Question

1. Determine the DTFS of the following signals. Write a MATLAB program to calculate the coefficients Ck for each of the signals. Plot a) t[n] vs n, and b) Ck vs 2k. (a) 5 points: t[n] = cos(0.3an) (b) 5 points: t[n] = 1 + cos(0.24an) + 3 sin(0.56n)

Determine the DTFS of (B) only.

Answer 1

x[n] = 1 + cos(0.24nn) + 3.sin(0.561)

We can write

12 = 0.2411 12 = 0.567

So

211 N = =Xm

Where m is the smallest integer so that is an integer

211 N, = 0.2471

25 N, =<xm

If m = 3, we get

NA = 25

21 N2 = -Xm

. 20 . N2 = 0,561

x=z= 'N

N2 = 25

Now to find the fundamental period of the signal.

N, 25 1 p N2 – 25 ia

So the fundamental period of the signal will be

No = pN, or qN

No = 1 x 25 or 1x 25

So

The fundamental period is

No = 25

Any periodic discrete time signal can be represented as

No- k=0

This can be in any one period of the signal.

For this signal

x[n] = ) Ckejklon IM

Or

k=-12

x[n] = 1 + cos(0.24nn) + 3.sin(0.561)

Using

eje + e-je cos(0)=9

eje - e-je sin(0)=

We can write

j2

1 3 x[n] = 1 +*230.24nn +5e-10.2471 + 10.56an e-10.567

We have shown that x[n] is periodic with a period of . So

No = 25

120 =

not nigatr []

မ မ မ မ မ | ယ + x- + {+r= (B) + I = [] ယ

+2.50 per veure at 1-07 4 ugerom + = []x

k=-12

x[n] = C-12 e-j12Non +....... +c-je-ion + Co + Cejlon + czejaNon +..........

Comparing, we get

Co = 1

C2 =

C-3

C-7 = به انح

So the DTFS coefficients are

Co = 1

C2 =

C-3

C-7 = به انح

corresponds to
ejklon
. The frequency is

So

corresponds to
ejohon
. So the corresponding frequency is
kΩ) = 0Ω) = 0

corresponds to
еј3 Поп
. So the corresponding frequency is
kΩ) = 3Ω = 3

corresponds to
ej7non
. So the corresponding frequency is
kΩ) = 7Ω = 7

corresponds to
e-j3non
. So the corresponding frequency is
kl. = -310 = -324 = -6

corresponds to
e-70n
. So the corresponding frequency is
kl. =-710 = 72 = -

Please note that the DTFS coefficients are periodic with the period of the signal. So here the DTFS coefficients will be periodic with a period of 25.

I have found the DTFS coefficients between the period -12 and 12. If you want to get the coefficients in the period 0 to 12, you just have to add 25 to the k value in ck.

For example

C-3 = C-3+25 = C22

C-7 = C-7+25 = C18

MATLAB Code to plot the signal and to plot the DTFS

clc;
clear all;
close all;

n = 0:24;
x = 1 + cos(0.24*pi*n) + 3*sin(0.56*pi*n);
stem(n,x,'linewidth',2);
grid on;
xlabel('Time Index, n');
ylabel('Amplitude');
title('x[n]');

X = 1/25*fft(x);
k = 0:24;
Wk = k*2*pi/25;
figure
stem(Wk,abs(X),'linewidth',2);
grid on;
xlabel('\Omega_k');
ylabel('|c_k|');
title('Plot of |c_k|')

The plots

[u]X Amplitude Time Index, n

-- - - - - Plot of