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I am trying few problems on the introductory part of DSP. One of the problem asks to calculate the steady state response of a system with impulse response $h[n] = (\frac{j}{2})^{n} u[n] $ to an input signal $x[n]=\cos(\pi n) u[n]$. However, I have no idea how to handle this complex $j$ in the convolution. I want someone to confirm if my solution is correct or not. If it is incorrect please tell me where I am getting wrong.

I start with calculating the Z transform of $h[n]$. $$\begin{align} H(z) &= \sum_{n=0}^{\infty}h[n] \, z^{-n} \\ &= \sum_{n=0}^{\infty}\left(\frac{j}{2}\right)^{n} \, z^{-n} \\ &= \frac{1}{1-\frac{j}{2}z^{-1}} \\ \end{align}$$ such that $|z| > \frac{1}{2}$.

Now the z transform of $x[n]$ is , $$\begin{align} X(z) &= \sum_{n=0}^{\infty}\cos(\pi n)z^{-n} \\ &= \sum_{n=0}^{\infty}\Big[\frac{e^{j\pi n}+e^{-j\pi n}}{2}\Big]z^{-n} \\ &= \frac{1}{2} \left(\frac{1}{1-e^{j\pi}z^{-1}}+\frac{1}{1-e^{-j\pi}z^{-1}}\right) \\ &=\frac{1}{1+z^{-1}} \\ \end{align}$$

The convolution output in the z domain is $Y(z)$. $$Y(z) = H(z)X(z) = \frac{2}{(2-jz^{-1})(1+z^{-1})}$$ Applying the final value theorem, I get, steady state output $y_{ss}$ as $$y_{ss} = \lim_{z \to 0} \, zY(z) = 0$$.

I definitely think that I am doing something wrong over here. Please help. I don't know how to handle this $j$ in the convolution operation like this, please help me in that too.

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  • $\begingroup$ I do not understand how it approaches to 0/0 form when z tends to infinity. I still think that the limit is 0. $\endgroup$ – Himanshu Sharma Oct 2 '18 at 12:21
  • $\begingroup$ I erased my erroneous comments, Matt’s answer is very good $\endgroup$ – Dan Boschen Oct 2 '18 at 14:58
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You can't apply the final value theorem here. First of all, the correct form of the final value theorem for the $\mathcal{Z}$-transform is

$$\lim_{n\to\infty}y[n]=\lim_{z\to 1}(z-1)Y(z)\tag{1}$$

Eq. $(1)$ assumes that the poles of $(z-1)Y(z)$ are inside the unit circle, which is not the case in your example. In your case, the limit $(1)$ does not exist.

What you have to do is find the frequency response of the system, which is just the transfer function evaluated on the unit circle $z=e^{j\omega}$ (if it converges on the unit circle). Note that the region of convergence (ROC) you've found does not make sense because $|z|$ must be a real number. The requirement for the sum to converge is

$$\left|\frac{j}{2z}\right|<1\quad\Longrightarrow\quad |z|>\frac12\tag{2}$$

Since $H(z)$ converges on the unit circle $|z|=1$, the frequency response is given by

$$H(e^{j\omega})=\frac{1}{1-\frac{j}{2}e^{-j\omega}}\tag{3}$$

An LTI system's response to a sinusoidal input $x[n]=\cos(\omega_0n)$ is

$$y[n]=|H(e^{j\omega_0})|\,\cos[\omega_0n+\phi(\omega_0)]\tag{4}$$

where $\phi(\omega)$ is the system's phase response:

$$H(e^{j\omega})=|H(e^{j\omega})|e^{j\phi(\omega)}\tag{5}$$

So in the limit, after all transients have died out, the system's (steady-state) response to a switched sinusoidal input $x[n]=\cos(\omega_0n)u[n]$ is given by $(4)$. What is left now is to determine the magnitude and phase of $H(e^{j\omega})$ given by $(3)$ and plug these values into Eq. $(4)$.

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  • $\begingroup$ Yes, I did a horrible mistake with the ROC ;p. By the way thanks, I understood your solution. $\endgroup$ – Himanshu Sharma Oct 3 '18 at 8:25

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