0
$\begingroup$

Suppose I use the ideal lowpass filter eq to attempt creating a IIR filter using it : $$ h[n] = \frac{\sin(\omega_c \, n)}{\pi \, n}, \qquad -\infty < n < \infty $$

Obviously, I can't really (unless I'm wrong) realize a filter with this impulse response because I don't have the finite difference equation to make the recursion happen.

But I'm told that it's also an unstable system (textbook solutions). My question is.. why?

The stability condition states that a system is stable if its impulse response satisfies: $$ \sum_{n=-\infty}^{\infty} \Big|h[n] \Big| < \infty $$

From what I see, if I look at the condition blindly, it does satisfy the stability criterion. Doesn't it? Then why it's unstable? Is it because of something like a discontinuity at $n=0$ (even if technically its equal to one)?

Thanks!

$\endgroup$
  • $\begingroup$ Hint: the harmonic series does not converge and so, since there can be values of $\omega_c$ such that $h(n)$ is of the form $\frac 1n$, there is no guarantee that the impulse response is stable. $\endgroup$ – Dilip Sarwate Aug 15 '18 at 19:34
1
$\begingroup$

No, the signal $$ h[n] = \frac{\sin(\omega_c \, n)}{\pi \, n}, \qquad -\infty < n < \infty $$

is not absolutely summable, as the sum

$$ \sum_{n=-\infty}^{\infty} \Big|h[n]\Big| = \sum_{n=-\infty}^{\infty} \left |\frac{\sin(\omega_c \, n)}{\pi \, n} \right| $$

diverges to infinity. Hence the absolute sum does not converge and therefore it's considered as an unstable system's impulse response. Yet, its square is absolutely summable:

$$ \sum_{n=-\infty}^{\infty} \Big|h[n]\Big|^2 = \sum_{n=-\infty}^{\infty} \left |\frac{\sin(\omega_c \, n)}{\pi \, n} \right|^2 $$ does not diverge. This kind of condition is stated as mean square convergence; i.e., a typical condition for discontinuous frequency responses that has finite jumps in its frequency response which should be continuous for all $\omega$ according to the Fourier theorem that would require uniform convergence at all frequency points $\omega$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.