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I have been given a vector: \begin{equation} v= \:\begin{pmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\end{pmatrix} \end{equation} my job is to find the power spectral density from this vector \begin{equation} \left|\Phi \:_{vv}\left(e^{jw}\right)\right|^2 \end{equation}

Can somebody please direct me in the right direction to get started on solving this problem? I know I am supposed to use the Z transform, then evaluate the Z transform at the unit circle hence the term \begin{equation}e^{j\omega }\end{equation} but this is a vector and there's no dependence on a discrete time index. Looking forward for your help. Any simple and convenient help is appreciated. Thank you very much in advance.

Update 1: If, \begin{equation} v\left(n\right)\:\:=\:\frac{1}{\sqrt{2}}x\left(n\right)+\frac{1}{\sqrt{2}}x\left(n-1\right) \end{equation} then: \begin{equation} V\left(e^{j\omega }\right)\:=\:\frac{1}{\sqrt{2}}X\left(e^{j\omega \:}\right)+\frac{1}{\sqrt{2}}e^{j\omega \:\:}X\left(e^{j\omega \:\:}\right) \end{equation} therefore: \begin{equation} \frac{V\left(e^{j\omega }\right)}{X\left(e^{j\omega }\right)}\:=\:H\left(e^{j\omega }\right)=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}e^{j\omega \:\:\:} \end{equation} How to prove the response is a low pass filter response?

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  • $\begingroup$ I assume the discrete time index is the position in the vector; meaning you have two samples in time of what is a constant signal (DC) $\endgroup$ – Dan Boschen Oct 4 '20 at 13:44
  • $\begingroup$ @DanBoschen so you mean \begin{equation} v=\:\frac{1}{\sqrt{2}}x\left(n\right)+\frac{1}{\sqrt{2}}x\left(n-1\right)\end{equation} is this the situation I have? $\endgroup$ – JordenSH Oct 4 '20 at 17:49
  • $\begingroup$ That's how I would read that. $\endgroup$ – Dan Boschen Oct 4 '20 at 19:31
  • $\begingroup$ @DanBoschen thank you. So following this, I would find a correlation matrix of v with itself. In this case, it would be a 2X2 matrix. Am I correct? $\endgroup$ – JordenSH Oct 4 '20 at 21:35
  • $\begingroup$ PSD is a single dimensional array: power versus frequency. Take the DFT which is a magnitude quantity, scale by N to normalize to the same units as the time axis, take the complex conjugate to convert to a power quantity and you have two bins of a "power spectral density". The power of each bin represents the equivalent noise bandwidth of one bin (PSD is typically power/Hz but in this case you don't know the time increment so you can only do power/bin) $\endgroup$ – Dan Boschen Oct 4 '20 at 21:43

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