CHAPTER 5. e h(t) = F 1 [H(f)] = F 1 F 1 j2πft

Transcription

1 CHAPER 5 Problem 5. : (a) aking the inverse Fourier transform of H(f), we obtain : [ ] [ ] e h(t) F [H(f)] F F jπf jπf jπf ( ) t sgn(t) sgn(t )Π where sgn(x) is the signum signal ( if x>, - if x<, and if x )andπ(x) isa rectangular pulse of unit height and width, centered at x. (b) he signal waveform, to which h(t) ismatched, is: ( ) ( t s(t) h( t) Π Π t ) h(t) ( ) t wherewehaveusedthesymmetryofπ with respect to the t axis. Problem 5. : (a) he impulse response of the matched filter is : { A h(t) s( t) ( t)cos(πf c( t)) t otherwise (b) he output of the matched filter at t is : g( ) h(t) s(t) t v τ A A [ v 3 A A h( τ)s(τ)dτ ( τ) cos (πf c ( τ))dτ v cos (πf c v)dv ( v πf c [ ( πf c 8 (πf c ) 3 8 (πf c ) 3 76 ) sin(4πf c v)+ v cos(4πf ] cv) 4(πf c ) ) sin(4πf c )+ cos(4πf ] c ) 4(πf c )

2 (c) he output of the correlator at t is : q( ) s (τ)dτ A τ cos (πf c τ)dτ However, this is the same expression with the case of the output of the matched filter sampled at t. hus, the correlator can substitute the matched filter in a demodulation system and vice versa. Problem 5.3 : (a) In binary DPSK, the information bit is transmitted by shifting the phase of the carrier by π radians relative to the phase in the previous interval, while if the information bit is then the phase is not shifted. With this in mind : Data : Phase θ :(π) π π π π π π Note : since the phase in the first bit interval is, we conclude that the phase before that was π. (b) We know that the power spectrum of the equivalent lowpass signal u(t) is: sin πf πf Φ uu (f) G(f) Φ ii (f) where G(f) A, is the spectrum of the rectangular pulse of amplitude A that is used, and Φ ii (f) is the power spectral density of the information sequence. It is straightforward to see that the information sequence I n is simply the phase of the lowpass signal, i.e. it is e jπ or e j depending on the bit to be transmitted a n (, ). We have : I n e jθn e jπan e jθ n e jπ k a k he statistics of I n are (remember that {a n } are uncorrelated) : E [I n ] E [ e jπ ] k a k k E [e jπa [ k ] k ejπ ej] k E [ I n ] E [ e jπ k a k e jπ k k] a [ E [I n+m In ] E e jπ n+m k a k e jπ ] n k a k E [ e jπ m kn+ a k] mkn+ E [e jπa k ] Hence, I n is an uncorrelated sequence with zero mean and unit variance, so Φ ii (f), and sin πf πf Φ uu (f) G(f) A Φ ss (f) [Φ uu(f f c )+Φ uu ( f f c )] 77

3 A sketch of the signal power spectrum Φ ss (f) is given in the following figure : Power spectral density of s(t) fc fc f Problem 5.4 : (a) he correlation type demodulator employes a filter : f(t) { t o.w } as given in Example 5--. Hence, the sampled outputs of the crosscorrelators are : r s m + n, m, where s,s A and the noise term n is a zero-mean Gaussian random variable with variance : σn N he probability density function for the sampled output is : p(r s ) p(r s ) πn e r N e (r A πn ) N Since the signals are equally probable, the optimal detector decides in favor of s if PM(r, s )p(r s ) >p(r s )PM(r, s ) 78

4 otherwise it decides in favor of s. he decision rule may be expressed as: PM(r, s ) PM(r, s ) e (r A ) r N e (r A )A s N > < s or equivalently : r s > < s A he optimum threshold is A. (b) he average probability of error is: P (e) P (e s )+ P (e s ) Q A A A N [ p(r s )dr + A πn e r N dr + N A p(r s )dr A e x dx + π ] Q [ SNR ] e (r A πn N A N ) dr π e x dx where SNR A N hus, the on-off signaling requires a factor of two more energy to achieve the same probability of error as the antipodal signaling. Problem 5.5 : Since {f n (t)} constitute an orthonormal basis for the signal space : r(t) N n r n f n (t), s m (t) 79

5 Nn s mn f n (t). Hence, for any m : C(r, s m ) r(t)s m(t)dt s m (t)dt Nn r n f n (t) N l s ml f l (t)dt Nn s mn f n (t) N l s ml f l (t)dt N n r n Nl s ml f n(t)f l (t)dt N n s mn Nl s ml f n(t)f l (t)dt N n r n s mn N n s mn where we have exploited the orthonormality of {f n (t)} : f n(t)f l (t)dt δ nl. he last form is indeed the original form of the correlation metrics C(r, s m ). Problem 5.6 : he SNR at the filter output will be : SNR y( ) E [ n( ) ] where y(t) is the part of the filter output that is due to the signal s l (t), and n(t) is the part due to the noise z(t). he denominator is : so we want to maximize : E [ n( ) ] E [z(a)z (b)] h l ( a)h l ( b)dadb N h l( t) dt SNR From Schwartz inequality : s l (t)h l ( t)dt Hence : and the maximum occurs when : s l(t)h l ( t)dt N h l( t) dt h l ( t) dt s l (t) dt SNR s l (t) dt E SNR max N N s l (t) h l ( t) h l(t) s l ( t) 8

