Topic: Discrete Fourier Transform

(This problem clarifies how zero-padding a signal changes its DFT.)

## Question

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]?

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

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

, 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

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

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

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

$X_k = \frac{1 - e^{-j \pi n \frac {1}{2} { (1 + k) } } \cdot n } {1 - e^{-j \pi n \frac {1}{2} { (1 + k) } }}$.

Instructor's comment: This line looks suspicious, ,don't you think? On the left-hand-side, you have a function of k, on the right-hand-side, you have a function of k and n. Don't you think it must be wrong? -pm

$X_k = 4$.

Instructor's comment: How did you get from the previous line to here???? -pm

x[n] can be rewritten like this:

$x[n]=(-j)^{n} = (e^{-j\frac{\pi}{2}})^{n} = e^{-j\frac{\pi}{2}n} = e^{j\frac{3\pi}{2}n}$

And then it makes the problem pretty straight forward.

\begin{align} X[k] &= \sum_{n=0}^{N-1} x[n] e^{\frac{-j 2 \pi k n}{N}} \\ &=\sum_{n=0}^{3} e^{j \frac{3\pi}{2} n} e^{-j 2 \frac{\pi}{4} k n} \ \ \text{N=4 due to periodicity of signal}\\ &=\sum_{n=0}^{3} e^{-j \frac{\pi}{2} n (k-3)} \\ &= 4 \delta (k-3) \ \ \text{by comparison to IDFT formula} \\ \end{align}

## Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale. Dr. Paul Garrett