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See the title. You could imagine that I am asking about an AM radio, but I will use a simpler example with a lock-in amplifier.

Let's say that I have a lock-in amplifier with a reference frequency of 330 Hz. If I take that signal and modulate it at 0.01 Hz and connect that to the input of the lock-in, then look at the DC output of, say, the X channel, I can see that modulation, even if the lock-in is AC coupled, with a cutoff frequency of 1 Hz (3 orders of magnitude higher).

I do not understand why this is. It seems to me to be the same as an AM radio. If you have a 500 kHz carrier frequency modulated by your signal, if you connect that to a bandpass filter, you still get your signal out, right?

So, to rephrase my question: why is the signal of interest not filtered out in both of the above cases?

Thanks!

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  • \$\begingroup\$ AM is multiplying sine waves. Filters are filtering out harmonics, which are additive components of Fourier series. Not multiplicative. \$\endgroup\$
    – Eugene Sh.
    May 12 '17 at 15:24
  • \$\begingroup\$ @EugeneSh. the referenced to harmonics is mistaken. The question concerns a bandpass filter, not a lowpass filter, and specifically asks why the modulated signal survives when the baseband signal is blocked. This is explained in the answers below. \$\endgroup\$ May 12 '17 at 16:48
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In the frequency domain, AM modulation ("multiplication") is the convolution of the modulated signal with the carrier. Thus, the Fourier Transform energy is at the carrier frequency, and has the bandwidth of the original signal.

So, as long as the width of your lock-in filter is as wide as the bandwidth of the original signal, you don't lose signal.

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  • \$\begingroup\$ Isn't convolution commutative? So why not say it's the convolution of the carrier with the baseband signal, and the energy is at the baseband frequency? Or put another way, in AM the carrier doesn't entirely go away, so why would the baseband entirely go away? \$\endgroup\$
    – Phil Frost
    May 12 '17 at 19:14
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If you have a 500 kHz carrier frequency modulated by your signal, if you connect that to a bandpass filter, you still get your signal out, right?

No. The original modulating waveform is filtered out. However the Amplitude Modulation process creates sidebands above and below the carrier frequency. If these are inside the bandwidth of the output filter then they will be preserved. When this is demodulated you get back (an approximation of) the baseband signal.

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  • \$\begingroup\$ Your "No." is not theoretically correct. if you could make a bandpass filter with steep enough skirts you could reject the carrier and extract the upper sideband. (\understatement ON)It's tough to do, though.(\understatement OFF) \$\endgroup\$ May 12 '17 at 18:45
  • \$\begingroup\$ And then it would be SSB, not AM. (tough, but not impossible. It's how we used to transmit 300 voice telephone channels via Microwave before PCM was developed - hundreds of LC filters that all had to be precisely tuned!). And it's still not 'your' signal. \$\endgroup\$ May 12 '17 at 23:09
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Let's say that I have a lock-in amplifier with a reference frequency of 330 Hz. If I take that signal and modulate it at 0.01 Hz and connect that to the input of the lock-in, then look at the DC output of, say, the X channel, I can see that modulation, even if the lock-in is AC coupled, with a cutoff frequency of 1 Hz (3 orders of magnitude higher).

A lock in amplifier actually demodulates the modulation (0.01 Hz) and presents it to you as seen on your X output. It doesn't just filter it. You cannot filter a plain ordinary AM waveform and hope to recover the baseband (modulation) without a demodulator.

Your understanding is flawed. Modulation doesn't create a baseband signal; it creates sidebands around the carrier hence, when you amplitude modulated 300 Hz with 0.01 Hz you would have sidebands at 299.99 Hz and 300.01 Hz. These easily pass through a 1 Hz filter.

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