Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 34 exercise 2.5
Author: Panagiotis
13
Jun
In exercise
[1, p. 34 exercise 2.5] we are asked to show that
the matrix
![\mathbf{R}=\sigma^2
\mathbf{I}+P_1\mathbf{e_1}\mathbf{e_1}^H+P_2\mathbf{e_2}\mathbf{e_2}^H](https://lysario.de/wp-content/cache/tex_ebb3c29b535afbfbcdb6f3d8498dcff2.png)
is circulant
when
![\mathbf{e_i}^H=\left[ {\begin{array}{*{20}c}
1 & {e^{-j2\pi f_i } } & {e^{-j2\pi (2f_i) } } & \cdots & {e^{-j2\pi (n - 1)f_i } }
\end{array}} \right], i=1,2](https://lysario.de/wp-content/cache/tex_38ce66bc9234a41f314618cd54d2cfd9.png)
.
Solution:
A matrix
![\mathbf{A}](https://lysario.de/wp-content/cache/tex_6c6404adc033dfed51422fdaf7fa0494.png)
is circulant when its elements are arranged in the following way:
Computing the dyad
![\mathbf{e_i}\mathbf{e_i}^H](https://lysario.de/wp-content/cache/tex_f3b6d6dfaa258d177f1c4b27e9190067.png)
(for
![i=1,2](https://lysario.de/wp-content/cache/tex_d24da99f976bd259b127a9bec9f3c4f8.png)
) results in
We observe that this is a
![n \times n](https://lysario.de/wp-content/cache/tex_50f17e5c11d610b19c0471830dc4dda1.png)
hermitian toeplitz matrix.
In order to be circulant the elements
![e^{j2\pi f_i}](https://lysario.de/wp-content/cache/tex_a3244d6661e607572be068a8c60d66be.png)
and
![e^{ - j2(n - 1)\pi f_i }](https://lysario.de/wp-content/cache/tex_d10b73a67c4dc79a4c57735bad1ce26a.png)
,
![e^{j2 \cdot 2\pi f_i}](https://lysario.de/wp-content/cache/tex_9998f34be642376a59d3f8234bf6af28.png)
and
![e^{ - j2(n - 2)\pi f_i }](https://lysario.de/wp-content/cache/tex_7ba1ac40b2012782811435ee0bb0814a.png)
, etc. have to be equal. Or in general that the following relation must hold:
Because
[1, p.22] ![f_i=\frac{l_i}{n}](https://lysario.de/wp-content/cache/tex_9623c667d34006fe1f5bbd7ec9d62c0a.png)
,
![\mathrm{i=1,2}](https://lysario.de/wp-content/cache/tex_4188db1d995453bc66c526c4a52b2848.png)
where
![l_i](https://lysario.de/wp-content/cache/tex_b39335b6584e8455ab4de3c86b439e21.png)
are distinct integers in the range
![[-n/2,n/2-1]](https://lysario.de/wp-content/cache/tex_24916734a0a5230271c84429761581eb.png)
for
![n](https://lysario.de/wp-content/cache/tex_7b8b965ad4bca0e41ab51de7b31363a1.png)
even or [-(n-1)/2, (n-1)/2] for
![n](https://lysario.de/wp-content/cache/tex_7b8b965ad4bca0e41ab51de7b31363a1.png)
odd we can rewrite the right hand side of (
2) as:
But
thus
which proofs that the dyad
![\mathbf{e_i}\mathbf{e_i}^H](https://lysario.de/wp-content/cache/tex_f3b6d6dfaa258d177f1c4b27e9190067.png)
is circulant because relation (
2) is satisfied. The matrix
![\mathbf{R}=\sigma^2 \mathbf{I}+P_1\mathbf{e_1}\mathbf{e_1}^H+P_2\mathbf{e_2}\mathbf{e_2}^H](https://lysario.de/wp-content/cache/tex_2f86b5d706c8273bb492e48953bb5338.png)
is circulant, as the product of a circulant matrix with a scalar is itself a circulant matrix, the identity matrix is a special form of a circulant matrix, and also the sum of circulant matrices is also a circulant matrix. QED.
[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.
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