Solution Manual Fundamentals of Communication Systems John G. Proakis Masoud Salehi Second Edition 2013 Chapter 2 Problem 2.1 (cid:16) (cid:16) (cid:17)(cid:17) 1. Π(2t+5)=Π 2 t+ 5 . ThisindicatesfirstwehavetoplotΠ(2t)andthenshiftittoleftby 2 5. Aplotisshownbelow: 2 Π(2t+5) (cid:54) 1 (cid:45) t −11 −9 4 4 2. (cid:80)∞ Λ(t−n)isasumofshiftedtriangularpulses. Notethatthesumoftheleftandrightside n=0 oftriangularpulsesthataredisplacedbyoneunitoftimeisequalto1,Theplotisgivenbelow x (t) ✻2 1 ✲ t −1 3. Itisobviousfromthedefinitionofsgn(t)thatsgn(2t)=sgn(t). Thereforex (t)=0. 3 4. x (t)issinc(t)contractedbyafactorof10. 4 1 0.8 0.6 0.4 0.2 0 −0.2 −0.4 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 3 Problem 2.2 1. x[n]=sinc(3n/9)=sinc(n/3). 1 0.8 0.6 0.4 0.2 0 −0.2 −0.4 −20 −15 −10 −5 0 5 10 15 20 (cid:18) (cid:19) 2. x[n]=Π n4−1 . If−1 ≤ n4−1 ≤ 1,i.e.,−2≤n≤10,wehavex[n]=1. 3 2 3 2 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 −20 −15 −10 −5 0 5 10 15 20 3. x[n]= n4u−1(n/4)−(n4 −1)u−1(n/4−1). Forn<0,x[n]=0,for0≤n≤3,x[n]= n4 and forn≥4,x[n]= n − n +1=1. 4 4 4 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 −5 0 5 10 15 20 Problem 2.3 x [n] = 1 and x [n] = cos(2πn) = 1, for all n. This shows that two signals can be different but 1 2 theirsampledversionsbethesame. Problem 2.4 Let x [n] and x [n] be two periodic signals with periods N and N , respectively, and let N = 1 2 1 2 LCM(N ,N ),anddefinex[n]=x [n]+x [n]. Thenobviouslyx [n+N]=x [n]andx [n+N]= 1 2 1 2 1 1 2 x [n],andhencex[n]=x[n+N],i.e.,x[n]isperiodicwithperiodN. 2 Forcontinuous-timesignalsx (t)andx (t)withperiodsT andT respectively,ingeneralwe 1 2 1 2 cannot find a T such that T = k T = k T for integers k and k . This is obvious for instance if 1 1 2 2 1 2 T =1andT =π. Thenecessaryandsufficientconditionforthesumtobeperiodicisthat T1 bea 1 2 T2 rationalnumber. Problem 2.5 Usingtheresultofproblem2.4wehave: 1. The frequencies are 2000 and 5500, their ratio (and therefore the ratio of the periods) is rational,hencethesumisperiodic. 2. Thefrequenciesare2000and 5500. Theirratioisnotrational,hencethesumisnotperiodic. π 3. Thesumoftwoperiodicdiscrete-timesignalisperiodic. 4. The fist signal is periodic but cos[11000n] is not periodic, since there is no N such that cos[11000(n+N)]=cos(11000n)foralln. Thereforethesumcannotbeperiodic. 