# Plotting |X(m)|, Xmagnitude of signal x(t)

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Jorge Cantu on 6 May 2015
Commented: Jorge Cantu on 6 May 2015
Hi everyone, I am trying to plot the frequency spectrum of the magnitude for the following signal x(t)=3+5sin(2*pi*4000*t)+2sin(2*pi*6000*t). However, the plot above does not look right. Can someone review my work and see if this is correct?
Here is the task, and the code.
(b) Obtain |X(m)|.
(c) Plot the one-sided spectrum with redundancy, |X(m)| vs frequency.
Code:
close all;
clear all;
t = 0:10:1000; % Time vector
fs = 16000;
xt=3+5*sin(2*pi*4000*t)+2*sin(2*pi*6000*t);
N = 7;
y = abs(fft(xt,N)); % Compute DFT of x
fax_bins = [0 : N-1]; %N is the number of samples in the signal
f_hertz = fax_bins*fs/N;
plot(f_hertz, y)
xlabel('Frequency (Bins)')
ylabel('Magnitude');
title('Double-sided Magnitude spectrum (bins)');
Jorge Cantu on 6 May 2015

Walter Roberson on 6 May 2015
When you use sin(2*pi*t) then each change of 1 in t corresponds to a full cycle of the sine wave. When you use sin(2*pi*4000*t) then each change in 1 in t corresponds to 4000 full cycles of the sine wave.
All of your t are integral, 0, 10, 20, and so on, so you are plotting at points corresponding to 0 full periods of sine, 10*4000 full periods of sine, 20*4000 full periods of sine, and so on. And each of those is of course going to come out exactly the same to within round-off error: all 0 to within round-off. Effectively the "xt" you calculate will be constant to within round-off error.
You need to figure out what you want "10" to represent for t. 10 seconds? If so then you get the problem I described, where you are always sampling at the same point on the wave.
Jorge Cantu on 6 May 2015
I added my plot above, I forgot to do that the first time, but have a look.