PLOT time domain & frequency domain of music signal

6 vues (au cours des 30 derniers jours)
sanbadgerdude
sanbadgerdude le 20 Avr 2017
Commenté : the cyclist le 20 Avr 2017
MATLAB PLOT Sinusoidal Music... Can someone help me with how to PLOT my signal for the following code in time domain and frequency domain? Apparently the way I am using PLOT is incorrect.
if true
% code
end
clc
clear all
fs = 8000;
C = 523.251;
F = 698.456;
t = 0:1/fs:0.6;
song = [F*t C*t];
y = sin(2*pi*song);
sound(y,fs);
figure
plot(t,y)

Réponses (1)

the cyclist
the cyclist le 20 Avr 2017
Modifié(e) : the cyclist le 20 Avr 2017
Try this instead
plot([t t+0.6+1/fs],y)
The problem with your code is that you used the t vector twice in generating the song, so y was double the length of t. I fixed that by extending the time vector in the plotting function.
You'll see that there is a discontinuity at t=0.6. The better way to have done this is to define one continuous time vector
t = 0:1/fs:1.2;
and then define your song over segments of that time vector. [But I was too lazy to do all that. :-) ]
  2 commentaires
sanbadgerdude
sanbadgerdude le 20 Avr 2017
I have actually extended my code to play several notes for different durations. I actually have a t1, t2, and t3 now, where
if true
% code
end
t1 = 0:1/fs:0.5;
t2 = 0:1/fs:1;
t3 = 0:1/fs:2;
Does this create more issues. I still could not get your solution to work.
the cyclist
the cyclist le 20 Avr 2017
Conceptually, this is pretty simple. You need a one-to-one correspondence between your t vector and your y vector, in order to plot them.
To make a simple example, you can do
plot([1 2 3 4],[2 3 5 7])
but you are trying to do
plot([1 2 3 4],[2 3 5 7 9 11 13 17])
which will give an error because those vectors do not have a one-to-one correspondence.
So, using your new t1 etc vectors, it looks like your full time sequence is something like
t = [t1 t2+max(t1) t3+max(t1)+max(t2)]
and then you can plot your y value against that time vector.

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