Digital Signal ProcessingSolutions ManualThomas J. CavicchiGrove City CollegeJohn Wiley & Sons, Inc. 2000 Digital Signal ProcessingSolutions ManualThomas J. CavicchiTable of Contents Begins on pageChapter 2......................................................1Chapter 3....................................................50Chapter 4..................................................120Chapter 5..................................................220Chapter 6..................................................283Chapter 7..................................................408Chapter 8..................................................494Chapter 9..................................................582Chapter 10................................................658Note from author to instructors:I will place corrections and/or improvements to these solutions on my Web page for this book.To view these, you will need the following login and password:password: kh7yb2q5 login: facultyIf you have corrections, comments, or suggestions, please email me at the address also shown on themain webpage for this book.1END-OF-CHAPTER PROBLEM SOLUTIONS FOR CHAPTER 2DIGITAL SIGNAL PROCESSINGTHOMAS J. CAVICCHI2.1) Consider the mass, whose position is labeled as xc(t), as one "node" and theother node, whose position is designated x1,c(t) to the right, as the node joiningBd and K. Then, at the mass node, the sum of the forces in the -xc(t) directionis equated with fc(t), directed to the right:Md2xc(t)/dt2 + Bdxc(t)/dt + B1d[xc(t) - x1,c(t)]/dt = fc(t), (*)while for the other node, summing the forces in the -x1,c direction,B1d[x1,c(t) - xc(t)]/dt + Kx1,c(t) = 0. (**)Applying the approach of Example 2.2, which by the way is really the "transformtechnique" in disguise, we note that (**) can be rewritten as{B1d/dt + K}x1,c(t) = B1dxc(t)/dt. Operating on (*) with {B1d/dt + K} and notingthat the derivative factors commute, we have{B1d/dt + K}{Md2/dt2 + (B + B1)d/dt}xc(t) - B1d/dt{B1d/dt + K}x1,c(t) = {B1d/dt + K}fc(t)or{B1d/dt + K}{Md2/dt2 + (B + B1)d/dt}xc(t) - B1d/dt{B1d/dt}xc(t) = {B1d/dt + K}fc(t)or{B1Md3/dt3 + [B1(B+B1)+KM-B12]d2/dt2 + K(B+B1)d/dt}xc(t) = {B1d/dt + K}fc(t)or{B1Md3/dt3 + [B1B+KM]d2/dt2 + K(B+B1)d/dt}xc(t) = {B1d/dt + K}fc(t)or{d3/dt3 + [B/M + K/B1]d2/dt2 + (K/M)(1+B/B1)d/dt}xc(t) = (1/M){d/dt + K/B1}fc(t).Thus, the coefficients for (2.20) area3 = aN = 1, a2 = B/M + K/B1, a1 = (K/M)(1+B/B1), b1 = 1/M, b0 = K/[MB1].2.2) Define iL,c(t) as the current ("downward") in L. Then vo,c(t) = LdiL,c(t)/dt. By KCL, is,c(t) = vo,c(t)/R + iL,c(t) + Cdvo,c(t)/dt. Take d/ dt o f th is s ec ond equation [in order to eliminate iL,c(t) by usingdiL,c(t)/dt = vo,c(t)/L]:dis,c(t)/dt = dvo,c(t)/R + diL,c(t)/dt + Cd2vo,c(t)/dt2, ordis,c(t)/dt = dvo,c(t)/R + vo,c(t)/L + Cd2vo,c(t)/dt2, ord2vo,c(t)/dt2 + (1/[RC])dvo,c(t)/dt + vo,c(t)/[LC] = (1/C)dis,c(t)/dt. (*)Thus a2 = aN = 1, a1 = 1/(RC), a2 = 1/(LC), b1 = 1/C, b0 = 0.From the theory of differential equations, we know that the output will involveterms having the form of the input and its derivatives up to order N = 2. We usehere the method of undetermined coefficients. Thus, letvo,c(t) = ocos(t/{LC}1/2) + sin(t/{LC}1/2), so2dvo,c(t)/dt = -[o/{LC}1/2]sin(t/{LC}1/2) + [/{LC}1/2]cos(t/{LC}1/2), andd2vo,c(t)/dt2 = -[o/(LC)]cos(t/{LC}1/2) - [/(LC)]sin(t/{LC}1/2).Substitution of these into (*) gives[-o/(LC) + (1/[RC])/{LC}1/2 + o/(LC)]cos(t/{LC}1/2)[-/(LC) + (1/[RC])o/{LC}1/2 + /(LC)]sin(t/{LC}1/2) = -[A/(C{LC}1/2)]sin(t/{LC}1/2).Equating co ef ficients of cos(t/{LC}1/2), we h a v e = 0, w hile equatingcoefficients of sin(t/{LC}1/2), we have-(1/[RC])o/{LC}1/2 = -A/[C{LC}1/2], or o = AR. Thusvo,c(t) = Arcos(t/{LC}1/2) = Ris,c(t).So, und er r es ona n ce , th ere i s zero phase shift; only scaling by R. Fromresonance theory (using the frequency response), we know that for = 1/{LC}1/2,the maximum output occurs. That is, Z() R.2.3) KVL around the outer loop gives disvs - Ls - iTRT = vL, (1) dtbut because Rs and Ls are in parallel, dis(iT - is)Rs = Ls (2a) dtor Ls disiT = + is, (2b) Rs dtso that (1) becomes RT disvs - Ls(1 + ) - RTis = vL, (3a) Rs dtor d( + a)is + bvL = bvs (3b) dt RT awhere a = , b = . By the definition of vL, Ls(1 + RT/Rs) RT diLiLRL + LL = vL (4a) dtor3 d( + e)iL - fvL = 0 (4b) dt RL ewhere e = , f = . By KCL and (2b), LL RL Ls dis dvLiT = + is = iL + C (5a) Rs dt dtor, using (3b) for dis/dt, Ls Ls dvL(1 - a )is + b ()(vs-vL) = iL + C (5b) Rs Rs dtor d( + g)vL + his + miL = gvs (5c) dt 1 1where g = , h = -Rsg, m = . RsC(1 + RT/Rs) CTo obtain a single differential equation relating vL to vs, we need to eliminate d d d2 dis and iL. Because {} = , the ( + o) factors can be treated dt dt dt2 dtalgebraically. Following the rules of algebra, "multiply" (operate on) (5c) by 1 d( + a) to get h dt 1 d2 d d m d { + (a + g) + ag}vL + ( + a)is + ( + a)iL h dt2 dt dt h dt(6) g d = ( + a)vs, h dtwhich when (3b) is substituted gives 1 d2 d m d g d ag (7) { + (a+g) + ag-bh}vL + ( + a)iL = { + - b}v. h dt2 dt h dt h dt h d m dSimilarly, "multiply" (7) by ( + e) and (4b) by ( + a) to give, dt h dtrespectively, 1 d3 d2 d { + (a+g+e) + (ag-bh+e{a+g}) + e(ag-bh) } vL h dt3 dt2 dt(8) m d d g d2 ag eg d ag+ ( + a)( + e) iL = { + ( - b + ) + e( - b)}vs h dt dt h dt2 h h dt h4I

