Fundamentals Of Communication Systems Solution Manual

599 Pages·4.565 MB·English

by  Proakis J.G., Salehi M.

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