k)+\phi_{i})
\cos(2\pi f_{\ell}n+\phi_{\ell}) = " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_1c3f4235f2ef8f08c13d6012695743ce.png" align="bottom" class="tex" alt="\sum\limits_{n=-M}^{M}\frac{A_{i}A_{\ell}}{2M+1} \cos(2 \pi (f_{i}-f_{\ell})n+2\pi f_{i}k+\phi_{i}-\phi_{\ell}) + " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_2401b627ce7ae906e1c377385d5d067e.png" align="bottom" class="tex" alt="\sum\limits_{n=-M}^{M}\frac{A_{i}A_{\ell}}{2M+1} \cos(2 \pi (f_{i} + f_{\ell})n+2\pi f_{i}k+\phi_{i}+\phi_{\ell}) " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eqeq319:autocorrsumsinusoids"> (5)</a></td></tr>
</table><br/>
We see that both distinct parts are of the form <img src="https://lysario.de/wp-content/cache/tex_f543bb690e9d48f373c97ca59baef9ad.png" align="bottom" class="tex" alt="\sum_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n+ 2\pi \frak{f}_{2} k + \Phi)" />, and thus it remains to simplify this relation. Again using a trigonometric formula <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19bib3">[3, p. 810]</a>:

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_1f3936db5ea57706dc65f496a900dd2f.png" align="bottom" class="tex" alt="\cos(\frak{A+B})=\cos\frak{A}\cos\frak{B}-\sin\frak{A}\sin\frak{B}" /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eq6"> (6)</a></td></tr>
</table><br/>
with <img src="https://lysario.de/wp-content/cache/tex_c87d0b2f18abaefd671c96742b1c1e82.png" align="bottom" class="tex" alt="\frak{A}= 2\pi \frak{f}_{1} n" /> and <img src="https://lysario.de/wp-content/cache/tex_37be91a33ef85b742887e9b54ecabcda.png" align="bottom" class="tex" alt="\frak{B}=2\pi \frak{f}_{2} k + \Phi" /> we can simplify the relation as:

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_0251a4d0a0e3e842a7c73dd0edb2eb09.png" align="bottom" class="tex" alt="\sum\limits_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n+ 2\pi \frak{f}_{2} k + \Phi) " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_de14b464b70b22569a527b33585a6984.png" align="bottom" class="tex" alt="\cos(2\pi \frak{f}_{2} k + \Phi)\sum\limits_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n) " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_da5e8ea7a4b6342d0d2e25d8775bc69b.png" align="bottom" class="tex" alt="-\sin(2\pi \frak{f}_{2} k + \Phi)\underbrace{\sum\limits_{n=-M}^{M}\sin(2\pi \frak{f}_{1} n)}_{=0, \textrm{ \scriptsize{odd symmetry of sine}}} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_de14b464b70b22569a527b33585a6984.png" align="bottom" class="tex" alt="\cos(2\pi \frak{f}_{2} k + \Phi)\sum\limits_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n) " /></td><td></td></tr>
</table><br/>

Furthermore it was shown in <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19bib2">[2]</a> that <img src="https://lysario.de/wp-content/cache/tex_df88ea80a2339657777417d7b5410258.png" align="bottom" class="tex" alt="\sum_{n=-M}^{M}\cos(2\pi (2 f_{0}))=\frac{\sin(\pi (2 f_{0})(2M+1))}{\sin(\pi(2f_{0})}" /> , for <img src="https://lysario.de/wp-content/cache/tex_b9cd7a5cec56faa63d062880f4658cc9.png" align="bottom" class="tex" alt="e^{j2\pi 2f_{0}}\neq 1" />, so we can further simplify the previous relation by:

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_e529da150cbf9a8336b84b7f62152a22.png" align="bottom" class="tex" alt="\sum\limits_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n+ 2\pi \frak{f}_{2} k + \Phi)=\cos(2\pi \frak{f}_{2} k + \Phi)\sum\limits_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n) " /></td><td></td></tr>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_12d1d9367838330026e6c09526f5e08e.png" align="bottom" class="tex" alt="= \cos(2\pi \frak{f}_{2} k + \Phi)\frac{\sin(\pi \frak{f}_{1}(2M+1))}{\sin(\pi\frak{f}_{1})}, \; e^{j2\pi\frak{f}_{1}}\neq 1" /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eqeq319:simplcossum"> (7)</a></td></tr>
</table><br/>
Thus finally we can simplify (<a href="#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eqeq319:autocorrsumsinusoids">5</a>) by:

<br/><table>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_b16bbb465f1bc587003747f2d5c93551.png" align="bottom" class="tex" alt="\sum\limits_{n=-M}^{M}\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}(n+k)+\phi_{i})
\cos(2\pi f_{\ell}n+\phi_{\ell}) = " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_d1e76671dd4cba29993f20a18dffb6a6.png" align="bottom" class="tex" alt="\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}k+\phi_{i}-\phi_{\ell}) \frac{\sin(\pi (f_{i}-f_{\ell})(2M+1))}{\sin(\pi (f_{i}-f_{\ell}))} + " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_f0d2f9a43a96ff05e088856872b20671.png" align="bottom" class="tex" alt="\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}k+\phi_{i}+\phi_{\ell}) \frac{\sin(\pi (f_{i}+f_{\ell})(2M+1))}{\sin(\pi (f_{i}+f_{\ell}))} " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eqeq319:autocorrsumsinusoids2"> (8)</a></td></tr>
</table><br/>

So finally the temporal autocorrelation (<a href="#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eqeq319:temporalautocorrelsum">4</a>) can be reduced using (<a href="#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eqeq319:autocorrsumsinusoids2">8</a>) to the following formula:


<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_ca6151c7d846302aeb171d16bcaf24ce.png" align="bottom" class="tex" alt="\hat{r}_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_e59a63236359bda116f01b9af852d056.png" align="bottom" class="tex" alt="\sum\limits_{i=1}^{P}\sum\limits_{\ell=1,\ell \neq i}^{P}\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}k+\phi_{i}-\phi_{\ell}) \frac{\sin(\pi (f_{i}-f_{\ell})(2M+1))}{\sin(\pi (f_{i}-f_{\ell}))} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_69eb11cb43b8842557668a9c30dbb747.png" align="bottom" class="tex" alt="+\sum\limits_{i=1}^{P}\sum\limits_{\ell=1,\ell \neq i}^{P}\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}k+\phi_{i}+\phi_{\ell}) \frac{\sin(\pi (f_{i}+f_{\ell})(2M+1))}{\sin(\pi (f_{i}+f_{\ell}))} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_9c65a2045cce144deae94222d8276882.png" align="bottom" class="tex" alt="+ \sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2} \cos(2\pi f_{i}k) " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_00ef85d820c183a7349b4c57baf51344.png" align="bottom" class="tex" alt="+\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2(2M+1)}\cos(2\pi f_{i}k+2\phi_{i})\frac{\sin(2 \pi f_{i}(2M+1))}{\sin(2\pi f_{i})}" /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eq9"> (9)</a></td></tr>
</table><br/>

Rearranging the terms on the right we see that the temporal autocorrelation function equals the ensemble autocorrelation function with an additional error <img src="https://lysario.de/wp-content/cache/tex_92e4da341fe8f4cd46192f21b6ff3aa7.png" align="bottom" class="tex" alt="\epsilon" />, which we have derived for the the case when <img src="https://lysario.de/wp-content/cache/tex_6e7bff522c4ab08fa907a1a1f1c89921.png" align="bottom" class="tex" alt="e^{(j2\pi f_{i})}\neq 1 \; , i=1,...,P " />:


<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_ca6151c7d846302aeb171d16bcaf24ce.png" align="bottom" class="tex" alt="\hat{r}_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_d233a8f2e65cecd7483b28f4ad77e595.png" align="bottom" class="tex" alt="\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2} \cos(2\pi f_{i}k) " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_e59a63236359bda116f01b9af852d056.png" align="bottom" class="tex" alt="\sum\limits_{i=1}^{P}\sum\limits_{\ell=1,\ell \neq i}^{P}\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}k+\phi_{i}-\phi_{\ell}) \frac{\sin(\pi (f_{i}-f_{\ell})(2M+1))}{\sin(\pi (f_{i}-f_{\ell}))} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_69eb11cb43b8842557668a9c30dbb747.png" align="bottom" class="tex" alt="+\sum\limits_{i=1}^{P}\sum\limits_{\ell=1,\ell \neq i}^{P}\frac{A_{i}A_{\ell}}{2M+1} \cos(2\pi f_{i}k+\phi_{i}+\phi_{\ell}) \frac{\sin(\pi (f_{i}+f_{\ell})(2M+1))}{\sin(\pi (f_{i}+f_{\ell}))} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_3c16dd2d972f594779cadc0e6b7087ed.png" align="bottom" class="tex" alt="+\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2(2M+1)}\cos(2\pi f_{i}k+2\phi_{i})\frac{\sin(2 \pi f_{i}(2M+1))}{\sin(2\pi f_{i})} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_31d6c64643fcd1591c423ab479644174.png" align="bottom" class="tex" alt="r_{xx}[k] +\epsilon " /></td><td></td></tr>
</table><br/>

At once we see again that when <img src="https://lysario.de/wp-content/cache/tex_2eff378d712981e7a93be3f8f2eb2b1a.png" align="bottom" class="tex" alt="M\rightarrow \infty" /> that the error goes to zero <img src="https://lysario.de/wp-content/cache/tex_c336fb3606a9c5136b99db10b2e31f5d.png" align="bottom" class="tex" alt="\epsilon \rightarrow 0" />. Thus the random process of the sum of sinusoids is also autocorrelation ergodic as long as <img src="https://lysario.de/wp-content/cache/tex_e6a0254ecd6dc76ee1799e33eaab4274.png" align="bottom" class="tex" alt="e^{(j2\pi f_{i})}\neq1 \; , i=1,...,P " />. It is also easy to recognize, by using the argumentation of the previous exercise <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19bib2">[2]</a> , that this is not true if <img src="https://lysario.de/wp-content/cache/tex_e6a0254ecd6dc76ee1799e33eaab4274.png" align="bottom" class="tex" alt="e^{(j2\pi f_{i})}\neq1 \; , i=1,...,P " /> does not hold. In this case parts of the the error of equation (<a href="#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19eqeq:319:directtempautocorr">3</a>) are proportional to <img src="https://lysario.de/wp-content/cache/tex_69691c7bdcc3ce6d5d8a1361f22d04ac.png" align="bottom" class="tex" alt="M" /> as was already observed in <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-19bib2">[2]</a> . QED.
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		<title>Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.18</title>
		<link>https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2</link>
		<comments>https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2#comments</comments>
		<pubDate>Fri, 24 Aug 2012 15:37:26 +0000</pubDate>
		<dc:creator>Panagiotis</dc:creator>
				<category><![CDATA[Kay: Modern Spectral Estimation, Theory and Application]]></category>
		<category><![CDATA[Solved Problems]]></category>
		<category><![CDATA[autocorrelation]]></category>
		<category><![CDATA[ergodic]]></category>

		<guid isPermaLink="false">https://lysario.de/?p=960</guid>
		<description><![CDATA[In [1, p. 62 exercise 3.18] we are asked to find the temporal autocorrelation function for the real sinusoidal random process of Problem [1, p. 62 exercise 3.16]: as . As a second step we are asked to determine if the random process autocorrelation ergodic. Solution: The sinusoidal process of Problem [1, p. 62 exercise [...]]]></description>
				<content:encoded><![CDATA[In <a href="https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2#lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2bib1">[1, p. 62 exercise 3.18]</a> we are asked to find the temporal autocorrelation function for the real sinusoidal random process of Problem <a href="https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2#lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2bib1">[1, p. 62 exercise 3.16]</a>:


<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_e2a95ef4fe5209acc420201c8b3143bb.png" align="bottom" class="tex" alt="\hat{r}_{xx}[k]=\frac{1}{2M+1}\sum\limits_{n=-M}^{M}x[n]x[n+k]" /></td><td></td></tr>
</table><br/>

as <img src="https://lysario.de/wp-content/cache/tex_2eff378d712981e7a93be3f8f2eb2b1a.png" align="bottom" class="tex" alt="M\rightarrow \infty" />. As a second step we are asked to determine if the random process autocorrelation ergodic.
<span id="more-960"></span>
<br /> <strong>Solution:</strong>
The sinusoidal process of Problem <a href="https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2#lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2bib1">[1, p. 62 exercise 3.16]</a> is given by <img src="https://lysario.de/wp-content/cache/tex_c2f33921dab216dfd22261344f21fadd.png" align="bottom" class="tex" alt="x[n]=A \cos(2\pi f_{0}n+\phi)" />, and the autocorrelation function was found to be equal to (see solution of exercise 3.16 <a href="https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2#lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2bib2">[2]</a>) <img src="https://lysario.de/wp-content/cache/tex_9f26223331739c3c8412b538256a496f.png" align="bottom" class="tex" alt="r_{xx}[k]=\frac{A^{2}}{2} \cos(2\pi f_{0}k)" />. The temporal autocorrelation is equal to :

<br/><table>
<tr><td><img src="https://lysario.de/wp-content/cache/tex_ca6151c7d846302aeb171d16bcaf24ce.png" align="bottom" class="tex" alt="\hat{r}_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_6fa0ae17af0dd7a64de7fd7979439993.png" align="bottom" class="tex" alt="\frac{1}{2M+1}\sum\limits_{n=-M}^{M}A^{2}\cos(2\pi f_{0}n+\phi)\cos(2\pi f_{0}(n+k)+\phi)" /></td><td></td></tr>
</table><br/>
The previous relation can be furthermore simplified by the property of trigonometric functions <a href="https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2#lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2bib3">[3, p. 810]</a> :

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_291b3ca6eef3af231c27b4d48ecca749.png" align="bottom" class="tex" alt="2\cos\frak{A}\cos\frak{B}=\cos(\frak{A-B})+\cos(\frak{A+B})" /></td><td><a name="lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2eq1"> (1)</a></td></tr>
</table><br/>
and thus the autocorrelation can be written as:

<br/><table>
<tr><td><img src="https://lysario.de/wp-content/cache/tex_ca6151c7d846302aeb171d16bcaf24ce.png" align="bottom" class="tex" alt="\hat{r}_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_87d8b6f9740f7d5e59084d3f5a8def55.png" align="bottom" class="tex" alt="\frac{A^{2}}{2(2M+1)}\sum\limits_{n=-M}^{M}\left(\cos(2\pi f_{0}k)+\cos(2\pi f_{0}(2n+k)+2\phi)\right)" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_8b9975cdf350106973269aa1b024c015.png" align="bottom" class="tex" alt="\frac{A^{2}}{2(2M+1)} (2M+1) \cos(2\pi f_{0}k) " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_b18f02278b96378f6371ea93202ee717.png" align="bottom" class="tex" alt="+ \frac{A^{2}}{2(2M+1)}\sum\limits_{n=-M}^{M}\cos(2\pi f_{0}(2n+k)+2\phi) " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_d6d6226a0004519f7a3b335129e23336.png" align="bottom" class="tex" alt="\frac{A^{2}}{2} \cos(2\pi f_{0}k)" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_303bc1c022305e79c74e2a1e18897ca5.png" align="bottom" class="tex" alt="+ \frac{A^{2}}{2(2M+1)}\sum\limits_{n=-M}^{M}\cos(4\pi f_{0}n+2\pi f_{0}k+2\phi)" /></td><td></td></tr>
<tr><td colspan="3"></td><td></td></tr>
</table><br/>
Again it is possible to further simplify the previous relation, this time by using the following trigonometric formula <a href="https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2#lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2bib3">[3, p. 810]</a>:

