Lysario - by Panagiotis Chatzichrisafis » Solved Problems https://lysario.de "ούτω γάρ ειδέναι το σύνθετον υπολαμβάνομεν, όταν ειδώμεν εκ τίνων και πόσων εστίν ..." Thu, 15 Apr 2021 17:28:24 +0000 de-DE hourly 1 Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 94 exercise 4.2 https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2 https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2#comments Tue, 24 Jun 2014 18:54:07 +0000 Panagiotis https://lysario.de/?p=1062 [1, p. 94 exercise 4.2] we are asked to consider the estimator
\hat{P}_{AVPER}(0)=\frac{1}{N}\sum\limits_{m=0}^{N-1}\hat{P}_{PER}^{m}(0) (1)

where
\hat{P}_{PER}^{m}(0)= x^{2}[m] (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 periodograms for each block are averaged. We are asked to find the mean and variance of \hat{P}_{AVPER}(0) and compare the result to that obtained in [2].
Solution: 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 U = \sum_{m=0}^{N-1}\frac{x^{2}[n]}{\sigma_{x}^{2}} is distributed according to a \chi^{2}(N) distribution [3, p. 682] with mean E\left\{U\right\}=N and variance Var\left\{U\right\}=2N. The average periodogram is obtained by \hat{P}_{PER}(0) =\frac{1}{N} U \cdot \sigma_{x}^{2} and thus the mean and the average is obtained by
E\left\{\hat{P}_{PER}(0)\right\} = \sigma_{x}^{2} E\left\{U\right\} = \sigma_{x}^{2}
Var\left\{\hat{P}_{PER}(0)\right\} = \sigma_{x}^{2} Var\left\{U\right\} = 2 \sigma_{x}^{2}.

Thus taking the average periodogram of one sample section has no statistical advantage for the result of the estimator. ]]> https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-2/feed 0 Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 94 exercise 4.1 https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-1 https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-1#comments Sat, 01 Feb 2014 16:40:19 +0000 Panagiotis https://lysario.de/?p=1050 [1, p. 94 exercise 4.1] we will show that the periodogram is an inconsistent estimator by examining the estimator at f=0, or
\hat{P}_{PER}(0)=\frac{1}{N}\left(\sum\limits_{n=0}^{N-1}x[n]\right)^{2}. (1)

If x[n] is real white Gaussian noise process with PSD
P_{xx}(f)=\sigma_{x}^{2} (2)

we are asked to find the mean an variance of hat{P}_{PER}(0). We are asked if the variance converge to zero as N \rightarrow \infty. The hint provided within the exercise is to note that
\hat{P}_{PER}(0)= \sigma_{x}^{2} \left(\sum\limits_{n=0}^{N-1}\frac{x[n]}{\sigma_{x}\sqrt{N}}\right)^{2} (3)

where the quantiiy inside the parenthesis is \sim N(0,1) Solution: Because Y = \sum_{n=0}^{N-1}\frac{x[n]}{\sigma_{x}\sqrt{N}} is distributed according to a normal distribution \sim N(0,1) the squared variable Y^{2}=\frac{\hat{P}_{PER}(0)}{\sigma_{x}^{2}} is distributed according to a \chi^{2}(1) with mean E\left\{Y\right\}=1 and variance Var\left\{Y\right\}= 2. From the previous relations we obtain the mean and the variance of the periodogram \hat{P}_{PER}(0) as
E\left\{\hat{P}_{PER}(0)\right\} = \sigma_{x}^{2} E\left\{Y\right\} = \sigma_{x}^{2}
Var\left\{\hat{P}_{PER}(0)\right\} = \sigma_{x}^{2} Var\left\{Y\right\} = 2 \sigma_{x}^{2}.

