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Related theory and graphs

2007-02-05 21:05:13 · 2 answers · asked by Mathew 1 in Science & Mathematics Engineering

2 answers

I assume that you are talking about modulation related to AM. Normal AM there is variations in the amplitude. It can be modeled as multiplication. S(t) x sin(2 pi F t)

where S(t) is the signal
F is Frequency
t is time

For AM S(t) is restricted to positive values only. This is due to the limitations of the devices used to do the multiplication.

In the time domain we are biasing S(t) with some DC to keep it always positive. In the frequency domain this zero frequency produces carrier F +/- zero of simply F. This is the carrier that is not suppressed. Now if the modulator were more complex to allow negative values for S(t) then this DC can be reduced or practically eliminated. This is known as suppressed carrier.

2007-02-05 21:33:06 · answer #1 · answered by Roy E 4 · 0 1

Normal amplitude modulation multiplies a carrier frequency (r.f.) by a modulating signal (audio) to get what looks like the carrier with the audio riding on top of it. In the frequency domain, for a single modulating signal, Fm and a carrier Fc, you get sidebands above and below the carrier. The sidebands are separated from the carrier by the frequency of the modulating signal. This modulated signal would contain frequency components at Fc-Fm, Fc, and Fc+Fm.

The way the power is distributed, the carrier has half the power and each sideband has 1/4 the total power. Since the carrier has no information, it can be removed by phasing or filtering which doubles the power in the sidebands which increases the signal by a factor of two (3 dB). Now if we filter out one of the sidebands (they both have the same info) we can increase the remaining one by another factor of two ( another 3 dB). So going to single sideband, we increase the talk power by 6 dB. To get the information back though, we must mix the received signal with a local carried of Fc.

http://en.wikipedia.org/wiki/Single_sideband

2007-02-05 23:22:25 · answer #2 · answered by Gene 7 · 0 0

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