Discrete-time system: divergent response to exponential input

Question

I’ve been given the following Difference Equation and tasked with finding the response to $x_0[n] = (-p)^n$ - I’ve managed to $h[n]$ but the convolution itself failed.(The given system is IAR). $$y[n] -py[n-1] = x[n]$$ So I know that an IAR DE is a causal LTI system, with $$H(z) = \frac{1}{1-pz^{-1}}$$ which I plugged into a chart and got: $$h[n] = (p)^n u[n]$$ Trying to do the convolution to get $y[n]$: $$y[n] = \sum_{k=-\infty}^{\infty} p^k u[k] \cdot p^{n-k} \cdot (-1)^{n-k} = (-p)^n \cdot \sum_{k=0}^{\infty} (-1)^k$$ which unless I’m mistaken is a divergent series…

Answer 1

We assume that the given system is causal and stable, i.e., the (real-valued) pole must satisfy $-1<p<1$. If the input is given by $x_0[n]=(-p)^n$, $-\infty < n < \infty$, then - as mentioned in a comment by Hilmar - the input is unbounded, and the output is also unbounded, which is confirmed by your result.

If, on the other hand, the input signal is assumed to be causal, i.e.,

$$x_0[n] = (-p)^nu[n]$$

then $x_0[n]$ as well as the system's response are bounded:

\begin{eqnarray*}y[n]&=&\sum_{k=-\infty}^{\infty}p^ku[k](-p)^{n-k}u[n-k]\\&=&(-p)^nu[n]\sum_{k=0}^n(-1)^k\\&=&(-p)^nu[n]\frac12\left[1+(-1)^n\right]\\&=&\frac12\left[p^n+(-p)^n\right]u[n]\end{eqnarray*}

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I would like to add some information concerning the claim that since $z_0^n$ is an eigenfunction of a discrete-time LTI system, the system's response should be $H(z_0)z_0^n$.

Note that for the given (causal and stable) system

$$H(z)=\sum_{k=-\infty}^{\infty}h[k]z^{-k}=\sum_{k=0}^{\infty}p^kz^{-k}=\frac{1}{1-pz^{-1}},\qquad |z|>|p|\tag{1}$$

For the given input signal we have $z_0=-p$, which doesn't satisfy $|z_0|>p$, so we can't evaluate $(1)$ to obtain $H(z_0)$ because the series doesn't converge for that value of $z_0$. Computing the response to an eigenfunction $z_0^n$ as $H(z_0)z_0^n$ is only valid if $z_0$ is inside the region of convergence of $H(z)$, but this is not the case for $z_0=-p$.

Answer 2

1.- Series y converges for p values satisfying abs(p)<1

p being real or complex.

y oscillates if abs(p)=1

y diverges when abs(p)>1

n cannot be extended to -Inf because :

• if y converges towards +Inf then y diverges towards -Inf and

• if y diverges towards +Inf then y converges towards -Inf .

2.- example convergence

p=.9-1j*.43
abs(p)

n=[1:100];
x=(-p).^n;

y=zeros(1,numel(n));
for k=2:1:numel(n)
    y(k)=x(k)+p*y(k-1);
end

figure(1);
subplot(3,1,1)
stem(n,real(y));
grid on;xlabel('n');title(['real(y) p = ' num2str(p) ' |p| = ' num2str(abs(p))])
subplot(3,1,2)
stem(n,imag(y));
grid on;xlabel('n');title(['imag(y) p = ' num2str(p) ' |p| = ' num2str(abs(p))])
subplot(3,1,3)
stem(n,abs(y));
grid on;xlabel('n');title(['|y| p = ' num2str(p) ' |p| = ' num2str(abs(p))])

3.- example divergence

p=.73-1j*.73
abs(p)

x=(-p).^n;

y=zeros(1,numel(n));
for k=2:1:numel(n)
    y(k)=x(k)+p*y(k-1);
end

figure(2);
subplot(3,1,1)
stem(n,real(y));
grid on;xlabel('n');title(['real(y) p = ' num2str(p) ' |p| = ' num2str(abs(p))])
subplot(3,1,2)
stem(n,imag(y));
grid on;xlabel('n');title(['imag(y) p = ' num2str(p) ' |p| = ' num2str(abs(p))])
subplot(3,1,3)
stem(n,abs(y));
grid on;xlabel('n');title(['|y| p = ' num2str(p) ' |p| = ' num2str(abs(p))])

4.- all even terms are null

Among other stability tests the one attempting to find lim(abs(a(k+1)/a(k))) cannot be used because all y even terms are all null ; n(1)=1