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Unformatted text preview: 563120 Ho) 2. {xiF Chapter 3 Solutions 3.11. (a) E, = (3)2 +2(2)2 +2(1)2 = 19
(b) E, = (3)2 + 2(2)2 +2(1)2 = 19
' (c) E, = 2(3)2 + 2(6)2 + 2(9)2 = 252
(d) E, = 2(2)2 + 2(4)2 = 40 3.12. (a)
3
_ ' 1 2 _ 1 2 2 2 _ E
P,  N0 + 1 “223M111! — 6 [2(1) +2(2) +31— 6 (b) P = —1— [2(1)2 + 2(2)2 + 2(3)”) = Z ‘ 12 3
(c) Nol N N l l 0. °  l 0. ° — l
P: = — n = —— =
Nougat} ()[a—l] No(a—1)
3 21 (a) The energy of z[—n] is E1 given by
E1 = Z FBI4"“2
Setting n — —m, we obtain
E1 = Z livlmll2 = Z liﬁi'mll2 = E:
mace tux—co
(b) The energy E2 of z[n — m] is given by
E2 = Z lzln — mu” = Z lzirll“ = E,
n=—oo r=—oo
(c) The energy E3 of zlm — n] is given by
E: = Z lzlm ‘nlla = Z :I=[1‘]2 = Z :I=[1‘]2 = E:
n=—oo r=oo r=—°o (d) The energy E4 of K z[n] is given by & E4 = i K:I:[‘rz.]}2 = K2 2 :I:['n]2 = KaE, "__m “3—” 3.23. Figure 83.23 shows all the signals. gem6,1 Figure 83.23 3.32. (a) Trivial
(b) Becaus
(c) Because n(n — 1) = d Because sing— : 0 for even 11 and u[n] + (1)"u
2 . follows. _
(e) Because cos? = O for odd n and u follows. e sin “4; =D~ for n = 0, the result follows. 0 for n = 0 and n: 1, the result follows.
[11] = 0 for odd n, the result [n] + (1)"+1 = 0 for even 11, the result . 3.33. Figure S3.33 shows all the signals. Figure S3.33
.. K 3.34. (a) z[n] = (n + 3) (u[n + 3] — u[n]) + (—n + 3) (u[n] — u[n e 4]) (b) z[n] =n(u[n] —u[n—4])+(—n+ 6) u —4 — —
(c) 2[n] = n (u.[n + 3] — u[n — 4]) ( [n l '14" 7])
(d) 2[n] = 2n (u[n + 2] — u[n]) + 2n (u[n] — u[n — 3]) In all four cases z[n] ma _ . , I y be represented by several other (sli htl d'ﬁ' :13212? instance, In case (a), we may also use 2[n] = (n + 3)g(u[1:,+ Mn _4] 115:;an r1] 7 uh;  Moreover because z[n] = O at n = 5:3, u[n + 3] and
ace . . . apply to Other “a; also’m u[n +2] and u[n — 3], respectively. Similar observations 3.35. (3.) e"°'5”=(0.6065)“, (b) em" a (1.6487)”, (G) 6"“ == (e'j”)" = (1)", (d) 6”” = (8”)“ = (1)"
Figure S3.3~5 shows locations of A and '1 in each case. By plotting these signals, we
observe that when '1 > 1 (A in RHP), the signal grows exponentially. When '7 < 1 (A
in LHP), the signal decays exponentially. When M = 1. (A on imaginary axis), the
signal amplitude is constant. ' Figure 83.35 3.36. (a) raw“)? = (F‘sW)" = (%)n
(b) e—(l—jw)n = (elejvr)n = 0%)."
(c) cum” = (e eh)" = (—e)" '
(d) eum" = (ea1*)" = (—e)"
(e) ram)" =‘(e1)“ eﬁ" = (macs sn — jsin en] (f) ea’ji)?‘ = (el)" e‘ji" = e“[cos ﬁn — jsin gin] 3.4—2. The net growth rate of the native population is 3.3 — 1.3 = 2% per year. Assuming the
immigrants enter at a uniform rate throughout the year, their birth and death rate
will be (3.3/2)% and (1.3/2)%, respectively of the immigrants at the end of the year.
