Test 2 Review Professor Deepa Kundur University of Toronto Professor Deepa Kundur (University of Toronto) Test 2 Review 1 / 57 Test 2 Review Reference: Sections: 4.1, 4.2, 4.3, 4.4, 4.6, 4.7, 4.8 of 5.1, 5.2, 5.3, 5.4, 5.5 6.1, 6.2, 6.3, 6.4, 6.5, 6.6 S. Haykin and M. Moher, Introduction to Analog & Digital Communications, 2nd ed., John Wiley & Sons, Inc., 2007. ISBN-13 978-0-471-43222-7. Professor Deepa Kundur (University of Toronto) Test 2 Review 2 / 57 Chapter 4: Angle Modulation Professor Deepa Kundur (University of Toronto) Test 2 Review 3 / 57 Angle Modulation I Consider a sinusoidal carrier: c (t ) = A c cos(2πf c t + φ c | {z } angle )= A c cos(θ i (t )) θ i (t ) = 2πf c t + φ c =2πf c t for φ c =0 f i (t ) = 1 2π d θ i (t ) dt = f c I Angle modulation : the message signal m(t ) is piggy-backed on θ i (t ) in some way. Professor Deepa Kundur (University of Toronto) Test 2 Review 4 / 57
15
Embed
Test 2 Review - University of Torontodkundur/course_info/316/KundurTest2Review... · Test 2 Review Professor Deepa Kundur University of Toronto Professor Deepa Kundur (University
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Test 2 Review
Professor Deepa Kundur
University of Toronto
Professor Deepa Kundur (University of Toronto) Test 2 Review 1 / 57
Professor Deepa Kundur (University of Toronto) Test 2 Review 42 / 57
Q: A signal g(t) with bandwidth 20 Hz is sampled at the Nyquist rate isuniformly quantized into L = 16, 777, 216 levels and then binary coded. (a) Howmany bits are required to encode one sample? (b) What is the number ofbits/second required to encode the audio signal?
A: (a) Number of bits per sample = log2(16777216) = 24 bits.A: (b) Number of bits per second = Number of bits per sample × Number ofsamples/second = 24× 2× 20 = 960 bits per second
Professor Deepa Kundur (University of Toronto) Test 2 Review 43 / 57
Q: The signal g(t) has a peak voltage of 8 V (i.e., ranges between ±8 V). Whatis the power of the uniform quantization noise eq(t)?Note: The power of quantization noise from a uniform quantizer is:
EQN =∆2
12Watts
where ∆ ≡ separation of quantization levels.
-1 10-2-3 2 3n
1
x [n]q x[n]
A: To compute ∆:
∆ =range of signal
L=
8− (−8)
16777216=
1
1048576=
1
220
EQN =∆2
12Watts =
(2−20)2
12=
2−40
12Watts
Professor Deepa Kundur (University of Toronto) Test 2 Review 44 / 57
Q: The original un-quantized signal has average power of 10 W. What is theresulting ratio of the signal to quantization noise (SQNR) of the uniformquantizer output in decibels?
A: It is given that ES = 10. Therefore,
SQNR = 10 log10
(ES
EQN
)= 10 log10
(10
2−40
12
)≈ 141.2 dB
Professor Deepa Kundur (University of Toronto) Test 2 Review 45 / 57
PCM: Transmission Path
PCM Data Sequence
ChannelOutput
Transmitter ReceiverTranmissionPathSO
URC
E
DES
TIN
ATIO
N
ChannelOutput
TranmissionLine
RegenerativeRepeater
TranmissionLine
RegenerativeRepeater
TranmissionLine
...PCM Data Shaped for Transmission
Decision-making
DeviceAmpli�er-Equalizer
TimingCircuit
DistortedPCMWave
RegeneratedPCMWave
Professor Deepa Kundur (University of Toronto) Test 2 Review 46 / 57
PCM: Regenerative RepeaterDecision-making
DeviceAmpli�er-Equalizer
TimingCircuit
DistortedPCMWave
RegeneratedPCMWave
t
t
t
t
t
THRESHOLD
BIT ERROR
“0” “0” “0”“1” “1” “1”
Original PCM Wave
Professor Deepa Kundur (University of Toronto) Test 2 Review 47 / 57
PCM: Receiver
Two Stages:
1. Decoding and Expanding:
1.1 regenerate the pulse one last time1.2 group into code words1.3 interpret as quantization level1.4 pass through expander (opposite of compressor)
2. Reconstruction:
2.1 pass expander output through low-pass reconstruction filter(cutoff is equal to message bandwidth) to estimate originalmessage m(t)
Professor Deepa Kundur (University of Toronto) Test 2 Review 48 / 57
Chapter 6: Baseband Data Transmission
Professor Deepa Kundur (University of Toronto) Test 2 Review 49 / 57
Baseband Transmission of Digital Data
ThresholdBinary input sequence
LineEncoder
0011100010Source
Threshold
Decision-Making Device
Sampleat time
DestinationTransmit-Filter G(f )
ChannelH(f )
Receive-�lter Q(f )
Outputbinary data
Threshold
Decision-Making Device
Sampleat time
Pulse Spectrum P(f )
bk = {0, 1} and ak =
{+1 if bk is symbol 1−1 if bk is symbol 0
s(t) =∞∑
k=−∞
akg(t − kTb)
x(t) = s(t) ? h(t)
y(t) = x(t) ? q(t) = s(t) ? h(t) ? q(t)
=∞∑
k=−∞
akg(t − kTb) ? h(t) ? q(t) =∞∑
k=−∞
akp(t − kTb)
Professor Deepa Kundur (University of Toronto) Test 2 Review 50 / 57
Baseband Transmission of Digital Data
ThresholdBinary input sequence
LineEncoder
0011100010Source
Threshold
Decision-Making Device
Sampleat time
DestinationTransmit-Filter G(f )
ChannelH(f )
Receive-�lter Q(f )
Outputbinary data
Threshold
Decision-Making Device
Sampleat time
Pulse Spectrum P(f )
∴ y(t) =∞∑
k=−∞
akp(t − kTb)
where p(t) = g(t) ? h(t) ? q(t)
P(f ) = G (f ) · H(f ) · Q(f ).
