0
$\begingroup$

Intro:

Consider a signal source S which emits a PAM modulated signal, bits encoded as +g(t) and -g(t) pulses.

The matched filter for S is f - f is a function of g.

Then S is fed to an FM modulator and yields a modulated signal, S', let's call the corresponding matched filter of the modulated signal f'.

Question:

If f is known how can f' be computed ?

$\endgroup$
2
$\begingroup$

Normally you wouldn't try to do the match filter directly on the received FM signal. Typically the receiver path would look something like this-

Received Signal -> Noise/Interference Filter -> FM Demod -> Matched Filter

$\endgroup$
  • $\begingroup$ I agree. In general, you want the matched filter to operate on a baseband signal. Also, the FM modulated pulses are unlikely to be orthogonal, so a matched filter doesn't help. $\endgroup$ – MBaz Jan 13 '16 at 15:27
  • $\begingroup$ Many thanks to Jim for the answer and to MBaz for pointing out the orthogonality issue ! Where can I read about this orthogonality issue ? Could you recommend a good book for that ? $\endgroup$ – jhegedus Jan 13 '16 at 18:26
  • 1
    $\begingroup$ There is no orthogonality issue that would affect the matched filter. The (baseband) signals are antipodal, not orthogonal. $\endgroup$ – Dilip Sarwate Jan 14 '16 at 7:51
  • 1
    $\begingroup$ @jhegedus The PAM signal can be expressed as $$\sum_{k=0}^{N}a_k g(t-kT),$$ where $a_k$ is either 1 or -1, and $T$ is the symbol rate. The data $a_k$ can only be recovered with a matched filter if the pulses $g(t-kT)$ form an orthonormal set for all integers $k$. When the PAM signal is frequency-modulated, the orthogonality of the pulses is lost. This is why the best approach is to get the signal back to baseband, and then apply the matched filter. Most books on digital communications (e.g. Sklar) cover this subject. $\endgroup$ – MBaz Jan 15 '16 at 16:18
  • $\begingroup$ Very interesting comment. Many thanks @MBaz. $\endgroup$ – jhegedus Jan 15 '16 at 18:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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