Modulation and Coding

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1 COMM1001 Modulation and Coding Dr. Wassim Alexan Spring 2018 Lecture 2

2 Bandwidth Efficiency Plane Fig. 1. Bandwidth efficiency plane (Sklar, Digital Communications, 2 nd edition) Wassim Alexan 2

3 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 E b / Can you explain this behavior? Wassim Alexan 3

4 Bandwidth Efficiency Plane For MPSK R = k R s = log 2 (M) R s W = 1 T s = R s For MFSK R / W = log 2(M) R s R s = log 2 (M) W = M T s = M R s R / W = log 2(M) R s M R s = log 2(M) M Wassim Alexan 4

5 R/W R/W for MPSK is Log 2 M 2 1 R/W for MFSK is Log 2 M M M Thus, for MPSK, R / W increases with M and for MFSK, R / W decreases with M. Moreover, for MFSK, having M = {2, 4} yields the following equal values of R / W M = 2 M = 4 R / W = log 2(2) 2 R / W = log 2(4) 4 = 1 2 = 1 2 Wassim Alexan 5

6 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-tonoise 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 Wassim Alexan 6

7 Exercise 1 Solution (a) (b) C = W log 2 (1 + SNR) = log 2 (101) 22, bits / s = kbits / s R s = R b / log 2 M = log = = 3, symbols / s = 3.2 ksymbols / s Wassim Alexan 7

8 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 Wassim Alexan 8

9 (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? Wassim Alexan 9

10 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. Wassim Alexan 10

11 Exercise 3 Consider that a 100 kbits/s data stream is to be transmitted on a voice-grade telephone circuit (with a bandwidth of 3 khz). Is it possible to approach error-free transmission with a SNR of 10 db? If it is not possible, suggest system modifications that might be made. Wassim Alexan 11

12 Exercise 3 Solution C = W log 2 (1 + SNR) = ( ) log 2 (11) = 10, 380 bits / s = 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 Wassim Alexan 12

13 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 E b / = 10 db, calculate the theoretically available capacity in the 3429-Hz bandwidth. (c) What is the required E b / that will enable a 3429-Hz bandwidth to have a capacity of 28.8 kbits/s? Wassim Alexan 13

14 Exercise 4 Solution (a) η = R / W = bits / s / Hz (b) We know from lecture 1 that 2 C/W = 1 + E b C W W 2C/W - 1 C = E b = 10 (c) C/ C E b = W 2C/W - 1 C = 10, by trial and error, C 20, 300 bits / s = = db Wassim Alexan 14

15 Exercise 5 Starting with Shannon s capacity theorem, C = W log 2 (1 + SNR), show that the theoretical limit for any combination of modulation and coding techniques is -1.6 db. Wassim Alexan 15

16 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 Shannon s capacity theorem states that S = E s T s = E s R s and N = W (1) Plugging (1) into (2), we get C = W log S N (2) C = W log E s R s W (3) Knowing that E s R s = k E b R b = E b C and dividing both sides of the equation by W, we can write (3) as Wassim Alexan 16

17 C W = log E b C W (4) Now, we let x = E b C W and express (4) as x E b / = log 2 (1 + x) (5) Dividing both sides by x / (E b / ) 1 = E b 1 x log 2(1 + x) (6) Using a characteristic of logarithmic functions (a log b = log b a ) 1 = E b log 2 (1 + x) 1/x (7) Wassim Alexan 17

18 Noting that as x 0, (1 + x) 1 x e, and thus 1 = E b log 2 e (8) Rearranging the terms E b = 1 log 2 e = = db -1.6 db (9) Wassim Alexan 18

19 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 E b /, 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 E b / values: E b db = (EIRP) dbw - (losses) db + (G / T) db/k log 10 (W) bit/s, where EIRP is the equivalent isotropic power and G / T is the satellite receiver figure of merit. Wassim Alexan 19

20 E b db = (EIRP) dbw - (losses) db + (G / T) db/k log 10 (W) bit/s Wassim Alexan 20

21 Exercise 6 Solution INTELSAT IV: E b db = (EIRP) dbw - (losses) db + (G / T) db/k log 10 (W) bit/s E b = log db, R / W = bits / s / Hz DSCS II: E b = log 10 ( ) 20.6 db, R / W = bits / s / Hz Wassim Alexan 21

22 DSCS II: E b db = (EIRP) dbw - (losses) db + (G / T) db/k log 10 (W) bit/s GAPSAT/MARISAT: E b = log db, R / W = bits / s / Hz E b = log N 10 (500) 3.6 db, R / W = bits / s / Hz Wassim Alexan 22

23 R/W 4 INTELSAT IV (13.6,4.58) DSCS II (21,1.44) GAPSAT (3.6,0.001) DSCS II (20.6,0.002) E b / 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 Wassim Alexan 23

Modulation and Coding Tradeoffs

0 Modulation and Coding Tradeoffs Contents 1 1. Design Goals 2. Error Probability Plane 3. Nyquist Minimum Bandwidth 4. Shannon Hartley Capacity Theorem 5. Bandwidth Efficiency Plane 6. Modulation and