Fourier transform of derivative expression with respect to time

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Hi, everybody. The Fourier transform of the derivative expression of a function with respect to time, for example dx(t)/dt, is jwX(w). Here is how w in the expression jw should be calculated.
  2 Commenti
Erkan
Erkan il 30 Gen 2024
In the articles, the jw expression is expressed as "complex time-harmonic angular frequency". How can I calculate the value of w at each time t within a for loop? Is it necessary to create a frequency domain (depending on the operating frequency)?
Paul
Paul il 30 Gen 2024
w is the independent variable of the Fourier transform. "Angular frequency" just means it has units of rad/sec (as opposed to Hz, which would be called "ordinary frequency"). With the default settings, Matlab's fourier and ifourier use w (rad/sec) as the independent variable in the frequency domain, see my example below.
w is not a quantity that's caculated as a function of time.
Perhaps if you explain a bit more about what you're trying to do, further clarification can be provided.

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Risposte (1)

Paul
Paul il 30 Gen 2024
Hi Erikan,
I'm not clear what you're asking about how j*w should be calculated. It's just part of the expression.
If you're asking how to input the Matlab expression for jw, that's just 1j*w
syms t w real
f(t) = exp(-t)*heaviside(t);
F(w) = fourier(f(t),t,w);
1j*w*F(w)
ans = 
[num,den] = numden(fourier(diff(f(t),t),t,w));
num/den
ans = 
  7 Commenti
Paul
Paul il 26 Feb 2024
The jw term in equation (15) is needed to develop plots like Figures 2 and 4. It does not seem to be needed for RK integration of equations (1)-(5).
The matrix on the LHS of (15) can be expressed as:
jw*eye(5) - A, where A is formed from the gamma_ij
We can from the matrix on the left-hand side of equation (15) as a a function of w by using a 3-D array, with the third dimension corresponding to w.
f = logspace(-3,2,100)*1e9; % Hz
f = 1×100
1.0e+11 * 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0001 0.0002 0.0002 0.0002 0.0002 0.0003 0.0003
w = 2*pi*f; % rad/sec, which I *think* is required for (15)
w = reshape(w,1,1,[]);
A = rand(5); % for example, I'm not going to type all the expressions for gamma_ij
M = 1j.*w.*eye(5) - A;
size(M)
ans = 1×3
5 5 100
Now you can use M to proceed with whatever needs to be done with it. A function like pagemldivide would probably come in handy.
Erkan
Erkan il 26 Feb 2024
Hi Paul, thanks for your answer but I think the steps you suggested cannot solve the problem. I'll need to do more research.

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