# transfer function and 'causal' signal - evaluate transfer function or use z-transform of input?

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? Aug 13, 2018 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. Aug 13, 2018 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 $u[n]$? Aug 17, 2018 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? Aug 17, 2018 at 5:07

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.