I have implemented a matched filter based on the Fourier Transform approach. In the real numbers domain that means that I use as the coefficients of my filter (B) the inverted time-samples of the signal that I'm trying to find and I compute:

real_output = IFFT(FFT(A).*FFT(B))

Where, if we assume the moment of exact match, A is an input signal of real value samples and B is the signal made from inverted time-samples of A (or I can also conjugate the spectrum of B and I get the required time inversion because the signal is real). In order to avoid aliasing I perform both FFT with length(A)+length(B)-1 samples. The output of this filter is always real and its maximum, points at the moment in which a match has occurred.

Now assume that instead of a real valued A we have a complex valued A. My questions are:

  1. Is it possible to apply a matched filter for complex domain signals?
  2. Would I have to invert the complex valued A to get the coefficients of my filter or should I do something else like conjugating them?
  3. How do I interpret the output of the filter, should I say a match happens if the absolute value of the output is maximum or just when the real part is maximized?
  4. In some experiments I have noticed that at the moment of a match the output of the filter is real if spectrum(B) is the complex conjugate of spectrum(A), is this correct?

2 Answers 2


Yes, it is possible (at least on paper or code, since complex signal don't exist physically) to apply a matched filter to complex signals.

This is one way to look at it that I think is illustrative. Assume that a pulse $p(t)$ is real and has energy 1, that is, $$\int_{-\infty}^\infty p^2(t) \, dt=1.$$ The filter matched to $p(t)$ has impulse response $p(-t)$. Then, if the filter's input is $ap(t)$ for real $a$, the matched filter's output (sampled at the appropriate time) is equal to $$ap(t)\star p(-t)=a.$$

If the pulse amplitude is complex, say $a+jb$ for real $a$ and $b$, then we can use the linearity of the convolution operation to calculate \begin{align} (a+jb)p(t) \star p(-t) &= \left(ap(t)+jbp(t)\right) \star p(-t) \\ &=ap(t)\star p(-t) + jbp(t)\star p(-t) \\ &= a+jb. \end{align}

One way to interpret the result is that the filter is matching the real part and the imaginary part independently (but "at the same time"). Recall that $j=\sqrt{-1}$ is orthogonal to the real line; this means that $a$ and $b$ don't "interfere" with each other during the convolution.

To summarize, the answers to your questions are:

  1. Yes.
  2. If you can define the system as using a real pulse $p(t)$, then the matched filter is $p(-t)$. If you must use a complex pulse, then the matched filter is $p^\star(-t)$.
  3. A "match" can be declared when the filter's ouput is close enough (in Euclidean distance sense) to the matched signal (see https://en.wikipedia.org/wiki/Maximum_a_posteriori_estimation for more details).
  4. The output of the filter should be real only when $b$ is zero.
  • $\begingroup$ Thanks for your answer, could you please answer the remaining enumerated questions? What do you mean by at least on paper or code? $\endgroup$
    – VMMF
    Apr 7, 2016 at 15:59
  • $\begingroup$ I've edited my answer; hopefully this answers your question better. $\endgroup$
    – MBaz
    Apr 14, 2016 at 16:59
  • $\begingroup$ @MBaz just for the "philosophical" argument: some say that "the changing magnetic field which is the source of the signal measured in MRI is a vector which we represent using complex notation". Could they be actually complex signals, that we cannot measure as complex? If not, should we really talk about "real" digital signals? $\endgroup$ Apr 14, 2016 at 19:02
  • $\begingroup$ @LaurentDuval I don't know, but if you could create higher-dimensional signals (beyond in-phase and quadrature) without time or frequency expansion, you'd revolutionize the field of communications. $\endgroup$
    – MBaz
    Apr 14, 2016 at 23:06

The answer by @MBaz is very complete. I am adding a shameless plug to the debate.

In a past work, I have been involved in the design of an algorithm aimed at finding, locally, the "best" matched filtering involving one signal $d$, and one template $x$ (and later several ones). Due to the noisiness and the non-stationnarity of the (seismic) data, we wanted to perform the matched filtering in a transformed domain. Skipping failures and other details, we ended up with a complex wavelet transform (you can think of windowed Fourier transforms).

And after some trials, we found that a "single" or unary coefficient matched filter was quite efficient. As weird it may seem (single-coefficient real filters are not so interesting), it worked because this coefficient was complex. The amplitude was interpreted as the amplitude correction, and the phase as (somehow) a sub-sample time shift. You can find the whole story in Adaptive multiple subtraction with wavelet-based complex unary Wiener filters, 2012. We later added a lag parameter to account for super-sample shifts. So you can check the derivations with a single complex coefficient quite easily, with the complex conjugation and stuff. So:

  1. Yes it does
  2. You should place yourself in the most general context of complex data for Fourier, correlation, etc. to avoid mistakes
  3. You have an amplitude and a shift term, but it is somehow tricky (sub-sample)
  4. Appparently so, but I am not as sure as @MBaz
  • $\begingroup$ Interesting! From what I see in the paper, the signal model is quite different from that used in communications, but it reminded me of the frequency-selective wireless channel, which has multiple reflections. $\endgroup$
    – MBaz
    Apr 14, 2016 at 23:03
  • $\begingroup$ @MBaz I appreciate you point to such topics in coommunications, a domain I have almost no idea about. I am sure there exist works which already have performed such short adaptive filtering in a local frequency domain. What I am still looking at is the deep interplay, at equivalent performance, between the transform redundancy and the adaptive filter length $\endgroup$ Apr 15, 2016 at 5:05
  • $\begingroup$ @LaurentDuval thank you very much for answering with an example of your professional experience $\endgroup$
    – VMMF
    Mar 14, 2018 at 14:24

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