Professor Deepa Kundur (University of Toronto) Test 2 Review 51 / 57
Baseband Transmission of Digital Data
Threshold
Decision-Making Device
Sampleat time
Transmit-Filter G(f )
ChannelH(f )
Receive-�lter Q(f )
ThresholdBinary input sequence
LineEncoder
0011100010Source Destination
Outputbinary data
Threshold
Decision-Making Device
Sampleat time
Pulse Spectrum P(f )
P(f ) = G(f )H(f )Q(f )
Professor Deepa Kundur (University of Toronto) Test 2 Review 52 / 57
Baseband Transmission of Digital Data
Threshold
Decision-Making Device
Sampleat time
Pulse Spectrum P(f )
P(f ) = G(f )H(f )Q(f )
yi = y(iTb) and pi = p(iTb)
yi =√Eai︸ ︷︷ ︸
signal to detect
+∞∑
k=−∞,k 6=i
akpi−k︸ ︷︷ ︸intersymbol interference
for i ∈ Z
To avoid intersymbol interference (ISI), we need pi = 0 for i 6= 0.
Professor Deepa Kundur (University of Toronto) Test 2 Review 53 / 57
The Nyquist Channel
I Minimum bandwidth channel
I Optimum pulse shape:
popt(t) =√E sinc(2B0t)
Popt(f ) =
{ √E
2B0−B0 < f < B0
0 otherwise, B0 =
1
2Tb
Note: No ISI.
pi = p(iTb) =√E sinc(2B0iTb)
√E sinc
(2 · 1
2TbiTb
)=√E sinc(i) = 0 for i 6= 0.
Disadvantages: (1) physically unrealizable (sharp transition in freq domain); (2)
slow rate of decay leaving no margin of error for sampling times.
Professor Deepa Kundur (University of Toronto) Test 2 Review 54 / 57
Raised-Cosine Pulse SpectrumI has a more graceful transition in the frequency domain
I more practical pulse shape:
p(t) =√E sinc(2B0t)
(cos(2παB0t)
1− 16α2B02t2
)
P(f ) =
√E
2B00 ≤ |f | < f1
√E
4B0
{1 + cos
[π(|f |−f1)(B0−f1)
]}f1 < f < 2B0 − f1
0 2B0 − f1 ≤ |f |
α = 1− f1B0
BT = B0(1 + α) where B0 =1
2Tband fv = αB0
Note: No ISI. ∵ pi = 0 for i 6= 0.Professor Deepa Kundur (University of Toronto) Test 2 Review 55 / 57
Raised-Cosine Pulse Spectrum
f (kHz)
A= sqrt(E)/2B
A/2
0
0B0-B 02B0-2B0B /2 0B /2 0B /2 0B /2
Raised-Cosine,α=0.5
Raised-Cosine,α=1
NyquistPulse, α=0
Nyquist PulseBandwidth
Trade-off: larger bandwidth than Nyquist pulse.Professor Deepa Kundur (University of Toronto) Test 2 Review 56 / 57
The Eye Pattern
Slope dictates sensitivityto timing error
Best samplingtime
Distortion at sampling time
NO
ISE
MA
RGIN
Time interval over which waveis best sampled.
ZERO-CROSSINGDISTORTION
Note: an “open” eye denotes a larger noise margin, lowerzero-crossing distortion and greater robustness to timing error. �
Professor Deepa Kundur (University of Toronto) Test 2 Review 57 / 57