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. ]]> https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-94-exercise-4-1/feed 0 Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.19 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 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 Fri, 17 May 2013 21:09:39 +0000 Panagiotis https://lysario.de/?p=925 [1, p. 62 exercise 3.19] we are asked to find for the multiple sinusoidal process
x[n]=\sum\limits_{i=1}^{P}A_{i}\cos(2\pi f_{i}n+\phi_{i})

the ensemble ACF and the temporal ACF as M\rightarrow \infty , where the \phi_{i}‘s are all uniformly distributed random variables on [0, 2 \pi) and independent of each other. We are also asked to determine if this random process is autocorrelation ergodic.
Solution: We note that the joint p.d.f of the uniformly random variables \phi_{i} is given by f(\boldsymbol{\phi})=\prod_{i=1}^{P}f(\phi_{i})=\frac{1}{(2\pi)^{P}}, with domain \mathbb{D}= \mathbb{A}^{P}, \; \mathbb{A} =[0, 2 \pi) we can proceed to calculate the ensemble autocorrelation function which is defined as:
r_{xx}[k]=E \left\{ x^{\star}[n] x[n+k] \right\}
=\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}
=\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}
+ \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}
=\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}
+ \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}
=\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}

As in exercise [2] we can use the trigonometric relation [3, p. 810] :
2\cos\frak{A}\cos\frak{B} = \cos(\frak{A-B}) + \cos(\frak{A+B}) (1)

(\frak{A}=2\pi f_{i}(n+k)+\phi_{i}, \frak{B}=2\pi f_{i}n+\phi_{i}) to further simplify the expression for the ensemble autocorrelation function:
r_{xx}[k]=\sum\limits_{i=1}^{P}\int_{0}^{2\pi} \frac{A_{i}^{2}}{(4\pi)}\cos(2\pi f_{i}k)d\phi_{i}
+\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}
=\sum\limits_{i=1}^{P}\int_{0}^{2\pi} \frac{A_{i}^{2}}{(4\pi)}\cos(2\pi f_{i}k)d\phi_{i}
+\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}
=\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2}\cos(2\pi f_{i}k) (2)

Having obtained the ensemble autocorrelation function we can proceed to obtain the temporal autocorrelation function, which we hope will be a good approximation \hat{r}_{xx}[k] to the ensemble autocorrelation function r_{xx}[k]. By definition the temporal autocorrelation function is given by
\hat{r}_{xx}[k]=\frac{1}{2M+1} \sum\limits_{n=-M}^{M} x[n+k]x^{\star}[n]
=\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
\cdot \left(\sum\limits_{\ell=1}^{P}A_{\ell}\cos(2\pi f_{\ell}n+\phi_{\ell}) \right)\Bigg]
=\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})
+ \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}) (3)

The second sum of the previous relation can be simplified by one of the derived formulas of [2], for e^{j2\pi f_{0}}\neq 1 :
\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) =
\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})}

Thus the temporal autocorrelation function may be expressed, as long as e^{j2\pi f_{i}}\neq 1, \; i=1,...,P as:
\hat{r}_{xx}[k]=\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})
+ \sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2} \cos(2\pi f_{i}k)
+\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})} (4)

In order to simplify the relation further, especially the first sum, we will again use (1), with \frak{A} =2\pi f_{i}(n+k)+\phi_{i} and \frak{B} = 2\pi f_{\ell}n+\phi_{\ell}, and thus :
\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}) =
\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}) +
\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}) (5)

We see that both distinct parts are of the form \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 [3, p. 810]:
\cos(\frak{A+B})=\cos\frak{A}\cos\frak{B}-\sin\frak{A}\sin\frak{B} (6)

with \frak{A}= 2\pi \frak{f}_{1} n and \frak{B}=2\pi \frak{f}_{2} k + \Phi we can simplify the relation as:
\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)
-\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}}}
=\cos(2\pi \frak{f}_{2} k + \Phi)\sum\limits_{n=-M}^{M}\cos(2\pi \frak{f}_{1} n)

Furthermore it was shown in [2] that \sum_{n=-M}^{M}\cos(2\pi (2 f_{0}))=\frac{\sin(\pi (2 f_{0})(2M+1))}{\sin(\pi(2f_{0})} , for e^{j2\pi 2f_{0}}\neq 1, so we can further simplify the previous relation by:
\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)
= \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 (7)

Thus finally we can simplify (5) by:
\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}) =
\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}))} +
\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}))} (8)

So finally the temporal autocorrelation (4) can be reduced using (8) to the following formula:
\hat{r}_{xx}[k]=\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}))}
+\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}))}
+ \sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2} \cos(2\pi f_{i}k)
+\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})} (9)