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_1f3936db5ea57706dc65f496a900dd2f.png" align="bottom" class="tex" alt="\cos(\frak{A+B})=\cos\frak{A}\cos\frak{B}-\sin\frak{A}\sin\frak{B}" /></td><td><a name="lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2eq2"> (2)</a></td></tr>
</table><br/>

<br/><table>
<tr><td><img src="https://lysario.de/wp-content/cache/tex_ca6151c7d846302aeb171d16bcaf24ce.png" align="bottom" class="tex" alt="\hat{r}_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_d6d6226a0004519f7a3b335129e23336.png" align="bottom" class="tex" alt="\frac{A^{2}}{2} \cos(2\pi f_{0}k)" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_5731ac3a1ae032bba973d51b053ee0fd.png" align="bottom" class="tex" alt="+ \frac{A^{2}}{2(2M+1)}\cos(2\pi f_{0}k+2\phi)\sum\limits_{n=-M}^{M}\cos(4\pi f_{0}n)" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_74a68a08d92eb76ba49b7c35d8dc53fe.png" align="bottom" class="tex" alt="- \frac{A^{2}}{2(2M+1)}\sin(2\pi f_{0}k+2\phi)\sum\limits_{n=-M}^{M}\sin(4\pi f_{0}n)" /></td><td></td></tr>
<tr><td colspan="3"></td><td></td></tr>
</table><br/>
Because the sinus function is odd <img src="https://lysario.de/wp-content/cache/tex_88d448d35d6c8e148b93ae79c5c10090.png" align="bottom" class="tex" alt="\sin(-x)=-\sin(x)" /> the symmetric sum <img src="https://lysario.de/wp-content/cache/tex_afc84c7b842c1f026c5c822c77df252c.png" align="bottom" class="tex" alt="\sum_{n=-M}^{M}\sin(4\pi f_{0}n)" /> equals zero. Generally the same is not true for the cosine sum. For <img src="https://lysario.de/wp-content/cache/tex_f4087d12e714129573fd158d06b8ff0a.png" align="bottom" class="tex" alt="f_{0}=1" /> for example the sum <img src="https://lysario.de/wp-content/cache/tex_c6be32f937fe4e7c253bd5b8d68c9184.png" align="bottom" class="tex" alt="\sum_{n=-M}^{M}\cos(4\pi f_{0}n)" /> equals <img src="https://lysario.de/wp-content/cache/tex_c221cd01d162dfa494e0f03152a96122.png" align="bottom" class="tex" alt="2M+1" /> and thus even for <img src="https://lysario.de/wp-content/cache/tex_2eff378d712981e7a93be3f8f2eb2b1a.png" align="bottom" class="tex" alt="M\rightarrow \infty" /> the temporal autocorrelation function will be in error to the true autocorrelation function.
In general it is true that <img src="https://lysario.de/wp-content/cache/tex_4eb94200fc6fa9f045ef8189f8843faf.png" align="bottom" class="tex" alt="- \frac{A^{2}}{2}\leq \frac{A^{2}}{2(2M+1)} \sum_{n=-M}^{M}\cos(4\pi f_{0}n)\leq \frac{A^{2}}{2}" />.
The error to the true autocorrelation function is equal to <img src="https://lysario.de/wp-content/cache/tex_6785ccfb207dc2d74bedc41ca7e26793.png" align="bottom" class="tex" alt="\epsilon=\frac{A^{2}}{2(2M+1)}\cos(2\pi f_{0}k+2\phi)\sum_{n=-M}^{M}\cos(4\pi f_{0}n)" />, and thus the process is in general not autocorrelation ergodic, because as we have shown for e.g. <img src="https://lysario.de/wp-content/cache/tex_f4087d12e714129573fd158d06b8ff0a.png" align="bottom" class="tex" alt="f_{0}=1" /> the temporal autocorrelation doesn&#8217;t even converge for large <img src="https://lysario.de/wp-content/cache/tex_69691c7bdcc3ce6d5d8a1361f22d04ac.png" align="bottom" class="tex" alt="M" />. That is:

<br/><table>
<tr><td><img src="https://lysario.de/wp-content/cache/tex_ca6151c7d846302aeb171d16bcaf24ce.png" align="bottom" class="tex" alt="\hat{r}_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_d6d6226a0004519f7a3b335129e23336.png" align="bottom" class="tex" alt="\frac{A^{2}}{2} \cos(2\pi f_{0}k)" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_5731ac3a1ae032bba973d51b053ee0fd.png" align="bottom" class="tex" alt="+ \frac{A^{2}}{2(2M+1)}\cos(2\pi f_{0}k+2\phi)\sum\limits_{n=-M}^{M}\cos(4\pi f_{0}n)" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_b0dd2042185ab85bec7332dff3ee8e01.png" align="bottom" class="tex" alt="r_{xx}[k]+\epsilon." /></td><td></td></tr>
</table><br/>

While it is true that there are frequencies <img src="https://lysario.de/wp-content/cache/tex_d43e51bee35b78083e05bcfc2119b318.png" align="bottom" class="tex" alt="f_{0}" /> for which the temporal autocorrelation doesn&#8217;t converge, there are also cases when the process is autocorrelation ergodic. To see this we have to further simplify the relation for the temporal autocorrelation. The sum <img src="https://lysario.de/wp-content/cache/tex_39591e99b1e067d9d36e63c1b88e812e.png" align="bottom" class="tex" alt="\sum_{n=-M}^{M}\cos(4\pi f_{0}n) " /> can be further simplified, by using <img src="https://lysario.de/wp-content/cache/tex_af3685bd4992acbe30633def26031653.png" align="bottom" class="tex" alt="\cos(4\pi f_{0}n) =\Re\{e^{(j4\pi f_{0}n)}\} " />,
and observing that the resulting sum is a geometric progression (<a href="https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2#lysario.desolved_problemssteven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2bib3">[3, p. 7]</a> <img src="https://lysario.de/wp-content/cache/tex_41ea55b8999c6eff858babc6e264288a.png" align="bottom" class="tex" alt="\sum\limits_{j=0}^{n}a_{0}r^{j}=a_{0}\frac{1-r^{n+1}}{1-r}, \; r &lt; 1 " />) with <img src="https://lysario.de/wp-content/cache/tex_8a1238e65218c9c54993ef5d21b9505b.png" align="bottom" class="tex" alt="a_{0}=1, r_{0}=e^{j4\pi f_{0}}" /> :

