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Fourier transform in continuous time

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Kelly Howard
Kelly Howard 2016-1-27
编辑： Santiago Vélez Jr. Casallas 2022-1-14
hi guys, i trying out an example from the textbook 'DSP using matlab'.
for t=0:100
a=-100:1
x=exp(a*t)
answer= fft(x)
end
is there any mistake in my code as i could not get the right answer using matlab

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回答（2 个）

Star Strider
Star Strider 2016-1-27
For continuous time signals, you have to use the Symbolic Math Toolbox:
syms a t w
FT = int(exp(a*t) * exp(j*w*t), t, 0, Inf)
FT =
limit(exp(t*a)*exp(t*w*1i), t, Inf)/(a + w*1i) - 1/(a + w*1i)
If ‘a’ is negative, this is much more tractable.
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Walter Roberson
Walter Roberson 2016-1-27
If you were to add the assumption a<0 then it might become more readable.
syms a t w
assume(a<0);
FT = int(exp(a*t) * exp(j*w*t), t, 0, Inf)
Kelly Howard
Kelly Howard 2016-2-1
let say i use a=log(0.3), i can't get any value out of it

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Santiago Vélez Jr. Casallas
编辑：Santiago Vélez Jr. Casallas 2022-1-14
The most direct way is using fourier(f), where "f" is your continuous signal written in symbolic type (sym). You mut first declare the symbols for the time and frequency with "syms", and then apply:
An additional tip:
--> for me this function doesn't work with euler exponential functions, unless you declare them properly. For example, with Laplace functions:
As you can see it doesn't give error,but the result is the very expresion I wrote.
This is because the euler function has especial treatments in fourier tranforms or the integral will not converge. In this example, the constant that acompanies variable "t" (in this case 5), and "t" itself, must be positive, you can find it in Laplace's theory. So you must specify this, or the integral that matlab does will just not converge:
I did it with abs(), but it can b done with assume() funtion as well,for example:
assume([t],'positive');
You can know further about fourier(t) function in here: https://la.mathworks.com/help/symbolic/sym.fourier.html

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