How is calculated the determinant of a matrix containing a fourier transform?

Hi everyone. Let's assume that there are two paired differential equations as shown in equations 1 and 2 and that these equations are solved, that is, the values ​​of A and B are known. However, to find the behavior of and in the frequency domain, the time derivatives of these equations are taken by taking the Fourier transform to obtain equations 3 and 4. Then, when the linear equation system is formed using equations 3 and 4, how is jw calculated when finding the determinant of the coefficient matrix of the system?

Sample number: N=1024 and Sample frequency: Fs=4kHz

(1)

+ (2)

(3)

(4)

(5)

[Erkan]

sorry, i rearranged equation 5

(5)

[Torsten]

I thought you want to reconstruct F_A(omega) and F_B(omega) given A(omega) and B(omega). So why do you need a determinant to do this ? You can simply solve equation (3) for F_A(omega) and equation (4) for F_B(omega) (after computing the Fourier Transforms of A(t) and B(t)).

[Erkan]

What I want to learn is how to calculate jw? w=2*pi*Fs/1024*(0:Fs/2)?

Answer 1

[Matt J]

N=1024; Fs=4000;

f=(0:(N-1-ceil((N-1)/2)))/N*Fs;

jw = 1j*2*pi*f;

determinant =  (jw-x2).*y1 - (jw-x1).*y2  ;

[Erkan]

thanks your answer. Do the jw frequencies in your answer represent the frequencies used in the FFT?

[Matt J]

thanks your answer.

You're welcome. Please Accept-click the answer if it resolves your quesiton.

Do the jw frequencies in your answer represent the frequencies used in the FFT?

Yes, the right-half spectrum.

[Erkan]

Well, isn't jw in your answer the same as w=2*pi*Fs/N*linspace(0,Fs/2,N/2) for the right-half spectrum?

[Matt J]

The spacing between the frequency samples needs to satisfy the equation

Let's check if your proposal satisfies this:

N=1024; Fs=4000;
w=2*pi*Fs/N*linspace(0,Fs/2,N/2);
dw=w(2)-w(1);

dw - 2*pi*Fs/N

ans = 71.5177

Nope.

Now let's check mine:

w=2*pi*(0:(N-1-ceil((N-1)/2)))/N*Fs;
dw=w(2)-w(1);

dw - 2*pi*Fs/N

ans = 0

[Erkan]

Got it. thanks

[Erkan]

Hi Matt, i have a question. As far as I know, the frequency range used in FFT is -Fs/2 to Fs/2. In this range, when jw=(-Fs/2:Fs/N:Fs/2)*2*pi), the number of samples of jw is 1025. But it should be 1024. How can I do this?

[Matt J]

The general formula for the (two-sided) discrete frequency axis is,

NormalizedAxis = -ceil((N-1)/2) : +floor((N-1)/2);

deltaOmega=2*pi*Fs/N; %frequency sample spacing

w=deltaOmega*NormalizedAxis;

[Erkan]

ok. Thank you very much