3
\$\begingroup\$

I am designing an AM receiver that has a carrier frequency of 2MHz with a bandwidth of 20kHz; i.e., the information signal has 10kHz bandwidth.

What I've designed is basically like this:

enter image description here

So I need a carrier signal in-phase with the carrier at the modulator. I have problems with obtaining this carrier signal from the received AM signal. I've designed some passive filters, but non of them had the enough quality factor/selectivity.

I'm looking for a circuit that can separate the carrier signal from the received AM signal, i.e., a heterodyned, Crystalline or Active (etc.) filter. I've tried to design (each) one of the above filters, but I couldn't come up with a working circuit. So I want a more detailed circuit of one of the above solutions. (one of them that can do the job suffices)

\$\endgroup\$
1
  • \$\begingroup\$ (ignoring the negative half wave) a simple diode is a device that implicitely multiplies your AM signal with a rectangular in-phase carrier. No need for an external carrier. \$\endgroup\$
    – Curd
    Jan 7, 2014 at 11:03

1 Answer 1

1
\$\begingroup\$

You can't demodulate full and proper DSBSC (double sideband, suppressed carrier) by trying to filter the carrier frequency but, you should be able to do this on your design because you are adding a dc offset to your modulation and this means that there will be carrier signal present that is "gettable" with filtering.

You should consider using a resonant tank circuit to achieve this because, your modulation method may over-modulate and reverse the phase of the carrier in the composite signal. The resonant tank will keep the coherent carrier present in the detector despite it reversing on parts of the composite signal. However, if you over-modulate too much this method produces rapidly diminishing results and that leaves you with fairly complex methods such as the Costas detector or a squaring detector, both shown below: -

enter image description here

enter image description here

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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