6 Problem 5.7 : [ ] N mr Re z(t)fm (t)dt (a) Define a m z(t)f m (t)dt. hen, N mr Re(a m ) [a m + a m ]. [ ] E(N mr )Re E (z(t)) fm (t)dt since, E [z(t)]. Also : E ( N mr ) E [ a m +(a m ) +a m a m 4 ] But E (a m)e [ z(a)z(b)f m(a)f m(b)dadb ], since E [z(a)z(b)] (Problem 4.3), and the same is true for E [ (a m )], since E [z (a)z (b)] Hence : E (Nmr ) E [ ] a ma m E [z(a)z (b)] fm (a)f m(b)dadb N f m(a) da EN (b) For m k : E [N mr N kr ] E [ a m+a m ] a k +a k E [ a ma k +a ma k +a ma k +a ma k 4 But, similarly to part (a), E [a m a k ]E[a m a k ], hence, E [N mrn kr ]E [ ] a ma k +a ma k 4. Now : E [a m a k] E [z(a)z (b)] fm(a)f k (b)dadb N f m (a)f k(a)da since, for m k, the waveforms are orthogonal. Similarly : E [a m a k], hence : E [N mr N kr ]. ] Problem 5.8 : (a) Since the given waveforms are the equivalent lowpass signals : E E s (t) dt A dt A / s (t) dt A dt A / 8

7 Hence E E E. Also :ρ E s (t)s (t)dt. (b) Each matched filter has an equivalent lowpass impulse response : h i (t) s i ( t). he following figure shows h i (t) : h (t) h (t) A A t t -A æ (c) h (t) s (t) A h (t) s (t) A / t A / t æ 8

8 (d) s (τ)s (τ)dτ A / t A s (τ)s (τ)dτ t æ (e) he outputs of the matched filters are different from the outputs of the correlators. he two sets of outputs agree at the sampling time t. (f) Since the signals are orthogonal (ρ ) the error probability for AWGN is P Q ( E N ), where E A /. Problem 5.9 : (a) he joint pdf of a, b is : p ab (a, b) p xy (a m r,b m i )p x (a m r )p y (b m i ) πσ e σ [(a m r) +(b m i ) ] φ tan b/a a u cos φ, b u sin φ he Jacobian of the transforma- a/ u a/ φ u, hence : b/ u b/ φ (b) u a + b, tion is : J(a, b) p uφ (u, φ) where we have used the transformation : { M m r + m i θ tan m i /m r u πσ e σ [(u cos φ m r) +(u sin φ m i ) ] u πσ e σ [u +M um cos(φ θ)] } { mr M cos θ m i M sin θ } 83

9 (c) p u (u) π p uφ (u, φ)dφ u u +M πσ e σ π π u u +M σ e σ π u u +M ( σ e σ I ) o um/σ e σ [ um cos(φ θ)] dφ e um cos(φ θ)/σ dφ Problem 5. : s(t)+z(t) s(t)+z(t) z(t) (a) U Re [ r(t)s (t)dt ],wherer(t) depending on which signal was sent. If we assume that s(t) wassent: [ ] [ ] U Re s(t)s (t)dt + Re z(t)s (t)dt E + N where E s(t)s (t)dt, and N Re [ z(t)s (t)dt ] is a Gaussian random variable with zero mean and variance EN (as we have seen in Problem 5.7). Hence, given that s(t) was sent, the probability of error is : P e P (E + N<A)P (N < (E A)) Q ( ) E A N E When s(t) is transmitted : U E + N, and the corresponding conditional error probability is : ( ) E A P e P ( E + N> A) P (N >(E A)) Q N E and finally, when is transmitted : U N, and the corresponding error probability is : ( ) A P e3 P (N >Aor N< A) P (N >A)Q N E (b) P e 3 (P e + P e + P e3 ) 3 [ Q ( ) ( )] E A A + Q N E N E 84

10 (c) In order to minimize P e : where we differentiate Q(x) x ( d dx f(x) g(a)da) df dx dp e da A E π exp( t /)dt with respect to x, using the Leibnitz rule : g(f(x)). Using this threshold : P e 4 ( ) 3 Q E N E 4 ( ) E 3 Q N Problem 5. : (a) he transmitted energy is : E s (t) dt A / E s (t) dt A / (b) he correlation coefficient for the two signals is : ρ s (t)s (t)dt / E Hence, the bit error probability for coherent detection is : P Q ( ) ( ) E E ( ρ) Q N N (c) he bit error probability for non-coherent detection is given by (5-4-53) : P,nc Q (a, b) e (a +b )/ I (ab) where Q (.) is the generalized Marcum Q function (given in (--3)) and : ) E a N ( ρ ( ) E N 3 ) E b N (+ ρ ( ) E N

11 Problem 5. : he correlation of the two signals in binary FSK is: ρ sin(π f) π f o find the minimum value of the correlation, we set the derivative of ρ with respect to f equal to zero. hus: and therefore : ϑρ ϑ f cos(π f)π π f π f tan(π f) sin(π f) (π f) π Solving numerically (or graphically) the equation x tan(x), we obtain x hus, π f f.75 and the value of ρ is.7. We know that the probability of error can be expressed in terms of the distance d between the signal points, as : P e Q d N where the distance between the two signal points is : d E b( ρ) and therefore : [ ] Eb ( ρ).7eb P e Q Q N N Problem 5.3 : (a) It is straightforward to see that : Set I : Four level PAM Set II : Orthogonal Set III : Biorthogonal 86

12 (b) he transmitted waveforms in the first set have energy : A or 9A. Hence for the first set the average energy is : E ( 4 A + ) 9A.5A All the waveforms in the second and third sets have the same energy : A.Hence : E E 3 A / (c) he average probability of a symbol error for M-PAM is (5--45) : ( ) (M ) P 4,PAM M Q 6E av 3 (M )N Q A N (d) For coherent detection, a union bound can be given by (5--5) : ) P 4,orth < (M ) Q (E s /N 3Q A N while for non-coherent detection : P 4,orth,nc (M ) P,nc 3 e Es/N 3 e A /4N?? (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 (t) andu 3 (t) (oru (t)/u 4 (t)). (f) he bit rate to bandwidth ratio for M-PAM is given by (5--85) : ( R W ) log M log 44 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 log M W M 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 87

