in my calculations i get to this solution:

\begin{aligned} H(e^{j*\omega}) & =\frac{1}{1-0.5 e^{-j\omega}} - \frac{0.5e^{-j\omega}}{1-0.5 e^{-j\omega}} &= 1\ \end{aligned}

Am i right that this is just an delta impulse?

\begin{aligned} h_2[n] & =\delta[n] \ \end{aligned}

  • $\begingroup$ Yes, what else could it be? $\endgroup$ – Raskolnikov Apr 21 '11 at 9:25
  • $\begingroup$ Just wanted to be sure. I am new to signal processing. Thank You. $\endgroup$ – madmax Apr 21 '11 at 10:36

Yes it is an impulse! There is a unique one-to-one mapping of discrete-time signals to their Fourier counterparts. In the case of the Kronecker-delta function, $\delta[n]$, the transform is 1 for all frequencies. Similarly, if we are given a '1' in the spectral domain, we can state without any doubt that $\delta[n]$ is the corresponding discrete-time signal.


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