# DFT

The questions below are due on Monday March 29, 2021; 10:00:00 PM.

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Note that this exercise is intended to be solved by hand, without the use of computation.

## 1) Part 1

Consider a signal x_1[n] = (-j)^n + e^{j\pi n/4}

Determine a closed-form expression for the DFT coefficients computed with N=8.
For example, {1\over 2}\delta[k-1] would be [0, 0.5, 0, 0, 0, 0, 0, 0].

Enter a python list of 16 literal values for the DFT coefficients.

## 2) Part 2

Consider a different signal x_2[n] that is zero outside the range 0\leq n < 6, whose corresponding DFT (computed with N=6) is given by:

X_2[k] = X_2[k+6] = \begin{cases} 1 & \text{if}~ k=0\\ j & \text{if}~ k=2\\ -j & \text{if}~ k=4\\ 0 & \text{otherwise} \end{cases}

Consider a new signal y_1[n] = 9-2x_2[n]. What are its DFT coefficients, Y_1[k]?
Try to arrive at your solution without explicitly determining x_2[n].

Enter your solution as a python list of 6 literal values. For example,
\begin{cases} j & \text{if }k=2\\ 0 & \text{otherwise} \end{cases}
would be [0, 0, 1j, 0, 0, 0].

Consider another new signal y_2[n] = 5(-1)^nx_2[n]. What are the DFT coefficients of y_2, Y_2[k]?
Try to arrive at your solution without explicitly determining x_2[n].

## 3) Part 3

Consider a signal x_3[n], which is nonzero only for n=0,1,2,3.

You know the values of the DTFT at two points only. In particular, you know the values X_3(\frac{\pi}{2}) and X_3(\frac{3\pi}{2}). You also know the value of \sum_m x_3[m] and \sum_m e^{j\pi m}x_3[m].

Check Yourself: Explain how you could determine x_3[n] from this information.

This question isn't graded, but it's really good quiz prep! So you should solve the problem on your own before looking at someone else's solution. Once you've done the problem, you can see the staff solution if you'd like by pressing Show/Hide below.

x_3[n] is non-zero for n=0,1,2,3. The DTFT of x_3[n] is:

\begin{aligned} X_3(\Omega) &= \sum_{n=-\infty}^{\infty} x_3[n] e^{-j \Omega n} \\ &= \sum_{n=0}^{3} x_3[n] e^{-j \Omega n} \\ \end{aligned}

Let X_3[K] be the DFT of x_3[n] with analysis window N=4. Then,

\begin{aligned} X_3 &= \frac{1}{4} \sum_{n=0}^{3} x_3[n] e^{-j \frac{2 \pi}{4} n \cdot 0} = \frac{1}{4} \sum_{n=0}^{3} x_3[n] \\ X_3 &= \frac{1}{4} \sum_{n=0}^{3} x_3[n] e^{-j \frac{2 \pi}{4} n} = \frac{1}{4} X_3\left(\frac{\pi}{2}\right) \\ X_3 &= \frac{1}{4} \sum_{n=0}^{3} x_3[n] e^{-j \frac{2 \pi}{4} n \cdot 2} = \frac{1}{4} \sum_{n=0}^{3} x_3[n] e^{-j \pi n} = \frac{1}{4} \sum_{n=0}^{3} x_3[n] e^{j \pi n} \\ X_3 &= \frac{1}{4} \sum_{n=0}^{3} x_3[n] e^{-j \frac{2 \pi}{4} n \cdot 3} = \frac{1}{4} X_3\left(\frac{3 \pi}{2}\right)\\ \end{aligned}

The right hand side of each of the equations are given as known in the problem. We can determine x_3[n] by applying the DFT synthesis equation using these four values.