I

x

x

x Ior which springbreaks.Q

Q

Linear spring regime: I kx.AI/x k in I kx. IxBI/x A{I/x k and m d d mf d ( + a)( + e)iL - ( + a) vL = 0, (9) h dt dt h dtwhich when substituted into (8) gives, upon multiplication by h, d3 d2 d d2 d (10){ + a2 + a1 + a0 } vL(t) = { b2 + b1 + b0 } vs(t) dt3 dt2 dt dt2 dtwherea2 = a+g+e, a1 = ag-bh+e(a+g)+fm, a0 = e(ag-bh)+fam,b2 = g, b1 = ag-bh+eg, b0 = e(ag-bh), (11)(a3 = 1 was tacitly enforced), which when circuit parameters are substituted andsimplified givesa2 = (LLLs + C[RLRTLs + RsRTLL + RsRLLs])/A,a1 = (RLLs + RsLL + RTLs + RsLs + RTRLRsC)/A, a0 = (RLRs + RTRs)/A(12)b2 = LsLL/A, b1 = (RsLL + RLLs)/A, b0 = RLRs/AwhereA = LLLsC(Rs + RT).(13)2.4) (a) Because K at operating point B is not equal to that at operating pointA, we must first find the operating point at which we are running the system, sothat we k n ow w hi ch l ine ar m od el applies. If K were independent of theoperating point, the system would be linear. See figure below. 5-2 -1 0 1 2-2-101200.20.40.60.81x1x2

I e xp-x1`2-2x2`2) Ix1(x1,x2) Ix2(x1,x2)x1x2x2x1Values oI linear approximation oI I Ior diIIerent x1, x2 all lie on the plane Iormed by the two tangents shown. I exp-x12 - 2x22}R

R

V

R

V

0.7 V~I

I

R

V

(b)2.5) To find the operating point, we replace the transistor by a simple modelvalid in forward bias, and the capacitors by open circuits, as shown in thefigure below.6The operating point calculation is as follows. By KCL,VB - VCC VB + + IB = 0. R2 R1By KVL, VB = 0.7 V + (+1)IBRE. Combining these,0.7 V + (+1)IBRE VCC - + IB = 0, or R1||R2 R2 (+1)RE VCC 0.7 VIB{ + 1} = - . R1||R2 R2 R1||R2Solving for IB, we obtain VCC 0.7 V - R2 R1||R2IB = = 10.93 A. (+1)RE 1 + R1||R2Also, IC = IB = 1.640 mA. Now that we have the operating point, we linearize the system about it. We usethe usual notation that, e.g., vBE = VBE + vbe, where vBE is the total base-emittervoltage, VBE is the operating-point base-emitter voltage, and vbe is the small-signal deviation of the base-emitter voltage about its operating point VBE. Inforward bias, the total base current is equal to qvBE/kT qVBE/kT qvbe/kTiB Is e = Is e - e qvbe/kT = IB e, (*) qVBE/kTwhere IB = Ise is the (approximate) operating point base current. Weexpand the exponential in (*) in a linear Taylor series:iB IB{1 + qvbe/(kT)} = IB + ib,where the small-signal base current ib isib = IBqvbe/(kT) = vbe/r

wherer

= kT/(qIB) = kT/(qIC) = 2.378 k.Notice that the linear model parameter (r

) depends STRONGLY on the operatingpoint (IB or IC). Substit ut ing the hybrid-pi model for the transistor, andassuming the capacitors are large enough to ignore their impedance at the signalfrequencies, we obtain the linear model shown below, valid for small-signalexcursions about the operating point determined above. 7

o

s o

in

v

s

1

2 L

c

s ob

bWe now easily obtainvo = -(RCRL)ib = -(RCRL)vs/r

, yielding for the voltage gain voAv = = -(RCRL)/r

= -12.61. vsIf we apply a test source vT to the output (and remove the load RL), with vs heldat zero, it is clear that vT sees the resistance Zo = RC = 167 , which is theoutput impedance. On the input side, vs sees Zin = R1R2r