13 Problem 5.4 : he following graph shows the decision regions for the four signals : B U C D A U A U > + U B U < U C U > + U D U < U C B W A D W æ As we see, using the transformation W U + U,W U U alters the decision regions to : (W >,W > s (t); W >,W < s (t); etc.). Assuming that s (t) was transmitted, the outputs of the matched filters will be : U E + N r U N r where N r,n r are uncorrelated (Prob. 5.7) Gaussian-distributed terms with zero mean and variance EN. hen : W E +(N r + N r ) W E +(N r N r ) will be Gaussian distributed with means : E [W ]E [W ]E, and variances : E [W ] E [W ]4EN. Since U,U are independent, it is straightforward to prove that W,W are independent, too. Hence, the probability that a correct decision is made, assuming that s (t) was transmitted is : P c s P [W > ] P [W > ] (P [W > ]) ( P [W < ]) ( Q ( )) E 4EN ( Q ( E N )) ( Q ( Eb N )) where E b E/ is the transmitted energy per bit. hen : ( ( )) ( )[ Eb Eb P e s P c s Q Q ( )] N N Q Eb N 88

14 his 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 5.5 : (a) he 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 { t s (τ)h(t τ)dτ Ae (t τ)/ dτ A ( e ) t/, t Ae (t τ)/ dτ A (e )e t/, t } (b) Asketchofv(t) is given in the following figure : v(t) t (c) y(t) v(t)cosπf c t, where f c >> /. Hence the maximum value of y corresponds to the maximum value of v, or y max y( )v max v( )A ( e ). (d) Working with lowpass equivalent signals, the noise term at the sampling instant will be : v N ( ) z(τ)h( τ)dτ he mean is : E [v N ( )] E [z(τ)] h( τ)dτ, and the second moment : E [ v N ( ) ] E [ z(τ)h( τ)dτ z (w)h( w)dw ] N h ( τ)dτ N ( e ) 89

15 he variance of the real-valued noise component can be obtained using the relationship Re[N] (N + N )toobtain:σ Nr E [ v N ( ) ] N ( e ) (e) he SNR is defined as : γ v max E [ v N ( ) ] A N e e + (the same result is obtained if we consider the real bandpass signal, when the energy term has the additional factor / compared to the lowpass energy term, and the noise term is σ Nr E [ v N ( ) ] ) (f) If we have a filter matched to s (t), then the output of the noise-free matched filter will be : v max v( ) s o (t) A and the noise term will have second moment : E [ v N ( ) ] E [ z(τ)s ( τ)dτ z (w)s ( w)dw ] N s ( τ)dτ N A giving an SNR of : γ v max E [ v N ( ) ] A N Compared with the result we obtained in (e), using a sub-optimum filter, the loss in SNR is equal to : ( ) e e+)(.95 or approximately.35 db Problem 5.6 : (a) Consider the QAM constellation of Fig. P5-6. Using the Pythagorean theorem we can find the radius of the inner circle as: a + a A a A he 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 + 3 A 9

16 (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) he average transmitted power of the PSK constellation is: P PSK 8 8 A P PSK A whereas the average transmitted power of the QAM constellation: P QAM ( 4 A 8 +4( + ) [ ] 3) +(+ 3) A P 4 QAM 8 he relative power advantage of the PSK constellation over the QAM constellation is: A gain P PSK P QAM 8 ( + ( + 3) )( ).597 db Problem 5.7 : (a) Although it is possible to assign three bits to each point of the 8-PSK signal constellation so that adjacent points differ in only one bit, (e.g. going in a clockwise direction :,,,,,,, ). this is not the case for the 8-QAM constellation of Figure P5-6. his is because there are fully connected graphs consisted of three points. o see this consider an equilateral triangle with vertices A, B and C. If, without loss of generality, we assign the all zero sequence {,,...,} to point A, then point B and C should have the form B{,...,,,,...,} C{,...,,,,...,} where the position of the in the sequences is not the same, otherwise BC. hus, the sequences of B and C differ in two bits. (b) Since each symbol conveys 3 bits of information, the resulted symbol rate is : R s symbols/sec 9

17 Problem 5.8 : For binary phase modulation, the error probability is [ ] Eb A P Q Q With P 6 we find from tables that A 4.74 A N N If the data rate is Kbps, then the bit interval is 4 and therefore, the signal amplitude is A Similarly we find that when the rate is 5 bps and 6 bps, the required amplitude of the signal is A. and A 6.73 respectively. N Problem 5.9 : (a) he PDF of the noise n is : p(n) λ e λ n where λ σ he optimal receiver uses the criterion : p(r A) p(r A) e λ[ r A r+a ] A > < A r A > < A he average probability of error is : P (e) P (e A)+ P (e A) λ 4 f(r A)dr + λ e λ r A dr + A e λ x dx + λ 4 e λa e A σ A f(r A)dr λ e λ r+a dr e λ x dx 9

18 (b) he variance of the noise is : Hence, the SNR is: σn λ λ and the probability of error is given by: For P (e) 5 we obtain: e λ x x dx e λx x dx λ! λ 3 λ σ SNR A σ P (e) e SNR ln( 5 ) SNR SNR db If the noise was Gaussian, then the probability of error for antipodal signalling is: [ ] Eb P (e) Q Q [ SNR ] N where SNR is the signal to noise ratio at the output of the matched filter. With P (e) 5 we find SNR 4.6 and therefore SNR db. hus the required signal to noise ratio is 5 db less when the additive noise is Gaussian. Problem 5. : he constellation of Fig. P5-(a) has four points at a distance A from the origin and four points at a distance A. hus, the average transmitted power of the constellation is: P a [ 4 (A) +4 ( A) ] 6A 8 he second constellation has four points at a distance 7A from the origin, two points at a distance 3A and two points at a distance A. hus, the average transmitted power of the second constellation is: P b [ 4 ( 7A) + ( 3A) +A ] 9 8 A Since P b <P a the second constellation is more power efficient. 93

