# Steady state value of a complex convolution

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.

• I do not understand how it approaches to 0/0 form when z tends to infinity. I still think that the limit is 0. – Himanshu Sharma Oct 2 '18 at 12:21
• I erased my erroneous comments, Matt’s answer is very good – Dan Boschen Oct 2 '18 at 14:58

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. $$(4)$$.

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