Solution Manual For Fundamentals Of Communication Systems, 2nd Edition

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Solution ManualFundamentals of Communication SystemsJohn G. ProakisMasoud SalehiSecond Edition2013

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Chapter 2Problem 2.11.Π(2t+5)=Π(2(t+52)). This indicates first we have to plotΠ(2t)and then shift it to left by52. A plot is shown below:6−114−94-tΠ(2t+5)12.∑∞n=0Λ(t−n)is a sum of shifted triangular pulses. Note that the sum of the left and right sideof triangular pulses that are displaced by one unit of time is equal to 1, The plot is given below✲✻tx2(t)−113. It is obvious from the definition of sgn(t)that sgn(2t)=sgn(t). Thereforex3(t)=0.4.x4(t)is sinc(t)contracted by a factor of 10.−1−0.8−0.6−0.4−0.200.20.40.60.81−0.4−0.200.20.40.60.813

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Problem 2.21.x[n]=sinc(3n/9)=sinc(n/3).−20−15−10−505101520−0.4−0.200.20.40.60.812.x[n]=Π(n4−13). If−12≤n4−13≤12, i.e.,−2≤n≤10, we havex[n]=1.−20−15−10−50510152000.10.20.30.40.50.60.70.80.913.x[n]=n4u−1(n/4)−(n4−1)u−1(n/4−1). Forn <0,x[n]=0, for 0≤n≤3,x[n]=n4andforn≥4,x[n]=n4−n4+1=1.4

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−50510152000.10.20.30.40.50.60.70.80.91Problem 2.3x1[n]=1 andx2[n]=cos(2π n)=1, for alln. This shows that two signals can be different buttheir sampled versions be the same.Problem 2.4Letx1[n]andx2[n]be two periodic signals with periodsN1andN2, respectively, and letN=LCM(N1, N2), and definex[n]=x1[n]+x2[n]. Then obviouslyx1[n+N]=x1[n]andx2[n+N]=x2[n], and hencex[n]=x[n+N], i.e.,x[n]is periodic with periodN.For continuous-time signalsx1(t)andx2(t)with periodsT1andT2respectively, in general wecannot find aTsuch thatT=k1T1=k2T2for integersk1andk2. This is obvious for instance ifT1=1 andT2=π. The necessary and sufficient condition for the sum to be periodic is thatT1T2be arational number.Problem 2.5Using the result of problem 2.4 we have:1.The frequencies are 2000 and 5500, their ratio (and therefore the ratio of the periods) isrational, hence the sum is periodic.2. The frequencies are 2000 and5500π. Their ratio is not rational, hence the sum is not periodic.3. The sum of two periodic discrete-time signal is periodic.4.The fist signal is periodic butcos[11000n]isnotperiodic, since there is noNsuch thatcos[11000(n+N)]=cos(11000n)for alln. Therefore the sum cannot be periodic.5

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Problem 2.61)x1(t)=e−tt >0−ett <00t=0=⇒x1(−t)=−e−tt >0ett <00t=0= −x1(t)Thus,x1(t)is an odd signal2)x2(t)=cos(120π t+π3)is neither even nor odd. We havecos(120π t+π3)=cos(π3)cos(120π t)−sin(π3)sin(120π t).Thereforex2e(t)=cos(π3)cos(120π t)andx2o(t)= −sin(π3)sin(120π t).(Note: This part can also be considered as a special case of part 7 of this problem)3)x3(t)=e−|t|=⇒x3(−t)=e−|(−t)|=e−|t|=x3(t)Hence, the signalx3(t)is even.4)x4(t)=tt≥00t <0=⇒x4(−t)=0t≥0−tt <0The signalx4(t)is neither even nor odd. The even part of the signal isx4,e(t)=x4(t)+x4(−t)2=t2t≥0−t2t <0= |t|2The odd part isx4,o(t)=x4(t)−x4(−t)2=t2t≥0t2t <0=t25)x5(t)=x1(t)−x2(t)=⇒x5(−t)=x1(−t)−x2(−t)=x1(t)+x2(t)Clearlyx5(−t)≠x5(t)since otherwisex2(t)=0∀t. Similarlyx5(−t)≠−x5(t)since otherwisex1(t)=0∀t. The even and the odd parts ofx5(t)are given byx5,e(t)=x5(t)+x5(−t)2=x1(t)x5,o(t)=x5(t)−x5(−t)2= −x2(t)6