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_c9275005e962430f74bccbce412acbd6.png" align="bottom" class="tex" alt="\sum\limits_{n=-M}^{M}\cos(4\pi f_{0}n)" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_a4f4b9362a3ea016733aef34238901e4.png" align="bottom" class="tex" alt="\Re\{\sum\limits_{n=-M}^{M}e^{j4\pi f_{0}n}\} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_436c2ae970a219191aa0ce8723cbcf10.png" align="bottom" class="tex" alt="\Re\{e^{-j4\pi f_{0}M}\sum\limits_{n=0}^{2M}e^{j4\pi f_{0}n}\} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_aa950a7cc88f30a79d53b557a327651a.png" align="bottom" class="tex" alt="\Re\{e^{-j4\pi f_{0}M}\frac{1-e^{j4\pi f_{0}(2M+1)}}{1-e^{j4\pi f_{0}}}\} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_8caea2e5192bbecac06e4b26fb0333b0.png" align="bottom" class="tex" alt="\Re\{\underbrace{e^{-j4\pi f_{0}M}\frac{e^{j2\pi f_{0}(2M+1)}}{e^{j2\pi f_{0}}}}_{=1}\cdot " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_82405ef837e998e1c62b3bd119acf2e2.png" align="bottom" class="tex" alt="\frac{e^{-j2\pi f_{0}(2M+1)}-e^{j2\pi f_{0}(2M+1)}}{e^{-j2\pi f_{0}}-e^{j2\pi f_{0}}}\} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_361e957accd34a5be587ef91b963c8a6.png" align="bottom" class="tex" alt="\Re\{\frac{\sin(2 \pi f_{0}(2M+1))}{\sin(2\pi f_{0})}\} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_90b473ecaccff4ffd7ef53f222a2a64e.png" align="bottom" class="tex" alt="\frac{\sin(2 \pi f_{0}(2M+1))}{\sin(2\pi f_{0})}" /></td><td></td></tr>
</table><br/>
Using this identity the error to the true autocorrelation function can be rewritten as:

<br/><table>
<tr><td><img src="https://lysario.de/wp-content/cache/tex_92e4da341fe8f4cd46192f21b6ff3aa7.png" align="bottom" class="tex" alt="\epsilon" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_70a394dbcbead9ae5bcfe3ab72edddcf.png" align="bottom" class="tex" alt="\frac{A^{2}}{2(2M+1)}\cos(2\pi f_{0}k+2\phi)\frac{\sin(2 \pi f_{0}(2M+1))}{\sin(2\pi f_{0})} \Rightarrow" /></td><td></td></tr>
<tr><td><img src="https://lysario.de/wp-content/cache/tex_18f2075faf346af745a59fc8ea396d98.png" align="bottom" class="tex" alt="|\epsilon|" /></td><td><img src="https://lysario.de/wp-content/cache/tex_de44c582df9d8d29dbbd70aca311c641.png" align="bottom" class="tex" alt="\leq" /></td><td><img src="https://lysario.de/wp-content/cache/tex_ffeaa8efdeb5d11bc009461f6a8b4daf.png" align="bottom" class="tex" alt="|\frac{A^{2}}{2(2M+1)}| \Rightarrow" /></td><td></td></tr>
<tr><td><img src="https://lysario.de/wp-content/cache/tex_0b88efe71bf5b13e4ec8f342e7fda629.png" align="bottom" class="tex" alt="\lim_{M\rightarrow \infty} |\epsilon|" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_36b8e1133a9d046fbd840f67896716b7.png" align="bottom" class="tex" alt="0," /></td><td></td></tr>
</table><br/>
Thus by only imposing that <img src="https://lysario.de/wp-content/cache/tex_3c70c5d3cd1f4a01c9416ea09b4f0312.png" align="bottom" class="tex" alt="e^{j4\pi f_{0}}\neq 1 " /> the process is autocorrelation ergodic.
<!-- The entry title is:Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.18The compare string is:LaTeX TesterAnd the comparison is: FALSE -->
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		</item>
		<item>
		<title>Scilab Startup Configuration on MacOsX</title>
		<link>https://lysario.de/scilab/scilab-startup-configuration-on-macosx</link>
		<comments>https://lysario.de/scilab/scilab-startup-configuration-on-macosx#comments</comments>
		<pubDate>Sat, 05 May 2012 14:20:52 +0000</pubDate>
		<dc:creator>Panagiotis</dc:creator>
				<category><![CDATA[Programming]]></category>
		<category><![CDATA[Scilab]]></category>

		<guid isPermaLink="false">https://lysario.de/?p=907</guid>
		<description><![CDATA[Sometimes it is useful to have several scripts (with standard definitions) loaded immediately into scilab. On MacOsX Scilab executes at startup the first lines of the file .scilab. The file has to be placed (if it does already exist) under the directory: /home/User/.Scilab/scilab-version If the file isn&#8217;t there just create one and call your scripts [...]]]></description>
				<content:encoded><![CDATA[Sometimes it is useful to have several scripts (with standard definitions) loaded immediately into scilab.
<br />
On MacOsX Scilab executes at startup <a href="https://help.scilab.org/docs/5.3.3/en_US/startup.html" target="_blank">the first lines of the file .scilab</a>. The file has to be placed (if it does already exist) under the directory:
<br />
<code>
/home/<img src="https://lysario.de/wp-content/cache/tex_524a50782178998021a88b8cd4c8dcd8.png" align="bottom" class="tex" alt="&lt;" />User<img src="https://lysario.de/wp-content/cache/tex_cedf8da05466bb54708268b3c694a78f.png" align="bottom" class="tex" alt="&gt;" />/.Scilab/<img src="https://lysario.de/wp-content/cache/tex_524a50782178998021a88b8cd4c8dcd8.png" align="bottom" class="tex" alt="&lt;" />scilab-version<img src="https://lysario.de/wp-content/cache/tex_cedf8da05466bb54708268b3c694a78f.png" align="bottom" class="tex" alt="&gt;" />
</code>
<br />
If the file isn&#8217;t there just create one and call your scripts with the definitions to be loaded into the scilab workspace.
<br /> 
<br />
For example you might want to load definitions of physical constants into the workspace that are defined within &#8216;physical_constants.sce&#8217;. For this purpose add a line like the following into the .scilab file: 
<br />
<br />
<code>exec('Users/<img src="https://lysario.de/wp-content/cache/tex_524a50782178998021a88b8cd4c8dcd8.png" align="bottom" class="tex" alt="&lt;" />User<img src="https://lysario.de/wp-content/cache/tex_cedf8da05466bb54708268b3c694a78f.png" align="bottom" class="tex" alt="&gt;" />/Development/scilab/Base/physical_constants.sce');</code>
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		</item>
		<item>
		<title>Book Review: Logicomix &#8211; An epic search for the truth</title>
		<link>https://lysario.de/allgemein/book-reviews/book-review-logicomix-an-epic-search-for-the-truth</link>
		<comments>https://lysario.de/allgemein/book-reviews/book-review-logicomix-an-epic-search-for-the-truth#comments</comments>
		<pubDate>Thu, 26 Apr 2012 20:45:44 +0000</pubDate>
		<dc:creator>Panagiotis</dc:creator>
				<category><![CDATA[Book Reviews]]></category>
		<category><![CDATA[Gödel]]></category>
		<category><![CDATA[Hilbert]]></category>
		<category><![CDATA[Logicomix]]></category>
		<category><![CDATA[Russel]]></category>

		<guid isPermaLink="false">https://lysario.de/?p=880</guid>
		<description><![CDATA[Last year i got a very different mathematical book as a gift. A comic novel from Doxiadis, Papadimitriou and Papadatos: Logicomix. It is easy to read and provides some insight of the mathematical quest in the area of logic of the last century. You&#8217;ll meet people like Bertrand Russel, David Hilbert, Gödel and many more [...]]]></description>
				<content:encoded><![CDATA[<iframe src="https://rcm-de.amazon.de/e/cm?lt1=_blank&#038;bc1=000000&#038;IS2=1&#038;bg1=FFFFFF&#038;fc1=000000&#038;lc1=0000FF&#038;t=netzspielwies-21&#038;o=3&#038;p=8&#038;l=as4&#038;m=amazon&#038;f=ifr&#038;ref=ss_til&#038;asins=1596914521" style="width:120px;height:240px;float:left; margin:10px 20px 0 0;" scrolling="no" marginwidth="0" marginheight="0" frameborder="0"></iframe> Last year i got a very different mathematical book as a gift. A comic novel from Doxiadis, Papadimitriou and Papadatos: Logicomix. It is easy to read and provides some insight of the mathematical quest in the area of logic of the last century. You&#8217;ll meet people like Bertrand Russel, David Hilbert, Gödel and many more outstanding persons. Some work results of the main actors are presented in that way that is accessible for a wide audience &#8211; according to the rules used in popular science media.  So even mathematical illiterate persons may enjoy the comic and due to the very nice illustrations i think even kids could enjoy it. The &#8220;showdown&#8221; is Gödel&#8217;s incompleteness theorem and i enjoyed the way the result was presented. The authors admit at the end of the book that the storyline is only loosely based on historical facts but actually this doesn&#8217;t reduce the value of the novel. Reading this comic was a nice way to relax during a cold winter&#8217;s day. Give it a try ! 



