Fourier transform of derivative expression with respect to time

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.

[Erkan]

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]

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.

Answer 1

[Paul]

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 = 

[Erkan]

Hi Paul, thanks for your answer but I think I didn't express my question fully. How should I create w (angular frequency) values ​​in jw expression?

[Paul]

Can you provide an example, either some math equations or some code, of what you're actually trying to do?

[Erkan]

hi Paul, I tried to explain the process I wanted to do in the picture. I hope I could explain. Note: differential equations are written symbolically.

[Paul]

So far, this isn't really a question about Matlab .....

What you call "matrix A" is properly a 2x2 matrix, not 2x3. We'd have

A(1,1) = 1j*w - x/t_a

A(1,2) = -y/t_b

A(2,1) = -m/t_d

A(2,2) = 1j*w - k/t_c

assuming x, y, m, k, t_a, t_b, t_c, and t_d are all constants.

I thought that the eqation highligted in the rectangle was going to be solved for A(w) and B(w) as functions of w, then those would be inverse transformed to solutions for A(t) and B(t)

I'm still not sure what is the purpose of the loop. It seems like this is important to understand.

What are Atheta(t,A,B) and Btheta(t,A,B)?

My apologies if I'm making this more complicated than it really is.

[Erkan]

Sorry for not being able to express my question fully. I hope this article will make the process I want to do more clear. The rate equations on the 2nd page of the article are calculated in the for loop containing the 4th runge-kutta method, and there is a jw expression in the matrix on the 3rd page of the article. I don't know how to calculate the w value here.

[Paul]

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]

Hi Paul, thanks for your answer but I think the steps you suggested cannot solve the problem. I'll need to do more research.