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

Obviously,I I can't really (unless I'm wrong) realize thisa 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 signalsystem (textbook solutions). My question is.. why?

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

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$n=0$ (even if technically its equal to one)?

Thanks!

Suppose I use the ideal lowpass filter eq to attempt creating a IIR filter using it  : $$ d(k) = \frac{sin(\omega_c \cdot k)}{\pi \cdot k}, -\infty < k < \infty $$

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

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

The stability condition states that a signal is stable if its impulse response satisfies: $$ \sum_{n=-\infty}^{\infty}|h(n)| < \infty $$

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!

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!

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Yannick
Yannick
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Simple question about Sinc()

Suppose I use the ideal lowpass filter eq to attempt creating a IIR filter using it : $$ d(k) = \frac{sin(\omega_c \cdot k)}{\pi \cdot k}, -\infty < k < \infty $$

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

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

The stability condition states that a signal is stable if its impulse response satisfies: $$ \sum_{n=-\infty}^{\infty}|h(n)| < \infty $$

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!