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Problem 2.7For the first two questions we will need the integralI=∫eaxcos2xdx.I=1a∫cos2x deax=1a eaxcos2x+1a∫eaxsin 2x dx=1a eaxcos2x+1a2∫sin 2x deax=1a eaxcos2x+1a2eaxsin 2x−2a2∫eaxcos 2x dx=1a eaxcos2x+1a2eaxsin 2x−2a2∫eax(2 cos2x−1) dx=1a eaxcos2x+1a2eaxsin 2x−2a2∫eaxdx−4a2IThus,I=14+a2[(acos2x+sin 2x)+2a]eax1)Ex=limT→∞∫T2−T2x21(t)dx=limT→∞∫T20e−2tcos2tdt=limT→∞18[(−2 cos2t+sin 2t)−1]e−2t∣∣∣∣T20=limT→∞18[(−2 cos2T2+sinT−1)e−T+3]=38Thusx1(t)is an energy-type signal and the energy content is 3/82)Ex=limT→∞∫T2−T2x22(t)dx=limT→∞∫T2−T2e−2tcos2tdt=limT→∞∫0−T2e−2tcos2tdt+∫T20e−2tcos2tdtBut,limT→∞∫0−T2e−2tcos2tdt=limT→∞18[(−2 cos2t+sin 2t)−1]e−2t∣∣∣∣0−T2=limT→∞18[−3+(2 cos2T2+1+sinT )eT]= ∞since 2+cosθ+sinθ >0. Thus,Ex= ∞since as we have seen from the first question the secondintegral is bounded.Hence, the signalx2(t)is not an energy-type signal.To test ifx2(t)is apower-type signal we findPx.Px=limT→∞1T∫0−T2e−2tcos2dt+limT→∞1T∫T20e−2tcos2dt7

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But limT→∞1T∫T20e−2tcos2dtis zero andlimT→∞1T∫0−T2e−2tcos2dt=limT→∞18T[2 cos2T2+1+sinT]eT>limT→∞1T eT>limT→∞1T (1+T+T2) >limT→∞T= ∞Thus the signalx2(t)is not a power-type signal.3)Ex=limT→∞∫T2−T2x23(t)dx=limT→∞∫T2−T2sgn2(t)dt=limT→∞∫T2−T2dt=limT→∞T= ∞Px=limT→∞1T∫T2−T2sgn2(t)dt=limT→∞1T∫T2−T2dt=limT→∞1T T=1The signalx3(t)is of the power-type and the power content is 1.4)First note thatlimT→∞∫T2−T2Acos(2π f t)dt=∞∑k=−∞A∫k+12fk−12fcos(2π f t)dt=0so thatlimT→∞∫T2−T2A2cos2(2π f t)dt=limT→∞12∫T2−T2(A2+A2cos(2π2f t))dt=limT→∞12∫T2−T2A2dt=limT→∞12A2T= ∞Ex=limT→∞∫T2−T2(A2cos2(2π f1t)+B2cos2(2π f2t)+2ABcos(2π f1t)cos(2π f2t))dt=limT→∞∫T2−T2A2cos2(2π f1t)dt+limT→∞∫T2−T2B2cos2(2π f2t)dt+ABlimT→∞∫T2−T2[cos2(2π (f1+f2)+cos2(2π (f1−f2)]dt=∞ + ∞ +0= ∞Thus the signal is not of the energy-type. To test if the signal is of the power-type we consider twocasesf1=f2andf1≠f2. In the first casePx=limT→∞1T∫T2−T2(A+B)2cos2(2π f1)dt=limT→∞12T (A+B)2∫T2−T2dt=12(A+B)28

