A composite signal is defined as: x(t) = sin(2*pi*10*t) + sin(2*pi *40*t) + sin(2*pi *60*t) + sin(2*pi *50*t). Separate the four frequencies without using any of the transform techniques.
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Amit Shukla
am 21 Nov. 2017
Bearbeitet: Walter Roberson
am 18 Dez. 2024
Which Filtering should i employ for this question? Almost all filtering techniques employ fft in one way or the other, Is there any other way?
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Birdman
am 21 Nov. 2017
My way is all about converting this expression to a equivalent transfer function and let you see the frequency peaks in Bode plot. The following code does the trick:
syms t
x=sin(2*pi*10*t)+sin(2*pi*40*t)+sin(2*pi*50*t)+sin(2*pi*60*t);
Xs=laplace(x);
XsTf=syms2tf(Xs);
opts=bodeoptions;
opts.FreqUnits='Hz';
bode(XsTf,opts)
The first two line defines the x expression symbolically. Then, I took the laplace transform of the expression and the converted it to transfer function(the attached syms2tf function does it). Afterwards, since I obtain a transfer function expression, I expect to see peaks at those frequencies in Bode Plot and the attached figure contains it. Hope this helps to you.
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Birdman
am 21 Nov. 2017
But you can not do it any other way around. You have already eliminated so much method by saying no transformation, but that transformation probably refers to fft and its similar transforms. I assume that this is a homework question and laplace transform is not something that can be thought as a transformation method.
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Walter Roberson
am 18 Dez. 2024
Bearbeitet: Walter Roberson
am 18 Dez. 2024
t = 0:0.001:1;
x1 = sin(2*pi*10*t) + sin(2*pi *40*t) + sin(2*pi *60*t) + sin(2*pi *50*t);
plot(t, x1)
x2 = @(f14) sin(2*pi*f14(1)*t) + sin(2*pi*f14(2)*t) + sin(2*pi*f14(3)*t) + sin(2*pi*f14(4)*t);
f = @(f14) sum((x2(f14)-x1).^2);
bestfreqs = fminsearch(f, [10.5 41 49 62])
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