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?
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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?
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???
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Risposte (5)
Rick Rosson
il 31 Mar 2013
Modificato: Rick Rosson
il 31 Mar 2013
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);
2 Commenti
Ankit Ghosh
il 20 Ago 2017
I think it is the other way round. May you please check if my understanding is wrong or you mistyped ?
Ankit Ghosh
il 20 Ago 2017
Modificato: Ankit Ghosh
il 20 Ago 2017
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.
RISET
il 15 Gen 2017
Modificato: RISET
il 15 Gen 2017
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)...
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Jayram Naykinde
il 25 Ott 2021
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)")
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azad Thanveer
il 2 Giu 2022
>> fm = 5000; >> t = 0: 0.01: 10; >> y = cos(2*pi*fm*t) >> plot(t,y)
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azad Thanveer
il 2 Giu 2022
fm = 5000;
>> t = 0: 0.01: 10;
>> y = cos(2*pi*fm*t)
>> plot(t,y)
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