0
$\begingroup$

From my studying difference equations and transfer functions, I understand that when a complex exponential input $x[n]=z_1^n$ is applied to an LTI system with transfer function $H(z)$, determining the output involves evaluating the transfer function at $z=z_1$, i.e. $H(z_1)$ which is a complex quantity that simply scales the input signal to produce the output. The textbook I have uses a sinusoid input to show the effect (the transfer function causes a magnitude and phase change to the input).

The answers to this question I asked a while ago made the point of recognizing the difference between $a^n$ and $a^n \cdot u[n]$. All of that was in the context of two sided signals and the bilateral z-transform.

In something I found online, in the context of causal signals and the unilateral z-transform, there was a question posed: "if $x[n]=(1/2)^n \cdot u[n] \to y[n]=\delta [n-2],$ what is the output for $x[n]=\cos(\frac{\pi}{3}n)?$", where the author determined the output by evaluating $H(z)$ determined from the first relationship at $z=e^{jn \pi /3}$. Since this is a 'causal' context, isn't that cosine really $x[n]=\cos(n\frac{\pi}{3}) \cdot u[n]$ and hence requires taking the z-transform of the input?

When can you just evaluate $H(z)$ at the value of interest, and when do you have to take the (unilateral) z-transform of the input?

$\endgroup$

2 Answers 2

1
$\begingroup$

For a discrete-time LTI system with transfer function $H(z)$, the response to $x[n]=z_1^n$ equals $y[n]=H(z_1)z_1^n$ if $z_1$ is inside the region of convergence of $H(z)$. This relationship holds whether or not the system is causal or stable.

From this relationship it follows that if the system is real-valued and stable, i.e., if its impulse response is real-valued and if $H(e^{j\omega})=|H(e^{j\omega})|e^{j\phi(\omega)}$ exists, the response to a sinusoidal input $x[n]=\cos(n\omega_0)$ is given by $y[n]=|H(e^{j\omega_0})|\cos[n\omega_0+\phi(\omega_0)]$.

I think your confusion comes from assuming something you call a "causal context". It doesn't matter if a system is causal or not for above relationships to hold. You can apply a sinusoidal input signal to a causal system, and as long as the system is real-valued and stable, the output will be given by the relationship mentioned above. If the input is a sinusoid switched on at a finite $n=n_0$, the system's response can be computed by solving the convolution sum or, equivalently, by using the $\mathcal{Z}$-transform.

$\endgroup$
4
  • $\begingroup$ So, where I am using the unilateral z-transform, a sinusoid or even an complex exponential signal such as $a^n$ is not considered switched on at n=0? In which case, only if I had a signal where the signal is turned on at $n=n_0$ where $n_0 >0$ would I use the z-transform of the input to determine the output. Is this correct? $\endgroup$
    – Westerley
    Aug 13, 2018 at 4:31
  • $\begingroup$ @Westerley: If you have a signal $z_1^n$ or $\cos(\omega_0n)$ then there's usually no restriction on the values of $n$. Otherwise one needs to state $n\ge 0$, or multiply the sequence with the unit step sequence $u[n]$. If the input is switched on (by specifically mentioning $n\ge 0$ or multiplication with $u[n]$) then you can use the (unilateral) Z-transform. You can always use convolution, no matter if the signal extends over all possible values of $n$ or just the non-negative values. $\endgroup$
    – Matt L.
    Aug 13, 2018 at 8:24
  • $\begingroup$ I think this is what confuses me: if I'm using the unilateral z-Transform, what's the difference betwen $ z_1^n $ and $ z_1^n \cdot u[n] $? Aren't they going to be the same? If I've determined a transfer function by taking the unilateral z-Transform of the impulse function and then apply an exponential signal, what will the difference between the exponential signal with and without the $u[n]$? $\endgroup$
    – Westerley
    Aug 17, 2018 at 4:51
  • $\begingroup$ More specifically, if I'm told that $x[n]=(1/2)^n \to y[n]$, I can determine the transfer function as $Y(z)/X(z)$, using the unilateral z-Transform to do so. But in that case, since I'm ignoring signal for $n<0$, what's the difference between $(1/2)^n$ and $(1/2)^n\cdot u[n]$? The 'switched on' signal and the eigenfunction signal look identical, no? $\endgroup$
    – Westerley
    Aug 17, 2018 at 5:07
0
$\begingroup$

I'm not sure that the notion of a "'causal' context" is a meaningful distinction.

The author is taking a shortcut between the z domain and the discrete time Fourier transform domain.

In your case you've got a system with a transfer function $H(z)$, and you've got a choice of two signals: $x_i(n) = \cos (\theta_0 n)$ or $x_f(n) = u(n)\left(\cos( \theta_0 n)\right)$.

The big difference here is that $x_i$ is of infinite extent in both directions, where $x_f$ at least has a finite beginning.

You can go one of two ways here.

One way is to recognize that if $H(z)$ is the transfer function of the system in the context of the z transform, then $H(e^{j\theta})$ is the transfer function of the system in the context of the discrete-time Fourier transform*. Then the response of the system to a sinusoid of infinite extent at frequency $\theta$ and phase $\phi$ is just $\phi H(e^{j\theta})$. This is nice, but you have to crank through all that Fourier stuff to prove it.

Another way is to start with $x_f(n) = u(n)\left(\cos( \theta_0 n)\right)$, take it's $z$ transform, multiply that by $H(z)$ and do all the partial fraction math**. You'll find that if $H(z)$ has an impulse response that settles out, you end up with $y(n) = \left | H(z) \right | \cos \left (\theta_0 n + \arg H(z) \right) + y_t(n)$, where the modes of $y_t(n)$ are the modes of $H(z)$, which means that over time it settles out and you're left with $\left | H(z) \right | \cos \left (\theta_0 n + \arg \left( H(z) \right) \right)$.

In other words, after the transient has settled, the responses to $x_i$ and $x_f$ are the same. Moreover, the response to the "uses the z transform" case is the same as the "uses the Fourier transform" case. If you're very careful about minding your p's and your q's you can observe that H(z) is a LTI system, and the "TI" in "LTI" means "time invariant". This means that you can use the z-domain analysis technique for a signal $x^{'}_f(n) = u(n + k) \cos (\theta_0 n)$. Then you can take the limit as $k \to \infty$ -- and get back to the discrete-time Fourer transform case.

* Not the DFT -- the DFT is of finite extent, while the DTFT is of infinite extent.

** You should work this out for yourself if you can, starting with examples of $H(z)$ that you know to be stable, and then if you're feeling your oats, with any general $H(z)$ that's assumed to be stable.

$\endgroup$

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.