= 2.31 k. Of course,in practical applications Zin is usually desired to be much larger, and Zo isdesired to be smaller, but this is only a one-transistor amplifier.The Thevenin-equivalent (linearized) model of the transistor amplifier is asshown in the final figure below.2.6) First note that Fr = frMg = 302.6 N.We first find the equilibrium constant operating point thrust force. Vmph = 50 mph corresponds to V = 80.47 km/h = 22.35 m/s.At the operating point,F - FR - cCDSV2 = 0. Solving for F, F = Fr + cCDSV2(*) = 302.6 + 0.5-1.2-0.4-2-22.352 548 N.Now let Af = f - F, Av = v - V. Then the equation of motion becomes (noting thatdV/dt = 0 because V is a constant),8Af + F - cCDS(Av+V)2 - Fr = MdAv/dt, orAf + F - cCDS(Av2 + 2VAv + V2) - Fr = MdAv/dt. Ignoring the second-order term,Af + F - cCDS(2VAv + V2) - Fr = MdAv/dt. Notice now that the terms F, Fr, and cCdSV2 cancel, due to (*). We are left withAf - cCDSVAv = MdAv/dt, ordAv/dt + (cCDSV/M)Av = Af/M. (**)(**) is the linearized differential equation, valid for small excursions aboutthe operating point (V, F).If we now apply a step function Af = 50 N, the final value of Av will be [settingAv = 0 in (**)]Avfinal = Af/(cCDSV) = 50/(1.23-0.4-2-22.35) = 2.27 m/s = 8.18 km/h = 5.08 mph.The time constant involved is, from (**), M/[cCDSV] = 1814/[1.23-0.4-2-22.35] = 82.5 s. Clearly, this is unacceptable in practical driving conditions, so we will needfar, far greater a Af to have reasonable acceleration. Of course, that willcause us to have to solve the nonlinear differential equation.2.7) y(1.5) = sin(3n)+0.2(1.5)3-5(1.5) + 2ln(1.5) = -4.6378.y' = 2ncos(2nx) + 0.6x2 - 2.5/x1/2 + 2/x.y'(1.5) = 2ncos(3n) + 0.6(1.5)2 - 2.5/(1.5) + 4/3 = -2n+1.35-2.0412+1.3333 = -5.6411.Thus, y(x) -4.6378 - 5.6411(x-1.5).For x very near 1.5 (first plot, on next page), the agreement is fairly good.With x ra n gi n g fa rt her fr om 1.5 (second plot), the linear approximationeventually fails.91.2 1.3 1.4 1.5 1.6 1.7 1.8-6.5-6-5.5-5-4.5-4-3.5-3-2.5xy and its linear approximationSolid exact y, ---- linear approximation of y.0.5 1 1.5 2 2.5-12-10-8-6-4-202xy and its Iinear approximationSoIid exact y. ---- Iinear approximation oI y. 100 5 10 1500.20.40.60.811.21.41.61.82Time t in sec.xc(t)Continuous-time input xc(t). t -ot' -ot -ot2.8) (a) sc(t) = - e dt' = -(-1/o)(e - 1)uc(t) = (/o)e uc(t). 0For o < 0, sc(t) begins at the negative value /o and exponentially goes toward- as t > . For o > 0, sc(t) be gi n s at the positive v a l u e /o andexponentially decays toward 0 as t > . 0 -ot'(b) sc(t) = e uc(t - t')dt'. -For t > 0: 0 -ot'sc(t) = e dt' = -(/o)(1 - 0) = -/o. -For t < 0: t -ot' -ot -otsc(t) = e dt' = -(/o)(e - 0) = -(/o)e. -Thus, sc(-) = 0, it exponentially rises to -/o at t = 0, and from there onremains fixed at -/o.2.9) (a) The plots are as shown below.110 5 10 15-0.0500.050.10.150.20.250.30.35Time (secs)AmplitudeSimulated output yc(t) for input xc(t).0 5 10 1500.20.40.60.811.21.41.61.82Time t in sec.xc(t)Continuous-time input xc(t-t0), where t0 = 5.(b) Again, the plots are as shown below. We indeed see that the result here isdelayed by t0 = 5 s relative to the result in part (a). We conclude what wealready knew from a knowledge of linear differential equations--that the systemis indeed time-invariant.120 5 10 15-0.0500.050.10.150.20.250.30.35Time (secs)AmplitudeSimulated output yc(t) for input xc(t-t0) where t0 = 50 5 10 15-3000-2500-2000-1500-1000-50005001000Time (secs)AmplitudeSimulated output yc(t) for inputs xc(t) and xc(t-t0) where t0 = 5.2.10) The plot is as shown below. Yes, time invariance still holds even when thesystem is unstable.132.11) (a) We have shown in Example 2.7 that (1 - e

