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I need to calculate the output power spectral density of the following digital filter

enter image description here

My calculations are as follows:

$y\left(n\right)\:=\:x\left(n-1\right)+d\left[x\left(n-1\right)+x\left(n\right)\right]$

$Y\left(z\right)\:=\:X\left(z\right)z^{-1}+d\left[X\left(z\right)z^{-1}+X\left(z\right)\right]$

$Y\left(z\right)\:=\:X\left(z\right)\left[\left(1+d\right)z^{-1}+d\right]$

$\frac{Y\left(z\right)}{X\left(z\right)}=\:H\left(z\right)=\left(1+d\right)z^{-1}+d$

Did I get this equation right? would somebody help me please because this diagram is kind of confusing for me.

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1 Answer 1

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A digital filter doesn't have power spectral density, a signal has. I guess what you want is the transfer function of this filter, i.e., $H(z)=Y(z)/X(z)$.

Let's add two intermediate signal $u(n)$ and $v(n)$ in the diagram.

enter image description here

We have:

$$ y(n) = v(n-1) + du(n) \tag{1} $$ $$ u(n)=v(n-1)+x(n)\tag{2} $$ $$ v(n) = x(n) -du(n) \tag{3} $$

Take Z-transform and we get $$ Y(z) = V(z)z^{-1} + dU(z) \tag{4} $$ $$ U(z) = V(z)z^{-1}+X(z)\tag{5} $$ $$ V(z)=X(z)-dU(z)\tag{6} $$

Furthermore, we can derive that

$$ V(z)=X(z)-d\Big[V(z)z^{-1} + X(z)\Big] $$

$$ \frac{V(z)}{X(z)} = \frac{1-d}{1+dz^{-1}} $$

$$ Y(z)=V(z)z^{-1} + d\Big[V(z)z^{-1}+X(z) \Big] =(1+d)V(z)z^{-1} + dX(z) $$

Thus, $$ H(z) = \frac{Y(z)}{X(z)} = (1+d) \frac{V(z)}{X(z)} z^{-1} + d = (1+d) \frac{(1-d)z^{-1}}{1+dz^{-1}} +d $$

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  • $\begingroup$ very impressive. Thank you very much. This for some reason was very confusing. I see how to solve something like that from now on. Basically to introduce new signals. $\endgroup$
    – JordenSH
    Apr 23, 2021 at 8:21
  • $\begingroup$ So if the signal, has an input power spectral density say $\sigma _x^2$ how would someone find the output power spectral density? I would appreciate it if you provide this update to the answer. $\endgroup$
    – JordenSH
    Apr 23, 2021 at 8:25
  • $\begingroup$ Power spectral density of input signal is $P_{xx} = X(e^{j\omega}) X^*(e^{j\omega})$. Since you already have the transfer function $H(z)$ and also the frequency response $H(e^{j\omega})$, just take $Y(e^{j\omega}) = H(e^{j\omega}) X(e^{j\omega})$ into account and you will see the answer. $\endgroup$
    – ZR Han
    Apr 23, 2021 at 8:32

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