Finding Impulse Response of Frequency Response

[Fizzill]

Homework Statement

Consider the frequency response H(e^jw) = e^j3w(5 + cos(w) + 2cos(w)).

Find the impulse response h[n].

Homework Equations

H(e^jw) = Ʃ(from k = 0 to M) bk * e^-jwk = Ʃ(from k = 0 to M) h[k] * e^-jwk

The Attempt at a Solution

So I'm not sure at all how to get about solving this for sure, I believe I did it but then again after going through my book and notes I wasn't able to find anything like it which makes me doubt my answer.

H(e^jw) = e^j3w(e^-j2w + 1/2e^-jw + 5 + 1/2e^jw + e^j2w)
= e^jw + 1/2e^j2w + 5e^j3w + 1/2e^j4w + e^j5w

which i believe would equate to

h[n] = ∂[n + 1] + 1/2∂[n + 2] + 5∂[n + 3] + 1/2∂[n + 4] + ∂[n + 5]

 

[KataKoniK]

Hi there,

Your attempt at the solution is partially correct. The frequency response H(e^jw) is a complex function, so it cannot be directly equated to the impulse response h[n], which is a real function. The correct approach to finding the impulse response is to first separate the real and imaginary parts of H(e^jw) and then apply the inverse discrete Fourier transform to each part separately.

So, we can rewrite H(e^jw) as:

H(e^jw) = e^j3w(5 + cos(w) + 2cos(w)) = 5e^j3w + e^j4w + 2e^j5w

Separating the real and imaginary parts, we get:

Re{H(e^jw)} = 5cos(3w) + cos(4w) + 2cos(5w)
Im{H(e^jw)} = 5sin(3w) + sin(4w) + 2sin(5w)

Now, applying the inverse discrete Fourier transform to each part, we get:

h[n] = Ʃ(from k = 0 to M) Re{H(e^jw)} * e^jwn = 5∂[n + 3] + ∂[n + 4] + 2∂[n + 5]
g[n] = Ʃ(from k = 0 to M) Im{H(e^jw)} * e^jwn = 5∂[n + 3] + ∂[n + 4] + 2∂[n + 5]

Therefore, the final impulse response would be:

h[n] = Re{h[n]} + Im{h[n]} = 10∂[n + 3] + 2∂[n + 4] + 4∂[n + 5]

I hope this helps clarify the process for finding the impulse response from a frequency response. Let me know if you have any further questions.