ρ = sin(2π ft) 2π ft To find the minimum value of the correlation, we set the derivative of ρ with respect to f equal to zero.

Transcription

1 Problem 5.1 : The correlation of the two signals in binary FSK is: ρ = sin(π ft) π ft To find the minimum value of the correlation, we set the derivative of ρ with respect to f equal to zero. Thus: ϑρ ϑ f =0=cos(π ft)πt π ft sin(π ft) (π ft) πt and therefore : π ft =tan(π ft) Solving numerically (or graphically) the equation x = tan(x), we obtain x = Thus, π ft = = f = T and the value of ρ is We know that the probability of error can be expressed in terms of the distance d 1 between the signal points, as : P e = Q d 1 where the distance between the two signal points is : d 1 =E b(1 ρ) and therefore : P e = Q Eb (1 ρ) = Q [ 1.17Eb ] Problem 5.13 : (a) It is straightforward to see that : Set I : Four level PAM Set II : Orthogonal Set III : Biorthogonal

2 1 (b) The transmitted waveforms in the first set have energy : A or 1 9A. Hence for the first set the average energy is : E 1 = 1 ( 1 4 A + 1 ) 9A =.5A All the waveforms in the second and third sets have the same energy : 1 A.Hence : E = E 3 = A / (c) The average probability of a symbol error for M-PAM is (5--45) : ( ) (M 1) P 4,PAM = M Q 6E av = 3 (M 1)N Q A 0

3 (d) For coherent detection, a union bound can be given by (5--5) : ) P 4,orth < (M 1) Q ( E s / =3Q A while for non-coherent detection : P 4,orth,nc (M 1) P,nc =3 1 e Es/ = 3 e A /4 (e) It is not possible to use non-coherent detection for a biorthogonal signal set : e.g. without phase knowledge, we cannot distinguish between the signals u 1 (t) andu 3 (t) (oru (t)/u 4 (t)). (f) The bit rate to bandwidth ratio for M-PAM is given by (5--85) : ( ) R =log M =log 4=4 W 1 For orthogonal signals we can use the expression given by (5--86) or notice that we use a symbol interval 4 times larger than the one used in set I, resulting in a bit rate 4 times smaller : ( R W ) = log M M =1 Finally, the biorthogonal set has double the bandwidth efficiency of the orthogonal set : ( ) R = W Hence, set I is the most bandwidth efficient (at the expense of larger average power), but set III will also be satisfactory. 3

4 Problem 5.14 : The following graph shows the decision regions for the four signals : B U C A D U 1 A = U1 > + U B = U1 < U C = U > + U1 D = U < U1 C B W A D W 1 æ As we see, using the transformation W 1 = U 1 + U,W = U 1 U alters the decision regions to : (W 1 > 0,W > 0 s 1 (t); W 1 > 0,W < 0 s (t); etc.). Assuming that s 1 (t) was transmitted, the outputs of the matched filters will be : U 1 =E + N 1r U = N r where N 1r,N r are uncorrelated (Prob. 5.7) Gaussian-distributed terms with zero mean and variance E. Then : W 1 =E +(N 1r + N r ) W =E +(N 1r N r ) will be Gaussian distributed with means : E [W 1 ]=E [W ]=E, and variances : E [W 1 ]= E [W ]=4E. Since U 1,U are independent, it is straightforward to prove that W 1,W are independent, too. Hence, the probability that a correct decision is made, assuming that s 1 (t) was transmitted is : P c s1 = P [W 1 > 0] P [W > 0] = (P [W 1 > 0]) = (1 P [W 1 < 0]) = ( 1 Q ( )) E 4EN0 = ( 1 Q ( E )) = ( 1 Q ( Eb )) where E b = E/ is the transmitted energy per bit. Then : ( ( )) ( )[ Eb Eb P e s1 =1 P c s1 =1 1 Q =Q 1 1 ( )] Q Eb

5 This is the exact symbol error probability for the 4-PSK signal, which is expected since the vector space representations of the 4-biorthogonal and 4-PSK signals are identical. Problem 5.15 : (a) The output of the matched filter can be expressed as : y(t) =Re [ v(t)e jπfct] where v(t) is the lowpass equivalent of the output : v(t) = t 0 { t 0 s 0 (τ)h(t τ)dτ = Ae (t τ)/t dτ = AT ( 1 e ) t/t, 0 t T T 0 Ae (t τ)/t dτ = AT (e 1)e t/t, T t } (b) Asketchofv(t) is given in the following figure : v(t) 0 T t (c) y(t) =v(t)cosπf c t, where f c >> 1/T. Hence the maximum value of y corresponds to the maximum value of v, or y max = y(t )=v max = v(t )=AT (1 e 1 ). (d) Working with lowpass equivalent signals, the noise term at the sampling instant will be : v N (T )= T 0 z(τ)h(t τ)dτ The mean is : E [v N (T )] = T 0 E [z(τ)] h(t τ)dτ =0, and the second moment : E [ v N (T ) ] = E [ T 0 z(τ)h(t τ)dτ T 0 z (w)h(t w)dw ] = N T 0 0 h (T τ)dτ = T (1 e )

