Communications II
Lecture 4: Effects of Noise on AM
Professor Kin K. LeungEEE and Computing Departments
Imperial College London© Copyright reserved
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Noise in Analog Communication Systems
• How do various analog modulation schemes perform in the presence of noise?
• Which scheme performs best?
• How can we measure its performance?
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We must find a way to quantify (=to measure) the performance of a modulation scheme.
We use the signal-to-noise ratio (SNR) at the output of the receiver:
N
S
PPSNR
outputreceivertheatnoiseofpoweraverageoutputreceivertheatsignalmessageofpoweraverage
0
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Model of an analog communication system
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PT : The transmitted power
Limited by: equipment capability, cost, government restrictions,interference with other channels, etc
The higher it is, the more the received power (PS ), the higher the SNR
For a fair comparison between different modulation schemes: PT should be the same for all
We use the baseband signal to noise ration SNRbaseband to calibrate theSNR values we obtain
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A Baseband Communication System
• It does not use modulation
• It is suitable for transmission over wires
• The power it transmits is identical to the message power: PT = P
• The results carry over to band-pass systems
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Average signal(=message) power:P =the area under the triangular curve
Assume:Additive, white noise with power spectral density PSD=N0/2
Average noise power at the receiver: PN = area under the straight line = 2WN0/2 = WN0
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SNR at the receiver output:
Note: Assume no propagation loss
Improve the SNR by:(a) increasing the transmitted power (PT ↑), (b) restricting the message bandwidth (W ↓), (c) making the receiver less noisy (N0 ↓).
WNPSNR T
baseband0
ST PP
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REVISION: Amplitude Modulation
General form of an AM signal:
A: the amplitude of the carrier
fc: the carrier frequency
m(t): the message signal
)2cos()]([)( tftmAts cAM
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Modulation index:
mp: the peak amplitude of m(t), i.e., mp = max |m(t)|
Amp
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Signal recovery
pmA 11) : use an envelope detector
2) Otherwise: use synchronous detection=product demodulation=coherent detection
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Synchronous detection
• Multiply the waveform at the receiver with a local carrier of the same frequency (and phase) as the carrier used at the transmitter:
• Use a LPF to recover (m(t) + A)/2 and finally m(t)
• Problem: At the receiver you need a signal perfectly synchronised with the transmitted carrier
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REVISION Amplitude Modulation: Double-sideband suppressed carrier (DSB-SC)
Signal recovery: With synchronous detection only
)2cos()()( tftAmts cDSBS
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Noise in DSB-SC
The true signal received is:
)2sin()()2cos()]()([)2sin()()2cos()()2cos()(
)2sin()()2cos()()()()()(
tftntftntAmtftntftntftAm
tftntftntstntstx
cscc
csccc
cscc
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)()(~ tntAmy c
For synchronous detection:
• multiply with 2cos(2πfct):
• Use a LPF to keep
)4sin()()]4cos(1)[()]4cos(1)[(
)4sin()()2(cos2)()2(cos)(
)()2cos(2)(22
tftntftntftAm
tftntftntftAm
txtfty
cs
ccc
csccc
c
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Signal power at the receiver output:
Power of the noise signal nc(t):
PAtmEAtmAEPS22222 )}({)}({
WNdfNPW
WN 00 2
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SNR at the receiver output:
To which transmitted power does this correspond?
WNPASNR
0
2
0 2
2
)}2(cos)({2
222
PA
tftmAEP cT
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So
Comparison with
Conclusion: a DSB-SC system provides no SNR performance gain over a baseband system.
SCDSBT SNRWN
PSNR 0
0
WNPSNR T
baseband0
basebandSCDSB SNRSNR
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Noise in standard AM, Synchronous Detection
Pre-detection signal:
)2sin()()2cos()]()([)2sin()()2cos()()2cos()]([
)()2cos()]([)(
tftntftntmAtftntftntftmA
tntftmAtx
cscc
csccc
c
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Signal Recovery:
)()(~ tntmAy c
• Multiply with 2 cos(2πfct):
• LPF
)4sin()()]4cos(1)[()]4cos(1)[()]4cos(1[)(
tftntftntftmtfAty
cscc
cc
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Signal power at the receiver output:
Noise power:
PtmEPS )}({ 2
PNPN 02
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SNR at the receiver output:
Transmitted power:
SNR of a baseband signal with the same transmitted power:
AMSNRWN
PSNR 0
0 2
222
22 PAPAPT
WNPASNRbaseband
0
2
2
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Thus:
Note:
Conclusion: the performance of standard AM with synchronous recovery is worse than that of a baseband system.
basebandAM SNRPA
PSNR
2
12 PAP
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Noise in standard AM, Envelope Detection
Phasor diagram of the signals present at an AM receiver
Ei(t): receiver output= y(t)
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Equation too complicated
Must use limiting cases to put it in a form where noise and message are added
22 )()]()([
)(ofenvelope)(
tntntmA
txty
sc
(95)
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1st Approximation: (a) Small Noise Case
Then
Then
)]([)( tmAtn
)]()([)( tntmAtn cs
)]()([)( tntmAty c
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Thus
And in terms of baseband SNR:
Valid for small noise only!
envSNRWN
PSNR 0
0 2
basebandenv SNRPA
PSNR
2
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)]([)( tmAtn
2nd Approximation: (b) Large Noise Case
Isolate the small quantity in (95):
)()()())((2
)()())((1)]()([
)()())((2)())((
)()]()([)(
22222
222
222
222
tntntntmA
tntntmAtntn
tntntmAtntmA
tntntmAty
sc
c
scsc
scc
sc
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where
: the envelope of the noise )()()( 22 tntntEn sc
)()()]([21)(
)()()()]([21)]()([)(
22
22222
tEtntmAtE
tntntntmAtntnty
n
cn
sc
csc
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From the phasor diagram: nc(t) = En(t) cosθn(t)
Then:
Use :1for2
11 xxx
)()(cos)]([21)()(
tEttmAtEty
n
nn
)(cos)]([)()(
)(cos)]([1)()(
ttmAtEtE
ttmAtEty
nn
n
nn
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Noise is multiplicative here!
No term proportional to the message!
Result: a threshold effect, as below some carrier power level (very low A), the performance of the detector deteriorates very rapidly.
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SSB modulation
Single (lower) sideband AM:
where is the Hilbert transform of m(t).is obtained by passing m(t) through a linear filter with transfer
function −jsgn(f ).and m(t) have the same power P .
The average power is A2P/4.
tftmAtftmAts ccSSB 2sin)(ˆ2
2cos)(2
)(
)(ˆ tm)(ˆ tm
)(ˆ tm
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Noise in SSB
Receiver signal x(t) = s(t) + n(t).
Apply a band-pass filter on the lower sideband. Using coherent detection:
After low-pass filtering,
)4sin()()(ˆ2
)4cos()()(2
)()(2
)2cos(2)()(
tftntmA
tftntmAtntmAtftxty
cs
ccc
c
)()(
2)( tntmAty c
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Signal power A2P/4
Noise power for nc(t) = that for band-pass noise = N0W
SNR at output
For a baseband system with the same transmitted power A2P/4
Conclusion: SSB achieves the same SNR performance as DSB-SC (and the baseband model) but only requires half the band-width.
WNPASNRSSB
0
2
4
WNPASNRbaseband
0
2
4
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Summary