COMM1001
Modulation and Coding
Dr. Wassim Alexan
Spring 2018
Lecture 2
Bandwidth Efficiency Plane
Fig. 1. Bandwidth efficiency plane (Sklar, Digital Communications, 2nd edition)
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Bandwidth Efficiency Plane
◆ For MFSK modulation, R /W decreases with increasing M
◆ The location of the MFSK points indicates that BFSK (M = 2) and QFSK (M = 4) have the same bandwidth efficiency, even though BFSK requires a higher value of Eb /N0
◆ Can you explain this behavior?
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Bandwidth Efficiency Plane
R = k Rs = log2(M) ·Rs
◆ For MPSK
W =1
Ts= Rs
R /W =log2(M) ·Rs
Rs= log2(M)
◆ For MFSK
W =M
Ts= M ·Rs
R /W =log2(M) ·Rs
M ·Rs=
log2(M)
M
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●
●
●
●
●
■ ■ ■ ■ ■
R/W for MPSK is Log2M
R/W for MFSK isLog2 M
M
2 4 8 16 32M
1
2
3
4
5R/W
Thus, for MPSK, R /W increases with M and for MFSK, R /W decreases with M. More-over, for MFSK, having M = {2, 4} yields the following equal values of R /W
◆ M = 2
R /W =log2(2)
2=
12
◆ M = 4
R /W =log2(4)
4=
12
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Exercise 1
Alphanumeric data are entered into a computer from a remote terminal through a voice-grade telephone channel. The channel has a bandwidth of 3.4 kHz and signal-to-noise ratio of 20 dB. The terminal has a total of 128 symbols. Assume that the symbols are equiprobable and the successive transmissions are statistically independent
◆ (a) Calculate the information capacity of the channel
◆ (b) Calculate the maximum symbol rate for which error-free transmission over the channel is possible
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Exercise 1 Solution
◆ (a) C = W log2(1 + SNR) = 3.4× 1000 log2(101) ≃ 22, 637.9 bits / s = 22.64 kbits / s
◆ (b)
Rs = Rb / log2 M =22.64× 1000
log2 128=
22.64× 10007
= 3, 234.3 symbols / s = 3.2 ksymbols / s
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Exercise 2
The following table has information regarding a number of M - ary modulation schemes.
◆ (a) Calculate the missing information (a and b) to complete the table
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◆ (b) Which modulation scheme would be best-suited for space communications and achieves a BER of 10-5, for the SNR range [11, 20] dB?
◆ (c) Which modulation scheme would be best-suited for a power-limited system and exhibits a spectral efficiency better than 1 bit/s/Hz?
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Exercise 2 Solution
◆ (a)
◆ (b) MFSK is well-suited for space communications, as it has a poor spectral efficiency. For the SNR range [11, 20] dB, BFSK would be the best-suited modulation scheme.
◆ (c) QPSK is the best-suited modulation scheme given these specifications.
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Exercise 3
Consider that a 100 kbits/s data stream is to be transmitted on a voice-grade tele-phone circuit (with a bandwidth of 3 kHz). Is it possible to approach error-free trans-mission with a SNR of 10 dB? If it is not possible, suggest system modifications that might be made.
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Exercise 3 Solution
C = W log2(1 + SNR) = (3× 1000 ) log2(11) = 10, 380 bits / s = 10.38 kbits / s
◆ Thus, error-free transmission is not possible at 100 kbits/s. To modify the system in order to reach this rate, basic trade-offs should be me made. Those include:
◆ Increasing the channel bandwidth
◆ Improving the SNR (either through increased power or the use of low-noise receivers)
◆ Using advanced modulation and coding techniques, e.g.: trellis-coded modulation
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Exercise 4
Consider a telephone modem operating at 28.8 kbits/s and uses trellis-coded QAM modulation.
◆ (a) Calculate the bandwidth efficiency of such a modem, assuming that the usable channel bandwidth is 3429 Hz.
◆ (b) Assuming AWGN and an available Eb /N0 = 10 dB, calculate the theoretically available capacity in the 3429-Hz bandwidth.
◆ (c) What is the required Eb /N0 that will enable a 3429-Hz bandwidth to have a capacity of 28.8 kbits/s?
