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In alternating current such as a signal in an antenna the electrical signal in the wire is switching it's polarity back and forth. How does the rectifier keep the other half out? Once you cut the second half out would there not be a "gap" in the electrical signal from the time period that it took to get to the other polarity? And although not enough room to ask in the original question , one is compelled to ask would not cutting the bottom half also cut into the "voice" part of the signal or is the voice part of the signal a duplicate up an down. By voice part I mean the signal that is somehow "mixed" in with a signal that has a much higher frequency that is called a carrier wave.

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  • \$\begingroup\$ Conversion from RF to baseband requires both rectification and storage of envelope voltage into a dielectric. So no gap. \$\endgroup\$ Commented Mar 7, 2017 at 17:42
  • \$\begingroup\$ @TonyStewart.EEsince'75 Isn't rectification enough? Either your ear or an audio transducer will inevitably provide envelope-energy storage. \$\endgroup\$
    – glen_geek
    Commented Mar 7, 2017 at 17:55
  • \$\begingroup\$ any reactance will do to suppress the carrier. \$\endgroup\$ Commented Mar 7, 2017 at 19:05

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Detection of Broadcast AM is exactly the same as rectification of the AC output from a transformer as used in many power supplies. The picture below shows half wave rectification: -

enter image description here

Once you cut the second half out would there not be a "gap" in the electrical signal from the time period that it took to get to the other polarity?

Yes, and that gap has a voltage close to 0 volts. You can see it in the rectified waveform above.

How does the rectifier keep the other half out?

This is what diodes do - they readily conduct current in one direction and block voltage when reverse biased. They operate like the chain-cog on the back wheel of most push bikes - you can pedal forward and apply power to the back wheel but, if you pedal backwards, barely nothing happens.

In AM, the carrier frequency is hundreds or thousands of kHz instaed of 50/60 Hz and the amplitude of the carrier is modulated in amplitude like this: -

enter image description here

Notice that the positive and negative envelopes are identical (but mirrored) in shape and the object is to "capture that shape: -

enter image description here

So, whereas a power supply rectifier and filter tries to produce a clean DC signal with no ripple, the detection of AM tries to produce a clean amplitude signal with little ripple (carrier content).

would not cutting the bottom half also cut into the "voice" part of the signal or is the voice part of the signal a duplicate up an down

There is a danger of this. If the original AM signal is overmodulated like this: -

enter image description here

This cannot be properly recovered to a clean audio wave by a simple diode rectifier and the output looks like the red line (envelope) below: -

enter image description here

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  • \$\begingroup\$ I accepted the answer but I guess I need to understand a little more on why the diode is even necessary. Sound waves are represented by complete sinusoidal waves. The rectification cuts them in half. So there must be enough of a wave to reproduce the exact sound wave, does't this mean it must look like the original sinusoidal wave? Maybe you can also expand how the carrier and signal are "mixed" . modulated. Also you mentioned a time constant in the output filter, could you share this with me please. Thank you \$\endgroup\$
    – Sedumjoy
    Commented Mar 8, 2017 at 16:52
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    \$\begingroup\$ Do you understand that without a diode, you can't convert an AC power waveform into a steady DC level (the act of rectification and smoothing. If you don't understand this you have to before moving onto the slightly more complex nature of AM. "Mixing" isn't adding, it's multiplication of A and B but with a DC offset on the modulator to ensure phase inversion doesn't take place. Try it out on a spreadsheet maybe? Get to grips with the AC power rectification and smoothing first because AM detection is pretty much that. \$\endgroup\$
    – Andy aka
    Commented Mar 8, 2017 at 18:35
  • \$\begingroup\$ The waves are multiplied. That is helpful. Why is it though that the signal cannot remain as such as an ac signal and not work to reproduce the sound in the speaker. Is there a problem with ac signal translating a coded sound wave sinusoidal wave into a speaker ? \$\endgroup\$
    – Sedumjoy
    Commented Mar 9, 2017 at 20:15
  • \$\begingroup\$ To transmit audio and prevent it clashing with someone else transmitting audio you need to segregate your audio by shifting it to part of theradio spectrum that is yours exclusively. The mixer does this and notice how a radio receiver has to be tuned to different stations. The tuning process also uses a mixer to differentiate the multitude of radio signals transmitted. \$\endgroup\$
    – Andy aka
    Commented Mar 10, 2017 at 8:41
  • \$\begingroup\$ A single transmission might occupy a few kHz of bandwidth around 1MHz but it won't leak into a different transmission at (say) 1.012 MHz. So now you have the concept of channels separated by a few kHz each exclusively able to use that part of the spectrum. An antenna perfectly tuned to 1MHz will still be pretty good across a broad section of many channels and this allows different channels to be tuned. \$\endgroup\$
    – Andy aka
    Commented Mar 10, 2017 at 8:54
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First: A crytstal radio works only with AM radio, i.e. the audio signal must be coded as the amplitude of the RF signal.

The diode does in deed cut the negative part of the RF signal, but those parts are very short and evened out by a capacitor after the diode (or just by the capacitance of the speaker).

That resulting signal is the amplitude of the RF signal and is the actual audio signal.

Since the radio frequency (carrier) is much higher (e.g. by factor 1000 or more) than the modulating signal (audio signal) the upper and lower halfes of the RF are "duplicates" (as you call it). So no information is lost by rectification.

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  • \$\begingroup\$ But the voice signal is a complete sine wave isn't it? How is that entire sinusoidal wave captured into the carrier signal? In fact, why is a diode even needed at all? \$\endgroup\$
    – Sedumjoy
    Commented Mar 8, 2017 at 16:47
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It does indeed only keep the upper cycles. But it does not matter. The carrier frequency is so much higher than the modulating amplitude frequency that it is unimportant as the resultant "chopping" after the diode is filtered out by the rest of the circuitry and by the inertia of the speaker.

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The idea of a diode "clipping" only one-half of the radio frequency waveform is not really correct for small signals. The diode is operating in its square-law region, where it is conducting pretty much all the time. However, it does conduct current at the top peak more than the bottom peak. The radio frequency wave sweeps over only a small section of the curved part of its current/voltage curve:
germanium diode current with no bias
The diode is most usually biased near the origin. But a DC bias might bring a diode to its most rapidly-changing curved section so that the power available in the radio signal is efficiently provided to the sound-producing transducer.
The amplitude-modulated wave shape provided by the antenna is nearly sinusoidal. Such a signal produces no sound in the audio transducer. The asymmetry provided by the diode's curve modifies a sinusoid by extending one peak, and stunting the other:
diode causes asymmetry to sine wave
For larger amplitudes, the asymmetry is more pronounced, and causes more current to flow in one direction through the transducer. This asymmetrical current produces sound.

Note that the diode direction makes no difference to the sound (for an unbiased diode). The amplitude-modulated radio wave is symmetrical in every way, so that its positive peak amplitude is matched by its negative peak amplitudes. Either positive peaks or negative peaks produce the same sound - you ears cannot tell the difference.

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