19 Problem 5. : he optimum decision boundary of a point is determined by the perpedicular bisectors of each line segment connecting the point with its neighbors. he decision regions for this QAM constellation are depicted in the next figure: O O O O O O O O O O O O O O O O æ Problem 5. : One way to label the points of the V.9 constellation using the Gray-code is depicted in the next figure. 94

20 O O O O O O O O O O O O O O O O æ Problem 5.3 : he transmitted signal energy is E b A where is the bit interval and A is the signal amplitude. Since both carriers are used to transmit information over the same channel, the bit SNR, E b N, is constant if A is constant. Hence, the desired relation between the carrier amplitudes and the supported transmission rate R is A c A s s Rc c R s With we obtain R c R s 3 3. A c A s.36 95

21 Problem 5.4 : (a) Since m (t) m 3 (t) the dimensionality of the signal space is two. (b) As a basis of the signal space we consider the functions: f (t) { t otherwise he vector representation of the signals is: f (t) m [, ] m [, ] m 3 [, ] (c) he signal constellation is depicted in the next figure : t <t otherwise (, ) ( (,,) ) (d) he three possible outputs of the matched filters, corresponding to the three possible transmitted signals are (r,r )( + n,n ), (n, + n )and(n, + n ), where n, n are zero-mean Gaussian random variables with variance N. If all the signals are equiprobable the optimum decision rule selects the signal that maximizes the metric (see 5--44): or since m i is the same for all i, C(r, m i )r m i m i C (r, m i )r m i hus the optimal decision region R for m is the set of points (r,r ), such that (r,r ) m > (r,r ) m and (r,r ) m > (r,r ) m 3.Since(r,r ) m r,(r,r ) m r and (r,r ) m 3 r, the previous conditions are written as r >r and r > r 96

22 Similarly we find that R is the set of points (r,r )thatsatisfyr >, r >r and R 3 is the region such that r < andr < r. he regions R, R and R 3 areshowninthenextfigure. R R R 3 (e) If the signals are equiprobable then: P (e m )P ( r m > r m m )+P ( r m > r m 3 m ) When m is transmitted then r [ + n,n ] and therefore, P (e m ) is written as: P (e m )P (n n > )+P (n + n < ) Since, n, n are zero-mean statistically independent Gaussian random variables, each with variance N, the random variables x n n and y n + n are zero-mean Gaussian with variance N. Hence: πn e x P (e m ) N dx + [ ] [ ] Q + Q Q N N πn When m is transmitted then r [n,n + ] and therefore: [ ] N P (e m ) P (n n > )+P(n < ) [ ] [ ] Q + Q Similarly from the symmetry of the problem, we obtain: [ ] [ ] P (e m )P(e m 3 )Q + Q Since Q[ ] is momononically decreasing, we obtain: [ ] [ ] Q <Q N N N 97 N N N e y N dy

23 and therefore, the probability of error P (e m ) is larger than P (e m )andp (e m 3 ). Hence, the message m is more vulnerable to errors. he reason for that is that it has both threshold lines close to it, while the other two signals have one of the their threshold lines further away. Problem 5.5 : (a) If the power spectral density of the additive noise is S n (f), then the PSD of the noise at the output of the prewhitening filter is S ν (f) S n (f) H p (f) In order for S ν (f) to be flat (white noise), H p (f) should be such that H p (f) S n (f) (b) Let h p (t) be the impulse response of the prewhitening filter H p (f). hat is, h p (t) F [H p (f)]. hen, the input to the matched filter is the signal s(t) s(t) h p (t). he frequency response of the filter matched to s(t) is S m (f) S (f)e jπft S (f)h p (f)e jπft where t is some nominal time-delay at which we sample the filter output. (c) he frequency response of the overall system, prewhitenig filter followed by the matched filter, is G(f) S m (f)h p (f) S (f) H p (f) e jπft S (f) S n (f) e jπft (d) he variance of the noise at the output of the generalized matched filter is σ S n (f) G(f) S(f) df S n (f) df At the sampling instant t t, the signal component at the output of the matched filter is Hence, the output SNR is y( ) Y (f)e jπf df S(f) S (f) S n (f) df s(τ)g( τ)dτ S(f) S n (f) df SNR y ( ) S(f) σ S n (f) df 98

24 Problem 5.6 : (a) he number of bits per symbol is k 48 R 48 4 hus, a 4-QAM constellation is used for transmission. he probability of error for an M-ary QAM system with M k,is ( ( P M ) [ ]) 3kEb Q M (M )N With P M 5 and k weobtain [ ] Eb Q N 5 6 E b N (b) If the bit rate of transmission is 96 bps, then k In this case a 6-QAM constellation is used and the probability of error is P M ( ( ) Q 4 [ ]) 3 4 Eb 5 N hus, [ ] 3 Eb Q 5 N 3 5 E b N (c) If the bit rate of transmission is 9 bps, then k In this case a 56-QAM constellation is used and the probability of error is P M ( ( ) Q 6 [ ]) 3 8 Eb 55 N With P M 5 we obtain E b N

25 (d) he 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. Problem 5.7 : Using the Pythagorean theorem for the four-phase constellation, we find: r + r d r d he radius of the 8-PSK constellation is found using the cosine rule. hus: d r + r r cos(45 o ) r d he average transmitted power of the 4-PSK and the 8-PSK constellation is given by: P 4,av d, P 8,av d hus, the additional transmitted power needed by the 8-PSK signal is: P log d ( db )d We obtain the same results if we use the probability of error given by (see 5--6) : [ P M Q γ s sin π ] M where γ s is the SNR per symbol. schemes, implies that In this case, equal error probability for the two signaling γ 4,s sin π 4 γ 8,s sin π 8 log γ 8,s γ 4,s log sin π 4 sin π db Since we consider that error ocuur only between adjacent points, the above result is equal to the additional transmitted power we need for the 8-PSK scheme to achieve the same distance d between adjacent points.

