Practice Problem

Compute the discrete Fourier transform of the discrete-time signal

$ x[n]= (-j)^n $.

How does your answer related to the Fourier series coefficients of x[n]?

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Answer 1

$ X_k = \sum_{n=0}^{N-1} x_n \cdot e^{-j 2 \pi \frac{k}{N} n} = \sum_{n=0}^{3} (-j)^n \cdot e^{-j 2 \pi \frac{k}{4} n} = 1 + (-j \cdot e^{-j \frac{\pi k}{2}} ) + (-1 \cdot e^{-j \frac{2\pi k}{2}} ) + (j \cdot e^{-j \frac{3\pi k}{2}} ) $

$ = 1 + (-j) \cdot (-j)^k + (-1) \cdot (1)^k + (j) \cdot (j)^k = (-j)^{k+1} + (j)^{k+1} = 0, -2, 0, 2 $

, when k = 0, 1 ,2 ,3. And it is periodic with K = 4.

Ouch... This is not right. since $ x[n] = (-j)^n = e^{((-j\pi/2) \cdot n )} $

it's fft should be only an impulse. And Matlab told me:

x = [1 -j -1 j];

fft(x)

ans =

    0     0     0     4

I'll fix it tomorrow. Or someone can point out my error?

instructor's comment: There is a much easier way to answer this question. Take a close look at the formula for the DFT and try to use a "comparison" approach. -pm

Answer 2

$ X_k = \sum_{n=0}^{N-1} x_n \cdot e^{-j 2 \pi \frac{k}{N} n} = \sum_{n=0}^{3} (-j)^n \cdot e^{-j 2 \pi \frac{k}{4} n} $.

$ X_k = \sum_{n=0}^{3} e^{-j \pi \frac{n}{2}} e^{-j 2 \pi \frac{k}{4} {n}} $.

$ X_k = \sum_{n=0}^{3} e^{-j \pi n \frac {1}{2} { (1 + k) } } ---- [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] $