PLOT time domain & frequency domain of music signal

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sanbadgerdude
sanbadgerdude 2017-4-20
评论： the cyclist 2017-4-20
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)

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回答（1 个）

the cyclist
the cyclist 2017-4-20
编辑：the cyclist 2017-4-20
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 个评论
sanbadgerdude
sanbadgerdude 2017-4-20
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 2017-4-20
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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