5 Problem 2.6 1)    e−t t >0  −e−t t >0 x1(t)= −et t <0 (cid:61)⇒x1(−t)= et t <0 =−x1(t)  0 t =0  0 t =0 Thus,x (t)isanoddsignal 1 (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) 2)x (t)=cos 120πt+ π isneitherevennorodd. Wehavecos 120πt+ π =cos π cos(120πt)− 2 3 3 3 (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17) sin π sin(120πt). Therefore x (t) = cos π cos(120πt) and x (t) = −sin π sin(120πt). 3 2e 3 2o 3 (Note: Thispartcanalsobeconsideredasaspecialcaseofpart7ofthisproblem) 3) x (t)=e−|t| (cid:61)⇒x (−t)=e−|(−t)| =e−|t| =x (t) 3 3 3 Hence,thesignalx (t)iseven. 3 4)    t t ≥0  0 t ≥0 x (t)= (cid:61)⇒x (−t)= 4 4  0 t <0  −t t <0 Thesignalx (t)isneitherevennorodd. Theevenpartofthesignalis 4  x (t)+x (−t)  t t ≥0 |t| x (t)= 4 4 = 2 = 4,e 2  −t t <0 2 2 Theoddpartis  x (t)−x (−t)  t t ≥0 t x (t)= 4 4 = 2 = 4,o 2  t t <0 2 2 5) x (t)=x (t)−x (t) (cid:61)⇒x (−t)=x (−t)−x (−t)=x (t)+x (t) 5 1 2 5 1 2 1 2 Clearly x (−t) ≠ x (t) since otherwise x (t) = 0 ∀t. Similarly x (−t) ≠ −x (t) since otherwise 5 5 2 5 5 x (t)=0∀t. Theevenandtheoddpartsofx (t)aregivenby 1 5 x (t)+x (−t) x (t) = 5 5 =x (t) 5,e 1 2 x (t)−x (−t) x (t) = 5 5 =−x (t) 5,o 2 2 6 Problem 2.7 (cid:82) ForthefirsttwoquestionswewillneedtheintegralI = eaxcos2xdx. (cid:90) (cid:90) 1 1 1 I = cos2x deax = eaxcos2x+ eaxsin2x dx a a a (cid:90) 1 1 = eaxcos2x+ sin2x deax a a2 (cid:90) 1 1 2 = eaxcos2x+ eaxsin2x− eaxcos2x dx a a2 a2 (cid:90) 1 1 2 = eaxcos2x+ eaxsin2x− eax(2cos2x−1)dx a a2 a2 (cid:90) 1 1 2 4 = eaxcos2x+ eaxsin2x− eax dx− I a a2 a2 a2 Thus, (cid:20) (cid:21) 1 2 I = (acos2x+sin2x)+ eax 4+a2 a 1) (cid:90) T (cid:90) T E = lim 2 x2(t)dx = lim 2 e−2tcos2tdt x T→∞ −T 1 T→∞ 0 2 (cid:12)T = lim 1(cid:104)(−2cos2t+sin2t)−1(cid:105)e−2t(cid:12)(cid:12)2 T→∞8 (cid:12)0 (cid:20) (cid:21) 1 T 3 = lim (−2cos2 +sinT −1)e−T +3 = T→∞8 2 8 Thusx (t)isanenergy-typesignalandtheenergycontentis3/8 1 2) (cid:90) T (cid:90) T E = lim 2 x2(t)dx = lim 2 e−2tcos2tdt x T→∞ −T 2 T→∞ −T 2 2   (cid:90)0 (cid:90) T = lim  e−2tcos2tdt+ 2 e−2tcos2tdt T→∞ −T 0 2 But, lim (cid:90)0 e−2tcos2tdt = lim 1(cid:104)(−2cos2t+sin2t)−1(cid:105)e−2t(cid:12)(cid:12)(cid:12)0 T→∞ −T T→∞8 (cid:12)−T 2 2 (cid:20) (cid:21) 1 T = lim −3+(2cos2 +1+sinT)eT =∞ T→∞8 2 since2+cosθ+sinθ >0. Thus,E =∞sinceaswehaveseenfromthefirstquestionthesecond x integral is bounded. Hence, the signal x (t) is not an energy-type signal. To test if x (t) is a 2 2 power-typesignalwefindP . x P = lim 1 (cid:90)0 e−2tcos2dt+ lim 1 (cid:90) T2 e−2tcos2dt x T→∞T −T T→∞T 0 2 7 ButlimT→∞ T1 (cid:82)0T2 e−2tcos2dt iszeroand 1 (cid:90)0 1 (cid:20) T (cid:21) lim e−2tcos2dt = lim 2cos2 +1+sinT eT T→∞T −T T→∞8T 2 2 1 1 > lim eT > lim (1+T +T2)> lim T =∞ T→∞T T→∞T T→∞ Thusthesignalx (t)isnotapower-typesignal. 