26 Problem 5.8 : For 4-phase PSK (M 4) we have the following realtionship between the symbol rate /,the required bandwith W and the bit rate R k / log M (see 5--84): W R log M R Wlog M W kbits/sec For binary FSK (M ) the required frequency separation is / (assuming coherent receiver) and (see 5--86): W M log M R R Wlog M M W kbits/sec Finally, for 4-frequency non-coherent FSK, the required frequency separation is /, 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 kbits/sec Problem 5.9 : We assume that the input bits, are mapped to the symbols - and respectively. terminal phase of an MSK signal at time instant n is given by he θ(n; a) π k a k + θ k where θ is the initial phase and a k is ± depending on the input bit at the time instant k. he following table shows θ(n; a) for two different values of θ (,π), and the four input pairs of data: {,,, }. θ b b a a θ(n; a) - - π - - π π - - π - π π - π π π

27 Problem 5.3 : (a) he envelope of the signal is s(t) s c (t) + s s (t) ( ) Eb πt cos + E ( ) b πt sin b b b b Eb b hus, the signal has constant amplitude. (b) he signal s(t) is equivalent to an MSK signal. synthesizing the signal is given in the next figure. A block diagram of the modulator for Serial data a n Serial / Parallel Demux a n cos( πt b ) π cos(πf c t) π + s(t) a n+ (c) A sketch of the demodulator is shown in the next figure. r(t) b ( )dt cos(πf cos( πt c t)) b ) π π b ( )dt t b hreshold Parallel to Serial t b hreshold

28 Problem 5.3 : h, L Based on (5-3-7), we obtain the 4 phase states : Θ s {,π/,π,3π/} he states in the trellis are the combination of the phase state and the correlative state, which take the values I n {±}. he transition from state to state are determined by θ n+ θ n + π I n and the resulting state trellis and state diagram are given in the following figures, where a solid line corresponds to I n, while a dotted line corresponds to I n. (θ n,i n ) (, ) (, ) (π/, ) (π/, ) (π, ) (π, ) (3π/, ) (π/, ) æ 3

29 (3π/,) (3π/, ) (π, ) (, +) (π, ) (, ) (π/, ) (π/, ) he treatment in Probl. 4.7 involved the terminal phase states only, which were determined to be {π/4, 3π/4, 5π/4, 7π/4}. We can easily verify that each two of the combined states, which were obtained in this problem, give one terminal phase state. For example (θ n,i n ) (3π/, ) and (θ n,i n ) (π, +), give the same terminal phase state at t (n +) : φ ((n +) ; I) θ n + θ(t; I) θ n +πhi n q( )+πhi n q( ) φ ((n +) ; I) θ n + π I n + π I 4 n 3π + π I 4 n 5π/4 or7π/4 Problem 5.3 : (a) (i) here are no correlative states in this system, since it is a full response CPM. Based on (5-3-6), we obtain the phase states : { Θ s, π 3, 4π } 3 4

30 (ii) Based on (5-3-7), we obtain the phase states : { Θ s, 3π 4, 3π, 9π 4 π 4,π,5π 4 7π 4, 8π 4 π, π 4 5π } 4 (b) (i) he combined states are S n (θ n,i n,i n ), where {,I n /n } take the values ±. Hence there are 3 combined states in all. (ii) he combined states are S n (θ n,i n,i n ), where {,I n /n } take the values ±. Hence there are 8 3 combined states in all. Problem 5.33 : A biorthogonal signal set with M 8 signal points has vector space dimensionallity 4. Hence, the detector first checks which one of the four correlation metrics is the largest in absolute value, and then decides about the two possible symbols associated with this correlation metric,based on the sign of this metric. Hence, the error probability is the probability of the union of the event E that another correlation metric is greater in absolute value and the event E that the signal correlation metric has the wrong sign. A union bound on the symbol error probability can be given by : P M P (E )+P(E ) But P (E ) is simply the probability of error for an antipodal signal set : P (E )Q ( ) E s N and the probability of the event E can be union bounded by : P (E ) 3[P ( C > C )] 3 [P (C >C )] 6P (C >C )6Q ( ) Es where C i is the correlation metric corresponding to the i-th vector space dimension; the probability that a correlation metric is greater that the correct one is given by the error probability for orthogonal signals Q ( ) E s N (since these correlation metrics correspond to orthogonal signals). Hence : ( ) ( ) Es Es P M 6Q + Q N (sum of the probabilities to chose one of the 6 orthogonal, to the correct one, signal points and the probability to chose the signal point which is antipodal to the correct one). N N Problem 5.34 : It is convenient to find first the probability of a correct decision. Since all signals are equiprob- 5

31 able, M P (C) i M P (C s i) All the P (C s i ), i,...,m are identical because of the symmetry of the constellation. By translating the vector s i to the origin we can find the probability of a correct decison, given that s i was transmitted, as : P (C s i ) d f(n )dn d f(n )dn... f(n N )dn N d where the number of the integrals on the right side of the equation is N, d is the minimum distance between the points and : f(n i ) n i πσ e σ e n πn i N Hence : P (C s i ) ( N ( f(n)dn) d ( [ ]) N d Q N d f(n)dn ) N and therefore, the probability of error is given by : P (e) P (C) ( Q [ M i ]) N d N ( [ ]) N d Q M N Note that since : N E s s m,i N ( d i i ) N d 4 the probability of error can be written as : ( [ ]) N Es P (e) Q NN Problem 5.35 : Consider first the signal : n y(t) c k δ(t k c ) k 6

