Chap. 6: Behavior of Analog Systems In the Presence of Noise

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1 Chap. 6: Behavior of Analog Systems In the Presence of Noise 6.1 Transmission Quality Measurement (figure of merit) A typical baseband system can be modeled as m(t) Transmitter H_p S T Channel H_c Channel noise n(t) S i,n(t) Receiver H_d S o,σ Figure 1: Baseband system. The system model is the message signal is m(t), with BW B and power E[m (t)] = ; transmit power is S T ; channel only introduces AWGN noise, n(t), with PSD S n (f) =N 0 /; the received signal power at receiver frontend is S i. 1 receiver output consists of signal with power S o, and noise component with power σ. For analog message signal m(t), the quality of the received signal is determined by S o /σ (SNR); S o /σ can be increased by increasing S T. For fair comparison and for easy convenience, we shall investigate SNR with a given S i. For this kind of basedband system, since there is no modulation, the receiver can be understood as a low pass filter with a BW B. Therefore, The received signal (message + noise) at the front end of the receiver: y i (t) =m(t)+n(t) The filtered output: y o (t) =m(t)+n LP (t) Noise power Signal to noise ratio: σ = B B S n (f)df = N 0 B = N 0B S o σ = E[m (t)] N 0 B = S i N 0 B = γ Observe that S o = S i, this figure γ will be our benchmark against which we will

2 compare other transmission (modulation) systems Bandpass System DSB-SC cosωc t n(t) cosωc t m(t) Bandpass filter f c ± B LPF k m(t) S i,σ i,y i (t) S o,σ o,y o (t) Transmitter Channel Receiver s(t) Figure : DSB-SC system. For general bandpass system (with modulation), Transmitted signal s(t) is the modulated waveform: s(t) =A(t) cos(ω c t + φ(t)) Channel: additive white Gaussian noise with PSD S n (f) =N 0 /. Receiver frontend is a BPF (with bandwidth appropriately chosen) to suppress out of 4

3 band AWGN noise y i (t) =s(t)+n c cos ω c t + n s sin ω c t Steps: 1. measure S i,σ i, and SNR i;. measure S o,σ o, and SNR o ; 3. express SNR o in terms of SNR i, and compare with the baseband case (γ). Check if the modulation wave transmission improves the SNR o. 5 DSB-SC cosωc t n(t) cosωc t m(t) Bandpass filter f c ± B LPF k m(t) S i,σ i,y i (t) S o,σ o,y o (t) Transmitter Channel Receiver s(t) Figure 3: DSB-SC system. Besides the specifications introduced above for general passband system, the low pass filter (LPF) bandwidth is B. We have following analysis: Channel input: s(t) = m(t) cos ω c t Channel output (also as the receiver frontend): m(t) cos ω c t + n(t) BPF output: y i (t) = m(t) cos ω c t + n i (t) = m(t) cos ω c t + n c (t) cos ω c t + n s (t)sinω c t 6

4 Demodulator output y o (t) = [ y i (t) cos ω c t] h LP F (t) = [m(t) cos ω c t + n c (t) cos ω c t + n s (t) cos ω c t sin ω c t] h LP F (t) = m(t)+ 1 n c (t) (1) Now, look at SNR at several stages: Step 1: (after BPF, i.e., input to the demodulator) S i = E[ m(t) cos ω c t] =E[m (t)]e[cos (ω c t)] = E[m (t)] = σ i = E[n i (t)] = N 0 B Step : (output of LPF) S o = E[m (t)] = = S i ( ) 1 σo = E[n c(t)] = 1 E[n i (t)] = 1 N 0B = N 0 B Step 3: SNR o = S o σ o = S i N 0 B = γ 7 For a fixed transmit power, the SNR at the demodulator output is the same for the baseband and the DSB-SC system. Theoretically, baseband and DSB-SC systems have identical capabilities. 8