<!-- The entry title is:Book Review: Logicomix &#8211; An epic search for the truthThe compare string is:LaTeX TesterAnd the comparison is: FALSE -->
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		</item>
		<item>
		<title>Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.17</title>
		<link>https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17</link>
		<comments>https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17#comments</comments>
		<pubDate>Thu, 19 Apr 2012 20:43:42 +0000</pubDate>
		<dc:creator>Panagiotis</dc:creator>
				<category><![CDATA[Kay: Modern Spectral Estimation, Theory and Application]]></category>
		<category><![CDATA[Solved Problems]]></category>
		<category><![CDATA[sample mean estimator]]></category>
		<category><![CDATA[variance]]></category>
		<category><![CDATA[white noise]]></category>

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		<description><![CDATA[In [1, p. 62 exercise 3.17] we are asked to verify that the variance of the sample mean estimator for the mean of a real WSS random process is given by [1, eq. (3.60), p. 58]. For the case when is real white noise we are asked to what the variance expression does reduce to. [...]]]></description>
				<content:encoded><![CDATA[In <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17bib1">[1, p. 62 exercise 3.17]</a> we are asked to verify that the variance of the sample mean estimator for the mean of a real WSS random process

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_1802f56fef4f03089a8b530f29ffae10.png" align="bottom" class="tex" alt="\frac{1}{2M+1}\sum\limits_{n=-M}^{n=M}x[n]" /></td><td></td></tr>
</table><br/>
is given by <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17bib1">[1, eq. (3.60), p. 58]</a>. For the case when <img src="https://lysario.de/wp-content/cache/tex_d3baaa3204e2a03ef9528a7d631a4806.png" align="bottom" class="tex" alt="x[n]" /> is real white noise we are asked to what the variance expression does reduce to. 
A hint that is given is to use the relationship from <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17bib1">[1, eq. (3.64), p. 59]</a>.  
<span id="more-845"></span>
<br /> <strong>Solution:</strong>
Let&#8217;s reproduce the corresponding equations cited in the problem statement, so <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17bib1">[1, eq. (3.60), p. 58]</a> is given by 

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_0b499c889ff801e3e4cfe1eb020d2cc6.png" align="bottom" class="tex" alt="\lim\limits_{M\rightarrow \infty}\frac{1}{2M+1} \sum\limits_{k=-2M}^{2M}\left(1-\frac{|k|}{2M+1}\right)c_{xx}[k]=0." /></td><td></td></tr>
</table><br/>
while <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17bib1">[1, eq. (3.64), p. 59]</a> is given by: 

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_e252ca41cef09873b0fe68f74a0a3cbb.png" align="bottom" class="tex" alt="\sum\limits_{m=-M}^{M}\sum\limits_{n=-M}^{M}g[m-n]=\sum\limits_{k=-2M}^{2M} \left(2M+1-|k|\right)g[k]" /></td><td></td></tr>
</table><br/>

First we note that the mean of the sample mean converges to the true mean of the WSS random process <img src="https://lysario.de/wp-content/cache/tex_d3baaa3204e2a03ef9528a7d631a4806.png" align="bottom" class="tex" alt="x[n]" />: 


<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_ab94d861246b3116e586683c6ea57e36.png" align="bottom" class="tex" alt="E\left\{ \hat{\mu}_{x} \right\}" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_f6bd9c48c4d56674f504638e16edb6c6.png" align="bottom" class="tex" alt="\frac{1}{2M+1}\sum\limits_{n=-M}^{M} E\left\{ x[n]\right\} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_71efd26075b22bf8b1fea5ee840747fa.png" align="bottom" class="tex" alt="\frac{1}{2M+1} (2M+1)\mu_{x} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_077d2e2d7390e157052459a1526e1e9e.png" align="bottom" class="tex" alt="\mu_{x}. " /></td><td></td></tr>
</table><br/>
From this we can derive the variance of the <img src="https://lysario.de/wp-content/cache/tex_40f3b3e31a1e46c12f0ea8fb7e891e52.png" align="bottom" class="tex" alt="\hat{\mu}_{x}" /> as: 

<br/><table>
<tr><td><img src="https://lysario.de/wp-content/cache/tex_a3742d54e5a4622349a3b6fe5fbb906b.png" align="bottom" class="tex" alt="E\left\{ (\hat{\mu}_{x}-\mu_{x})^{2}\right\}" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_2294f8b24e693791e39e19fe1162805e.png" align="bottom" class="tex" alt="E\left\{\hat{\mu}_{x}^{2}\right\}-2\mu_{x} E\left\{\hat{\mu}_{x}\right\}+\mu_{x}^{2} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_da19eb7009facdab9c96329036d3f54c.png" align="bottom" class="tex" alt="E\left\{\hat{\mu}_{x}^{2}\right\}-\mu_{x}^{2}" /></td><td></td></tr>
<tr><td colspan="3"></td><td></td></tr>
</table><br/>

The squared sample mean can be written as: 

<br/><table>
<tr><td><img src="https://lysario.de/wp-content/cache/tex_c2ccfad2fdfae1c2736391565a901fe2.png" align="bottom" class="tex" alt="\hat{\mu}_{x}^{2}" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_ca3e0ddb764d6a3c9097a1432b47b7d3.png" align="bottom" class="tex" alt="\left(\frac{1}{2M+1}\right)^{2} \sum\limits_{n=-M}^{n=M} x[n] \sum\limits_{l=-M}^{l=M}x[l]" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_d971b37c82939f61543d9d5801d45e5b.png" align="bottom" class="tex" alt="\left(\frac{1}{2M+1}\right)^{2}\sum\limits_{n=-M}^{n=M}\sum\limits_{l=-M}^{l=M}x[n]x[l], \; \textnormal{replacing } l=n+k \Rightarrow" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_524140ec37a2340a41b30b949aac8bf2.png" align="bottom" class="tex" alt="\left(\frac{1}{2M+1}\right)^{2}\sum\limits_{n=-M}^{n=M}\sum\limits_{k=-M-n}^{k=M-n}x[n]x[n+k]." /></td><td></td></tr>
</table><br/>
From the previous relation we can derive that the mean squared sample mean is given by: 

