# Hilbert transform of low pass and high pass signal

Why is the Hilbert transform of the product of a low pass signal and a high pass signal equal to the product of the low pass signal itself with the Hilbert of the high pass signal?

• This sounds like a homework problem? If so, please show what you have done to solve this and we can help you where you are having trouble--- do you know where to start? Mar 9 '17 at 14:57
• @DanBoschen I think that the problem as stated is too hard to be assigned as homework... I hope I'm not just giving away a homework answer, though.
– MBaz
Mar 9 '17 at 15:44
• well, the answer is obvious if the lowpass signal and highpass signal have no non-zero spectrum in common. but no practical lowpass and highpass signals are that. Mar 9 '17 at 16:07

If $s(t)$ is a baseband signal and $x(t)$ is highpass, then under certain conditions, the Hilbert transform of $s(t)x(t)$ is $s(t)\hat{x}(t)$, where $\hat{x}(t)$ is the Hilbert transform of $x(t)$. This is known as Bedrosian's identiy (or theorem).

The easiest case to consider is when the high-pass signal is a sinusoid.

The frequency response of a Hilber transformer is $$H(j\omega)=\begin{cases}-j,\text{ if }\omega>0 \\+j,\text{ if }\omega<0.\end{cases}$$

Note that the Hilbert transform of $\cos(\omega_0t)$ is $\sin(\omega_0t)$.

Now consider a band-limited baseband signal $s(t)$ with Fourier transform $S(j\omega)$. The spectrum of $s(t)\cos(\omega_0t)$ is $$\frac12 S(\omega-\omega_0) + \frac12 S(\omega+\omega_0).$$ The Hilbert transform of $s(t)\cos(\omega_0t)$ is then $$\frac{-j}{2} S(\omega-\omega_0) + \frac{j}{2} S(\omega-\omega_0)=\frac{-j}{2}\left(S(\omega-\omega_0) - S(\omega+\omega_0)\right),$$ which hapens to be the spectrum of $s(t)\sin(\omega_0t)$.

In conclusion, the Hilbert transform of the baseband signal $s(t)$ times the passband signal $\cos(\omega_0t)$ is the original baseband signal times the Hilbert transform of the passband signal.

This idea applies whenever you're finding the Hilbert transform of a signal whose spectrum is divided into two distinct bands which do not overlap; in other words, one band consists of only positive frequencies, and the other band is completely negative frequencies. This means that the lowest frequency component of the high-pass signal has to be larger than the bandwidth of $s(t)$. You can find more about the Bedrosian identity here, and a couple of proofs here.