I've been trying to solve the following problem, but I noticed that I get two different answers by means of two different techniques:

Find the impulse response, $h(n)$, of the following system, described in frequency: $Y(e^{j\omega}) = e^{-j\omega}X(e^{j\omega}) + \frac{dX(e^{j\omega})}{d\omega}$

Try #1: I found the difference equation for the system and iterated while making $x(n) = \delta(n)$:

$y(n) = x(n-1)+\frac{n}{j}x(n)$

$y(0) = 0 + 0$

$y(1) = 0 + 1$

$y(2) = 0 + 0$

And so on. Which yields $h(n) = \delta(n-1)$.

I found this same result by noticing that $X(e^{j\omega}) = 1$ for $x(n) = \delta(n)$, which gives: $Y(e^{j\omega}) = e^{-j\omega}$ and by using the time shift property we find that $h(n) = \delta(n-1)$.

Try #2: I used the inversion formula for $Y(e^{j\omega}) = e^{-j\omega}$:

$y(n) = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{-j\omega}e^{j\omega n}d\omega$ which yields $y(n) = \frac{sin(\pi(n-1))}{\pi(n-1)}$

Did I miss something? What is the correct result?


1 Answer 1


Your second try yielded a normalized sinc function. That function has a zero-crossing whenever $\pi(n-1)$ ia a nonzero multiple of $\pi$, which is for every $n\ne 1$ (assuming n is an integer) and is equal to 1 when $\pi(n-1)=0$.
Therefore, both solutions are the same.

  • $\begingroup$ Thanks. I should have plotted the function before posting a question. $\endgroup$
    – Thiago
    Commented Jun 8, 2014 at 20:41

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