Rearranging the terms on the right we see that the temporal autocorrelation function equals the ensemble autocorrelation function with an additional error \epsilon, which we have derived for the the case when e^{(j2\pi f_{i})}\neq 1 \; , i=1,...,P :
\hat{r}_{xx}[k]=\sum\limits_{i=1}^{P} \frac{A_{i}^{2}}{2} \cos(2\pi f_{i}k)
\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}))}
+\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}))}
+\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})}
=r_{xx}[k] +\epsilon

At once we see again that when M\rightarrow \infty that the error goes to zero \epsilon \rightarrow 0. Thus the random process of the sum of sinusoids is also autocorrelation ergodic as long as e^{(j2\pi f_{i})}\neq1 \; , i=1,...,P . It is also easy to recognize, by using the argumentation of the previous exercise [2] , that this is not true if e^{(j2\pi f_{i})}\neq1 \; , i=1,...,P does not hold. In this case parts of the the error of equation (3) are proportional to M as was already observed in [2] . QED. ]]> 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/feed 0 Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.18 https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2 https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2#comments Fri, 24 Aug 2012 15:37:26 +0000 Panagiotis https://lysario.de/?p=960 [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]:
\hat{r}_{xx}[k]=\frac{1}{2M+1}\sum\limits_{n=-M}^{M}x[n]x[n+k]

as M\rightarrow \infty. 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 3.16] is given by 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 [2]) r_{xx}[k]=\frac{A^{2}}{2} \cos(2\pi f_{0}k). The temporal autocorrelation is equal to :
\hat{r}_{xx}[k]=\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)

The previous relation can be furthermore simplified by the property of trigonometric functions [3, p. 810] :
2\cos\frak{A}\cos\frak{B}=\cos(\frak{A-B})+\cos(\frak{A+B}) (1)

and thus the autocorrelation can be written as:
\hat{r}_{xx}[k]=\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)
=\frac{A^{2}}{2(2M+1)} (2M+1) \cos(2\pi f_{0}k)
+ \frac{A^{2}}{2(2M+1)}\sum\limits_{n=-M}^{M}\cos(2\pi f_{0}(2n+k)+2\phi)
=\frac{A^{2}}{2} \cos(2\pi f_{0}k)
+ \frac{A^{2}}{2(2M+1)}\sum\limits_{n=-M}^{M}\cos(4\pi f_{0}n+2\pi f_{0}k+2\phi)

Again it is possible to further simplify the previous relation, this time by using the following trigonometric formula [3, p. 810]:
\cos(\frak{A+B})=\cos\frak{A}\cos\frak{B}-\sin\frak{A}\sin\frak{B} (2)


\hat{r}_{xx}[k]=\frac{A^{2}}{2} \cos(2\pi f_{0}k)
+ \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)
- \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)

Because the sinus function is odd \sin(-x)=-\sin(x) the symmetric sum \sum_{n=-M}^{M}\sin(4\pi f_{0}n) equals zero. Generally the same is not true for the cosine sum. For f_{0}=1 for example the sum \sum_{n=-M}^{M}\cos(4\pi f_{0}n) equals 2M+1 and thus even for M\rightarrow \infty the temporal autocorrelation function will be in error to the true autocorrelation function. In general it is true that - \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 \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. f_{0}=1 the temporal autocorrelation doesn’t even converge for large M. That is:
\hat{r}_{xx}[k]=\frac{A^{2}}{2} \cos(2\pi f_{0}k)
+ \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)
=r_{xx}[k]+\epsilon.

While it is true that there are frequencies f_{0} for which the temporal autocorrelation doesn’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 \sum_{n=-M}^{M}\cos(4\pi f_{0}n) can be further simplified, by using \cos(4\pi f_{0}n) =\Re\{e^{(j4\pi f_{0}n)}\} , and observing that the resulting sum is a geometric progression ([3, p. 7] \sum\limits_{j=0}^{n}a_{0}r^{j}=a_{0}\frac{1-r^{n+1}}{1-r}, \; r < 1 ) with a_{0}=1, r_{0}=e^{j4\pi f_{0}} :
\sum\limits_{n=-M}^{M}\cos(4\pi f_{0}n)=\Re\{\sum\limits_{n=-M}^{M}e^{j4\pi f_{0}n}\}
=\Re\{e^{-j4\pi f_{0}M}\sum\limits_{n=0}^{2M}e^{j4\pi f_{0}n}\}
=\Re\{e^{-j4\pi f_{0}M}\frac{1-e^{j4\pi f_{0}(2M+1)}}{1-e^{j4\pi f_{0}}}\}
=\Re\{\underbrace{e^{-j4\pi f_{0}M}\frac{e^{j2\pi f_{0}(2M+1)}}{e^{j2\pi f_{0}}}}_{=1}\cdot
\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}}}\}
=\Re\{\frac{\sin(2 \pi f_{0}(2M+1))}{\sin(2\pi f_{0})}\}
=\frac{\sin(2 \pi f_{0}(2M+1))}{\sin(2\pi f_{0})}