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Iff1≠f2thenPx=limT→∞1T∫T2−T2(A2cos2(2π f1t)+B2cos2(2π f2t)+2ABcos(2π f1t)cos(2π f2t))dt=limT→∞1T[A2T2+B2T2]=A22+B22Thus the signal is of the power-type and iff1=f2the power content is(A+B)2/2 whereas iff1≠f2the power content is12(A2+B2)Problem 2.81.Letx(t)=2Λ(t2)−Λ(t), thenx1(t)=∑∞n=−∞x(t−4n). First we plotx(t)then by shiftingit by multiples of 4 we can plotx1(t).x(t)is a triangular pulse of width 4 and height 2from which a standard triangular pulse of width 1 and height 1 is subtracted. The result is atrapezoidal pulse, which when replicated at intervals of 4 gives the plot ofx1(t).✲✻tx1(t)12−26−62.This is the sum of two periodic signals with periods 2πand 1.Since the ratio of the twoperiods is not rational the sum is not periodic (by the result of problem 2.4)3. sin[n]is not periodic. There is no integerNsuch that sin[n+N]=sin[n]for alln.Problem 2.91)Px=limT→∞1T∫T2−T2A2∣∣∣ej(2π f0t+θ)∣∣∣2dt=limT→∞1T∫T2−T2A2dt=limT→∞1T A2T=A2Thusx(t)=Aej(2π f0t+θ)is a power-type signal and its power content isA2.2)Px=limT→∞1T∫T2−T2A2cos2(2π f0t+θ) dt=limT→∞1T∫T2−T2A22dt+limT→∞1T∫T2−T2A22cos(4π f0t+2θ) dtAsT→ ∞, the there will be no contribution by the second integral. Thus the signal is a power-typesignal and its power content isA22.9

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3)Px=limT→∞1T∫T2−T2u2−1(t)dt=limT→∞1T∫T20dt=limT→∞1TT2=12Thus the unit step signal is a power-type signal and its power content is 1/24)Ex=limT→∞∫T2−T2x2(t)dt=limT→∞∫T20K2t−12dt=limT→∞2K2t12∣∣∣∣T /20=limT→∞√2K2T12= ∞Thus the signal is not an energy-type signal.Px=limT→∞1T∫T2−T2x2(t)dt=limT→∞1T∫T20K2t−12dt=limT→∞1T2K2t12∣∣∣∣T /20=limT→∞1T2K2(T /2)12=limT→∞√2K2T−12=0SincePxis not bounded away from zero it follows by definition that the signal is not of the power-type(recall that power-type signals should satisfy 0< Px<∞).Problem 2.10Λ(t)=t+1,−1≤t≤0−t+1,0≤t≤10,o.w.u−1(t)=1t >01/2t=00t <0Thus, the signalx(t)=Λ(t)u−1(t)is given byx(t)=0t <01/2t=0−t+10≤t≤10t≥1=⇒x(−t)=0t≤ −1t+1−1≤t <01/2t=00t >0The even and the odd part ofx(t)are given byxe(t)=x(t)+x(−t)2=12Λ(t)xo(t)=x(t)−x(−t)2=0t≤ −1−t−12−1≤t <00t=0−t+120< t≤101≤t10

