# What will the impulse response of a matched filter look like if the input is complex?

If $$h(t)$$ is the impulse response of a filter matched to a signal $$s(t)$$, I read that $$h(t) = ks(t_o - t)$$.

But what if the signal is complex? I went through the derivation of the matched filter and saw that just like how the inner product of two vectors is maximum if the vectors are "parallel". I think $$H(\omega)$$ has to be proportional to $$S^*(\omega)e^{-\omega t_0}$$ for the SNR to be maximized. This is a generalization of the Cauchy–Schwarz inequality that I don't fully understand.

So from this we can say, $$H(\omega) = S^*(\omega)e^{-\omega t_0}$$

To get RHS,

frequency shifting, $$s(t-t_o) \iff e^{-j\omega t_0}S(\omega)$$

taking conjugate on both sides, $$s^*(t-t_o) \iff e^{j\omega t_0}S^*(-\omega)$$

scaling with $$-1$$, $$s^*(t_o-t) \iff e^{-j\omega t_0}S^*(\omega)$$

So can I conclude $$h(t) = s^*(t_o-t)$$??

To apply this general formula to a specific example, consider this question from the 2013 exam of ISRO,  From what I learned, can I say that the answer is option D?

The impulse response for the given complex signal $$s(t)$$ is complex conjugate time reversal of $$s(t)$$, i.e. the real part of $$s(t)$$ simply gets time reversed (flips across y axis) so A B or D, while the imaginary part is flipped across both the y and x axis. C or D.

$$h(t) = s^*(t_o-t)$$

$$h(t) = (x(t_o-t)+iy(t_o-t))^*$$

$$h(t) = x(t_o-t)-iy(t_o-t)$$

So my question is, what is the impulse response of a matched filter if the input is complex and why? Like please point out any mistakes in my logic, I especially would love to gain some intuition on why $$H(\omega)= kS^*(\omega)e^{-\omega t_0}$$

I think it is difficult to gain intuition for the result of a matched filter in the time domain. Instead look at it from the frequency domain. \begin{align} Y(\omega) &=& H(\omega)S(\omega) \\ & = & e^{-j\omega t_0}S^*(\omega) S(\omega)\\ & = & e^{-j\omega t_0} ||S(\omega)||^2 \end{align}
In the case where the envelope of $$S(\omega)$$ is rectangular (as in your example) then $$||S(\omega)||^2$$ is equal to a rectangular pulse. The output then be a time delayed version of a sinc() pulse. The wider the frequency range of $$S(\omega)$$ then the narrower the main pulse of the sinc() pulse.
• The impulse response for a matched filter will change depending on which input signal it's matched to right? When $s(t)$ rectangular like in the question, I thought $S(\omega)$ would be sinc() in shape. So what's the answer to the question? Is it D? – Aditya Feb 8 at 1:23
• If you change the signal you are matching to, i.e $S(\omega)$ then the matched filter will change as well. The sinc() appears at the matched filter output when you apply the signal you are trying to detect to the input - it isn't the impulse response. The matched filter is the time reversed complex conjugate - so both the real and imaginary components are reversed and then take the negative of the imaginary, so yes D. Note, to make the filter causal, it is usually time delayed. – David Feb 8 at 14:31