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Considering this general 1st order transfer function

$$ H(z) = \frac{b_0 + b_1z^{-1}}{1-az^{-1}} $$

How to find (analytically) the transient and steady-state responses?

With steady-state response I mean response to a sinusoid. I'm particularly interested in the case when $a=\pm1$.

Conceptually I have some difficulties grasping how to deal with the fact that the z-transform for a causal sinusoid only exists for $|z|>1$ (explained here) and that for the case when $a=\pm1$ the z-transform does not converge for $|z|=1$. So the 'straight-forward' procedure of multiplying $H(z)$ with the z-transform of a causal sine and then inverse z-transform the result seems to not be possible/valid. I'm confused.

Edit:

simple example

fs = 48000;
f0 = 1000; 
T = 3/f0;
N = round(T*fs);
t = linspace(0,T,N);
x = sin(2*pi*f0*t);

bd = [1 0];
ad = [1 -1];
yi = filter(bd, ad, x);
figure;
plot(t,x,'b'); hold on;
plot(t,yi,'r')

blue: input, red: output

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The region of convergence (ROC) of the $\mathcal{Z}$-transforms of a step-modulated sinusoid

$$x[n]=\sin(\omega_0n)u[n]\tag{1}$$

is the region $|z|>1$. The ROC of the transfer function

$$H(z)=\frac{b_0+b_1z^{-1}}{1- z^{-1}}\tag{2}$$

also equals $|z|>1$ (assuming a causal system). Consequently, multiplying the two transforms doesn't pose any problem, and the result also converges for $|z|>1$.

As an example, let's compute the response $y[n]$ of the system $(2)$ to the input $(1)$. For the sake of simplicity we assume $b_0=1$ and $b_1=0$ in $(2)$. First, we compute the response to

$$\tilde{x}[n]=e^{j\omega_0n}u[n]\tag{3}$$

The $\mathcal{Z}$-transform of $(3)$ is

$$\tilde{X}(z)=\frac{1}{1-e^{j\omega_0}z^{-1}}\tag{4}$$

The $\mathcal{Z}$-transform of the response $\tilde{y}[n]$ is given by

$$\begin{align}\tilde{Y}(z)&=\tilde{X}(z)H(z)\\&=\frac{1}{1-e^{j\omega_0}z^{-1}}\cdot \frac{1}{1-z^{-1}}\\&=\frac{A}{1-e^{j\omega_0}z^{-1}}+\frac{A^*}{1-z^{-1}}\tag{5}\end{align}$$

with

$$A=\frac{1}{1-e^{-j\omega_0}}=H(e^{j\omega_0})\tag{6}$$

The inverse $\mathcal{Z}$-transform of $(5)$ is

$$\tilde{y}[n]=H(e^{j\omega_0})e^{j\omega_0n}u[n]+H^*(e^{j\omega_0})u[n]\tag{7}$$

The response to the step-modulated sinusoid $(1)$ is easily obtained from $(7)$ by taking its imaginary part:

$$\begin{align}y[n]&=\textrm{Im}\big\{\tilde{y}[n]\big\}\\&=\big|H(e^{j\omega_0})\big|\sin\big(\omega_0n+\arg\left\{H(e^{j\omega_0})\right\}\big)u[n]\\&\qquad -\textrm{Im}\big\{H(e^{j\omega_0})\big\}u[n]\tag{8}\end{align}$$

The first term in $(8)$ is the steady-state response, and the second term is the transient response, which doesn't decay because of the system's pole at $z=1$.

So the output has a DC component due the imaginary part of $H(e^{j\omega_0})$. Note that this is no contradiction with the linearity of the system, because the input signal is not a single spectral line but it has a continuous spectrum extending down to DC due to the sinusoid being switched on at $n=0$. This DC value of the input's Fourier transform triggers the system's eigenfrequency, which is a DC component. Note that this always happens with LTI systems: the output may contain oscillations at frequencies that are only determined by the system, not by the input signal. However, unlike in the given example, usually these transients decay because most of the time we consider asymptotically stable systems.

I slightly modified your Matlab/Octave script and added the analytical result $(8)$ for comparison. The analytical result is identical (up to numerical accuracy) to the result obtained by filtering:

fs = 48000;
f0 = 1000; 
w0 = 2*pi*f0/fs;
N = 200;
n = 0:N-1;
x = sin(w0*n);

bd = [1 0];
ad = [1 -1];
yi = filter(bd, ad, x);
figure;
plot(n,x,'b'); hold on;
plot(n,yi,'r')

% analytical computation
A = 1 / ( 1 - exp( -1i*w0 ) );
y2 = abs(A) * sin( w0*n + angle(A) ) - imag(A);

plot(n,y2,'k.'), hold off
legend('input','response by filtering','response by analytical computation')

enter image description here

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  • $\begingroup$ ah, so I should have focused on the Fourier transform instead of the z-transform. Is there any chance that you can explain/derive or provided a reference to the added pole expressions for $H_{-1}$ and $H_{1}$? Can one say that the delta function in the $H_{1}$ expression change the magnitude response but not the phase response? Would you call the systems $H_{-1}$ and $H_{1}$ linear? The reason for this question is that if $\sin\big(n\omega_0\big)$ is input to the $H_{1}$ system a DC offset is added so in some sense intuitively this conflicts with the notion of linearity.. $\endgroup$ – bjarne Oct 8 at 9:36
  • $\begingroup$ your answer is really good and I'm going to accept it and upvote (although it seems I can't right now). I just need to digest your answer to understand exactly how it works. $\endgroup$ – bjarne Oct 8 at 9:38
  • $\begingroup$ also I'm a little confused about your expression (4). Would you call that the steady-state response? what is the transient response then? $\endgroup$ – bjarne Oct 8 at 9:40
  • $\begingroup$ Do you think it feasible to multiply the Fourier transform of a sine to the $H_1$ system and analytically inverse Fourier transform the result? $\endgroup$ – bjarne Oct 8 at 9:46
  • $\begingroup$ hmm, I'm a little bit confused. Does the system have a transient response? I will try to add a small Matlab snippet for the $H_1$ system (with $b_0=1$ and $b_1=0$) which illustrates that there is a DC offset added to the output for a sine input (I somehow hope that the transient response can explain this). $\endgroup$ – bjarne Oct 8 at 9:58

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