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Exercise 4 Solution
◆ (a)
η = R /W =28.8× 1000
3429≃ 8.4 bits / s /Hz
◆ (b) We know from lecture 1 that
2C/W = 1 +Eb
N0
C
W→
W2C/W - 1
C=
Eb
N0= 10
3429 2C/3429 - 1
C= 10, by trial and error, C ≈ 20, 300 bits / s
◆ (c)
Eb
N0=
W2C/W - 1
C=
342928.8× 1000
28.4 - 1 = 40.1 ≈ 16 dB
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Exercise 5
Starting with Shannon’s capacity theorem, C = W log2(1 + SNR), show that the theoreti-cal limit for any combination of modulation and coding techniques is -1.6 dB.
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Exercise 5 Solution
We start the derivation with the equations of the signal and noise powers relating to the transmission of a symbol over an AWGN channel
S =Es
Ts= Es Rs and N = N0 W (1)
Shannon’s capacity theorem states that
C = W log2 1 +S
N (2)
Plugging (1) into (2), we get
C = W log2 1 +Es Rs
N0 W (3)
Knowing that Es Rs = k Eb Rb = Eb C and dividing both sides of the equation by W , we can write (3) as
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C
W= log2 1 +
Eb C
N0 W (4)
Now, we let x = Eb
N0 C
W and express (4) as
x
Eb /N0= log2(1 + x) (5)
Dividing both sides by x / (Eb /N0)
1 =Eb
N0
1x
log2(1 + x) (6)
Using a characteristic of logarithmic functions (a log b = log ba)
1 =Eb
N0log2(1 + x)1/x (7)
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Noting that as x → 0, (1 + x)1x → e , and thus
1 =Eb
N0log2 e (8)
Rearranging the terms
Eb
N0=
1log2 e
= 0.693 = -1.59267 dB ≃ -1.6 dB (9)
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Exercise 6
The table in the next slide characterizes four different satellite-to-earth-terminal links. For each link assume that the space loss is 196 dB. For each link, plot an operating point on the bandwidth efficiency plane, R /W versus Eb /N0, and characterize the link according to one of the following descriptions: bandwidth limited, severely bandwidth limited, power limited, and severely power limited
◆ The following relation will help you calculate Eb /N0 values:
Eb
N0
dB= (EIRP)dBW - (losses)dB + (G /T)dB/K + 228.6 - 10 log10(W)bit/s,
where EIRP is the equivalent isotropic power and G /T is the satellite receiver figure of merit.
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Eb
N0
dB= (EIRP)dBW - (losses)dB + (G /T)dB/K + 228.6 - 10 log10(W)bit/s
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Exercise 6 Solution
Eb
N0
dB= (EIRP)dBW - (losses)dB + (G /T)dB/K + 228.6 - 10 log10(W)bit/s
◆ INTELSAT IV:
Eb
N0= 22.5 - 196 + 40.7 + 228.6 - 10 log10165× 106 ≈ 13.6 dB,
R /W =165× 106
36× 106≈ 4.58 bits / s /Hz
◆ DSCS II:
Eb
N0= 28 - 196 + 10 + 228.6 - 10 log10(100× 103) ≈ 20.6 dB,
R /W =100× 103
50× 106≈ 0.002 bits / s /Hz
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Eb
N0
dB= (EIRP)dBW - (losses)dB + (G /T)dB/K + 228.6 - 10 log10(W)bit/s
◆ DSCS II:
Eb
N0= 28 - 196 + 39 + 228.6 - 10 log1072× 106 ≈ 21 dB,
R /W =100× 103
50× 106≈ 1.44 bits / s /Hz
◆ GAPSAT/MARISAT:
Eb
N0= 28 - 196 - 30 + 228.6 - 10 log10(500) ≈ 3.6 dB,
R /W =500
500× 103 ≈ 0.001 bits / s /Hz
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●
●
●
●
6 12 18 24 30Eb/N0
0.10.5
2
4
R/W
INTELSAT IV (13.6,4.58)
DSCS II (21,1.44)
DSCS II (20.6,0.002)
GAPSAT (3.6,0.001)
◆ INTELSAT IV (13.6,4.58): Severely bandwidth limited
◆ DSCS II (21,1.44): bandwidth limited
◆ DSCS II (20.6,0.002): Power limited
◆ GAPSAT (3.6,0.001): Severely power limited
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