How Matched filter is transformed due to frequency modulation?

Question

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 ?

Jim Clay · Accepted Answer · 2016-01-13 14:31:12Z

2

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

answered Jan 13, 2016 at 14:31

Jim Clay

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$\begingroup$ 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
Commented Jan 13, 2016 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
Commented Jan 13, 2016 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
Commented Jan 14, 2016 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
Commented Jan 15, 2016 at 16:18
$\begingroup$ Very interesting comment. Many thanks @MBaz. $\endgroup$
– jhegedus
Commented Jan 15, 2016 at 18:03

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