)/o, 0 t Tyc(t) = e-(e

- 1)/o, t > T 0, otherwise. The code for plotting yc(t) for 0 t 10 s is as shown below.It is the solid curve in the plots.(b) Let the maximum number of terms to be kept in the sum approximating theintegral be Nintmax. We will run our sum from m = 0 to the smaller of Nintmax - 1(giving us Nintmax terms in our sum) ans [t/At], the greatest integer of t/At. By the rectangular rule estimate of the convolution integral, we have -o(t-)yc(t) = e uc(t-) [uc() - uc(-T)]d - min(T,t) -o(t-) min(Nintmax-1,[t/At]) -o(t - mAt) = e d At- Z e. 0 m=0The code for plotting is again given in the program below, and theresults are plotted using O stems. See the code for explanationof discrepancies for Nintmax = 10. (c) We will use the transfer function representation to provideas specifying input to "lsim." -(o+s)t 1Hc(s) = e dt = for Re{s} > -Re{o}. 0 s + oNoting the boxcar form of the input, we write the code belowfor using lsim to do the integration for us (it chooses its own"At" which may be variable within the same simulation). The results are plottedin the plot below using * stems. It of course provides the same, accurateestimate regardless of the At (Nintmax) parameter for part (b).The complete Matlab code follows.clg; T=2; Nt=120; alf=0.8; t=linspace(0,5*T,Nt);A=exp(alf*T)-1;% Exact solution (NOTICE NO "FOR" LOOPS!):% Note that we use "plot" because we are plotting a continuous-time signal.% (Although we are really plotting only every dt, we think of yc(t),% the exact solution, as "continuous"--Matlab draws straight lines between% our calculated points.y=(1-exp(-alf.*t)).*(tT);y=y./alf;plot(t,y,'r')xlabel('t (s)')ylabel('yc(t) and estimates')hold on; pause141 . 2 * s t e m s = l s i m r e s u l t ; d t = 0 . 2 s , N i n t m a x = 1 0 . S o l i d = e x a c t s o l . , O s t e m s = r e c t a n g . r u l e e s t . ;% Rectangular rule estimate:Nintmax=100;%Nintmax=10;dt=T/Nintmax; N1=Nintmax-1;% At t=0, the integral runs from 0 to 0, so we know y(0) = 0.% Note that array index=0 is not permitted in Matlab.yest(1)=0;% Remember that i is the index corresponding to evaluation time t,% while m below is the integration index corresponding to tau.for i=2:Nt m=[0:min(N1,floor(t(i)/dt))]; sterms=exp(-alf.*(t(i)-m.*dt)); yest(i)=sum(sterms);endyest=yest.*dt;% Note that we use "stem" because we are plotting estimates of% yc(t) at only discrete times.stem(t,yest)pause% Case of too-small Nintmax (e.g., Nintmax = 10):% Reason for ratchet appearance on way up and none on way down:% On way up, as you are "waiting" for t(i)/dt to become% large enough to include another term in the sum,% the exponent becomes larger negative for each term due% to the increase in t(i); hence yest becomes smaller and smaller% until a new sample is included--i.e., t(i)/dt exceeds a new% integer. On the way down the curve, the exact result goes down% too (note that now as t(i) > T, the same number of points is% used for each integration--Nintmax-1),so no ratchet appearance), making theestimate% "more exact" by "luck"!% "lsim" estimate:num=1; den=[1 alf]; u=t 2, the upper limit on the integral will be A = 2.For t e [1, 2], we thus have A -0.5(t-) -0.5(t+5) A+5 0.5vyc(t) = (+5)e d = e v e dv, 1 6where in the latter expression we set v = +5, so d = dv and t- = t+5-v. In the above integral, set u = v, du = dv, dw = e0.5vdv, w = 2e0.5v. Thus 0.5v A+5 -0.5(t+5) ve A+5 0.5vyc(t) = e { - 2 e dv } 0.5 6 6 -0.5(t+5) 0.5(A+5) 3 0.5(A+5) 3 = e { 2[(A+5)e - 6e ] - 4[e - e ] }. -0.5(t+5) 0.5(A+5) 3 = 2e { (A+3)e - 4e }. (*)This is our general expression. Now we set A = t: 0.5(1-t)yc(t) = 2{(t+3) - 4e }, t e [1, 2].As a check, we should, and do obtain yc(1) = 0, for there is zero "area" yet whenthe time-reversed hc(t) just begins overlapping xc(). The other end value is -0.5yc(2) = 2{(5 - 4 e }, for reference below.For t > 2, we merely set A = 2 in the general expression (*): -0.5(t+5) 3.5 3yc(t) = 2e { 5e - 4e )}, t > 2. -0.5As a check, we should, and do obtain yc(2) = 2{(5 - 4 e }, in agreement withour expression for t e [1, 2].There will be such continuity in the output as long as there are no impulsefunctions in xc(t) and/or hc(t).(b) The plots are as shown below.170 1 2 3 4 5 6 7 8 9 1001234567Time thc(t). xc(t). yc(t)Solid yc(t) xc(t)*hc(t).-.-. hc(t) exp(-0.5t)uc(t); --- xc(t) t5 Ior 1t2 and 0 otherwise.-1-1-4-4x

