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

  • 2
    $\begingroup$ How did you define the Hilbert transform ? Note it is a linear operator so it is enough to look at how it acts on $e^{i \omega t}$ $\endgroup$
    – reuns
    Feb 13, 2017 at 12:09
  • $\begingroup$ $\hat x = x(t)*\frac{1}{\pi t}$ $\endgroup$
    – Shine Sun
    Feb 13, 2017 at 23:21

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.