# Fourier transform of derivative expression with respect to time

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##### 2 Comments

Paul
on 30 Jan 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.

### Answers (1)

Paul
on 30 Jan 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)

[num,den] = numden(fourier(diff(f(t),t),t,w));

num/den

##### 7 Comments

Paul
on 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

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)

Now you can use M to proceed with whatever needs to be done with it. A function like pagemldivide would probably come in handy.

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