()h

(t)t1-3 1h

(t-)t1t3 t-1th

(t) is nonzero for t 0,so the system is noncausal.2.13) Refer to the pictures shown below. 18-7 -6 -5 -4 -3 -2 -1 0-8-7-6-5-4-3-2-10Time t (s)yc(t)Result of continuous-time convolution.t + 3 < -4: t < -7. For this case, yc(t) = 0. t+3 (t+3)2-(-4)2 t2+6t-7-4 < t + 3 < -1: -7 < t < -4: yc(t) = d = = . -4 2 2 -1 -4 < t < -3: yc(t) = d = -15/2. -4 -1 (-1)2-(t-1)2 -4 < t -1 < -1: -3 < t < 0: yc(t) = d = = t(1 - t/2). t-1 2 t > 0: yc(t) = 0. xc(t) begins at t = -3, and yc(t) begins at t = -7 > systemnoncausal. Check endpoints: yc(-4) = -7.5 = yc(-3).The plot of yc(t) is as shown in the figure below.19-T -T T t1Eor generality. let`s also look at howa non-symmetrical h

(t) would be handled.t - -T. or T t.t - T. or -T t.These identiIications show that h

(t - ) is time-reversed and shiItedto the right by t seconds (iI t is positive; to the leIt iI t is negative).Tx

()1h

(t)We use y

(t) ' x

()h

(t-)d.-88Special cases:No overlap for T t -T. or t -2T;yc(t) 0 for this range of t.Txc()1hc(t - )-T -Tt TtTxc()1hc(t - )-T -TtTtEor -T Tt T. or -2T t 0.yc(t) 1|T t - (-T)] 2T t.Notice that yc(-2T) 0 and yc(0) 2T.Txc()1hc(t - )-T Tt-TtEor -T -Tt T. or 0 t 2T.yc(t) 1|T - (-T t )] 2T - t.Notice that yc(2T) 0 and yc(0) 2T.tyc(t)-2T 2TSimilarly. for t ~ 2T. there is noverlap. s yc(t) 0.02T2.14) See the graphical solution below.202.15) (a) ac(t) = A[cos(t)cos(U) - sin(t)sin(U)] andbc(t) = B[cos(t)cos(o) - sin(t)sin(o)].Thuscc(t) = A[cos(t)cos(U) - sin(t)sin(U)] + B[cos(t)cos(o) - sin(t)sin(o)] = [Acos(U)+Bcos(o)]cos(t) - [Asin(U)+Bsin(o)]sin(t) = Ccos(t+v) = [Ccos(v)]cos(t) - [Csin(v)]sin(t).Equating coefficients of cos(t),Ccos(v) = Acos(U)+Bcos(o). (*)Equating coefficients of sin(t),Csin(v) = Asin(U)+Bsin(o). (**)Adding (*)2 and (**)2 and using sin2(x) + cos2(x) = 1,C2 = {Acos(U)+Bcos(o)}2 + {Asin(U)+Bsin(o)}2 = A2 + B2 + 2AB[cos(U)cos(o) + sin(U)sin(o)] = A2 + B2 + 2ABcos(U-o), orC = [A2 + B2 + 2ABcos(U-o)].Taking the ratio (**)/(*) gives Asin(U)+Bsin(o)tan(v) = . Acos(U)+Bcos(o)(b) Using the phasor approach, which is of course mathematically equivalent tothe above, we writecc(t) = Ccos(t+v) = Re{Cejt} where C = Cejac(t) = Acos(t+U) = Re{Aejt} where A = Aejand bc(t) = Bcos(t+o) = Re{Bejt} where B = Bej.Thuscc(t) = Re{Aejt + Bejt} = Re{(A + B)ejt} = Re{Cejt},where (using Euler's formula below)C = A + B = Acos(U)+Bcos(o) + j[Asin(U)+Bsin(o)] Asin(U)+Bsin(o) = {[Acos(U)+Bcos(o)]2 + [Asin(U)+Bsin(o)]2}1/2 /_tan-1{} Acos(U)+Bcos(o) Asin(U)+Bsin(o) = [A2 + B2 + 2ABcos(U-o)]1/2 /_tan-1{} Acos(U)+Bcos(o) = C /_U and thus the results are as in part (a) done without using phasors.However, the amount of thought and effort involved is reduced by using phasors.212.16) Differential equations approach:Write vo,c(t) = Bcos(t+o) = Bcos{t+U + (o-U)} = B[cos(t+U)cos(o-U) - sin(t+U)sin(o-U)] (*) = Ldi/dt where i is the "downward" current in L d vs,c(t)-vo,c(t) d = L { } + C {vs,c(t)-vo,c(t)} dt R dtord2 1 d 1 d2 1 d vo,c(t) + vo,c(t) + vo,c(t) = vs,c(t) + vs,c(t).dt2 RC dt LC dt2 RC dtSubstituting (*) and vs,c(t) = Acos(t+U) into this differential equation, we have 1 B{( - 2)[cos(t+U)cos(o-U) - sin(t+U)sin(o-U)] LC + [-sin(t+U)cos(o-U) - cos(t+U)sin(o-U)]} RC = -A[2cos(t+U) + sin(t+U)]. RCEquating coefficients of cos(t+U), we have 1 B[( - 2)cos(o-U) - sin(o-U)] = -A2, (**) LC RCand equating coefficients of sin(t+U), we have 1 B[(- + 2)sin(o-U) - cos(o-U)] = -A/(RC). (***) LC RCTaking the ratio (**)/(***), we obtain 1 - 2 - tan(o-U) LC RC = RC, or 1 (- + 2)tan(o-U) - LC RC 1 1 1[RC(- + 2) + ]tan(o-U) = - 2 + 2 = , LC RC LC LCor 1 n (****)o-U = tan-1{ } = - tan-1{[(2LC - 1)RC + L/R]}. [(2LC - 1)RC + L/R] 2As a final expression for the phase o, we may solve (****) for o as no = U + - tan-1{[(2LC - 1)RC + L/R]}. 2 To obtain the magnitude B, we may use (***), slightly rewritten as follows:22 1 | (