Using this identity the error to the true autocorrelation function can be rewritten as:
\epsilon=\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
|\epsilon|\leq|\frac{A^{2}}{2(2M+1)}| \Rightarrow
\lim_{M\rightarrow \infty} |\epsilon|=0,

Thus by only imposing that e^{j4\pi f_{0}}\neq 1 the process is autocorrelation ergodic. ]]> https://lysario.de/solved_problems/steven-m-kay-modern-spectral-estimation-theory-and-applicationsp-62-exercise-3-18-2/feed 0 Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.17 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 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 Thu, 19 Apr 2012 20:43:42 +0000 Panagiotis https://lysario.de/?p=845 [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
\frac{1}{2M+1}\sum\limits_{n=-M}^{n=M}x[n]

is given by [1, eq. (3.60), p. 58]. For the case when 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 [1, eq. (3.64), p. 59].
Solution: Let’s reproduce the corresponding equations cited in the problem statement, so [1, eq. (3.60), p. 58] is given by
\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.

while [1, eq. (3.64), p. 59] is given by:
\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]

First we note that the mean of the sample mean converges to the true mean of the WSS random process x[n]:
E\left\{ \hat{\mu}_{x} \right\}=\frac{1}{2M+1}\sum\limits_{n=-M}^{M} E\left\{ x[n]\right\}
=\frac{1}{2M+1} (2M+1)\mu_{x}
=\mu_{x}.

From this we can derive the variance of the \hat{\mu}_{x} as:
E\left\{ (\hat{\mu}_{x}-\mu_{x})^{2}\right\}=E\left\{\hat{\mu}_{x}^{2}\right\}-2\mu_{x} E\left\{\hat{\mu}_{x}\right\}+\mu_{x}^{2}
=E\left\{\hat{\mu}_{x}^{2}\right\}-\mu_{x}^{2}

The squared sample mean can be written as:
\hat{\mu}_{x}^{2}=\left(\frac{1}{2M+1}\right)^{2} \sum\limits_{n=-M}^{n=M} x[n] \sum\limits_{l=-M}^{l=M}x[l]
=\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
=\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].

From the previous relation we can derive that the mean squared sample mean is given by:
E\left\{\hat{\mu}_{x}^{2}\right\}=\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\}}
=\left(\frac{1}{2M+1}\right)^{2}\sum\limits_{n=-M}^{n=M}\sum\limits_{k=-M-n}^{k=M-n}r_{xx}[k] (1)
=\left(\frac{1}{2M+1}\right)^{2}\sum\limits_{n=-M}^{n=M}\sum\limits_{k=-2M}^{2M}r_{xx}[k]
- \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]
- \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] (2)

In the previous relations the factors (1-\delta(n-M))=1-\delta_{nM} and (1-\delta(n+M))=1-\delta_{n(-M)} denote that for n=M the second summand, while for n=-M the third summand degenerates to zero. The three double sums can be further simplified:
\sum\limits_{n=-M}^{n=M}\sum\limits_{k=-2M}^{2M}r_{xx}[k] =(2M+1)\sum\limits_{k=-2M}^{2M}r_{xx}[k]. (3)

