2 AM Receiver Received Signal X s(t) is the transmitted signal W-W1H(f)ff1f2-f1-f2ReceivedSignals(t) is the transmitted signalLet M(t) be a random process representing the information bearing signal. m(t) will denote a sample function of M(t). M(t) is assumed zero mean WSS with autocorrelation function RM(τ) and power spectral density SM(f). M(t) is assumed a low-pass signal or a baseband signal with spectral content limited to W Hz, i.e.,and the signal power isnW(t) is a sample of zero mean, white noise with power spectral density N0/2.The received signal after the ideal BPF iswhere n(t) is narrow-band noise.
3 Noise in the ReceiverFor DSB-SC Amplitude Modulation, And the PSD of n(t) is From our work on bandpass processes, n(t) can be broken into in-phase and quadrature components where nc(t) and ns(t) are uncorrelated processes, i.e. Furthermore, and given byfc -Wfc+W-fcSN(f)fc-fc -W-fc+WW-WN0f
4 Evaluation of DSB-SCRecall that the transmitted signal is The received signal, r(t), after the ideal BPF filter is where the ideal bandpass filter H(f) at the receiver’s input is and W is the bandwidth of the information process m(t). Find the demodulated r(t):
5 DSB-SC SNR at OutputAssuming synchronous demodulation (e.g., ), find y(t). Power of the received signal at the receiver’s output: Power of the noise at the receiver’s output:
6 DSB-SC SNR at InputTo transmit out signal m(t), we used a transmitter with power equal to the power of s(t) given by This was also considered the power of the signal at the receiver’s input. The noise power at the receiver input calculated for the message bandwidth is
7 Conventional AM (DSB)Recall the transmitted signal is where |m(t)| ≤ 1. If we follow the methodology in DSB-SC assuming synchronous modulation and θ = 0 without any loss of generality, then the output of the low-pass filter is The dc component, Ac/2, is not part of the message and must be removed. The output after a dc blocking device is: Find the SNR at the receiver’s output and input.
8 SSBRecall the transmitted signal is The received signal is Again assuming synchronous demodulation with perfect phase, the output after the LPF is Find:
9 ExampleThe message process M(t) is a stationary process with the autocorrelation functionIt is also known that all the realizations of the message process satisfy the condition max |m(t)|=6. It is desirable to transmit this message to a destination via a channel with 80-dB attenuation and additive white noise with power-spectral density Sn(f) = N0/2 = W/Hz, and achieve a SNR at the modulator output of at least 50 dB. What is the required transmitter power and channel bandwidth if the following modulations schemes are employed?DSB-SCSSBConventional AM with modulation index equal to 0.8.
10 Angle Modulation Effect of additive noise on modulated FM signal Amplitude Modulation vs Angular ModulationImportance of zero-crossing -> instantaneous frequencyApproximateBlock diagram of the receiver
11 ReceiverBandpass FilterAngle DemodulatorLowpass FilterBandpass filters limits noise to bandwidth of modulated signaln(t) is bandpass noiseOr, in Phasor formwhere11
13 Solve for SNR Found demodulated signal y(t) Composed of signal and additive noiseAssumption: m(t) is a sample function of a zero mean stationary Gaussian process with autocorrelation function RM(τ).What about ?Recall:
15 Noise Power Spectrum at Demodulated Output PMFM
16 Noise and Signal Power at Output (LPF) Noise Power at LPFTo Determine Power Out of Signal, recallSignal Power at LPF
17 SNR for Angle Modulation Therefore, SNRoutUsing Modulation IndexesAnd denotingThen
18 SNRout / SNRb Proportional to modulation index squared Increasing improves SNR gain at the expense of bandwidth expansionThe maximum possible SNR gain improvement is exponential as can be shown using Shannon theoryWe cannot increase without limit sent at some point our results will not be valid since they are only approximate resultsFM, like any other nonlinear modulation technique, exhibits a threshold effect and performance. Above certain SNRb , the SNRout is proportional to 2 SNRb . Below the threshold, SNRout maybe worse than SNRb.
19 SNRout / SNRbIn AM, increasing Ac increases SNRout since the received message is proportional to Ac. Here, increasing Ac also increases SNRout but through a different mechanism. Here increasing Ac reduces the amount of noise that affects the message signal.To compensate for the high noise PSD at high frequencies and FM, the PSD of the signal is pre-emphasized in the transmitter to increase its immunity to noise at high frequencies and it is then the emphasized at the receivers output.It can be shown that at threshold, we have for FM systems Given the received power of the modulated signal, this relation gives us the max which ensures that the system works above threshold. Another restricting factor results from Carlson’s rule Then, give and receive power and channel bandwidth
20 ExampleConsider an FM broadcast system with parameter and Assuming find the output SNR and calculate the improvement (in dB) over the baseband system.