I was looking for a straightforward proof for Bedrosian's theorem which says the Hilbert transform of the baseband signal times the passband signal is the original baseband signal times the Hilbert transform of the passband signal in which by assuming $x(t)$ is a passband signal and and $m(t)$ is a baseband signal, then we have : $$c(t)=\widehat{x(t)m(t)}=\widehat {x(t)}m(t)$$ Now for the sake of proof since we know Fourier transform of Hilbert equals to : $$\hat X(f)=-j\operatorname{sgn}(f)X(f)$$ Then we say : $$\hat c(f)=-j\operatorname{sgn}(f) [x(f) \star m(f)]$$ Now can we say( because $x(t)$ is a passband signal) $X(f)$ is only in positive f part and $\operatorname{sgn}(f)$ value is constant and bring it in? and then we have: $$[-j\operatorname{sgn}(f)x(f)] \star m(f)=\widehat {x(f)}\star m(f)$$ And so we can say

$$\hat c(f)=\widehat {x(f)}\star m(f)$$ $$\hat c(t)=\widehat{x(t)m(t)}=\widehat {x(t)}m(t)$$


1 Answer 1


One easy way to see the validity of the theorem is to consider the corresponding analytic signals. Let a signal $s(t)$ be defined by


where $m(t)$ is a low pass (message) signal, and $c(t)$ is a bandpass (carrier) signal. Note that we don't need to assume that $c(t)$ is sinusoidal. Let $\hat{s}(t)$ denote the Hilbert transform of $s(t)$, and $\hat{c}(t)$ is the Hilbert transform of $c(t)$.

The equality claimed by Bedrosian's theorem


is equivalent to


where $s_a(t)$ and $c_a(t)$ denote the analytic signals corresponding to $s(t)$ and $c(t)$, respectively. Analytic signals are complex-valued signals having only positive frequency components. The analytic signal is given by the original signal plus its Hilbert transform as the imaginary part:


In the frequency domain, the signal $m(t)c_a(t)$ is centered at the (positive) center frequency of the bandpass signal $c(t)$. Its bandwidth equals the bandwidth of $c(t)$ plus twice the bandwidth of the lowpass signal $m(t)$. It has only positive frequency components if the spectra of $m(t)$ and $c(t)$ do not overlap. In that case it equals the analytic signal $s_a(t)$, as claimed in Eq. $(3)$. Otherwise, $m(t)c_a(t)$ has also negative frequency components and $(3)$ is not satisfied.

Note that $(2)$ and $(3)$ are also valid if $c(t)$ is a highpass signal, as long as the spectra of $m(t)$ and $c(t)$ don't overlap.

Also take a look at this related answer and the figure therein.

  • $\begingroup$ Thanks , so i suppose my proof was wrong , right ? $\endgroup$
    – Elias
    Commented Apr 17, 2021 at 20:39
  • $\begingroup$ @Elias: Your argument wasn't very clear to me. $\endgroup$
    – Matt L.
    Commented Apr 17, 2021 at 21:09
  • $\begingroup$ @MattL. Did you mean $s(t)$ instead of $x(t)$ in your second paragraph? $\endgroup$ Commented Jun 9, 2022 at 2:52
  • $\begingroup$ @DanBoschen: Yes, I did. Corrected, thx. $\endgroup$
    – Matt L.
    Commented Jun 9, 2022 at 15:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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