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_1ef1b28b5ad1a096d209456fcc4b9f41.png" align="bottom" class="tex" alt="E\left\{\hat{\mu}_{x}^{2}\right\}" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_317ff3a81fbf73676bcc6860ce3ca115.png" align="bottom" class="tex" alt="\left(\frac{1}{2M+1}\right)^{2}\sum\limits_{n=-M}^{n=M}\sum\limits_{k=-M-n}^{k=M-n}\ensuremath{E\left\{x[n]x[n+k]\right\}}  " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_1f50033434ff7d80f9c93940bed3b407.png" align="bottom" class="tex" alt="\left(\frac{1}{2M+1}\right)^{2}\sum\limits_{n=-M}^{n=M}\sum\limits_{k=-M-n}^{k=M-n}r_{xx}[k]  " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:usehint"> (1)</a></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_199ece4161bcb7662515f993c1ca24a5.png" align="bottom" class="tex" alt="\left(\frac{1}{2M+1}\right)^{2}\sum\limits_{n=-M}^{n=M}\sum\limits_{k=-2M}^{2M}r_{xx}[k] " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_3209dd09953589843c62c97d7baa3911.png" align="bottom" class="tex" alt="- \left(\frac{1}{2M+1}\right)^{2}\sum\limits_{n=-M}^{n=M}(1-\delta(n-M))\sum\limits_{k=-2M}^{-M-n-1}r_{xx}[k]" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_3c579113779788f576074a579b80514e.png" align="bottom" class="tex" alt="- \left(\frac{1}{2M+1}\right)^{2}\sum\limits_{n=-M}^{n=M}(1-\delta(n+M))\sum\limits_{k=M-n+1}^{2M}r_{xx}[k]  " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:smwhl"> (2)</a></td></tr>
</table><br/>
In the previous relations the factors <img src="https://lysario.de/wp-content/cache/tex_266e916c3ce19a5649a07dd988e7ad96.png" align="bottom" class="tex" alt="(1-\delta(n-M))=1-\delta_{nM}" /> and <img src="https://lysario.de/wp-content/cache/tex_15b15eec897c8ace28946d4b1cc53c39.png" align="bottom" class="tex" alt="(1-\delta(n+M))=1-\delta_{n(-M)}" /> denote that for <img src="https://lysario.de/wp-content/cache/tex_093754a4657deb413176628af0be6247.png" align="bottom" class="tex" alt="n=M" /> the second summand, while for <img src="https://lysario.de/wp-content/cache/tex_5a37f5ba50abaff1fc41c8b87dab2892.png" align="bottom" class="tex" alt="n=-M" /> the third summand degenerates to zero. 
 
The three double sums can be further simplified:


<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_de180dd0158180fe708e9aa11872d161.png" align="bottom" class="tex" alt="\sum\limits_{n=-M}^{n=M}\sum\limits_{k=-2M}^{2M}r_{xx}[k] =(2M+1)\sum\limits_{k=-2M}^{2M}r_{xx}[k]. " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:smnd1"> (3)</a></td></tr>
</table><br/>
 
The second double sum can be written by: 

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_b5ba37e14cc35676ccfaa1c6d203399e.png" align="bottom" class="tex" alt="(1-\delta_{nM})\sum\limits_{n=-M}^{M}\sum\limits_{k=-2M}^{-M-n-1}r_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_4a33f504d780a2ddfebc0ac04bc949d6.png" align="bottom" class="tex" alt="\underbrace{0}_{n=M}+\underbrace{r_{xx}[-2M]}_{n=M-1} + " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_c42a8ecbeb8ef251cdc4173a20d909ad.png" align="bottom" class="tex" alt="+\underbrace{r_{xx}[-2M]+r_{xx}[-2M+1]}_{n=M-2} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_9efc22c952d8ba18f0f1d7a3e10741b3.png" align="bottom" class="tex" alt="+...+\underbrace{\sum\limits_{k=-2M}^{-1}r_{xx}[k]}_{n=-M} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_b3fe520c1fec030c5fb82812c9152ab1.png" align="bottom" class="tex" alt="2M r_{xx}[-2M] +" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_ef55e8b0feec659d86412a1551750fe0.png" align="bottom" class="tex" alt="+ (2M-1) r_{xx}[-2M+1] + " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_2a6ddcd33c2ec41d3961fa19b81d3d1b.png" align="bottom" class="tex" alt="+ ... + r_{xx}[-1]  " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_3c3bf44215e7f043c48984b18514ca43.png" align="bottom" class="tex" alt="\sum\limits_{u=-2M}^{0} |u|r_{xx}[u], " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:smnd2"> (4)</a></td></tr>
</table><br/>
while similar to the previous derivation the third double sum can be simplified to: 

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_0e7da7f7d8383cd538a8c4d55aa38b7d.png" align="bottom" class="tex" alt="(1-\delta_{n(-M)})\sum\limits_{n=-M}^{M}\sum\limits_{k=-2M}^{-M-n-1}r_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_c04a93614a798ab765890b2f6da0aea7.png" align="bottom" class="tex" alt="\underbrace{0}_{n=-M}+\underbrace{r_{xx}[-2M]}_{n=-M+1} + " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_d37c169cbfee62c2f0408837cc7a8c88.png" align="bottom" class="tex" alt="+\underbrace{r_{xx}[-2M]+r_{xx}[-2M+1]}_{n=-M+2} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_6d68dbbc5d985140e199b177c88782a9.png" align="bottom" class="tex" alt="+...+\underbrace{\sum\limits_{k=1}^{2M}r_{xx}[k]}_{n=M} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_07709c02aa6fdbc02fd77154fb10e5ae.png" align="bottom" class="tex" alt="2M r_{xx}[2M] " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_aa5fb06576916ff6ed70852e7625adad.png" align="bottom" class="tex" alt="+(2M-1) r_{xx}[2M-1] " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_2e495564a0f6a3f00c442225e0e958b8.png" align="bottom" class="tex" alt="+ ... + r_{xx}[1]  " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_3b041e4aa32b3a552c7ac74e8dcc392c.png" align="bottom" class="tex" alt="\sum\limits_{u=0}^{2M} |u|r_{xx}[u], " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:smnd3"> (5)</a></td></tr>
</table><br/>
Using (<a href="#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:smnd1">3</a>),(<a href="#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:smnd2">4</a>) and (<a href="#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:smnd3">5</a>) in (<a href="#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:smwhl">2</a>) we derive the relationship: 

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_1ef1b28b5ad1a096d209456fcc4b9f41.png" align="bottom" class="tex" alt="E\left\{\hat{\mu}_{x}^{2}\right\}" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_967a7b3547edaa3815d92924db58f948.png" align="bottom" class="tex" alt="\left(\frac{1}{2M+1}\right)\sum\limits_{k=-2M}^{2M}r_{xx}[k] " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_1bbf4430e01236349b29d7b91ebdbbab.png" align="bottom" class="tex" alt="-\left(\frac{1}{2M+1}\right)^{2} \left(\sum\limits_{u=-2M}^{0} |u|r_{xx}[u]+\sum\limits_{u=0}^{2M} |u|r_{xx}[u]\right) " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_8f3d9b1dfb237857f98c171926bbe659.png" align="bottom" class="tex" alt="\left(\frac{1}{2M+1}\right)\sum\limits_{k=-2M}^{2M}r_{xx}[k] -\left(\frac{1}{2M+1}\right)^{2} \sum\limits_{u=-2M}^{2M} |u|r_{xx}[u] " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_8d2f90d7c42d4b0cffd2c702a17f4a97.png" align="bottom" class="tex" alt="\frac{1}{2M+1} \sum\limits_{k=-2M}^{2M}\left(1-\frac{1}{2M+1}|k|\right)r_{xx}[k] " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:relautocorr"> (6)</a></td></tr>
</table><br/>
Observe that we could have replaced in (<a href="#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:usehint">1</a>) the sum <img src="https://lysario.de/wp-content/cache/tex_6a637b736fb66c3b0eadc5bf548acef2.png" align="bottom" class="tex" alt="\sum_{k=-M-n}^{M-n}r_{xx}[k]" /> by its equivalent form <img src="https://lysario.de/wp-content/cache/tex_3f5b522e59c60cd506578271c6d3c1f9.png" align="bottom" class="tex" alt="\sum_{k=-M}^{M}r_{xx}[k-n]" /> and applied relation <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17bib1">[1, eq. (3.64), p. 59]</a> of the hint. Then we would have come to the same result. In the approach taken in this solution we also derived the relation provided by the hint. 
Now relating the autocovariance to the autocorrelation function: 