32 he signal y(t) has duration n c and its matched filter is : g(t) n y( t) y(n c t) c k δ(n c k c t) k n n c n i+ δ((i ) c t) c n i+ δ(t (i ) c ) i i that is, a sequence of impulses starting at t and weighted by the mirror image sequence of {c i }. Since, n n s(t) c k p(t k c )p(t) c k δ(t k c ) k k the Fourier transform of the signal s(t) is: n S(f) P (f) c k e jπfkc k and therefore, the Fourier transform of the signal matched to s(t) is: H(f) S (f)e jπf S (f)e jπfnc n P (f) c k e jπfkc e jπfnc P (f) k n i P (f)f[g(t)] c n i+ e jπf(i )c hus, the matched filter H(f) can be considered as the cascade of a filter,with impulse response p( t), matched to the pulse p(t) and a filter, with impulse response g(t), matched to the signal y(t) n k c k δ(t k c ). he output of the matched filter at t n c is (see 5--7) : n s(t) c k p (t k c )dt k n c c k k wherewehaveusedthefactthatp(t) is a rectangular pulse of unit amplitude and duration c. Problem 5.36 : he bandwidth required for transmission of an M-ary PAM signal is W R log M Hz 7

33 Since, we obtain R 8 3 samples sec 8 bits sample 6 KHz M 4 W.667 KHz M 8 8KHz M bits sec Problem 5.37 : (a) he inner product of s i (t) ands j (t) is n n s i (t)s j (t)dt c ik p(t k c ) c jl p(t l c )dt k l n n c ik c jl p(t k c )p(t l c )dt k l n n c ik c jl E p δ kl k l n E p c ik c jk k he quantity n k c ik c jk is the inner product of the row vectors C i and C j. Since the rows of the matrix H n are orthogonal by construction, we obtain s i (t)s j (t)dt E p hus, the waveforms s i (t) ands j (t) are orthogonal. n k c ik δ ij ne p δ ij (b) Using the results of Problem 5.35, we obtain that the filter matched to the waveform n s i (t) c ik p(t k c ) k can be realized as the cascade of a filter matched to p(t) followed by a discrete-time filter matched to the vector C i [c i,...,c in ]. Since the pulse p(t) is common to all the signal waveforms s i (t), we conclude that the n matched filters can be realized by a filter matched to p(t) followed by n discrete-time filters matched to the vectors C i, i,...,n. 8

34 Problem 5.38 : (a) he optimal ML detector (see 5--4) selects the sequence C i that minimizes the quantity: n D(r,C i ) (r k k E b C ik ) he metrics of the two possible transmitted sequences are and w D(r,C ) (r k k w D(r,C ) (r k k E b ) + E b ) + n kw+ n kw+ (r k (r k + E b ) E b ) Since the first term of the right side is common for the two equations, we conclude that the optimal ML detector can base its decisions only on the last n w received elements of r. hat is or equivalently n kw+ (r k E b ) n kw+ n kw+ C (r k + E b ) > < C r k C > < C (b) Since r k E b C ik + n k, the probability of error P (e C )is n P (e C ) P E b (n w)+ n k < P n kw+ kw+ n k < (n w) he random variable u n kw+ n k is zero-mean Gaussian with variance σu (n w)σ. Hence Eb (n w) x P (e C ) exp( π(n w)σ π(n w)σ )dx Q Eb (n w) σ 9 E b

35 Similarly we find that P (e C )P(e C ) and since the two sequences are equiprobable Eb (n w) P (e) Q σ (c) he probability of error P (e) is minimized when E b(n w) σ is maximized, that is for w. his implies that C C and thus the distance between the two sequences is the maximum possible. Problem 5.39 : r l (t) s l (t)e jφ + z(t). Hence, the output of a correlator-type receiver will be : r r l(t)s l (t)dt ( sl (t)e jφ + z(t) ) s l (t)dt e jφ s l(t)s l(t)dt + z(t)s l(t)dt e jφ E + n c + jn s E cos φ + n c + j (E sin φ + n s ) where n c Re [ z(t)s l (t)dt],n s Im [ z(t)s l (t)dt]. Similarly for the second correlator output: r r l(t)s l(t)dt ( sl (t)e jφ + z(t) ) s l(t)dt e jφ s l(t)s l (t)dt + z(t)s l (t)dt e jφ Eρ + n c + jn s e jφ E ρ e ja + n c + jn s E ρ cos (φ a )+n c + j (E ρ sin (φ a )+n s ) where n c Re [ z(t)s l(t)dt ],n s Im [ z(t)s l(t)dt ]. Problem 5.4 : n ic Re [ z(t)s li(t)dt ],n is Im [ z(t)s li(t)dt ], i,. he variances of the noise terms