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Problem 2.111) Suppose thatx(t)=x1e(t)+x1o(t)=x2e(t)+x2o(t)withx1e(t),x2e(t)even signals andx1o(t),x1o(t)odd signals. Then,x(−t)=x1e(t)−x1o(t)so thatx1e(t)=x(t)+x(−t)2=x2e(t)+x2o(t)+x2e(−t)+x2o(−t)2=2x2e(t)+x2o(t)−x2o(t)2=x2e(t)Thusx1e(t)=x2e(t)andx1o(t)=x(t)−x1e(t)=x(t)−x2e(t)=x2o(t)2) Letx1e(t),x2e(t)be two even signals andx1o(t),x2o(t)be two odd signals. Then,y(t)=x1e(t)x2e(t)=⇒y(−t)=x1e(−t)x2e(−t)=x1e(t)x2e(t)=y(t)z(t)=x1o(t)x2o(t)=⇒z(−t)=x1o(−t)x2o(−t)=(−x1o(t))(−x2o(t))=z(t)Thus the product of two even or odd signals is an even signal. Forv(t)=x1e(t)x1o(t)we havev(−t)=x1e(−t)x1o(−t)=x1e(t)(−x1o(t))= −x1e(t)x1o(t)= −v(t)Thus the product of an even and an odd signal is an odd signal.3) One trivial example ist+1 andt2t+1.Problem 2.121)x1(t)=Π(t)+Π(−t). The signalΠ(t)is even so thatx1(t)=2Π(t). . . . . . . . . . . . . . . . . .12121211

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2)x2(t)=Λ(t)·Π(t)=0,t <−1/21/4,t= −1/2t+1,−1/2< t≤0−t+1,0≤t <1/21/4,t=1/20,1/2< t. . . . . . . . ...........14−121213)x3(t)=∑∞n=−∞Λ(t−2n)......−3−13114)x4(t)=sgn(t)+sgn(1−t). Note thatx4(0)=1,x4(1)=1.........0215)x5(t)=sinc(t)sgn(t). Note thatx5(0)=0.12

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−1−0.8−0.6−0.4−0.200.20.40.60.81−4−3−2−101234Problem 2.131) The value of the expressionsinc(t)δ(t)can be found by examining its effect on a functionφ(t)through the integral∫∞−∞φ(t)sinc(t)δ(t)=φ(0)sinc(0)=sinc(0)∫∞−∞φ(t)δ(t)Thus sinc(t)δ(t)has the same effect as the function sinc(0)δ(t)and we conclude thatx1(t)=sinc(t)δ(t)=sinc(0)δ(t)=δ(t)2) sinc(t)δ(t−3)=sinc(3)δ(t−3)=0.3)x3(t)=Λ(t) ?∞∑n=−∞δ(t−2n)=∞∑n=−∞∫∞−∞Λ(t−τ)δ(τ−2n)dτ=∞∑n=−∞∫∞−∞Λ(τ−t)δ(τ−2n)dτ=∞∑n=−∞Λ(t−2n)13

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4)x4(t)=Λ(t) ? δ′(t)=∫∞−∞Λ(t−τ)δ′(τ)dτ=(−1) ddτΛ(t−τ)∣∣∣∣τ=0=Λ′(t)=0t <−112t= −11−1< t <00t=0−10< t <1−12t=101< t5)x5(t)=cos(2t+π3)δ(3t)=13cos(2t+π3)δ(t)=13cos(π3)δ(t). Hencex5(t)=16δ(t).6)x6(t)=δ(5t) ? δ(4t)=15δ(t) ?14δ(t)=120δ(t)7)∫∞−∞sinc(t)δ(t)dt=sinc(0)=18)∫∞−∞sinc(t+1)δ(t)dt=sinc(1)=0Problem 2.14The impulse signal can be defined in terms of the limitδ(t)=limτ→012τ(e−|t|τ)Bute−|t|τis an even function for everyτso thatδ(t)is even. Sinceδ(t)is even, we obtainδ(t)=δ(−t)=⇒δ′(t)= −δ′(−t)Thus, the functionδ′(t)is odd. For the functionδ(n)(t)we have∫∞−∞φ(t)δ(n)(−t)dt=(−1)n∫∞−∞φ(t)δ(n)(t)dtwhere we have used the differentiation chain ruleddt δ(k−1)(−t)=dd(−t) δ(k−1)(−t) ddt (−t)=(−1)δ(k)(−t)14

