Generate trapezoidal waveform from a square waveform using convolution

Hello,

I try to generate a trapezoidal waveform from a square waveform using convolution using rectpuls() and fft() but I don't manage to obtain the good result with correct amplitude (i.e. same as the original signal). Could you please help me ?

Thank you !

Example:

t = 0:10e-6:10;
signal = 270*square(t);

trise=100e-6;
x = rectpuls(t-trise/2,trise);
conv = ifft(fft(signal).*fft(x));

figure
plot(t,signal,'k-+'); hold on
plot(t,conv,'r-s'); hold on
legend('signal', 'conv')
xlabel('Time (s)')
ylabel('Signal (-)')

Answer 1

[William Rose]

@Victor Dos Santos,

[edit correct capitalization errors]

Your convolution worked and it did produce a trapezoidal signal. You just need to zoom in more on the transitions to see the trapezoidal shape. See zoomed-in plot below.

If you want wider transitions, then you need to make the signal x have a wider pulse. In its present form, the pulse in x is only 0.0001 time units long.

[Victor Dos Santos]

@William Rose @Abhijeet Thanks for your answers, I know the convolution worked but as you can see the amplitude of the convolution signal does not match the square one.

The answer @Abhijeet gived me to correct the amplitude didn't work:

t = 0:10e-6:10;

amplitude = 270;

signal = amplitude*square(t);

trise = 30e-6;
x = amplitude*rectpuls(t-trise/2,trise);

conv =  (1/length(signal))*ifft(fft(signal).*fft(x));

figure
plot(t,signal,'k-+'); hold on
plot(t,conv,'r-s'); hold on
legend('signal', 'conv')
xlabel('Time (s)')
ylabel('Signal (-)')

[William Rose]

@Victor Dos Santos, if you want the convolution to have the same amplitude as the original signal, then divide by the sum of the valyues in x:

t=0:.01:10; signal=270*square(t);
trise=0.3; x=rectpuls(t-trise/2,trise);
conv=ifft(fft(signal).*fft(x))/sum(x);
figure
subplot(211), plot(t,signal,'-r.',t,conv,'-b.'); legend('signal','conv')
subplot(212), plot(t,x,'-g.'); legend('x')

Or you can normalize x before using it in the convolution, so that the area under the curve in x is one:

x=x/sum(x);

Then do the convolution, and you won't need to divide by sum(x):

conv=ifft(fft(signal).*fft(x));

[Victor Dos Santos]

Thank you @William Rose, that's what I was looking for !

[William Rose]

@Victor Dos Santos, you're welcome.