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?

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.

  • 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? – Westerley Aug 13 at 4:31
  • @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. – Matt L. Aug 13 at 8:24
  • 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 $u[n]$? – Westerley Aug 17 at 4:51
  • 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? – Westerley Aug 17 at 5:07

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