Question

3. For this problem, we consider a signal made of the sum of three simultaneously present sinusoids (t) sin(2m fit) sin(2"f2t) +sin(2m fst) at frequencies of fl = 3 Hz, f2 = 7 Hz, and f3-11 Hz. (a) Plot the signal as a function of time. Label the axis as amplitude and time in seconds. 〉Ou (b) The signal is to be sampled by a detector. According to the Nyquist Sampling Theorem can pick a range that covers several periods of the sinusoids. discussed in class, what is the minimum (lowest) sampling frequency that will allow these three sinusoids to be correctly 'reconstructed' from the samples (i.e. from the sampled signal)? (c) Plot two sinusoids at twice the Nyquist frequency and at half the Nyquist frequency on top of the curve in part (a). Use only two different symbols instead of connected curves in the plot of the two sinusoids. Describe which symbol represents oversampling and undersampling of the signal in part (a). What would happen to the reconstructed result if the sampling frequency is below the Nyquist frequency? I need the MATLAB code for A, B and C labeled please

Answer 1

solution:

Execute main.m 1 clc 2 clear 3 close all 4 7 -E(t) sin (2*pi*f1*t) sin(2*pi*f2*t) sin(2*pi*f3*t); s + + 9 %time range 0-3000s and 3000 graph points 10 t linspace (0, 3000,1000); 12 %(a) 13 figure 14 plot (t,s (t)) 15 xlabel('time [s]') 16 ylabel('amplitude') 17 title( 's(t) sin(2*på*3*t) + sin(2*på*7*t) + sin(2*på#11*t)') 18 19 %(b) 20%{ 21 Nyquist says sampling frequency fs of a signal should be greater or to twice the highest equal frequency 2 component in the sampled signal. (in this case fs 2f3) if the original 23 signal is to be recovered from the sampled signa] 25 Therefore, the lowest(minimum) sampling frequency fs2*f3 2 1122Hz 26 27 %} 29 30 %(c) 31 %{ 32 33 from (b), sampling frequency fs 22Hz 34 Nyquist frequency fn-half of sampling frequency(1/2)*fs (1/2)*22 11Hz 35

37 Frequency fT, at Twice of Nyquist frequency is given by 38 fT-2 fn -2*1122Hz 39 40 Frequency fH, at Half of Nyquist frequency is given by 1 fH(1/2)*fn(1/2)11-5.5Hz 43 %} 46 47 48 A-1; 49 %A 1; 50 51 ST-@(t) A*sin (2 pi fT t) 52 SH(t) A*sin(2*pi*fH*t); 53 54 %using a sammler scale of 100points and shorter time range 0-1000 for better visual 55 t linspace(0,1000,100) 56 57 figure 58 plot (t,s (t)) 59 hold on 60 plot (t,sT(t),'o r) 61 hold on 62 plot (t, sH(t), o g') 63 legend('s', ST, SH' 64 xlabel ('time [s]') 65 ylabel('amplitude') 66 title(['amplitude of sT and sH is num2str (A)]) 67 68 69 %{

70 oversampling and undersampling 71 ST - Over-sampling 72 SH - Under-sampling | 73 % 74 75 % 76 sampling a signal below the nyquist frequency causes aliasing where 77 higher frequency components masquerade as though they are lower frequency components 78 which causes distortion and makes it impossible to re-construct/recover the original signal from the 9 sampled signal 80%)
output screen shot:

3000 2500 2000 1500 -3 1000 500

time [s] 1 * 1000 200 400 600 800

1 * 1000 200 400 600 800

*FY 8 1 200 400 800 1000

0 .2 -3 400 200 600 800 1000

1000 200 400 600 800

01 이。 0 .2 -3 400 600 200 800 1000
sample program:

clc
clear
close all

f1 = 3; f2 = 7; f3 = 11; %Hz

s = @(t) sin(2*pi*f1*t) + sin(2*pi*f2*t) + sin(2*pi*f3*t);

%time range 0-3000s and 3000 graph points
t = linspace(0,3000,1000);

%(a)
figure
plot(t,s(t))
xlabel('time [s]')
ylabel('amplitude')
title('s(t) = sin(2*pi*3*t) + sin(2*pi*7*t) + sin(2*pi*11*t)')

%(b)
%{
Nyquist says sampling frequency fs of a signal should be greater or equal to twice the highest frequency
component in the sampled signal. (in this case fs > = 2*f3) if the original
signal is to be recovered from the sampled signal

Therefore, the lowest(minimum) sampling frequency fs = 2*f3 = 2*11 = 22Hz

%}

%(c)
%{

from (b), sampling frequency fs = 22Hz
Nyquist frequency fn = half of sampling frequency = (1/2)*fs = (1/2)*22 = 11Hz

Frequency fT, at Twice of Nyquist frequency is given by
fT = 2*fn =2*11 = 22Hz

Frequency fH, at Half of Nyquist frequency is given by
fH = (1/2)*fn =(1/2)*11 = 5.5Hz

%}

fT = 22; fH = 5.5; %Hz

A = 1;
%A = 1;

sT = @(t) A*sin(2*pi*fT*t);
sH = @(t) A*sin(2*pi*fH*t);

%using a sammler scale of 100points and shorter time range 0-1000 for better visual
t = linspace(0,1000,100);

figure
plot(t,s(t))
hold on
plot(t,sT(t),'o r')
hold on
plot(t,sH(t),'o g')
legend('s','sT','sH')
xlabel('time [s]')
ylabel('amplitude')
title(['amplitude of sT and sH is ' num2str(A)])

%{
oversampling and undersampling
sT - Over-sampling
sH - Under-sampling
%}

%{
sampling a signal below the nyquist frequency causes aliasing where
higher frequency components masquerade as though they are lower frequency components
which causes distortion and makes it impossible to re-construct/recover the original signal from the
sampled signal
%}

sample output:

s(t) sin(2*pi*3*t) sin(2*pi*7*t) sin (2*pi* 11*t) 3 0 -2 1500 2500 500 1000 2000 3000 time [s]

amplitude of sT and sH is 1 3 O ST O SH 0 -2 0 100 200 300 400 500 600 700 800 900 1000 time [s]

amplitude of sT and sH is 3 O ST 0 -3 0 100 200 300 400 500 600 700 800 900 1000 time [s]