# Fourier series coefficients of Hilbert transform

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 $e^{i \omega t}$ Feb 13, 2017 at 12:09
• $\hat x = x(t)*\frac{1}{\pi t}$ Feb 13, 2017 at 23:21

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!