LC - 1)RC L/R| 1 B[(2LC - 1)sin(o-U) - cos(o-U)] = -A/(RC), LC RCyielding -A -AL/RB = = R L (2LC - 1)sin(o-U) - cos(o-U) (2LC - 1)sin(o-U) - cos(o-U) L R -A(L/R)/cos(o-U) = . (2LC - 1)tan(o-U) - L/RUsing tan(o-U) from the first expression in (****) above, -A(L/R)/cos(o-U)B = 2LC - 1 - L/R [(2LC - 1)RC + L/R] -A(2L/R)[(2LC - 1)RC + L/R]/cos(o-U) = 2LC - 1 - (2L/R)[(2LC - 1)RC + L/R] -A(2L/R)[(2LC - 1)RC + L/R]/cos(o-U) = (2LC - 1)(1 - 2LC) - (L/R)2 A(2L/R)[(2LC - 1)RC + L/R]/cos(o-U) = . (2LC - 1)2 + (L/R)2Now use the following right triangle, for cos(o-U): From first expr. in (****), where the hypotenuse, by the Pythagorean theorem, is {(2LC-1)2(RC)2+(L/R)2+1+22LC(2LC-1)}1/2. [(2LC - 1)RC + L/R]Thus cos(o-U) = . {(2LC-1)2(RC)2 + (L/R)2 + 1 + 22LC(2LC - 1)}Thus our expression for B becomes A(L/R){(2LC - 1)2(RC)2 + (L/R)2 + 1 + 22LC(2LC - 1)}1/2B = . (2LC - 1)2 + (L/R)223Performing long division in the hopes of simplification, (RC)2 + 1 _________________________________________________(2LC - 1)2 + (L/R)2 (2LC-1)2(RC)2+(L/R)2+1+22LC(2LC-1) (2LC-1)2(RC)2+(2LC)2 -(2LC)2+2(2LC)2+(L/R)2+1-22LC.The preceding line simplifies to (2LC)2-22LC+1+(L/R)2 = (2LC-1)2 + (L/R),which justifies the "+1" in the quotient with zero remainder. Hence,{(2LC-1)2(RC)2+(L/R)2+1+22LC(2LC-1)}1/2 1 + (RC)2 = {}1/2. (2LC - 1)2 + (L/R)2 (2LC - 1)2 + (L/R)2Thus we obtain L 1 + (RC)2B = A {}1/2. R (2LC - 1)2 + (L/R)2Phasor approach (MUCH EASIER!):Define phasors jU joVs = Ae and Vo = Beso that vs,c(t) = Re{Vsejt} and vo,c(t) = Re{Voejt}.Then jo jLVo = Be = Vs R/[jC] jL + R+1/[jC] jL = Vs R jL + 1 + jRC jL(1 + jRC) = Vs R + jL(1+jRC) 1 + (RC)2 = AL {}1/2 /_ U+tan-1(1/[-RC]) - tan-1{L/[R-2LRC]} (R-2LRC)2 + (L)2or, using the identity tan-1(a) + tan-1(b) = tan-1{(a+b)/(1-ab)}, L 1 + (RC)2 -1/(RC) - L/[R-2LRC]Vo = A {}1/2 /_ U+tan-1 }. R (2LC - 1)2 + (L/R)2 1 - L/{(R-2LRC)RC}Multiplying numerator and denominator of the inverse tangent argument byRC [R-2LRC], simplifying the resulting numerator, and then multiplyingnumerator and denominator by -1/R and simplifying that result, we obtain24 L 1 + (RC)2 1Vo = A {}1/2 /_ U+tan-1 }, R (2LC - 1)2 + (L/R)2 [(2LC-1)RC + L/R]which is i n e xa ct a gr ee me n t wi th the results of the differential equationsapproach, but by using FAR less effort.2.17) L and C have their v and i related by a first-order differential equation.So when these first-order differential equations are inserted into the KVL, KCLequations and they are combined into one, the result is an Nth-order differentialequation relating vc(t) and ic(t). Thus, for any RLC network, the total voltagevc(t) across the network and current ic(t) into it satisfy a linear, constant-coefficient differential equation N d

vc(t) M d

ic(t) Z a

= Z b

.=0 dt =0 dt

If we let vc(t) = Vcos(t+U) and view it as the excitation of the network, thenic(t) has the formic(t) = Icos(t+o). This is because the particular (steady-state) solution involves the excitationfunction and all of its derivatives up to order N. Thus in this case ic(t) mustbe sinusoidal of frequency, but different amplitude and phase, of course.Writing phasor expressionsvc(t) = Re{Vejt} and ic(t) = Re{Iejt} where V = Vej and I = Iej, we have N d

Re{Vejt} M d

Re{Iejt} Z a

= Z b

=0 dt =0 dt

or N d

Vejt M d

Iejt Z a

Re{} = Z b

Re{}=0 dt =0 dt

or N M Z a

Re{(j)

