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131B_1_AMnoise

131B_1_AMnoise - signal power A 2 I 1 a 2 R m H LM The...

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NOISEANALYSISFORAMPLITUDEMODULATION WestudytheeffectofNoiseinAMsystems Wehaveseenthatthesignaltransmiited overthechannel canbeexpressed: s H t L = A H 1 + am H t LL Sin @ 2 Π f c t + Φ D ThechannelismodelledasanadditiveGaussian noisechannelsothatthesignalreceived canbeexpressed: r H t L = s H t L + n H t, Ω L where n H . L iswhiteGaussianwithspectraldensity N.Henceusingthenarrowbandnoiserepresentationwehave r H t L = H A H 1 + am H t LL Cos @ Φ D + n I H t, Ω L L Sin @ 2 Π f c t D + I A H 1 + am H t LL Sin @ Φ D + n q H t, Ω LM Cos @ 2 Π f c t D wherewemodeltheunknowncarrierphase Φ arandomvariableuniform @ 0, 2 Π D , independentofthenoise. ThenSin @ Φ D andCos @ Φ D areuncorrelatedsothat eachtermisuncorrelatedwiththeotherandthecovariance functionofr H t L is R r H t L = A 2 I 1 + a 2 R m H t LM Cos @ 2 Π f c t D + R n H t L Cos @ 2 Π f c t D where Aisthereceivedcarrieramplitude, R n H t L = 2WNSin @ Π 2 Wt D H Π 2Wt L andhenceNoisepower R n H 0 L = 2W N whereNisthespectraldensityofthechannelnoise, and
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Unformatted text preview: signal power A 2 I 1 + a 2 R m H LM The ratio of received signal power to noise power H SNR L is : H S ± N L b = A 2 I 1 + a 2 R m H LM ² H 2 W N L If we demodulate with a phase coherent demodulator and low pass filter we would get A H 1 + a m H t LL + n 1 H t, Ω L and we remove the DC component with a DC block, we get Aa m H t L + n 1 H t, Ω L and the output SNR is A 2 a 2 R m H L± 2 W N = I A 2 a 2 R m H L ² A 2 I 1 + a 2 R m H LMM H S ± N L b = Η H S ± N L b where Η < 1, is the modulation efficiency Example : For speech signals R H L ~ 0.1 And a ~ 0.8-0.9, so that Η ~ .075 or a loss of 11 db because of the power needed to transmit the carrier. 2 AMnoise.nb...
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