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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…

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    $\begingroup$ Are you sure that the input signal isn't causal, i.e., $x_0[n]=(-p)^nu[n]$? $\endgroup$
    – Matt L.
    Commented Jan 14, 2023 at 16:54
  • $\begingroup$ imgur.com/a/CRAXY9X < a photo of the question I’m trying to solve. It’s given that the system is IAR, the first subquestion wanted me to find the p values for which it’s BIBO stable, and the second one wanted the response to $x[n]$ $\endgroup$ Commented Jan 14, 2023 at 16:59
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    $\begingroup$ The system is BIBO stable for $|p| < 1$ so in this case the output should be bounded if the input is bounded. HOWEVER the input as written is unbounded, and hence the output will be as well. Matt L's comment is a really good one! $\endgroup$
    – Hilmar
    Commented Jan 14, 2023 at 17:43
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    $\begingroup$ Sorry, I can't read that. I'm guessing that there is some explicit or implicit assumption in here that's missing. $\endgroup$
    – Hilmar
    Commented Jan 14, 2023 at 18:32
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    $\begingroup$ @Lonimous: I added some information to my answer concerning your last comment. $\endgroup$
    – Matt L.
    Commented Jan 15, 2023 at 12:14

2 Answers 2

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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*}


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 $z_0=-p$.

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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))])

enter image description here

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))])

enter image description here

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

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