<br/><table>
<tr><td><img src="https://lysario.de/wp-content/cache/tex_be7c8f1bd829ab6d117b7a827ab99d91.png" align="bottom" class="tex" alt="c_{xx}[k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_23cf961d22540c3ae331aad2cd2a439f.png" align="bottom" class="tex" alt="E\left\{(x[n]-\mu_{x})(x[n+k]-\mu_{x})\right\}" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_470c3a6b3ff20865fb940fc06cee07db.png" align="bottom" class="tex" alt="\ensuremath{E\left\{x[n]x[n+k]\right\}}-\ensuremath{E\left\{x[n]\right\}}\mu_{x}-\mu_{x}\ensuremath{E\left\{x[n+k]\right\}}+\mu_{x}^{2}" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_d73a97a1a82050231e1c3ed02af674e0.png" align="bottom" class="tex" alt="r_{xx}[k]-\mu_{x}^{2}" /></td><td></td></tr>
</table><br/>
and replacing the autocorrelation in (<a href="#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:relautocorr">6</a>) by <img src="https://lysario.de/wp-content/cache/tex_df64c393064cb5c4f3055c9655b56073.png" align="bottom" class="tex" alt="r_{xx}[k]=c_{xx}[k]+\mu_{x}^{2}" /> we obtain the variance of the sample mean as: 

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_0763198cc2f8e78f70e0b8390d01d976.png" align="bottom" class="tex" alt="E\left\{\hat{\mu}^{2}_{x}\right\}-\mu_{x}^{2}" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_f77d0ff6751906e26c2d2c77c26b1722.png" align="bottom" class="tex" alt="\frac{1}{2M+1}\sum\limits_{k=-2M}^{2M}\left(1-\frac{|k|}{2M+1}\right)(c_{xx}[k]+\mu_{x}^{2}) - \mu_{x}^{2} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_9aba1dc9147fcf40537b965cd62a66dd.png" align="bottom" class="tex" alt="\frac{1}{2M+1}\sum\limits_{k=-2M}^{k=2M}\left(1-\frac{|k|}{2M+1}\right)c_{xx}[k] " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_4c7756ffb50abc1f0f07db3096b04e47.png" align="bottom" class="tex" alt="+ \frac{\mu_{x}^{2}}{2M+1}\sum\limits_{k=-2M}^{2M}\left(1-\frac{|k|}{2M+1}\right) -\mu_{x}^{2} " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:varsamplmean1"> (7)</a></td></tr>
</table><br/>
Noting that <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17bib2">[2, p. 125]</a>: <img src="https://lysario.de/wp-content/cache/tex_2fb2960ed085d1a9ff41d277f4f1178f.png" align="bottom" class="tex" alt="\sum_{k=0}^{N}k=\frac{N (N+1)}{2}" /> we can simplify the second part of the previous equation by: 

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_7b86a8d8534a0cbbf58844790d5753a3.png" align="bottom" class="tex" alt="\sum\limits_{k=-2M}^{2M}\left(1-\frac{|k|}{2M+1}\right)" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_41440cf888557774fe07881385292b8e.png" align="bottom" class="tex" alt="4M+1 -\frac{2}{2M+1} \sum\limits_{k=0}^{2M}k " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_deb74eb16ad72ce33fa861a04318f94b.png" align="bottom" class="tex" alt="4M+1 -\frac{2}{2M+1}\frac{2M (2M+1)}{2} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_50855bcb53fc863b0f0660504215a53e.png" align="bottom" class="tex" alt="4M+1-2M " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_c221cd01d162dfa494e0f03152a96122.png" align="bottom" class="tex" alt="2M+1" /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eq8"> (8)</a></td></tr>
</table><br/>
So finally we can rewrite the variance of the sample (<a href="#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eqeq317:varsamplmean1">7</a>) mean as : 

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_0763198cc2f8e78f70e0b8390d01d976.png" align="bottom" class="tex" alt="E\left\{\hat{\mu}^{2}_{x}\right\}-\mu_{x}^{2}" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_d7a30028ac4ed6db6d87fb92eb2a4a14.png" align="bottom" class="tex" alt="\frac{1}{2M+1}\sum\limits_{k=-2M}^{k=2M}\left(1-\frac{|k|}{2M+1}\right)c_{xx}[k]" /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_7215ee9c7d9dc229d2921a40e899ec5f.png" align="bottom" class="tex" alt=" " /></td><td><img src="https://lysario.de/wp-content/cache/tex_75c74180f772cbf882e954654a7e424b.png" align="bottom" class="tex" alt="+ \frac{\mu_{x}^{2}}{2M+1}(2M+1)-\mu_{x}^{2} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_d7a30028ac4ed6db6d87fb92eb2a4a14.png" align="bottom" class="tex" alt="\frac{1}{2M+1}\sum\limits_{k=-2M}^{k=2M}\left(1-\frac{|k|}{2M+1}\right)c_{xx}[k]" /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17eq9"> (9)</a></td></tr>
</table><br/>
which is the relation of the sample mean used in <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-17bib1">[1, eq. (3.60), p. 58]</a>. QED.
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		<title>Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.16</title>
		<link>https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-16</link>
		<comments>https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-16#comments</comments>
		<pubDate>Fri, 13 Apr 2012 17:58:22 +0000</pubDate>
		<dc:creator>Panagiotis</dc:creator>
				<category><![CDATA[Kay: Modern Spectral Estimation, Theory and Application]]></category>
		<category><![CDATA[Solved Problems]]></category>
		<category><![CDATA[Complex Sinoid]]></category>
		<category><![CDATA[Random process]]></category>
		<category><![CDATA[Wide Sense Stationary]]></category>

		<guid isPermaLink="false">https://lysario.de/?p=839</guid>
		<description><![CDATA[In [1, p. 62 exercise 3.16] we are asked to show that the random process (1) , where is uniformly distributed on , is WSS by finding its mean and ACF. Using the same assumptions we are asked to repeat the exercise for a single complex sinusoid (2) Solution: The mean of the random process [...]]]></description>
				<content:encoded><![CDATA[In <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-16#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-16bib1">[1, p. 62 exercise 3.16]</a> we are asked to show that the random process 
<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_2601f4a14f4a21685c3519fd9aad5951.png" align="bottom" class="tex" alt="x[n]= A \cos(2\pi f_{0}n+\phi)" /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-16eq1"> (1)</a></td></tr>
</table><br/> , where <img src="https://lysario.de/wp-content/cache/tex_1ed346930917426bc46d41e22cc525ec.png" align="bottom" class="tex" alt="\phi" /> is uniformly distributed on <img src="https://lysario.de/wp-content/cache/tex_f771f932c7bc8de334071aa4c63281a0.png" align="bottom" class="tex" alt="(0,2\pi)" />, is WSS by finding its mean and ACF. Using the same assumptions we are asked to repeat the exercise for a single complex sinusoid 

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_b5f22b91c82ff2c5d69179471bd04442.png" align="bottom" class="tex" alt="x[n]=A\exp[j(2\pi f_{0}n+\phi)]." /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-16eq2"> (2)</a></td></tr>
</table><br/>
<span id="more-839"></span> 
<br /> <strong>Solution:</strong>
The mean of the random process is given by: 