The second double sum can be written by:
(1-\delta_{nM})\sum\limits_{n=-M}^{M}\sum\limits_{k=-2M}^{-M-n-1}r_{xx}[k]=\underbrace{0}_{n=M}+\underbrace{r_{xx}[-2M]}_{n=M-1} +
+\underbrace{r_{xx}[-2M]+r_{xx}[-2M+1]}_{n=M-2}
+...+\underbrace{\sum\limits_{k=-2M}^{-1}r_{xx}[k]}_{n=-M}
=2M r_{xx}[-2M] +
+ (2M-1) r_{xx}[-2M+1] +
+ ... + r_{xx}[-1]
=\sum\limits_{u=-2M}^{0} |u|r_{xx}[u], (4)

while similar to the previous derivation the third double sum can be simplified to:
(1-\delta_{n(-M)})\sum\limits_{n=-M}^{M}\sum\limits_{k=-2M}^{-M-n-1}r_{xx}[k]=\underbrace{0}_{n=-M}+\underbrace{r_{xx}[-2M]}_{n=-M+1} +
+\underbrace{r_{xx}[-2M]+r_{xx}[-2M+1]}_{n=-M+2}
+...+\underbrace{\sum\limits_{k=1}^{2M}r_{xx}[k]}_{n=M}
=2M r_{xx}[2M]
+(2M-1) r_{xx}[2M-1]
+ ... + r_{xx}[1]
=\sum\limits_{u=0}^{2M} |u|r_{xx}[u], (5)

Using (3),(4) and (5) in (2) we derive the relationship:
E\left\{\hat{\mu}_{x}^{2}\right\}=\left(\frac{1}{2M+1}\right)\sum\limits_{k=-2M}^{2M}r_{xx}[k]
-\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)
=\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]
=\frac{1}{2M+1} \sum\limits_{k=-2M}^{2M}\left(1-\frac{1}{2M+1}|k|\right)r_{xx}[k] (6)

Observe that we could have replaced in (1) the sum \sum_{k=-M-n}^{M-n}r_{xx}[k] by its equivalent form \sum_{k=-M}^{M}r_{xx}[k-n] and applied relation [1, eq. (3.64), p. 59] 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:
c_{xx}[k]=E\left\{(x[n]-\mu_{x})(x[n+k]-\mu_{x})\right\}
=\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}
=r_{xx}[k]-\mu_{x}^{2}

and replacing the autocorrelation in (6) by r_{xx}[k]=c_{xx}[k]+\mu_{x}^{2} we obtain the variance of the sample mean as:
E\left\{\hat{\mu}^{2}_{x}\right\}-\mu_{x}^{2}=\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}
=\frac{1}{2M+1}\sum\limits_{k=-2M}^{k=2M}\left(1-\frac{|k|}{2M+1}\right)c_{xx}[k]
+ \frac{\mu_{x}^{2}}{2M+1}\sum\limits_{k=-2M}^{2M}\left(1-\frac{|k|}{2M+1}\right) -\mu_{x}^{2} (7)

Noting that [2, p. 125]: \sum_{k=0}^{N}k=\frac{N (N+1)}{2} we can simplify the second part of the previous equation by:
\sum\limits_{k=-2M}^{2M}\left(1-\frac{|k|}{2M+1}\right)=4M+1 -\frac{2}{2M+1} \sum\limits_{k=0}^{2M}k
=4M+1 -\frac{2}{2M+1}\frac{2M (2M+1)}{2}
=4M+1-2M
=2M+1 (8)

So finally we can rewrite the variance of the sample (7) mean as :
E\left\{\hat{\mu}^{2}_{x}\right\}-\mu_{x}^{2}=\frac{1}{2M+1}\sum\limits_{k=-2M}^{k=2M}\left(1-\frac{|k|}{2M+1}\right)c_{xx}[k]
+ \frac{\mu_{x}^{2}}{2M+1}(2M+1)-\mu_{x}^{2}
=\frac{1}{2M+1}\sum\limits_{k=-2M}^{k=2M}\left(1-\frac{|k|}{2M+1}\right)c_{xx}[k] (9)

which is the relation of the sample mean used in [1, eq. (3.60), p. 58]. QED. ]]> 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/feed 0 Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.16 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 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 Fri, 13 Apr 2012 17:58:22 +0000 Panagiotis https://lysario.de/?p=839 [1, p. 62 exercise 3.16] we are asked to show that the random process
x[n]= A \cos(2\pi f_{0}n+\phi) (1)

, where \phi is uniformly distributed on (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
x[n]=A\exp[j(2\pi f_{0}n+\phi)]. (2)