2 3) (cid:90) T (cid:90) T (cid:90) T E = lim 2 x2(t)dx = lim 2 sgn2(t)dt = lim 2 dt = lim T =∞ x T→∞ −T 3 T→∞ −T T→∞ −T T→∞ 2 2 2 (cid:90) T (cid:90) T P = lim 1 2 sgn2(t)dt = lim 1 2 dt = lim 1T =1 x T→∞T −T T→∞T −T T→∞T 2 2 Thesignalx (t)isofthepower-typeandthepowercontentis1. 3 4) Firstnotethat (cid:90) T2 (cid:88)∞ (cid:90)k+21f lim Acos(2πft)dt = A cos(2πft)dt =0 T→∞ −T k− 1 2 k=−∞ 2f sothat (cid:90) T (cid:90) T lim 2 A2cos2(2πft)dt = lim 1 2 (A2+A2cos(2π2ft))dt T→∞ −T T→∞2 −T 2 2 (cid:90) T = lim 1 2 A2dt = lim 1A2T =∞ T→∞2 −T T→∞2 2 (cid:90) T E = lim 2 (A2cos2(2πf t)+B2cos2(2πf t)+2ABcos(2πf t)cos(2πf t))dt x 1 2 1 2 T→∞ −T 2 (cid:90) T (cid:90) T = lim 2 A2cos2(2πf t)dt+ lim 2 B2cos2(2πf t)dt+ 1 2 T→∞ −T T→∞ −T 2 2 (cid:90) T AB lim 2 [cos2(2π(f +f )+cos2(2π(f −f )]dt 1 2 1 2 T→∞ −T 2 = ∞+∞+0=∞ Thusthesignalisnotoftheenergy-type. Totestifthesignalisofthepower-typeweconsidertwo casesf =f andf ≠f . Inthefirstcase 1 2 1 2 (cid:90) T P = lim 1 2 (A+B)2cos2(2πf )dt x 1 T→∞T −T 2 (cid:90) T = lim 1 (A+B)2 2 dt = 1(A+B)2 T→∞2T −T 2 2 8 Iff ≠f then 1 2 (cid:90) T P = lim 1 2 (A2cos2(2πf t)+B2cos2(2πf t)+2ABcos(2πf t)cos(2πf t))dt x 1 2 1 2 T→∞T −T 2 (cid:34) (cid:35) 1 A2T B2T A2 B2 = lim + = + T→∞T 2 2 2 2 Thusthesignalisofthepower-typeandiff =f thepowercontentis(A+B)2/2whereasiff ≠f 1 2 1 2 thepowercontentis 1(A2+B2) 2 Problem 2.8 1. Let x(t)=2Λ(cid:16)t(cid:17)−Λ(t), then x (t)=(cid:80)∞ x(t−4n). First we plot x(t) then by shifting 2 1 n=−∞ it by multiples of 4 we can plot x (t). x(t) is a triangular pulse of width 4 and height 2 1 fromwhichastandardtriangularpulseofwidth1andheight1issubtracted. Theresultisa trapezoidalpulse,whichwhenreplicatedatintervalsof4givestheplotofx (t). 1 x (t) ✻1 1 ✲ t −6 −2 2 6 2. This is the sum of two periodic signals with periods 2π and 1. Since the ratio of the two periodsisnotrationalthesumisnotperiodic(bytheresultofproblem2.4) 3. sin[n]isnotperiodic. ThereisnointegerN suchthatsin[n+N]=sin[n]foralln. Problem 2.9 1) Px = lim 1 (cid:90) T2 A2(cid:12)(cid:12)(cid:12)ej(2πf0t+θ)(cid:12)(cid:12)(cid:12)2dt = lim 1 (cid:90) T2 A2dt = lim 1A2T =A2 T→∞T −T T→∞T −T T→∞T 2 2 Thusx(t)=Aej(2πf0t+θ) isapower-typesignalanditspowercontentisA2. 