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,

image part 1

image part 2

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 $H(\omega)= kS^*(\omega)e^{-\omega t_0}$


1 Answer 1


The impulse response doesn't change because it is the output of the filter when the input is an impulse.

When the complex input is applied to a complex filter the output is still determined by the convolution of the input and the impulse response of the filter. The addition/integration and multiplication operations that take place inside a convolution operation have to account for complex numbers.

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.

If you are trying to gain insight into how a matched filter is derived - one insight I can provide is that you have to understand that the Signal to Noise ratio (SNR) is defined differently at the input of the filter, than the SNR that we are trying to maximize at the output. At the output we are interested in the maximum value of the output signal, while at the input the SNR is measured using the signal's energy.

  • $\begingroup$ 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? $\endgroup$
    – Aditya P
    Commented Feb 8, 2019 at 1:23
  • 1
    $\begingroup$ 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. $\endgroup$
    – David
    Commented Feb 8, 2019 at 14:31

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