Fourier series coefficients of Hilbert transform

Question

Suppose signal $x(t)$ is periodic with period T. Then $x(t)$ can be represented by its Fourier series representation $$x(t)=\sum_{k=-\infty} ^\infty{} X_ke^{j2\pi kt/T}$$ Let the fourier series representation of $$y(t)=\hat x(t)=\sum_{k=-\infty} ^\infty{} Y_ke^{j2\pi kt/T}$$Where $\hat x(t)$ is Hilbert transform of $x(t)$.Express the Fourier series coefficients $Y_k$ in terms of the coefficient $X_k$

Can someone help me!!?Thanks a lot

How did you define the Hilbert transform ? Note it is a linear operator so it is enough to look at how it acts on $e^{i \omega t}$ — reuns, Feb 13 '17 at 12:09

score 1 · Accepted Answer · answered May 17 '20 at 23:07

Hilbert transform of x(t) is x(t-T/4). We know that for a shift of $t_0$, the Fourier series coefficients get multiplied with $e^{-j2\pi\frac{t_0}{T}}$. In this case coefficients $X_k$ will get multiplied with $e^{-jk\frac{\pi}{2}}$, i.e $Y_k = X_k\cdot e^{-jk\frac{\pi}{2}}$.

I think your had concerns about Hilbert transform of a signal rotating the exponential Fourier series coefficients in the clockwise by $90^\circ$ in the positive frequency axis and anti-clockwise by $90^\circ$ in the negative frequency axis, but it appears like it is not the case for Fourier series!

However except for this definition, the above answer satisfies all the other established properties of Hilbert transform and Fourier series

Fourier series coefficients of Hilbert transform

1 Answers1