2) P = lim 1 (cid:90) T2 A2cos2(2πf t+θ)dt = lim 1 (cid:90) T2 A2dt+ lim 1 (cid:90) T2 A2 cos(4πf t+2θ)dt x 0 0 T→∞T −T T→∞T −T 2 T→∞T −T 2 2 2 2 AsT →∞,thetherewillbenocontributionbythesecondintegral. Thusthesignalisapower-type signalanditspowercontentis A2. 2 9 3) (cid:90) T (cid:90) T P = lim 1 2 u2 (t)dt = lim 1 2 dt = lim 1 T = 1 x T→∞T −T −1 T→∞T 0 T→∞T 2 2 2 Thustheunitstepsignalisapower-typesignalanditspowercontentis1/2 4) (cid:90) T (cid:90) T (cid:12)T/2 Ex = Tli→m∞ −2T x2(t)dt =Tli→m∞ 02 K2t−12dt =Tli→m∞2K2t12(cid:12)(cid:12)(cid:12)0 2 √ = lim 2K2T12 =∞ T→∞ Thusthesignalisnotanenergy-typesignal. (cid:90) T (cid:90) T Px = lim 1 2 x2(t)dt = lim 1 2 K2t−12dt T→∞T −T T→∞T 0 2 = lim 12K2t12(cid:12)(cid:12)(cid:12)T/2 = lim 12K2(T/2)12 = lim √2K2T−12 =0 T→∞T (cid:12)0 T→∞T T→∞ SinceP isnotboundedawayfromzeroitfollowsbydefinitionthatthesignalisnotofthepower-type x (recallthatpower-typesignalsshouldsatisfy0<P <∞). x Problem 2.10    t+1, −1≤t ≤0  1 t >0 Λ(t)= −t+1, 0≤t ≤1 u−1(t)= 1/2 t =0  0, o.w.  0 t <0 Thus,thesignalx(t)=Λ(t)u−1(t)isgivenby    10/2 tt =<00  t+01 t−≤1≤−1t <0 x(t)= (cid:61)⇒x(−t)=  −t0+1 0t ≥≤1t ≤1  10/2 tt =>00 Theevenandtheoddpartofx(t)aregivenby x(t)+x(−t) 1 x (t) = = Λ(t) e 2 2  x(t)−x(−t)  −t02−1 t−≤1≤−1t <0 xo(t) = = 0 t =0 2  −t02+1 10<≤tt ≤1 10 Problem 2.11 1)Supposethat x(t)=x1(t)+x1(t)=x2(t)+x2(t) e o e o withx1(t),x2(t)evensignalsandx1(t),x1(t)oddsignals. Then,x(−t)=x1(t)−x1(t)sothat e e o o e o x(t)+x(−t) x1(t) = e 2 x2(t)+x2(t)+x2(−t)+x2(−t) = e o e o 2 2x2(t)+x2(t)−x2(t) = e o o =x2(t) 2 e Thusx1(t)=x2(t)andx1(t)=x(t)−x1(t)=x(t)−x2(t)=x2(t) e e o e e o 2)Letx1(t),x2(t)betwoevensignalsandx1(t),x2(t)betwooddsignals. Then, e e o o y(t)=x1(t)x2(t) (cid:61)⇒ y(−t)=x1(−t)x2(−t)=x1(t)x2(t)=y(t) e e e e e e z(t)=x1(t)x2(t) (cid:61)⇒ z(−t)=x1(−t)x2(−t)=(−x1(t))(−x2(t))=z(t) o o o o o o Thustheproductoftwoevenoroddsignalsisanevensignal. Forv(t)=x1(t)x1(t)wehave e o v(−t)=x1(−t)x1(−t)=x1(t)(−x1(t))=−x1(t)x1(t)=−v(t) e o e o e o Thustheproductofanevenandanoddsignalisanoddsignal. 3)Onetrivialexampleist+1and t2 . t+1 Problem 2.12 1)x (t)=Π(t)+Π(−t). ThesignalΠ(t)isevensothatx (t)=2Π(t) 1 1 2 1 . . . . . . . . . . . . . . . . . . 1 1 2 2 11