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I'm trying to understand the idea behind filtering the carrier wave from the AM signal to obtain the modulating signal.

There is some explanation in this source: http://reviseomatic.org/help/2-modulation/Amplitude%20Modulation.php

which mentions:

A diode is used to pass the radio frequency signal in one direction only.

A low pass RC filter is used to pass the audio frequencies and decouple or smooth out the radio frequencies.

Two ideas confuses me:

Why is diode used before using the RC filter? Why not just using the RC low-pass filter to filter out the audio signal?

Another thing I wonder is that the plots for AM modulation showing side bands.. They are not the frequency domain of the AM signal right? I mean the following confuses me because it doesnt show the audio signal freq. in x axis. as if there is no audio freq. component.

enter image description here

edit:

Below simulates an AM 1 MHz carrier with a 500Hz modulating signal:

AM in time domain: enter image description here

AM in FFT: enter image description here

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  • \$\begingroup\$ The audio frequency (more strictly the information frequency)is the difference between the carrier and the sidebands. \$\endgroup\$
    – Peter Smith
    Sep 22, 2016 at 12:42
  • \$\begingroup\$ so this plot is not FFT of the AM signal right? \$\endgroup\$
    – user16307
    Sep 22, 2016 at 13:51
  • \$\begingroup\$ Not sure which plot you are looking at, but a plot of the FFT of an AM-modulated RF signal will not show any audio frequency component, only a carrier and sidebands with the audio encoded in their difference. At some intermediate stage in a transmitter you might still see a baseband audio frequency signal that hasn't been filtered out yet, but that won't exactly carry vary far on the air - and if your goal were to transmit data that way, why bother with the RF parts at all? \$\endgroup\$
    – Chris Stratton
    Sep 22, 2016 at 15:04
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    \$\begingroup\$ You have overmodulated the carrier. For 100% amplitude modulation Vm should be half the amplitude of Vc, with an offset of half Vc. This produces a modulation envelope that follows the full peak to peak waveform of Vm. \$\endgroup\$
    – Bruce Abbott
    Sep 22, 2016 at 17:34

2 Answers 2

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The main body of this answer was made before the OP added an edit section to his question - that edit section is inapplicable because the type of modulation (although AM) is full double sideband suppressed carrier modulation and diode detectors will be useless if trying to recover decent audio or linear base-band signals.

A rather idealized but useful picture of an AM modulated signal. Note that the carrier is 500 kHz and the "audio" is in fact a 100 kHz sine wave: -

enter image description here

The spectrum shows that the content of the modulated carrier comprises three frequencies ranging from 400 kHz to 600 kHz (a total width of 200 kHz centred at 500 kHz). If the "audio" were a 1 kHz sine wave then the total width would be 2 kHz ranging from 499 kHz to 501 kHz.

So, the picture above, depicts the output of a modulator where a 100 kHz base-band sine wave is modulating a 500 kHz carrier. Note that the "finished" spectrum does not contain any base-band signal of 100 kHz.

Why not just using the RC low-pass filter to filter out the audio signal?

Because there is no base-band part of the spectrum any more - base-band and carrier are sitting up at about 500 kHz. If every radio transmission also contained base-band frequencies then the lower end of the radio spectrum would be a mess. The whole point is that the base-band spectrum moves up and straddles the carrier frequency.

A carrier frequency does just what it says - it "carries" the base-band rather like a shopping bag carries your groceries so, if the carrier were 500 MHz, the spectrum around 500 MHz would range from 499.90 MHz to 500.10 MHz i.e. still a 200 kHz width.

Demodulation by a diode envelope detector: -

enter image description here

Think of the diode demodulator as just a plain ordinary 50 Hz AC half wave rectifier. The output of the diode has a load and a smoothing capacitor. That is all it is, plain and simple. If you adjusted the amplitude of the AC with (say) a variac, the DC output level would rise and fall as you turned the control knob on the variac. The action of moving the knob modulates the AC amplitude. A diode rectifier IS an envelope detector. Here's probably a more accurate picture to consider: -

enter image description here

Forget about it having the title of RF detector and imagine it's a simple mains transformer fed from a variac.

