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!

• 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. – Dilip Sarwate Aug 15 '18 at 19:34

No, the signal $$h[n] = \frac{\sin(\omega_c \, n)}{\pi \, n}, \qquad -\infty < n < \infty$$
$$\sum_{n=-\infty}^{\infty} \Big|h[n]\Big| = \sum_{n=-\infty}^{\infty} \left |\frac{\sin(\omega_c \, n)}{\pi \, n} \right|$$
$$\sum_{n=-\infty}^{\infty} \Big|h[n]\Big|^2 = \sum_{n=-\infty}^{\infty} \left |\frac{\sin(\omega_c \, n)}{\pi \, n} \right|^2$$ 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 $\omega$.