Solution: The mean of the random process is given by:
\mu_{x}[n]=\int_{0}^{2\pi}A\cos(2\pi f_{0}n+\phi)\frac{1}{2\pi} d\phi
=\frac{A}{2\pi}\Big[\sin(2\pi f_{0}n+\phi)\Big]_{\phi=0}^{\phi=2\pi}
=0

and is thus independent of the sampling variable. Let the p.d.f of \phi be f_{pdf}(\phi)=\frac{1}{2\pi} then the ACF of the random process is given by
r_{xx}[n,k]=E\{x^{\star}[n]x[n+k]\}
=\int_{0}^{2\pi}x^{\star}[n]x[n+k]f_{pdf}d\phi
=\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
=\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 (3)

From the trigonometric formula [2, p. 810 rel 21.2-12]:
\cos\mathfrak{A} \cos\mathfrak{B}=\frac{cos(\mathfrak{A-B})+cos(\mathfrak{A+B})}{2} (4)

by setting \mathfrak{A}=2\pi f_{0}(n+k)+\phi and \mathfrak{B}=2\pi f_{0}n+\phi we obtain because \int_{0}^{2\pi}cos(4\pi f_{0}n+ 2\pi f_{0}k+2 \phi)d\phi =0:
r_{xx}[n,k]=\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
=\frac{A^{2}}{2}\cos(2\pi f_{0}k)
=r_{xx}[k].

Thus it is shown that the random process x[n] is WSS. Repeating the same for a single complex sinusoid x[n]=Ae^{j2\pi f_{0}n+\phi} we obtain for the mean that it is equal to zero and independent of n:
\mu_{x}[n]=E\left\{x[n]\right\}
=\int_{0}^{2\pi}\frac{A}{2\pi}e^{j(2\pi f_{0}n+\phi)}d\phi
=\frac{A}{2\pi j}e^{j2\pi f_{0}n}\left[e^{j\phi}\right]_{0}^{2\pi}
=0. (5)

The autocorrelation of the complex sinusoid is obtained by:
r_{xx}[n,k]=E\left\{x[n+k]x^{\star}[n]\right\}
=\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
=A^{2}e^{j2\pi f_{0}k}
=r_{xx}[k],

which again is independent of 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. ]]> 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/feed 0 Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.15 https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-15 https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-15#comments Sat, 24 Mar 2012 16:03:14 +0000 Panagiotis https://lysario.de/?p=826 [1, p. 62 exercise 3.15] it is requested to verify the ACF and PSD relationships given in [1, p. 53 eq. (3.50)] and [1, p. 54 eq. (3.51)].
Solution: Starting from the first relationship of [1, p. 53 eq. (3.50)] and the definition of the cross correlation function we can derive
r_{xy}[n,k]=E\{x^{\ast}[n]y[n+k]\}
=E\{x^{\ast}[n]\sum\limits_{l}x[n+k-l]h[l]\}
=\sum\limits_{l}E\{x^{\ast}[n]x[n+k-l]\}h[l]

Using the starting assumption that x[n] is WSS we can derive the final result, that the cross correlation is also independent of the observation instance n and depends only on the lag k:
r_{xy}[n,k]=r_{xy}[k]=\sum\limits_{l}r_{xx}[k-l]h[l]

Similar the second relation can be obtained by:
r_{yx}[n,k]=E\{y^{\ast}[n]x[n+k]\}
=E\{\sum\limits_{l}x^{\ast}[n-l]h^{\ast}[l]x[n+k]\}
=\sum\limits_{l}E\{x^{\ast}[n-l]x[n+k]\}h^{\ast}[l]
=\sum\limits_{l}r_{xx}[k+l]h^{\ast}[l]

and setting l^{\prime} = -l:
r_{yx}[n,k]=\sum\limits_{l^{\prime}}r_{xx}[k-l^{\prime}]h^{\ast}[-l^{\prime}]
=r_{xx}[k]\star h^{\ast}[-k] =r_{yx}[k]

The third equation can be derived by:
r_{yy}[n,k]=E\{y^{\ast}[n]y[n+k]\}
=E\{\sum\limits_{l}h^{\ast}[l]x^{\ast}[n-l]\sum\limits_{\lambda}h[\lambda]x[n+k-\lambda] \}
=\sum\limits_{l}\sum\limits_{\lambda}h^{\ast}[l]h[\lambda]E\{x^{\ast}[n-l]x[n+k-\lambda] \}
=\sum\limits_{l}h^{\ast}[l]\sum\limits_{\lambda} h[\lambda]r_{xx}[k+l-\lambda] (1)