The population p[n] at the beginning of the kth year is p[n — 1] plus the net increase
in the native population plus i[n  1], the immigrants entering during (71. — 1)st year plus the net increase in the immigrant population for the year (n — 1). 3.3 — 1.3 . 3.3  1.3 .
pin] = pin—11+ mo shuns—11+ 2x100 zln—ll .—_ 1.02p[n — 1] + 1.01i[n — 1] or p[n] — 1.02p[n — 1] = 1.01i[n — 1] or p[n + 1] — I.02p[n] = _1.01i[n] 3.44. The input a:[n] =: u.[n], which has a constant value of unity for all n 2 0. Also
. yin]. — yin 1' 1} ;. Tulnl. .Hence.the diﬁerence between two successive output values
is always constant of value T. Clearly y[n] must be a ramp with a possible constant
component. Thus
y[n1=(nT+ c)u[n1 To ﬁnd the value of unknown constant c, we let n = 0 and obtain
ylol = 6 But from the input equation yIn]  y[n 4 1] = Tu[n], we ﬁnd y[0] = T [remember that
y[ 1] = 0]. Hence yIn] = (n + 1)Tu[n] : nTu[n] for T —+ 0 3.46. Thelnode equation at the kth node is 2'1 + £2 + £3 = 0, or v[n 1] u[n] v['n.+1] v[n] _ m = R + R (112 0
Therefore 
a(v[n — 11+ vln + 11 — Zvinl) — ulnl = 0
01'
u[1_7.+1] — (2+3 v[n]+u['n.1] =0
that is u[n+2]— (2+l1;)v[n+1]+v[n]=0 3.48. Notice, y1[n] = 6]n] +6[n— 1] +26[n2]. Furthermore, 2:2[11] = zl[n—1] —2:1:1[n—2].
Since the system is LTI, ' y2[n] = y1[n — 1] — 2y1[n — 2].
MATLAB is used to plot the result. >> y1 '= inline(’(n=0)+(n==1)+2*(n=2)’); n  [2:8];
>> stem(n,y1(n1)2*y1(n2),’k'); axis([~2 8 4.5 4.5]);
>> x1a.be1(’n’); ylabel(’y.2[n] ’); Figure 33.48: Plot of y2[n] = 91 [n — 1] — 2y1[n — 2]. 3.49. Using the sifting property, this system operation is rewritten as yIn]
0.5 (:i:[n] + :i:[—n]). ~ (3) This system extracts the even portion of the input. (b) Yes, the system is BIBO stable. If theinput is bounded, then the output is neces sarily bounded. That is, if ]z[n] 5 Ms < 00, then ]y[n] = l0.5 (:1:[n] + z]—n]) S
 9,5(]r[n}l+]z{‘m]ﬁsM¢ <ool~ ~~      ~   ~~   ~ ' (c) Yes, the system is linear. Let y1[n] = 0.5(zlin] + z1[n]) and yarn] =
0.5(zg[n] +Jng[—n.]). Applying 0.21nt + bzr2[n] to the system yields y[n] =
0.5 (0.271][n] + 62:2[11] + (an [—n] + b22{n])) = 0.5a(:z:1[n] +21[—n]) +0.5b(:rg[n] +
Sal"ll = agilnl ! byzlnl ' (d) No, the system is not memoryless. For example, at time n = 1 the output
y[1] = 0.5(z[1] + z[—1]) depends on a. past value of the input, z[—1]. (e) No, the system is not causal. For example, at timen = 1 the output y[—1] =
O.5(a:[~1] + :z:[1]) depends on a future value of the input, :r[1]. (f) No, the system is not timeinvariant. For example, let the input be 2:1[11] =
u[n+ 10] u_[n— 11]. Since this input isslready even, the output is just the input,
y1[n] =e 2:1[n], Shifting by a. nonzero integer N, the signal 1:2[12] = 1:1[n  N] is
not even, and the output is y2[n] aé y1[n — N] = 2:1[12 — N]. Thus, the system
cannot be timeinvariant. ' . ' 3.51. (3.) yin + 1] = 05ylnl (1) Setting a = —1 andsubstitutinggL—ll .=.10, yields y_[0] '= 0.5(10) = 5 Setting 1: =3: 0, and substituting y[0] = 5, yields y[1] = 0.5(5) = 2.5 Setting n a: 1 in (1), and substituting y[1] = 2.5, yields 3.5—4. ‘ y[2] = o.5(2.5) = 1.25
Mn 1+ 2] = 3y[n + 1)  2y[n] + :{n + 21+ 3:1:[n + 11+ 3:1:[n] Setting n = —2, and substituting y[—1] = 3, y[—2] = 2, :c[—1] = x[—2] = 0, :z:[0] = 1, yields y[0] = —3(3) — 2(2) + 1 + 3(0) +3(0) = —12 Setting n = —1, and substitnt‘nrg 11.10] = —12., y[—1] = 3, z[—1] = 0, 2.10] = 1, z[1] = 3, yields y[1] = —3(—12) — 2(3) + 3 + 3(1) + 3(0) = 36 Proceeding along same lines, we obtain 3.65. (a) (b) y[2] = —3(36) — 2(12) + 9 + 3(3) + 3(1) = —63 By deﬁnition, any element in the Fibonacci sequence is the sum of the previous