36 are: E [n c n c ] E [ Re [ z(t)s l (t)dt] Re [ z(t)s l (t)dt]] 4 E [ ( z(t)s l(t)dt + z (t)s l (t)dt ) ] 4 E [z(a)z (t)] s l (t)s l (a)dtda N 4 s l(t)s l(t)dt N E wherewehaveusedtheidentityre [z] (z + z ), and the fact from Problem 5.7 (or 4.3) that E [z(t)z(t + τ)], E[z (t)z (t + τ)]. Similarly : E [n c n c ]E [n s n s ]E [n s n s ]N E where for the quadrature noise term components we use the identity : Im[z] (z j z ). he covariance between the in-phase terms for the two correlators is : E [n c n c ] E [( 4 z(t)s l(t)dt + z (t)s l (t)dt )( z(t)s l(t)dt + z (t)s l (t)dt )] because the 4 crossterms that are obtained from the above expressions contain one of : E [z(t)z(t + τ)] E [z (t)z (t + τ)] s l (t)s l(t)dt s l (t)s l(t)dt Similarly : E [n c n s ]E[n s n s ]E[n s n c ]. Finally, the covariance between the in-phase and the quadrature component of the same correlator output is : E [n c n s ] E [( 4j z(t)s l (t)dt + z (t)s l (t)dt )( z(t)s l (t)dt z (t)s l (t)dt )] ( 4j ( + 4j 4j ( E [z(a)z(t)] s l (a)s l (t)dtda + E [z (a)z(t)] s l (a)s l(t)dtda E [z(a)z(t)] s l(a)s l(t)dtda + E [z (a)z (t)] s l (a)s l dtda) E [z (a)z(t)] s l (a)s l(t)dtda ) E [z (a)z (t)] s l (a)s ldtda ) Similarly : E [n c n s ]. he joint pdf is simply the product of the marginal pdf s, since these noise terms are Gaussian and uncorrelated, and thus they are also statistically independent : p(n c,n c,n s,n s )p(n c )p(n c )p(n s )p(n s )

37 where, e.g : p(n c ) 4πNoE exp( n c /4N E). Problem 5.4 : he first matched filter output is : Similarly : r r r l (τ)h ( τ)dτ r l (τ)h ( τ)dτ r l (τ)s l ( ( τ))dτ r l (τ)s l (τ)dτ r l (τ)s l ( ( τ))dτ r l (τ)s l (τ)dτ which are the same as those of the correlation-type receiver of Problem From this point, following the exact same steps as in Problem 5.39, we get : r E cos φ + n c + j (E sin φ + n s ) r E ρ cos (φ a )+n c + j (E ρ sin (φ a )+n s ) Problem 5.4 : (a) he noncoherent envelope detector for the on-off keying signal is depicted in the next figure. t r(t) t ( )dτ cos(πf ct) π t ( )dτ r c t r s ( ) ( ) r + V hreshold Device (b) If s (t) is sent, then the received signal is r(t) n(t) and therefore the sampled outputs r c, r s are zero-mean independent Gaussian random variables with variance N. Hence, the random variable r r c + r s is Rayleigh distributed and the PDF is given by : p(r s (t)) r r e σ σ r e r N N

38 If s (t) is transmitted, then the received signal is : Eb r(t) cos(πf c t + φ)+n(t) b Crosscorrelating r(t) by cos(πf ct) and sampling the output at t,resultsin r c E b b r(t) cos(πf ct)dt E b cos(πf c t + φ)cos(πf c t)dt + b (cos(πf ct + φ)+cos(φ)) dt + n c E b cos(φ)+n c n(t) cos(πf ct)dt where n c is zero-mean Gaussian random variable with variance N. Similarly, for the quadrature component we have : r s E b sin(φ)+n s he PDF of the random variable r rc + r s E b + n c + n s follows the Rician distibution : p(r s (t)) r ( ) r +Eb r Eb σ e σ I r ( ) e r +E b σ N r Eb I N N (c) For equiprobable signals the probability of error is given by: V P (error) p(r s (t))dr + p(r s (t))dr V Since r> the expression for the probability of error takes the form P (error) V V p(r s (t))dr + V ( r r +E b r Eb σ e σ I σ p(r s (t))dr ) dr + r r V σ e σ dr he optimum threshold level is the value of V that minimizes the probability of error. However, when E b N the optimum value is close to: E b and we will use this threshold to simplify the analysis. he integral involving the Bessel function cannot be evaluated in closed form. Instead of I (x) we will use the approximation : I (x) ex πx 3

39 which is valid for large x, that is for high SNR. In this case : V ( ) r r +E b r Eb σ e σ I dr σ E b r πσ e (r Eb ) /σ dr E b his integral is further simplified if we observe that for high SNR, the integrand is dominant in the vicinity of E b and therefore, the lower limit can be substituted by. Also r πσ E b πσ and therefore : Finally : E b r πσ E b e (r Eb ) /σ dr P (error) [ ] Q Eb + N [ ] Q Eb N E b Q [ Eb N E b + e E b 4N πσ e (r Eb ) /σ dr ] r e N r N dr Problem 5.43 : (a) D Re ( V m V m ) where Vm X m + jy m. hen : D Re ((X m + jy m )(X m jy m )) X m X m + Y m Y m ( ) X ( m+x m Xm Xm ) + ( Ym+Y m ) ( ) Ym Y m (b) V k X k + jy k ae cos(θ φ)+jae sin(θ φ)+n k,real + N k,imag. Hence : U Xm+X m, E(U )ae cos(θ φ) U Ym+Y m, E(U )ae sin(θ φ) U 3 Xm X m, E(U 3 ) U 4 Ym Y m, E(U 4 ) 4

40 hevarianceofu is : E [U E(U )] E [ (N m,real + N m,real ) ] E [Nm,real ] EN, and similarly : E [U i E(U i )] EN, i, 3, 4. he covariances are (e.g for U,U ): cov(u,u )E[(U E(U )) (U E(U ))] E [ (N 4 m,r + N m,r )(N m,i + N m,i ) ], since the noise components are uncorrelated and have zero mean. Similarly for any i, j : cov(u i,u j ). he condition cov(u i,u j ), implies that these random variables {U i } are uncorrelated, and since they are Gaussian, they are also statistically independent. Since U 3 and U 4 are zero-mean Gaussian, the random variable R distribution; on the other hand R U + U has a Rice distribution. U 3 + U 4 has a Rayleigh (c) W U + U, with U,U being statistically independent Gaussian variables with means ae cos(θ φ), ae(sin θ φ) and identical variances σ EN. hen, W follows a non-central chi-square distribution with pdf given by (--8): p(w ) ( ) e (4aE +w )/4EN a I w,w 4EN N Also, W U 3 + U 4, with U 3,U 4 being zero-mean Gaussian with the same variance. Hence, W follows a central chi-square distribution, with pfd given by (--) : p(w ) 4EN e w /4EN,w (d) P b P (D <) P (W W < ) P (w >w w )p(w )dw ( w p(w )dw ) p(w )dw e w /4EN p(w )dw ψ(jv) vj/4en ( jvσ ) exp ( jv4a E jvσ ) vj/4en e a E/N where we have used the characteristic function of the non-central chi-square distribution given by (--7) in the book. 5