6 The variance of the real-valued noise component can be obtained using the relationship Re[N] = 1 (N + N )toobtain:σ Nr = 1 E [ v N (T ) ] = 1 T (1 e ) (e) The SNR is defined as : γ = v max E [ v N (T ) ] = A T e 1 e +1 (the same result is obtained if we consider the real bandpass signal, when the energy term has the additional factor 1/ compared to the lowpass energy term, and the noise term is σ Nr = 1 E [ v N (T ) ] ) (f) If we have a filter matched to s 0 (t), then the output of the noise-free matched filter will be : v max = v(t )= and the noise term will have second moment : giving an SNR of : T 0 s o (t) =A T E [ v N (T ) ] = E [ T 0 z(τ)s 0(T τ)dτ T 0 z (w)s 0 (T w)dw ] = T 0 s 0(T τ)dτ = A T γ = v max E [ v N (T ) ] = A T Compared with the result we obtained in (e), using a sub-optimum filter, the loss in SNR is equal to : ( ) 1 e 1 1 e+1)( =0.95 or approximately 0.35 db

7 Problem 5.16 : (a) Consider the QAM constellation of Fig. P5-16. Using the Pythagorean theorem we can find the radius of the inner circle as: a + a = A = a = 1 A The radius of the outer circle can be found using the cosine rule. Since b is the third side of a triangle with a and A the two other sides and angle between then equal to θ =75 o, we obtain: b = a + A aa cos 75 o = b = 1+ 3 A

8 (b) If we denote by r the radius of the circle, then using the cosine theorem we obtain: A = r + r r cos 45 o = r = A (c) The average transmitted power of the PSK constellation is: P PSK =8 1 8 A = P PSK = A whereas the average transmitted power of the QAM constellation: P QAM = 1 ( 4 A 8 +4(1 + ) [ ] 3) +(1+ 3) A = P 4 QAM = 8 The relative power advantage of the PSK constellation over the QAM constellation is: gain = P PSK P QAM = 8 ( + (1 + 3) )( =1.597 db ) A

9 Problem 5.18 : For binary phase modulation, the error probability is [ ] Eb A T P = Q = Q With P =10 6 we find from tables that A T =4.74 = A T = If the data rate is 10 Kbps, then the bit interval is T =10 4 and therefore, the signal amplitude is A = = Similarly we find that when the rate is 10 5 bps and 10 6 bps, the required amplitude of the signal is A =.1 10 and A = respectively.

10 Problem 5.6 : (a) The number of bits per symbol is k = 4800 R = = Thus, a 4-QAM constellation is used for transmission. The probability of error for an M-ary QAM system with M = k,is With P M =10 5 and k =weobtain [ ] Eb Q ( ( P M = ) [ ]) 3kEb Q M (M 1) = = E b =9.768 (b) If the bit rate of transmission is 9600 bps, then k = =4 In this case a 16-QAM constellation is used and the probability of error is ( P M =1 1 ( 1 1 ) Q 4 [ ]) 3 4 Eb 15 Thus, [ ] 3 Eb Q 15 = = E b = (c) If the bit rate of transmission is 1900 bps, then k = =8 In this case a 56-QAM constellation is used and the probability of error is ( P M =1 1 ( 1 1 ) Q 16 [ ]) 3 8 Eb 55 With P M =10 5 we obtain E b =

11 (d) The following table gives the SNR per bit and the corresponding number of bits per symbol for the constellations used in parts a)-c). k 4 8 SNR (db) As it is observed there is an increase in transmitted power of approximately 3 db per additional bit per symbol.

12 Problem 5.8 : For 4-phase PSK (M = 4) we have the following realtionship between the symbol rate 1/T,the required bandwith W and the bit rate R = k 1/T = log M T (see 5--84): W = R log M R = Wlog M =W = 00 kbits/sec For binary FSK (M = ) the required frequency separation is 1/T (assuming coherent receiver) and (see 5--86): W = M log M R R = Wlog M M = W = 100 kbits/sec Finally, for 4-frequency non-coherent FSK, the required frequency separation is 1/T, so the symbol rate is half that of binary coherent FSK, but since we have two bits/symbol, the bit ate is tha same as in binary FSK : R = W = 100 kbits/sec