By setting \rho[k+l]=\sum_{\lambda} h[\lambda]r_{xx}[k+l-\lambda] =h[k+l]\ast r_{xx}[k+l] we can rewrite (1) by
r_{yy}[n,k]=\sum\limits_{l}h^{\ast}[l] \rho[k+l]
=h^{\ast}(-k)\star \rho(k)
=h^{\ast}(-k)\star h[k] \ast r_{xx}[k] (2)

which proves the last equation of [1, p. 53 eq. (3.50)] . We can now proceed to prove the relations for [1, p. 54 eq. (3.51)]
P_{xy}(z)=\mathcal{Z}\{r_{xy}[k]\}
=\mathcal\{r_{xx}[k]\star h[k]\}
=H(z)R_{xx}(z) (3)

The second relation P_{yx}(z) can similar be proven by:
P_{yx}(z)=\mathcal{Z}\{r_{yx}[k]\}
=\mathcal{Z}\{h^{\ast}[-k]\star r_{xx}[k]\}
=\mathcal{Z}\{h^{\ast}[-k]\}R_{xx}(z)
=R_{xx}(z)\sum\limits_{k=-\infty}^{\infty}h^{\ast}[-k]z^{-k}

By change of variables z=(u^{-1})^{\ast} and k=-l for the sum of the previous equation can be written as:
P_{yx}(z)=R_{xx}(z)\sum\limits_{l=-\infty}^{\infty}h^{\ast}[l](u^{-l})^{\ast}
=R_{xx}(z)(\sum\limits_{l=-\infty}^{\infty}h[l]u^{-l})^{\ast}
=R_{xx}(z)(H(u))^{\ast}
=R_{xx}(z) \left[H(u=\frac{1}{z^{\ast}})\right]^{\ast}
=R_{xx}(z) \left[H(\frac{1}{z^{\ast}})\right]^{\ast}

The last equation proves the second part of [1, p. 54 eq. (3.51)]. By similar reasoning the third relation can also be shown to be
P_{yy}(z)=\mathcal{Z}\{r_{yy}[k]\}
=\mathcal{Z}\{h^{\ast}[-k]\star h[k]\star r_{xx}[k]\}
=\mathcal{Z}\{h^{\ast}[-k]\}\mathcal{Z}\{ h[k]\}\mathcal{Z}\{r_{xx}[k]\}
=\left[H(\frac{1}{z^{\ast}})\right]^{\ast}H(z)R_{xx}(z),

which concludes the proof. QED. ]]> https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-15/feed 0 Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.14 https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-14 https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-14#comments Sun, 04 Mar 2012 07:34:41 +0000 Panagiotis https://lysario.de/?p=811 [1, p. 62 exercise 3.14] it is requested to proof prove that the autocorrelation matrix given by [1, eq. 3.46] is also positive semidefinite. This shall be done by usage of the definition of the semidefinite property of the ACF in [1, eq. 3.45].
Solution: A matrix \mathbf{A} is positive semidefinite if \mathbf{x}^{H}\mathbf{A}\mathbf{x}\geq 0. And the autocorrelation matrix given by ([1, eq. 3.46]) is:
\mathbf{R}_{xx}=E\left\{\mathbf{x}\mathbf{x}^{H}\right\} (1)
=\left[\begin{array}{cccc}r_{xx}[0] & r_{xx}[-1]& \cdots&r_{xx}[-(M-1)] \\r_{xx}[1] &r_{xx}[0] &\cdots&r_{xx}[-(M-2)] \\ \vdots &\vdots & \ddots& \vdots \\ r_{xx}[M-1]&r_{xx}[M-2]&... &r_{xx}[0]\end{array}\right] (2)