5 6.3: SSB-SC An SSB-SC system is shown below. Suppose we consider USB. The passband of the BPF is [f c,f c + B] with a bandwidth B; LPF is the same as DSB-SC with bandwidth B m(t) cosω c t SSB filter n(t) Bandpass filter LSB cosω c t LPF k m(t) S i,σ i,y i (t) S o,σ o,y o (t) Transmitter Channel Receiver s(t) Figure 4: DSB-SC system. Channel input (for USB) S USB (t) = m(t) cos ω c t m h (t)sinω c t E[S USB(t)] = 1 E[m (t)] + 1 E[m h(t)] = E[m (t)] = 9 Channel output: BPF output: Demodulator output y(t) =S USB (t) =m(t) cos ω c t m h (t)sinω c t + n(t) y i (t) =m(t) cos ω c t m h (t)sinω c t + n c (t) cos ω c t + n s (t)sinω c t y o (t) = [y i (t) cos ω c t] h LP F (t) = {[m(t)+n c (t)] cos ω c t +[n s (t) m h (t)] sin ω c t}{ cos ω c t} h LP F (t) = {[m(t)+n c (t)] + [m(t)+n c (t)] cos ω c t +[n s (t) m h (t)] sin ω c t} h LP F (t) = m(t)+n c (t) Now, look at SNR at several stages: Step 1: (after BPF) S i = E[S SSB] =E[m (t)] = σ i = ()(B) N 0 = N 0B 10

6 Step : (output of LPF) S o = E[m (t)] = σ o = E[n c(t) =E[n i (t)] = N 0 B Step 3 We have SNR o = S o σ o = S i N 0 B = γ SNR DSB SC = SNR SSB SC = SNR baseband = γ AM: Coherent/synchronous AM Coherent AM detection is identical to DSB-SC in every respect except for the additional and redundant carrier term. For re-emphasize, the BPF passband is [f c B,f c + B] with a BW of B. The LPF BW is B. AM signal is given as: Signal power: φ AM = [A + m(t)] cos ω c t S i =( ) E[(A + m(t)) ]E[cos ω c t]=a + E[m (t)] = A + Demodulation: φ AM cos ω c t = [A + m(t)] cos ω c t =[A + m(t)] + [A + m(t)] cos(ω c t) LPF output = m(t) (with DC removed). Therefore, S o = E[m (t)] = 1

7 The noise after BPF is n i (t). Before it is fed to LPF, the noise is ni (t) cos(ω c t) = n c cos (ω c t)+ n s cos(ω c t)sin(ω c t) = 1 n c (t)+ 1 n c (t) cos(ω c t)+ 1 n s (t)sin(ω c t) After LPF: n o (t) = 1 n c (t) σ o = 1 E[n c(t)] = 1 E[n i (t)] = 1 N 0B = N 0 B Therefore, SNR = S o σ o = N 0 B = A + S i N 0 B = A + γ Observe that A m p =[m(t)] max and for maximum SNR at A = m p [SNR] max = where t = m p/ 1, which leads to m γ = 1 p + t +1 γ SNR γ/ 13 Therefore, SNR AM SNR DSB/SSB 3dB. 14

8 6.5 AM: Envelope Detection For ED, the transmit part and the receiver front end BPF is the same as the case for coherent AM in Chap.6.4. We begin the analysis from the signal at the output of BPF (input to the demodulator): y i (t) =[A + m(t)] cos ω c t + n i (t) =[A + m(t)+n c (t)] cos ω c t + n s (t)sinω c t The signal power S i = E[(A + m(t)) ]E[cos (ω c t)] = A + To find the envelope of y i (t), we rewrite y i (t) in the form y i (t) = [A + m(t)+n c (t)] + n s cos(ω c t +Θ i (t)) = E i (t) cos(ω c t +Θ i (t)) We only consider small noise case A + m(t) >> n(t) A + m(t) >> n c /n s Then E i (t) A + m(t)+n c (t) 15 The DC component A is blocked by the envelope detector, therefore, y o (t) =m(t)+n c (t) Therefore, the signal power S o = E[m (t)] =. The noise power and SNR = S o σ o σ o = E[n c(t)] = E[n i (t)] = N 0 B = N 0 B = A + S i N 0 B = A + γ Therefore, when noise is small, the performance of the envelope detector is identical to that of the synchronous detector. 16

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