Vejt} = Z b

Re{(j)

Iejt}=0 =0 or N M Re{[ Z a

(j)

V - Z bm (j)mI]ejt} = 0. (*) =0 m=0 The only way that (*) can hold FOR ALL t is if the coefficient of ejt is zero,for Wejt is a rotating vector in the complex plane whose real part is zero onlytwice per cycle, not for all t.Thus we obtain M Z bm(j)m V m=0Z() = , which is a function of and a constant with respect I N to time. Thus, the impedance is also. Z a

(j)

=0250 1 2 3 4 500.20.40.60.81Time t (s)xc(t)Square-wave generated using "rem" function of Matlab.2.18) In order for xc(t) to be T-periodic in t, it must satisfy xc(t) = xc(t+T)for any t. Equivalently in this case,xc(t+T) = A1cos{1(t+T) + U1} + A2cos{2(t+T) + U2} = A1cos{1t+U1 + 1T} + A2cos{2t+U2 + 2T} = A1[cos(1t+U1)cos(1T) - sin(1t+U1)sin(1T)] + A2[cos(2t+U2)cos(2T) - sin(2t+U2)sin(2T)] = xc(t) for all t if and only if:(i) cos(1T) = 1, (ii) cos(2T) = 1, (iii) sin(1T) = 0, (iv) sin(2T) = 0.Of course, if (i) and (ii) are satisfied then (iii) and (iv) will also be. (i) and (ii) are satisfied for, respectively, 1T = 2n and 2T = 2nm where andm must be integers.Taking the ratio of these conditions, we obtain 1/2 = /m, which says that 1/2must be a rational number (i.e., a ratio of integers). For the smallest (, m)for which the condition holds, we obtain T = 2n/1 (or, equivalently, 2nm/2).Thus, e.g., if 1 = n/2 rad/s and 2 = 10n rad/s, then 1/2 = 1/20, so period T= 2n-1/[n/2] = 4 s. But if 1 = n/2 rad/s and 2 = 2-n, then 1/2 = 1/(22),which is not a rational number, so xc(t) is aperiodic.Notice that the condition is independent of the amplitudes and phases.Moreover, this result generalizes for the sum of ANY two periodic functions: Theratio of their periods must be a rational number.2.19) We write x = rem(t+T1/2,T) N, w e e xpe c t y ( n ) to b e neg at iv e because the larger h(n) values aremultiplying the negative values of x(n) in the convolution sum. Gradually, y(n)will then go to zero. Although technically it takes "forever" for y(n) to go toexactly zero, we expect it to be nearly zero roughly N samples after h(n) isnearly zero. Thus the plot for a near 1 is as shown in the figure below. Fora nearly zero, h(n) is like a Kronecker delta, so we expect y(n) x(n) = asingle sawtooth waveform. Thus, h(n)a=0 = o(n), and h(n)a=1 = u(n).28y(n)nRough sketch oI expected result oI convolution.NAssumed: h(n) declines slowly with n. (c) We have L 2 n, n e [0, N-1]y(n) = Z (1 - m) an-m, where L = N-1, n N . m=0 N-1 0, otherwiseThus L 2y(n) = an Z (1 - m)a-m m=0 N-1 1 - a-(L+1) 2 1/a + La-(L+2) - (L+1)a-(L+1) = an { } 1 - a-1 N-1 (1 - 1/a)2or (1-a-(L+1))(1-a-1) - [2/(N-1)]{a-1+La-(L+2)-(L+1)a-(L+1)}y(n) = an { } (1 - 1/a)2 1-a-1-a-(L+1)+a-(L+2) - [2/(N-1)]{a-1+La-(L+2)-(L+1)a-(L+1)} = an { } (1 - 1/a)2 1-(1+2/[N-1])a-1+{-1+2(L+1)/(N-1)}a-(L+1)+{1-2L/(N-1)}a-(L+2) = an { } (1 - 1/a)2or N-1 - (N+1)a-1 + (2L-N+3)a-(L+1) + (N-2L-1)a-(L+2)y(n) = an { } (N-1)(1 - 1/a)2Now, specializing for n e [0, N-1], with L = n: N-1 - (N+1)a-1 + (2n-N+3)a-(n+1) + (N-2n-1)a-(n+2)y(n) = an { } (N-1)(1 - 1/a)2and for n N, with L = N-1: N-1 - (N+1)a-1 + (N+1)a-N - (N-1)a-(N+1)y(n) = an { }. (N-1)(1 - 1/a)2The plots for a = 0.95 are as shown below; they agree with the general picture290 20 40 60 80 100-2-1.5-1-0.500.511.5Time index ny(n)O stems = result using conv routine; * stems = closed-form expression; a = 0.4.0 20 40 60 80 100-8-6-4-202468Time index ny(n)O stems = result using conv routine; * stems = closed-form expression; a = 0.95.in (b), though not exactly because the picture in (b) essentially assumed h(n)= u(n), which is not exactly true; thus, e.g., y(n) reaches zero before n = N.Also, the plots for a = 0.4 are as finally shown below. They also agree with thegeneral conclusion that y(n) x(n) for this case. Note that in all cases, theresult using "conv" agrees essentially exactly with that using the closed-formformula derived above.302.24) (a) The differential equation is obtained by balancing the forces in the"x" (vertical) direction:Md2x/dt2 + Bdx/dt + Kx = Mg, where g is the acceleration of gravity.Under backward difference, the differential equation becomes(M/At2)[x(n) - 2x(n-1) + x(n-2)] + (B/At)[x(n) - x(n-1)] + Kx(n) = Mg.Collecting terms,{M/At2 + B/At + K}x(n) - {(2M/At + B)/At}x(n-1) + {M/At2}x(n-2) = Mg, or 2M + BAt M MgAt2x(n) - x(n-1) + x(n-2) = . M + BAt + KAt2 M + BAt + KAt2 M + BAt + KAt2With the given parameter values, we obtainx(n) -0.7895x(n-1) + 0.3509x(n-2) = 0.0344 (for n > 0; RHS = 0, o.w. when we hold the mass in place at x = 0).Following the directions in the problem statement, letting N = (5 s)/At, weobtain the results shown in the figures below. With At = 0. 