Let \mathbf{\alpha}=\left[\begin{array}{ccc} \alpha[0]&\cdots & \alpha[M-1]\end{array}\right]^{T} and \mathbf{x}=\left[\begin{array}{ccc} x[0]&\cdots & x[M-1]\end{array}\right]^{T} then from [1, eq. 3.45] we obtain:
E\left\{\left| \sum\limits_{n=0}^{M-1}\alpha^{\ast}[n]x[n] \right|^{2}\right\}\geq0
E\left\{\left| \mathbf{\alpha}^{H}\mathbf{x} \right|^{2}\right\}\geq0
E\left\{ \mathbf{\alpha}^{H}\mathbf{x}\mathbf{x}^{H}\mathbf{\alpha} \right\}\geq0
\mathbf{\alpha}^{H}E\left\{\mathbf{x}\mathbf{x}^{H}\right\}\mathbf{\alpha}\geq0
\mathbf{\alpha}^{H}\mathbf{R}_{xx}\mathbf{\alpha}\geq0

Thus \mathbf{R}_{xx} is positive semidefinite. QED. ]]> https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-14/feed 0 Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.13 https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-13 https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-13#comments Thu, 23 Feb 2012 17:14:49 +0000 Panagiotis https://lysario.de/?p=803 [1, p. 62 exercise 3.13] it is requested to prove that the PSD of a real WSS random process is a real even function of frequency. Solution: The PSD of a WSS process is given by the Fourier transform of its autocorrelation function:
P_{xx}(f)=\sum\limits_{k=-\infty}^{\infty}r_{xx}[k]e^{-j2\pi f k}. (1)

We note that for a real process x[n], r_{xx}[k] will be also real thus we obtain the relations:
r^{\ast}_{xx}[k]=r_{xx}[-k]=r_{xx}[k]. (2)

First let us prove that for a real WSS this function is even. Having the previous relations in mind:
P_{xx}(-f)=\sum\limits_{k=-\infty}^{\infty}r_{xx}[k]e^{j2\pi f k}
=\sum\limits_{k=-\infty}^{\infty}r_{xx}[-k]e^{j2\pi f k}

Let u=-k then
P_{xx}(-f)=\sum\limits_{u=-\infty}^{\infty}r_{xx}[u]e^{-j2\pi f u}
=P_{xx}(f) (3)

Thus we have shown that the PSD is symmetric P_{xx}(f)=P_{xx}(-f). The next step is to show that the PSD is also real. We know that the real part of a complex function is given as one half of the sum of the complex function with its complex conjugate form, that is Re\left\{P_{xx}(f)\right\}=\frac{1}{2}\left(P_{xx}(f)+P^{\ast}_{xx}(f)\right). Let’s determine the complex conjugate of the PSD:
P_{xx}^{\ast}(f)=\sum\limits_{k=-\infty}^{\infty}r^{\ast}_{xx}[k]e^{j2\pi f k}
=\sum\limits_{k=-\infty}^{\infty}r_{xx}[-k]e^{j2\pi f k}=P_{xx}(-f)=P_{xx}(f) (4)

Again relation (2) was used in order to obtain the last result. The real part of the PSD is thus given by
Re\left\{P_{xx}(f)\right\}=\frac{1}{2}\left(P_{xx}(f)+P^{\ast}_{xx}(f)\right)
=Re\left\{P_{xx}(f)\right\}=\frac{1}{2}\left(P_{xx}(f)+P_{xx}(f)\right)
=P_{xx}(f),

because of (4). So the real part of the PSD P_{xx}(f) is the PSD itself, or stating it in another way: the PSD of a WSS process is real. Thus together with (3) we have shown that the PSD of a real WSS random process is a real even function of frequency. QED. ]]> https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-13/feed 0 Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 62 exercise 3.12 https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-12 https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-12#comments Mon, 20 Feb 2012 17:51:12 +0000 Panagiotis https://lysario.de/?p=796 [1, p. 62 exercise 3.12] we are asked to prove that for a WSS random process
r_{xx}[-k]=r^{\ast}[k]. (1)


Solution: Using the definition of the autocorrelation [1, eq. (3.40,p52)]:
r_{xx}^{\ast}[k]=E\left\{x[n+k]x[n]^{\ast}\right\}^{\ast}
=E\left\{x[n+k]^{\ast}x[n] \right\}
=r_{xx}[-k]

QED. ]]> https://lysario.de/solved_problems/steven-m-kay-%e2%80%9cmodern-spectral-estimation-%e2%80%93-theory-and-applications%e2%80%9dp-62-exercise-3-12/feed 0