2
$\begingroup$

I am having trouble on the following exercise and I can't figure out where I am doing something wrong:

Given an LTI system described by the following difference equation: $$y(n)=x(n)+2x(n-2)+y(n-1)$$ find the frequency response $H(e^{i\omega})$ and then find the impulse response $h(n)$ using a DTFT.

My attempt:

The difference equation becomes: $y(n)-y(n-1)=x(n)+2x(n-2)\overset{\mathcal{Z-}\text{transform}}{\Longrightarrow}Y(z)-z^{-1}Y(z)=X(z)+2z^{-2}X(z)$ and because $H(z)=\frac{Y(z)}{X(z)}$, $H(z)=\frac{1+2z^{-2}}{1-z^{-1}}=\frac{z^2+2}{z^2-z}=\frac{z^2+2}{z(z-1)}$ and because (here I am not so sure) the system is causal, it must be that we require $|z|>1$. Even if the ROC wasn't $|z|>1$, $|z|=1$ is a pole so it can't be in the ROC. But, the exercise is asking for a DTFT, which is the $\mathcal{Z-}$transform $z=e^{i\omega}$, so the $\mathcal{Z}-$transform on the unit circle $|z|=1$. The integral for $h(n)$ would then be $$h(n)=\frac{1}{2\pi i}\oint_{|z|=1}{\frac{z^2+2}{z(z-1)}z^{n-1}dz}$$ which obviously diverges because $|z|=1$ is a pole. Am I doing anything wrong? Because it seems to me that the problem is impossible with a DTFT. There is, however, a solution with an inverse $\mathcal{Z}-$transform. Thanks, in advance.

$\endgroup$

1 Answer 1

1
$\begingroup$

You're absolutely right, the given system doesn't have a frequency response, at least not as a conventional function. There's a pole on the unit circle, so the system is not stable (actually, it is marginally stable).

However, the impulse response is very easy to determine. You could either iterate through the given difference equation with $x[n]=\delta[n]$ (assuming zero initial condition), or you could rewrite $H(z)$ as

$$H(z)=\frac{1}{1-z^{-1}}+\frac{2z^{-2}}{1-z^{-1}}\tag{1}$$

Recognizing the first term on the left-hand side of $(1)$ as the $\mathcal{Z}$-transform of the unit step sequence $u[n]$, one arrives quickly at the following expression for $h[n]$:

$$h[n]=u[n]+2u[n-2]\tag{2}$$

where I've assumed causality of the system.

Since the unit step sequence does have a DTFT (if we allow generalized functions, i.e., a Dirac delta), we can write the frequency response of the given system as an expression with a Dirac delta impulse. But this is not something that you would try to derive from first principles, at least not as an engineer.

$\endgroup$
1
  • 1
    $\begingroup$ Thank you! I had actually gone about calculating the integral using the residue theorem and got some pretty weird $(1-n)-$ order derivatives, but these were easily discarded with the use of the $\Gamma$ function. I did get the same expression as you did, but I wondered why the problem was specifically asking for the frequency response, when there was none. I am a mathematics grad student helping an engineer friend, so I have been studying these systems for the past few days and you have been extremely helpful! $\endgroup$ Jun 17 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.