41 Problem 5.44 : v(t) { } sin πt k [I k u(t k b )+jj k u(t k b b )] where u(t) b, t b. Note, o.w. that u(t b )sin π(t b) cos πt, b t 3 b b b Hence, v(t) may be expressed as : v(t) [ I k sin π(t k b) jj k cos π(t k ] b) k b b he transmitted signal is : Re [ v(t)e jπfct] k [ I k sin π(t k b) cos πf c t + J k cos π(t k ] b) sin πf c t b b (a) (k+) b hreshold X k b ()dt Detector Sampler Î k t (k +) b Input sin π b cosπf c t (k+) b hreshold X k b ()dt Detector Sampler Î k t (k +) b cos π b sinπf c t æ (b) he offset QPSK signal is equivalent to two independent binary PSK systems. Hence for coherent detection, the error probability is : ) P e Q (γ b,γ b E b, E b u(t) dt N (c) Viterbi decoding (MLSE) of the MSK signal yields identical performance to that of part (b). 6

42 (d) MSK is basically binary FSK with frequency separation of f /. For this frequency separation the binary signals are orthogonal with coherent detection. Consequently, the error probability for symbol-by-symbol detection of the MSK signal yields an error probability of P e Q ( γ b ) which is 3dB poorer relative to the optimum Viterbi detection scheme. For non-coherent detection of the MSK signal, the correlation coefficient for f / is : sin π/ ρ.637 π/ From the results in Sec we observe that the performance of the non coherent detector method is about 4 db worse than the coherent FSK detector. hence the loss is about 7 db compared to the optimum demodulator for the MSK signal. Problem 5.45 : (a) For n repeaters in cascade, the probability of i out of n repeaters to produce an error is given by the binomial distribution ( ) n P i p i ( p) n i i However, there is a bit error at the output of the terminal receiver only when an odd number of repeaters produces an error. Hence, the overall probability of error is P n P odd ( ) n p i ( p) n i i iodd Let P even be the probability that an even number of repeaters produces an error. hen P even ( ) n p i ( p) n i i ieven and therefore, ( ) n n P even + P odd p i ( p) n i (p + p) n i i One more relation between P even and P odd can be provided if we consider the difference P even P odd. Clearly, P even P odd ( ) n p i ( p) n i ( ) n p i ( p) n i ieven a ieven i ( n i i iodd ) ( p) i ( p) n i + ( p p) n ( p) n 7 iodd ( n i ) ( p) i ( p) n i

43 where the equality (a) followsfromthefactthat( ) i is for i even and wheni is odd. Solving the system P even + P odd P even P odd ( p) n we obtain P n P odd ( ( p)n ) (b) Expanding the quantity ( p) n,weobtain ( p) n n(n ) np + (p) + Since, p we can ignore all the powers of p which are greater than one. Hence, P n ( +np) np 6 4 Problem 5.46 : he overall probability of error is approximated by (see 5-5-) [ ] Eb P (e) KQ hus, with P (e) 6 and K, we obtain the probability of each repeater P r Q [ ] E b N 8. he argument of the function Q[ ] that provides a value of 8 is found from tables to be Eb 5.6 N Hence, the required E b N is 5.6 /5.7 N Problem 5.47 : (a) he antenna gain for a parabolic antenna of diameter D is : ( πd G R η λ ) 8

44 If we assume that the efficiency factor is.5, then with : λ c f m D m we obtain : G R G db (b) he effective radiated power is : EIRP P G G 6.6 db (c) he received power is : P R P G G R ) db 55.3 dbm ( 4πd λ Note that : dbm log ( actual power in Watts 3 ) 3+log (power in Watts ) Problem 5.48 : (a) he antenna gain for a parabolic antenna of diameter D is : ( πd G R η λ ) If we assume that the efficiency factor is.5, then with : λ c f m and D m we obtain : G R G db (b) he effective radiated power is : EIRP P G db 9

45 (c) he received power is : P R P G G R ) db 67. dbm ( 4πd λ Problem 5.49 : he wavelength of the transmitted signal is: λ m 9 he gain of the parabolic antenna is: ( πd G R η λ ).6 ( ) π db.3 he received power at the output of the receiver antenna is: P R P G G R (4π d λ ) ( ) db Problem 5.5 : (a) Since 3 K, it follows that N k W/Hz If we assume that the receiving antenna has an efficiency η.5, then its gain is given by : ( ) πd G R η db λ Hence, the received power level is : P R P G G R (4π d λ ) ( db 8.5 ) (b) If E b N db, then R P ( ) R Eb N N Mbits/sec

46 Problem 5.5 : he wavelength of the transmission is : λ c f m If MHz is the passband bandwidth, then the rate of binary transmission is R b W 6 bps. Hence, with N 4. W/Hz we obtain : P R N R b E b N he transmitted power is related to the received power through the relation (see 5-5-6) : P R P G G R (4π d P P ( R 4π d ) λ ) G G R λ Substituting in this expression the values G.6, G R 5, d 36 6 and λ.75 we obtain P dbw