two. Thus, 'fIn] = f [n  1] + f [n — 2]. Written in standard form, this yields flnl  fln 11fln 21=0 This is a somewhat unusual system in the fact that it has no input. In the lingo
of signalls and systems, f [n] is a zeroinput response that is completely driven by
the amtiliary conditions. The characteristic equation is 1’ my — 1 = O. This yields twocharacteristic roots _ 1+¢5 1\/§
2 2 Since one characteristic root is in the righthalf planeI the system is not stable. To determine a particular Fibonaccinumber, it is convenient to determine a
closed form expression for f Since f [n] is a zeroinput response, it has form
f[n] =‘c1'yi‘ + cry; The stadiumequations yie'l'cl 'f[1] = 0. = c171 + 0272 and f[2] =4; 1 = ‘cnfcz'yg. Solving yields c; {Ti$3133 = 3%11 1% 0.2764 and 2 5
__ 5 1 ~
Ca _ '12 '11‘11 ~ 0'7236' MATIiAB is used to solve for the requested values of f [n]. >> gammal  (1+sqrt(5))/2; gamma2  (lrsqrt(5))/23
>> c1' = gamma2/(gamma1*gamma2* (gamma2gamma1));
>> c2 = gammal/(gamma1*gamma2*(gamma2—gamma1)) ; '
>> £510 = c1*gamma1'50+c2*gamma2‘50 ' £50 ‘1 7.7787e+009 >> £11000 3 ci*gamma1‘1000+c2*gamma2‘1000 ﬂood“ = 2.6864e+208 :3 1.618 and 72 = a: —0.618. 71= Thus, 1\/5
2 f5—1
2V5 1+ 2 50 50
f[50]= «5) + ) z7.7787(109) (/5+1‘ 1000
1 3%) z 2.6864(10208) ( 3.71. (3.) The chatatteristic equation is '1 + 2 a1 = 2, b1 i: 1. Therefore h[n] = $6.211] + c(2)" (1) We need one value of h[n] to determine c. This is determined by iterative solution of .
(E + 2)h[n] = 6[n] or h[n + 1] + 2h[n] = 6[n] 100 Setting n = —1, and substituting h[—1] = 6[—1] = 0, yields
h[0] = 0  Setting = 0 in Eq. (1) and using h[0] = 0 yields 1 1
0=— '=_....
2+6 =# c 2 Therefoub 1 1
wu=gw§emww (b) The chairmeristic root is —2, b; = 0, a1 = 2. Therefore
' ‘ m=ww m
We nee4 one value of h[n] to determine c. This is done by solving iteratiwa
h[n + 11+ 2h[n] = ah + 1] '
Setting t; = —1,'a.nd substituting h[—1] = o, 6[0] = 1, yields
h[0]_ = 1
Setting in = 0 in Eq. (2) and using h[0] = 0 yields 1=c and — 7  ‘h[11}=(—2)”u[n}““ 3.7;3. (E2  6E + 25)y[n] = (2E2 — 4E)z['n.] The characterﬁtic roots are Seijo‘gza, b2 = 0. Therefore h[n] = c(5)" cos(0.923n + 0)u['n.] of h[n] to determine c and 0. This is done by solvixig iteratively (2) ; (1)
.We need two vtiues h[n] — 6h[n — 1] + 25h[n — 21 = 26[n] — 46['n. — 1]
Setting n = 0 {yields h[0] — 6(0) + 25(0) = 2(1) — 4(0) => h[0] = 2 in (2) yields V hm  6(2) + 25(0) = 2(0) — 4 => h[1] = 8 Setting 1?. = 0) 1 in (1) and substituting h[0] = 2, h[1] = 8 yields Setting n —‘= 1 2 = ccosﬁ
3} s = 5c cos(0.923 + a) = 3.017est — 3.987csin9
Solution of thiese two equations yields _ . ~ _ _ _ _  c = 2.061
9 = —0.244tad ‘1 ccose = 2
‘3 csinﬁ = ~0.493]. and ‘
'h[n] = 2.061(5)" cos(0.923n — 0.244)u[n] 3.81. (2)"u[n —11*e'“u[n+11
i z '(—2)"‘u[m  1]e"(“"“)u[n  m  1] i
\
m=—oo 102 However, u.[m‘— 1] = 0 for m. < 1 and u[n —m+ 1] = 0 for m > 11+ 1. Hence the summation 1 We can also yin] = 6"" 2 (2e)"‘ innits may be restricted for 1 5 m 5 n + 1. "+1 e_n [(—2e)""’2 + 2e] % m=1 —2e1 _ 252 ﬁ+1_ —(n+1)
;  2e+1[( 2) e 1‘4"] dbtain this answer by using the convolution Table and the shift property of convolution. If we advance impulse response h.[n] by one unit and delay the input by
one unit, the convolution remains unchanged according to the shift property. Hence
we should obtain the convolution by using h[”1=(2)""1u[n] and 4n] =e—(n1)u[n] Thus the desired convolution is given by yln] (2)"+1u[n] * e'("'1)u[n]
= —2e {(2)"u[n] III e'"u[n]} From the convolution Table, we obtain W uln] uln] _ n+1_ —(n+1)
1 _2e [( 2) ‘3 262 _.
