# Find the impulse response of an LTI system? why there is a 3/2 in the (b) problem? Does the impulse response even exist?

• well, the $|\omega| \le \pi$ is a little bit of a problem because $$e^{-j3 \pi /2} \ne e^{+j3 \pi /2}$$ the problem should maybe have left the value of $H(\omega)$ ambiguous for $\omega= \pm \pi$. should maybe have said $|\omega| < \pi$ in 1(b). – robert bristow-johnson Oct 5 '17 at 5:39
• BTW, if this is homework, it should be tagged as such. my recommendation is to tell us what it appears to be for problem 1.(a) and then extend that guess for 1.(b), but tell us what you might think is odd about it in 1.(b). i presume "$H(\omega)$" means the DTFT of impulse response $h[n]$. – robert bristow-johnson Oct 5 '17 at 5:46
• @robertbristow-johnson Ya, they probably should have stated it as $-\pi \le \omega \lt \pi$ to allow for it to be $2\pi$-periodic and not have matching values at the extremities. – Peter K. Oct 5 '17 at 9:23
• but @PeterK., if $h[n]$ is real, we know that $H(\pm\pi)$ must be the same and be real. i think the only possibility is for $H(\pm\pi)=0$. – robert bristow-johnson Oct 5 '17 at 16:42
• @robertbristow-johnson Agreed, but who said its real?! 😜 – Peter K. Oct 5 '17 at 20:44

For the case a-) I assume you are trained enough in DSP to see that : $$\mathcal{ F } \{ \delta[n-d] \} = e^{-j \omega d }$$ for all integer $d$.

Hence given a frequency response of $H(\omega) = e^{-j3\omega }$ it's apparent that the corresponding impulse response is $$h_a[n] = \delta[n-3]$$

The problem in case b-) is in the fact that it suggests a non-integer amount of shift $\alpha$ of the unit impulse $\delta[n]$ so as to obtain $\delta[n-\alpha]$ which would then have a corresponding frequency response $H_b(\omega) = e^{ -j \omega \alpha }$. But this makes no sense in the domain of discrete-time sequences which cannot be shifted by non-integer amounts which might force you to argue that the corresponding impulse response does not even exist.

The solution requires an investigation of the relation between continuous-time and discrete-time signals through sampling as the other answer outlines. Instead here I put a shorthand result.

First observe that for any integer $d$: $$\delta[n-d] = \frac{ \sin(\pi(n-d)) }{\pi(n-d) } = \text{sinc}(n-d)$$ is satisfied.

The righthand side is a sampled (and therefore discrete) sinc pulse whose continuous equivalent is $$\text{sinc}(x) = \frac{\sin(\pi x)}{\pi x}$$

Furthemore observe that the discrete-time sinc pulse $\sin(\pi n) /{ \pi n}$ is actually the impulse response of an ideal lowpass filter with a cutoff frequency of $\omega_c = \pi$ which is indeed an allpass filter with a frequency response of: $$H_{apf}(\omega) = 1 ~~~,~~~ \text{for all } ~~~|\omega| < \pi$$

As a consequence, the frequency response of a shifted sinc pulse $\frac{ \sin(\pi(n-d)) }{\pi(n-d) }$ will be:

$$H_s(\omega) = e^{-j \omega d }$$ for any $d$, integer or real, according to the time-shift property of the Fourier transform.

Hence it can be concluded by the inverse Fourier transform relation of the last statement that the impulse response $h_b[n]$ associated with the frequency response $$H_b(\omega) = e^{-j \omega (3/2) }$$ is $$h_b[n] = \frac{ \sin(\pi(n-3/2)) }{\pi(n-3/2) }$$

Note that $h_b[n]$ cannot be obtained from $\frac{ \sin(\pi n) }{\pi n}$ by just shifting it $3/2$ samples which is not possible. Therefore you should obtain the samples of the impulse response $h_b[n]$ by evaluating its formula for each n.