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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