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Thus, ifn=2l(even)∫∞−∞φ(t)δ(n)(−t)dt=∫∞−∞φ(t)δ(n)(t)dtand the functionδ(n)(t)is even. Ifn=2l+1 (odd), then(−1)n= −1 and∫∞−∞φ(t)δ(n)(−t)dt= −∫∞−∞φ(t)δ(n)(t)dtfrom which we conclude thatδ(n)(t)is odd.Problem 2.15x(t) ? δ(n)(t)=∫∞−∞x(τ)δ(n)(t−τ) dτThe signalδ(n)(t)is even ifnis even and odd ifnis odd. Consider first the case thatn=2l. Then,x(t) ? δ(2l)(t)=∫∞−∞x(τ)δ(2l)(τ−t) dτ=(−1)2ld2ldτ2lx(τ)∣∣∣∣τ=t=dndtnx(t)Ifnis odd then,x(t) ? δ(2l+1)(t)=∫∞−∞x(τ)(−1)δ(2l+1)(τ−t) dτ=(−1)(−1)2l+1d2l+1dτ2l+1x(τ)∣∣∣∣τ=t=dndtnx(t)In both casesx(t) ? δ(n)(t)=dndtnx(t)The convolution ofx(t)withu−1(t)isx(t) ? u−1(t)=∫∞−∞x(τ)u−1(t−τ)dτButu−1(t−τ)=0 forτ > tso thatx(t) ? u−1(t)=∫t−∞x(τ)dτProblem 2.161) Nonlinear, since the response tox(t)=0 is noty(t)=0 (this is a necessary condition for linearityof a system, see also problem 2.21).2) Nonlinear, if we multiply the input by constant−1, the output does not change. In a linear systemthe output should be scaled by−1.15

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3) Linear, the output to any input zero, therefore for the inputαx1(t)+βx2(t)the output is zerowhich can be considered asαy1(t)+βy2(t)=α×0+β×0=0. This is a linear combination of thecorresponding outputs tox1(t)andx2(t).4) Nonlinear, the output tox(t)=0 is not zero.5) Nonlinear. The system is not homogeneous for ifα <0 andx(t) >0 theny(t)=T [αx(t)]=0whereasz(t)=αT [x(t)]=α.6) Linear. For ifx(t)=αx1(t)+βx2(t)thenT [αx1(t)+βx2(t)]=(αx1(t)+βx2(t))e−t=αx1(t)e−t+βx2(t)e−t=αT [x1(t)]+βT [x2(t)]7) Linear. For ifx(t)=αx1(t)+βx2(t)thenT [αx1(t)+βx2(t)]=(αx1(t)+βx2(t))u(t)=αx1(t)u(t)+βx2(t)u(t)=αT [x1(t)]+βT [x2(t)]8) Linear. We can write the output of this feedback system asy(t)=x(t)+y(t−1)=∞∑n=0x(t−n)Then forx(t)=αx1(t)+βx2(t)y(t)=∞∑n=0(αx1(t−n)+βx2(t−n))=α∞∑n=0x1(t−n)+β∞∑n=0x2(t−n))=αy1(t)+βy2(t)9) Linear. Assuming that only a finite number of jumps occur in the interval(−∞, t]and that themagnitude of these jumps is finite so that the algebraic sum is well defined, we obtainy(t)=T [αx(t)]=N∑n=1αJx(tn)=αN∑n=1Jx(tn)=αT [x(t)]whereNis the number of jumps in(−∞, t]andJx(tn)is the value of the jump at time instanttn,that isJx(tn)=lim→0(x(tn+)−x(tn−))Forx(t)=x1(t)+x2(t)we can assume thatx1(t),x2(t)andx(t)have the same number of jumpsand at the same positions. This is true since we can always add new jumps of magnitude zero to thealready existing ones. Then for eachtn,Jx(tn)=Jx1(tn)+Jx2(tn)andy(t)=N∑n=1Jx(tn)=N∑n=1Jx1(tn)+N∑n=1Jx2(tn)so that the system is additive.16

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