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  • \$\begingroup\$ But when I do FFT in LTspice for an AM signal it shows base band signal as well in freq domain. Do you mean that Figure 2 is not FFT of the AM signal? \$\endgroup\$
    – user16307
    Sep 22, 2016 at 13:56
  • \$\begingroup\$ i edited mt question, FFT of an AM with 500Hz baseband. It shows the baseband signal as a content of the AM signal. An RC set for 1000Hz low pass filter would pass the base band or not? Is base band not 500Hz anymore? \$\endgroup\$
    – user16307
    Sep 22, 2016 at 14:10
  • \$\begingroup\$ Lets say the carrier is 1MHz and audio signal is 500Hz. So in freq. domain we dont have the baseband anymore(unlike in mt edit)? Audio signal as a component of the AM is lost and it shifted next to the carrier? So the RC should be with a cut off freq: 1MHz - 500Hz? \$\endgroup\$
    – user16307
    Sep 22, 2016 at 14:17
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    \$\begingroup\$ Are you talking about all that mushy stuff at about -160 dB down? \$\endgroup\$
    – Andy aka
    Sep 22, 2016 at 15:03
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    \$\begingroup\$ In the picture you added you show full mathematical multiplication - this is not what AM broadcast stations transmit and this CANNOT be properly converted to audio by a diode detector. Some people will call that sort of modulation "over-modulation" or suppressed carrier modulation. AM broadcast only lightly modulates and retains the shape of the audio superimposed on the carrier. If you inspected the peak at 1 MHz you would only see the 500 Hz displaced sidebands and NO carrier. \$\endgroup\$
    – Andy aka
    Sep 22, 2016 at 16:40
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A diode is used to pass the radio frequency signal in one direction only. That is a confusing explanation, a better one would be:

A diode is used to rectify the signal, resulting in a signal representing the amplitude of the radio signal.

This signal is then filterer to "average out" the signal so that the radio signal (RF) is removed.

Why is diode used before using the RC filter? Why not just using the RC low-pass filter to filter out the audio signal?

Because the audio signal is only present after rectification, i.e. after the diode. Filtering the radio (RF) signal would remove all signal completely !

Another thing I wonder is that the plots for AM modulation showing side bands.. They are not the frequency domain of the signal right?

Of course they are present. A time domain or frequency domain representation of the same signal is still the same signal. So a 1000 kHz RF signal AM modulated with a 1 kHz signal creates a spectrum as in your last picture. Sum and difference frequencies will be present so 1000 kH + and - 1 kHz so there will be components at 999, 1000 and 1001 kHz.

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  • \$\begingroup\$ im sorry but i dont inderstand both still.. "Because the audio signal is only present after rectification, i.e. after the diode. Filtering the radio (RF) signal would remove all signal completely!" why audio signal is present after rectification only?.. "Of course they are present" but when I observe FFT of an AM signal I see the base band where it supposed to be not shifted nearby the carrier. So the plot looks different than FFT of AM. Isnt it? \$\endgroup\$
    – user16307
    Sep 22, 2016 at 14:01
  • \$\begingroup\$ i edited mt question, FFT of an AM with 500Hz baseband. It shows the baseband signal. But it wouldn't show in the side band plots. \$\endgroup\$
    – user16307
    Sep 22, 2016 at 14:08
  • \$\begingroup\$ why audio signal is present after rectification only? Where is the audio signal ? It is in the peaks of the RF signal. How do you detect those peaks ? With a diode. I think you not seeing the sidebands has more to do with how you use the FFT. Try to keep the ratio Frf/Fbaseband limited to for example 100. Simulate an integer number of periods of the baseband signal. \$\endgroup\$
    – Bimpelrekkie
    Sep 22, 2016 at 14:10
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    \$\begingroup\$ Don't talk about AM as if it is a signal, it is not. Call it: AM Modulated carrier. Indeed that will have no 500 Hz component because that 500 Hz is used to modulate that 100 kHz carrier. Filtering that will leave nothing as this signal has no low frequency (500Hz) components. Only after demodulation in this case amplitide detection with a peak detector (one could use a diode and an RC filter) does the LF signal of 500 Hz re-appear. In your line of thought you skipped the demodulation it is a required step ! \$\endgroup\$
    – Bimpelrekkie
    Sep 22, 2016 at 17:35
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    \$\begingroup\$ This is the basis for all wireless communication, you cannot transmit a baseband signal (containing data or sound) like that with small antennas. You could with large antennas but it is inconvenient. A high frequency carrier is needed. For example for Wifi a 2.5 GHz or 5 GHz carrier frequency is used. 2.5 and 5 GHz can be easily transmitted and received using small antennas. Then after receiving the signal is demodulated and that restores the low frequency baseband signal ready for further processing. \$\endgroup\$
    – Bimpelrekkie
    Sep 22, 2016 at 17:39

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