Question

Calculate DFT of x (n) = {1, 0, 1, 0} *

a) X (k) = {2, 0, 2, 0}

b) X (k) = {1, 0, 1, 0}

c) X (k) = {2, 0, 1, 0}

d) none

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Answer 1

Answer:

Step-by-step explanation:

1.1 Compute the DFT of the 2-point signal by hand (without a calculator or

computer).

x = [20, 5]

1.2 Compute the DFT of the 4-point signal by hand.

x = [3, 2, 5, 1]

1.3 The even samples of the DFT of a 9-point real signal x(n) are given by

X(0) = 3.1,

X(2) = 2.5 + 4.6 j,

X(4) = −1.7 + 5.2 j,

X(6) = 9.3 + 6.3 j,

X(8) = 5.5 − 8.0 j,

Determine the missing odd samples of the DFT. Use the properties of the

DFT to solve this problem.

1.4 The DFT of a 5-point signal x(n), 0 ≤ n ≤ 4 is

X(k) = [5, 6, 1, 2, 9], 0 ≤ k ≤ 4.

A new signal g(n) is defined by

g(n) := W−2 n

5 x(n), 0 ≤ n ≤ 4.

What are the DFT coefficients G(k) of the signal g(n), for 0 ≤ k ≤ 4?

1.5 Compute by hand the circular convolution of the following two 4-point

signals (do not use MATLAB, etc.)

g = [1, 2, 1, −1]

h = [0, 1/3, −1/3, 1/3]

1.6 What is the circular convolution of the following two sequences?

x = [1 2 3 0 0 0 0];

h = [1 2 3 0 0 0 0];

1.7 What is the circular convol

Answer 2

Concept:

The discrete Fourier transform (DFT) is a complex-valued frequency function that converts a finite sequence of equally-spaced function samples into a same-length sequence of equally-spaced discrete-time Fourier transform samples.

Given:

$x(n)=\{1,0,1,0\}$

Find:

The DFT of $x(n)$.

Solution:

$X(0)=\sum_{n=0}^{3} x(n)\\X(0)=x(0)+x(1)+x(2)+x(3)\\X(0)=1+0+1+0\\X(0)=2$$X(1)=\sum_{n=0}^{3} x(n)e^{-j\omega n}\\X(1)=x(0)+x(1)e^{-j\frac{2 \pi}{4} }+x(2)e^{-2j\frac{2 \pi}{4} }+x(3)e^{-3j\frac{2 \pi}{4} }\\X(1)=1+0+1*(-1)+0\\X(1)=0$

$X(2)=\sum_{n=0}^{3} x(n)e^{-2j\omega n}\\X(2)=x(0)+x(1)e^{-2j\frac{2 \pi}{4}}+x(2)e^{-4j\frac{2 \pi}{4}}+x(3)e^{-6j\frac{2 \pi}{4}}\\X(2)=1+0+1*1+0\\X(2)=2$

$X(3)=\sum_{n=0}^{3} x(n)e^{-3j\omega n}\\X(3)=x(0)+x(1)e^{-3j\frac{2 \pi}{4}}+x(2)e^{-6j\frac{2 \pi}{4}}+x(3)e^{-9j\frac{2 \pi}{4}}\\X(3)=1+0+1*(-1)+0\\X(3)=0$

Hence, the DFT of $x(n)$ is $X(k)= \{2,0,2,0\}$. Option(a) is correct.

Hence, the DFT of $x(n)$ is $X(k)= \{2,0,2,0\}$. Option(b) is incorrect.

Hence, the DFT of $x(n)$ is $X(k)= \{2,0,2,0\}$. Option(c) is incorrect.

Hence, the DFT of $x(n)$ is $X(k)= \{2,0,2,0\}$. Option(d) is incorrect.

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