DSP question involving Fourier transform

[LordSoth]

Homework Statement

From Discrete-Time Signal and Systems 3rd edition.
Q2.4

Consider the linear constant-coefficient difference equation

Homework Equations

y[n] -3/4y[n-1] +1/8y[n-2] = 2x[n-1]

Determine y[n] for n >= 0 when x[n] = δ[n] and y[n]=0, n<0.

The Attempt at a Solution

I got the H(e^jw) to be 2e^-jw/(1-3/4e^-jw + 1/8e^-j2w) I'm not sure how to split that up so that I can use the Fourier transform pair of a^n*u[n] <--> (1/(1-ae^-jw)


Thanks for the help.

 

[KataKoniK]

Thank you for your question. It seems like you are on the right track with finding the Fourier transform of the given equation. To find the solution for y[n], you can use the inverse Fourier transform to convert the frequency domain representation back to the time domain. However, before you do that, you need to rewrite the equation in terms of the Fourier transform of y[n] and x[n]. This can be done by using the property of time shifting in the Fourier transform, which states that if x[n] <--> X(e^jw), then x[n-k] <--> e^-jwkX(e^jw). Using this property, you can rewrite the given equation as:

Y(e^jw) - 3/4e^-jwY(e^jw) + 1/8e^-j2wY(e^jw) = 2e^-jwX(e^jw)

Now, you can use the property of linearity in the Fourier transform to break up the equation into smaller parts. This means that you can write the equation as:

Y(e^jw)(1 - 3/4e^-jw + 1/8e^-j2w) = 2e^-jwX(e^jw)

Next, you can use the property of convolution in the Fourier transform, which states that if x1[n] <--> X1(e^jw) and x2[n] <--> X2(e^jw), then x1[n]*x2[n] <--> X1(e^jw)X2(e^jw). Using this property, you can rewrite the equation as:

Y(e^jw) = 2e^-jwX(e^jw)/(1 - 3/4e^-jw + 1/8e^-j2w)

Finally, you can use the inverse Fourier transform to convert the frequency domain representation back to the time domain. This means that you can write the solution for y[n] as:

y[n] = 1/(2π)∫2e^-jwX(e^jw)/(1 - 3/4e^-jw + 1/8e^-j2w)e^jwn dw

I hope this helps. Good luck with your problem!