Continuous Time Fourier Series of h(t)e^(-4|t|)

[ryukyu]

1. Determine the Continuous Time Fourier Series representation of the response where:

h(t)=e^(-4*|t|)

and

x(t)=summation n=-inf to inf of dirac(t-n) The attempt at a solution:

I used Laplace transforms to find the frequency response:
H(s) = 4/(16-s^2)

from what I recall isn't the x(t) simply going to return the h(t) since the shift n is from -inf to inf?

If so then the output response would be x(t) convolved with h(t) which is messy work to do, so I'm assuming things are easier kept in the frequency domain...?

 

[Valkarie]

From what I recall the Fourier series representation is the following,X(jω) = ∑n=-∞ to ∞ X(n)δ (ω-2πn) And H(jω) = 4/(16-ω^2)So the Continuous Time Fourier Series representation of the output would be:Y(jω) = X(jω)*H(jω) = 4/16 * ∑n=-∞ to ∞ X(n)δ (ω-2πn)

 

[Mene]

Yes, you are correct. Since x(t) is a summation of delta functions, it can be written as the impulse response of an ideal low-pass filter with cutoff frequency 1/2π. This means that the Fourier transform of x(t) is simply H(jω), where ω is the frequency variable. Therefore, the Fourier series representation of the response would be H(jω) multiplied by the Fourier series of h(t).

In this case, since h(t) is an even function, the Fourier series of h(t) would only have cosine terms. Plugging in H(jω) = 4/(16+jω^2), we get:

H(jω) = 4/[16 + (ω^2 - jω)^2]
= 4/[16 + ω^4 + 2ω^2(-jω)]
= 4/[16 + ω^4 + 2ω^3]
= 4/[16 + (ω^2)^2 + 2(ω^2)^2]
= 4/[16 + 3(ω^2)^2]

Therefore, the Fourier series representation of the response would be:

x(t) convolved with [4/[16 + 3(ω^2)^2]]cos(ωt)

This can be simplified further using trigonometric identities, but it may not be necessary depending on the specific problem.