= 23 + 1 B'WH " 3 MD] "M which odnﬁrlhs earlier result. 3.83. Here let us delay a:[n] by one unit and advance hfn] by one unit to obtain y[n]. y[n] 3"“ulnl * [2"1+ 3(5)"+1]u[n1
1‘  3"+1u[n] * 2""1u[n] + 3(3)"+1u[n] * (—5)"+1u[n] ﬂ 3mm *2"u[n1} —45 {3"uln1* (—5)"u[nl} ‘ [3"+1 — 2n+1]u[n]  45t[w] urn] 99 n n n
[#3) —3(2) ~225(5) 114"] 1 8 3311.  (a) l
M
a
.E.
.3.
*
A
P
C"
v
a
5—.
.3. WI] (b) ]
z[n] = 2("_3)u[n] = 2'32"u[n] = é2"ub1]
From 12115 regult in part (a), it follows that yln] = gm“ — (05)"+11u[n1 = 112t2"+1 —(05)"+‘1u[n1 "(6) z[n] = 2"u[n  2] = 4{2("'2)u[n — 2]} Note th 2("")u[n — 2] is the same as the input 2"u[n] in part (a) delayed by 2 erefore from the shift property of the convolution, its response will be units.
the s as in part (a) delayed by 2 units. The input here is 4{2<"2>u{n — 21}.
Therefo; yln = 4%['/"‘+1'2  (0.5)"H‘21uln — 2] = 28“  (05)";‘luln  2] 3.815. The equation describing this situation is [see Eq. (3.9b)]
(E —— a.)y[n] = Ezr[n] a, :1 + r = 1.01 The initial con 'tion y[— 1] = 0. Hence there is only zerostate component. The input
is 500u[n] — 15 6[n — 4] because at n = 4, instead of depositing the usual $500, she
withdraws $10qp. To ﬁnd Mn], we solve iteratively ‘ (E — a)h[n] = E6[n] or h[n + 1] — ah[n] = + 1] Setting n = ~i, and substituting h['— 1] = O, 6[0] = 1, yields
' h[0] = 1 Also, the chardoteristic root is a. and b0 = 0. Therefore I h[n] = ca"u[n] Setting n = O and substituting Mo] = 1 yields l 1 = c Therefore
kin} = (a)“u[nl = (1.0mm) The (zerostatip) response is “[11] = (1.01)'3u[n]u[n1 ‘ l = (1.01)“u[n] at {5001411)  15006 [n  4]} = 500(1.01)"u[nl * u[‘n] — 1500(1.01)"“u[n — 4] = '%[(1.01)“+1 — 11141;} — 1500(1.01)"4u[n — 4}
= 50000[(1.01)"+1 — 1]u[n] — 1500(1.01)"4u[n — 4] 3.846. This problem is identical to the savings account problem with negative initial deposit
(loan). If M is the initial loan, then y[0] = —M. If y[n] is the loan balance, then {see Eq. (3.9b)]
y[n+1]ay[n]=x[n+1] a=1+r (E  GM"! = Exlnl The characteristic root is a, and the impulse response for this system is found in Prob.. , 3.8«15 to be
h[n] = a"'u.[n] This proble can be solved in two ways. _ First math d: We may consider the loan of M dollars as an a negative input —M6[n].
