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how do Create a message signal m(t) = cos(2πfmt), fm = 5 KHz. and Plot the signal both in time domain and the magnitude of its spectrum in frequency domain?

조회 수: 12 (최근 30일)
how do Create a message signal m(t) = cos(2πfmt), fm = 5 KHz. and Plot the signal both in time domain and the magnitude of its spectrum in frequency domain?
My code is below for time domain
>> fm = 5000;
>> t = 0: 0.01: 10;
>> y = cos(2*pi*fm*t)
>> plot(t,y)
I just get a straight line???

답변 (5개)

Rick Rosson
Rick Rosson 2013년 3월 31일
편집: Rick Rosson 2013년 3월 31일
The sampling rate that you are using is 100 samples per second, whereas the carrier frequency of the message signal is 5,000 hertz. According to the Nyquist Sampling Theorem, you need a sampling rate that is at leat twice the highest frequency that you want to represent. So, in this example, you need a sampling rate of at least 10,000 samples per second.
Let's try 800,000 samples per second, and see if it helps:
Fs = 800e3;
dt = 1/Fs;
t = (0:dt:0.002-dt)';
Fm = 5000;
y = cos(2*pi*Fm*t);
figure;
plot(t,y);
  댓글 수: 2
Ankit Ghosh
Ankit Ghosh 2017년 8월 20일
I think it is the other way round. May you please check if my understanding is wrong or you mistyped ?
Ankit Ghosh
Ankit Ghosh 2017년 8월 20일
편집: Ankit Ghosh 2017년 8월 20일
fs = 1e3;
dt = 1/fs;
t = (0:0.001:10)';
fm = 2e3;
sig = sin(2*pi*fm/fs*t);
plot(t,sig);
title('sin wave with 1KHz sampling rate')
This is working perfectly fine for me as per the text book equations.

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RISET
RISET 2017년 1월 15일
편집: RISET 2017년 1월 15일
as fm =5000 ; so (fm*t)=an integer; for cos(2*pi*n) value is always=0; so try to change it to some other values like 4999 or 3479 or 23 or 5 ;yes for fm = 5947; t = 0: 0.01:1; y = cos(2*pi*fm*t); plot(t,y) u will see the graphs cosine nature is lost it is due to the t intervals u have taken .try to change them as small as possible ....like fm = 5947; t = 0: 0.001:.1; y = cos(2*pi*fm*t); plot(t,y) u will see what u want....{i have not maintained the desired frequency but try to conceptualise what nyquist rate leading towards)...

Jayram Naykinde
Jayram Naykinde 2021년 10월 25일
perform Sampling of a signal
clc;
clear all;
close all;
f0=3;
fs=10;
T=1/f0;
t=0:0.001:5*T;
x_t=cos(2*pi*3*t);
Ts=1/fs;
n=0:Ts:5*T;
x_n=cos(2*pi*f0*n);
subplot(2,1,1)
plot(t,x_t,'-.');
xlabel('Time')
ylabel("x(t)")
subplot(2,1,2) stem(n,x_n,"filled"); xlabel('Time') ylabel("x(n)")

azad Thanveer
azad Thanveer 2022년 6월 2일
>> fm = 5000; >> t = 0: 0.01: 10; >> y = cos(2*pi*fm*t) >> plot(t,y)

azad Thanveer
azad Thanveer 2022년 6월 2일
fm = 5000;
>> t = 0: 0.01: 10;
>> y = cos(2*pi*fm*t)
>> plot(t,y)

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