# Power Spectral Density of a Filter

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

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.

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.

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

• 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. Apr 23, 2021 at 8:21
• 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. Apr 23, 2021 at 8:25
• 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. Apr 23, 2021 at 8:32