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I am having trouble understanding how this ring modulator works i.e. how it multiplies the input & carrier to give us appropriate output? Any help or guide will bedeeply appreciated. Reference - Wikipedia Reference - http://en.wikipedia.org/wiki/Ring_modulation

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  • \$\begingroup\$ Have you tried reading your linked article's text? \$\endgroup\$ – EM Fields May 13 '15 at 10:35
  • \$\begingroup\$ Yes I have gone through the explaination they give there but to my mind it's not clear how the bridge is causing the two signals to get multiplied - also, the theory there says that for this part of cycle this is on & so on... but how excatly voltages are getting multiplied - this I AM NOT able to concieve. \$\endgroup\$ – Madhusudana May 13 '15 at 10:39
  • \$\begingroup\$ They're kind of cheating when they say "multiply" without explaining that when a diode is switched OFF and is, therefore, blocking a signal, that's equivalent to multiplying the signal by zero, and when it's switched ON and the signal is allowed to pass, that's equivalent to multiplying the signal by 1. \$\endgroup\$ – EM Fields May 13 '15 at 10:54
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The use of "multiply" in this instance refers to that when a diode is switched OFF and is, therefore, blocking a signal, that's equivalent to multiplying the signal by zero, and when it's switched ON and the signal is allowed to pass, that's equivalent to multiplying the signal by 1.

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  • \$\begingroup\$ Oh! is it that? I thought that two signals are getting algebrically multiplied. Anyway can you give some reference from where I can read how to multiply two signals as it actually happens in amplitude modulation. Thanx! \$\endgroup\$ – Madhusudana May 13 '15 at 11:06
  • \$\begingroup\$ @Madhusudana: Sure. Try this \$\endgroup\$ – EM Fields May 13 '15 at 12:46

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