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_73318898fec221d69b0aff4e5811d7a7.png" align="bottom" class="tex" alt="\mu_{x}[n]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_bf5033bb2646a01c29735d200c64c6c1.png" align="bottom" class="tex" alt="\int_{0}^{2\pi}A\cos(2\pi f_{0}n+\phi)\frac{1}{2\pi} d\phi " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_49226f2b7649baf325f8830498022eba.png" align="bottom" class="tex" alt="\frac{A}{2\pi}\Big[\sin(2\pi f_{0}n+\phi)\Big]_{\phi=0}^{\phi=2\pi} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_2e5751b7cfd7f053cd29e946fb2649a4.png" align="bottom" class="tex" alt="0 " /></td><td></td></tr>
</table><br/>
and is thus independent of the sampling variable. 
Let the p.d.f of <img src="https://lysario.de/wp-content/cache/tex_1ed346930917426bc46d41e22cc525ec.png" align="bottom" class="tex" alt="\phi" /> be <img src="https://lysario.de/wp-content/cache/tex_34dc939c37467e0aa504e549f153b95d.png" align="bottom" class="tex" alt="f_{pdf}(\phi)=\frac{1}{2\pi}" /> then the ACF of the random process is given by

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_a9ccd043d218f2a731575e10120e2cc6.png" align="bottom" class="tex" alt="r_{xx}[n,k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_1cf0768778603d3632ee055806d1401d.png" align="bottom" class="tex" alt="E\{x^{\star}[n]x[n+k]\} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_6e245aad02736b947e6c5047a914a526.png" align="bottom" class="tex" alt="\int_{0}^{2\pi}x^{\star}[n]x[n+k]f_{pdf}d\phi " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_aa45956ce2e51b6a07ac71eb2041b459.png" align="bottom" class="tex" alt="\int_{0}^{2\pi}A \cos(2\pi f_{0}n+\phi)A \cos(2\pi f_{0}(n+k)+\phi)\frac{1}{2\pi}d\phi  " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_1e695b4dd9e37c7e7a71d121e1511478.png" align="bottom" class="tex" alt="\frac{A^{2}}{2\pi}  \int_{0}^{2\pi}\cos(2\pi f_{0}n+\phi) \cos(2\pi f_{0}(n+k)+\phi)d\phi " /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-16eqeq316:autocorr"> (3)</a></td></tr>
</table><br/>
From the trigonometric formula <a href="https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-16#lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-16bib2">[2, p. 810 rel 21.2-12]</a>: 

<br/><table>
<tr><td colspan="3"><img src="https://lysario.de/wp-content/cache/tex_ab0a19f23468a3618f72e7c666cce274.png" align="bottom" class="tex" alt="\cos\mathfrak{A} \cos\mathfrak{B}=\frac{cos(\mathfrak{A-B})+cos(\mathfrak{A+B})}{2}" /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-16eq4"> (4)</a></td></tr>
</table><br/>
by setting <img src="https://lysario.de/wp-content/cache/tex_337f2e5f84a5f0318e838dfc2e555b2e.png" align="bottom" class="tex" alt="\mathfrak{A}=2\pi f_{0}(n+k)+\phi" /> and <img src="https://lysario.de/wp-content/cache/tex_fd3fea1716754c59240fabff57d0d82a.png" align="bottom" class="tex" alt="\mathfrak{B}=2\pi f_{0}n+\phi" /> we obtain because <img src="https://lysario.de/wp-content/cache/tex_fae66be44b2955b7f824de5016c30784.png" align="bottom" class="tex" alt="\int_{0}^{2\pi}cos(4\pi f_{0}n+ 2\pi f_{0}k+2 \phi)d\phi =0" />: 

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_a9ccd043d218f2a731575e10120e2cc6.png" align="bottom" class="tex" alt="r_{xx}[n,k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_e6067f6d5657c504f074f86e4d85629c.png" align="bottom" class="tex" alt="\frac{A^{2}}{2} \cos(2\pi f_{0}k)+ \frac{A^{2}}{4\pi}\int_{0}^{2\pi}cos(4\pi f_{0}n+ 2\pi f_{0}k+2 \phi)d\phi " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_0bc79bbb92459529b2e7338055c4dbab.png" align="bottom" class="tex" alt="\frac{A^{2}}{2}\cos(2\pi f_{0}k) " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_072e56c7dccf2213e22ca14c6c4ab92a.png" align="bottom" class="tex" alt="r_{xx}[k]." /></td><td></td></tr>
</table><br/>
Thus it is shown that the random process <img src="https://lysario.de/wp-content/cache/tex_d3baaa3204e2a03ef9528a7d631a4806.png" align="bottom" class="tex" alt="x[n]" /> is WSS. 
Repeating the same for a single complex sinusoid <img src="https://lysario.de/wp-content/cache/tex_595b02f1c2a03c9a28998fa3afacf385.png" align="bottom" class="tex" alt="x[n]=Ae^{j2\pi f_{0}n+\phi}" /> we obtain for the mean that it is equal to zero and independent of <img src="https://lysario.de/wp-content/cache/tex_7b8b965ad4bca0e41ab51de7b31363a1.png" align="bottom" class="tex" alt="n" />: 

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_73318898fec221d69b0aff4e5811d7a7.png" align="bottom" class="tex" alt="\mu_{x}[n]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_bf177921e1bb3730949af17df7a7d0d1.png" align="bottom" class="tex" alt="E\left\{x[n]\right\} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_1e005239d6014496866063139a14a37f.png" align="bottom" class="tex" alt="\int_{0}^{2\pi}\frac{A}{2\pi}e^{j(2\pi f_{0}n+\phi)}d\phi " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_5ef99fdceba9955fc192a1f856e5b703.png" align="bottom" class="tex" alt="\frac{A}{2\pi j}e^{j2\pi f_{0}n}\left[e^{j\phi}\right]_{0}^{2\pi} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_4b6c6e6106b8933b98cdbbb388209294.png" align="bottom" class="tex" alt="0." /></td><td><a name="lysario.desolved_problemssteven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-16eq5"> (5)</a></td></tr>
</table><br/>
The autocorrelation of the complex sinusoid is obtained by: 

<br/><table>
<tr> <td><img src="https://lysario.de/wp-content/cache/tex_a9ccd043d218f2a731575e10120e2cc6.png" align="bottom" class="tex" alt="r_{xx}[n,k]" /></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_11ce8b20d6e1c2ea048b2c969885fb6a.png" align="bottom" class="tex" alt="E\left\{x[n+k]x^{\star}[n]\right\} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_1256b66611c843505706c81a9f3e1773.png" align="bottom" class="tex" alt="\frac{A^{2}}{2\pi}\int_{0}^{2\pi}e^{j(2\pi f_{0}(n+k)+\phi)}e^{-j(2\pi f_{0}n+\phi)}d\phi " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_3db0485881ef3906483a0f7bf0353b9e.png" align="bottom" class="tex" alt="A^{2}e^{j2\pi f_{0}k} " /></td><td></td></tr>
<tr><td></td><td><img src="https://lysario.de/wp-content/cache/tex_43ec3e5dee6e706af7766fffea512721.png" align="bottom" class="tex" alt="=" /></td><td><img src="https://lysario.de/wp-content/cache/tex_e547e774a269350debc3a433e8c75f49.png" align="bottom" class="tex" alt="r_{xx}[k], " /></td><td></td></tr>
</table><br/>
which again is independent of <img src="https://lysario.de/wp-content/cache/tex_7b8b965ad4bca0e41ab51de7b31363a1.png" align="bottom" class="tex" alt="n" /> and thus the complex sinusoid is also wide sense stationary. 
We see that in this case, the real part of the complex sinusoid equals the double of the autocorrelation of the real sinusoid. 

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