The monthl payment of P starting at n = 1 also is an input. Thus the total input is
a:[n] = M n] + Pu[n  1] with zero initial conditions. Because u[n] = 6[n] +u['n— 1],
s the input in a more convenient form as x[n] = —(M + P)6[n] + Pu[n). we can expr The loan balance (response) y[n] is Alsoa=1+r yln] = hlnl = {—Ma"+P[ a: :1:[n] n = N, the loan balance is zero. Therefore 01' M: rM —Ma”u[n] — P [a” — 1L y[N] = _MaN+P[ 1. yaw = ca“u[n] aN—l = a"u[n] a: + P)6[n] + Pu[n]}
= (M + P)a"u[n] + Pa"u[n] * u[n] —(M + P)a”u[n] + P u[n] an+1 _ 1 ____i—] u[n] a— : :11]u[n1 and a — 1 = r where r is the interest rate per dollar per month. At
=o rM 1+aN = 1+(1+r)N Second method: In this approach, the initial condition is y[0] = —M, and the
input is z[n] =9 Pu[n — 1] because the monthly payment of P starts at n = 1. The
characteristic root is a, and The zero—input response is Setting n = 0, and substituting yo[0] = —M, yields c = —M and The zero—state response Iy[n] is yo[n] = —Ma“u[n] Here we use shift property of convolution. If we let The shift property yields The total balance is yo[n] + y[n] 111 Pa"u[n] a: u[n — 1] = :r[n — I] = P [ yoln] + W] = Ma"u[n] + p [ a an+l _ 1 an an —1 a— —1 —1 y[n] = h[n] a: :1:[n] = h[n]‘* Pu['n.— 1] = Pa”u[n] 2:: u[n  1] wln] = awn] « uIn] = [a—‘_1—]u[nl 11w” 1.1M . 3.826. For n > 1, u[n] = u[n  1] = 1. Therefore a“,— 1 al Loan balance = Ma“,+ P [ J n > 1 which conﬁrms the result obtained by the ﬁrst method. From here on the procedure
is identical to tihat of the ﬁrst method. (a) Actually, tem. 0 plus Wh
before e yln] = id only a ﬁrstorder difference equation is necessary to describe this sys
reﬁll n, the amount of sugar y[n] is equal to the amount added ﬁn]
ver was still in the mug. Since Joe drinks 2/3 of his'cup of coffee
reﬁll, onethird of the sugar from the previous cup remains. Thus, — 1]/3 + z[n]. Thus, a1 = 1/3,a2 =0,bo= 1,bl=0, and bg= . In stande form, the diﬂerence equation is y[n] — y[n — 1]/3 = z[n]. (b) Since .10? adds two teaspoons of sugar each time he ﬁlls his cup, (0) (d) The tot a:[n] = 2u[n]. solution to a difference equation is the sum of the zeroinput response and the erastate response. Since Joe starts with a clean mug, y[—_1] = 0 and the zer put response is necessarily zero, yo[n] = 0. Thus, the total solution is ' just the erostate solution. Tolobtai the herostate solution, the "impulse—response Mn] is needed. In this 
case, h[ = ﬁg +37o[n]u[n], where gob] = c7". To determine c, input a:[n] = 6[n] into the Thus, That is, teaspoo
One we at the s
reﬁll. T riginal difference equation to yield h[n] — h[n — 1] / 3 = 6[11]. Since hIn] h[0] — h[—1]/3 = h[0] = 6{0] = 1 = :70[0}u[o] = c. Thus, 14:] (= 3)“"u[n].
1_ ~ n+1 is causal
The zertigstate solution is a:['n.] * h[n] = (Egg) 2(3)'“) u[n] = 2Wufn] =
(3 — 3" u [n] . an] =' (3 — 3") ulnl . __ . _ n =
5121;0th — is; <3 3 l 3
after many cups of coffee, Joe's mug reaches a steadystate of three
of sugar. ystate value of three and then add two teaspoons of sugar at each
t is,
:rIn] = 2u[n] + 6 The addbd 6[n] “jump starts” the sugar content of the ﬁrst cup to the steadystate value. If Joe dtsires a steady value of two teaspoons, which is twothirds the original steadys ate value, the input is simply scaled by 2/3. That is, the steady level Mn} = 2u[n] is achieved using the input 2[n] = gum] +‘§5[n]. to make Joe’s coffee remain a constant‘for all non—negative n is begin ~ . ...
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