00 01 s = 0. 1 m se c, the discrete-time solution is completelyindistinguishable from the continuous-time solution. For At = 0.001 s (1 ms), we just begin to see small disagreement. For At = 0.01 s (10 ms), the disagreement is severe, and for At = 0.1 s, thediscrete-time results are complete garbage. To interpret these results, note that the damped natural frequency at which theoutput oscillates is, from undergraduate analysis of second-order systems, Fd ={K/M - (B/2M)2}1/2/(2n) = 2 Hz, T = 0.5 s. [In this case, we have extremely smalldamping (damping ratio = 0.1), so the damped natural frequency undamped naturalfre qu enc y. ] Th e num be r of samples per period is thus T/At, wh ic h for therespective values of At (At values in ascending order) 5000, 500, 50, and 5samples/period. We conclude in this case that for highly accurate results with this first-orderapproximation of the derivative, we should have at least 500 points per period.This large num be r is d ue t o th e extremely low order of the derivativeapproximation.310 0.5 1 1.5 2 2.5 3 3.5 4 4.5 500.020.040.060.080.10.12Time (secs)AmplitudeSolid contin-time step soln: ---- discrete-time soln (points conn w/lines): dt 0.0001 s.0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 500.020.040.060.080.10.12Time (secs)AmpIitudeSoIid contin-time step soIn; ---- discrete-time soIn (points conn w/Iines); dt 0.001 s.320 0.5 1 1.5 2 2.5 3 3.5 4 4.5 500.020.040.060.080.10.12Time (secs)AmplitudeSolid contin-time ste soln; ---- discrete-time soln (points conn w/lines); dt 0.01 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 500.020.040.060.080.10.12Time (secs)AmplitudeSolid contin-time step soln; ---- discrete-tim soln (points con w/lines); dt 0.1 s.332 3 -1 1 018n4 -1 20 nx(n)*h(n)x(n)*h(n)23451-1-2(a)44 5 65 6 78 93 17 8(b)2.25) See the figures below. N-1 N-1 1 - a-min(n+1,N) 2.26) y(n) = Z an-mu(n-m) = an u(n) Z a-mu(n-m) = an u(n). m=0 m=0 1 - a-1 n-3 1 - n-22.27) y(n) = 5 Z u(k)ku(n-3-k) = 5 Z ku(n-3-k) = 5 Z k = 5 , n 3 k=- k=0 k=0 1 - (zero o.w.) = 10(1 - 22-n)u(n-3).2.28) Because the system is moving-average, the answer is obtained by inspection.Let x(n) = o(n). Then h(n) = o(n) - o(n-1) + 3o(n-6). 342.29) Set x(n) = o(n), so y(n) = h(n). Then, for n = 0, a0h(0) + a1h(-1) = b0, or, because h(-1) = 0, h(0) = b0/a0.For n 1, we havea0h(n) = -a1h(n-1), or h(n) = -(a1/a0)h(n-1). With n = 1, this becomesh(1) = -(a1/a0)(b0/a0). In general, h(n) = [-a1/a0]n(b0/a0)u(n).For more complicated systems, this approach becomes unwieldy, and we are thankfulfor transform methods!2.30) In Problem 2.29, let b0 = a0 = 1 and a1 = -0.7. Then for the "plant," y(n) - 0.7y(n-1) = w(n), where w(n) = A[x(n) - y(n)].Soy(n) - 0.7y(n-1) = A[x(n) - y(n)], or(A+1)y(n) - 0.7y(n-1) = Ax(n), or 0.7 Ay(n) - y(n-1) = x(n). A+1 A+1Following the result of Problem 2.29, we directly obtain A 0.7 nh(n) = ( ) u(n). A+1 A+1In or de r th at th is sy st em b e stable, we know that h(n) must be absolutelysummable, which from the text we know occurs if the geometric term dies away asn > . Thus we must have 0.7 < 1 for stability, or A+1 > 0.7, or A+1A > -0.3 union with A < -1.7. Numerically, this is verified using, e.g., Matlab's control systems toolbox.2.31) Assuming the student has correctly written the easy reversal and zero-padding m-files, the code for "conv1" is as shown below.The convolution formula for the appropriate limits is Nx-1y(n) = Z x(n-m) h(m), where Nx is the length of the array x. m=0Note that the length of y is Nx+Nh-1, which for Nx=4,Nh=5 is Ny=8. Thus n rangesfrom 0 to Nx+Nh-2 = 7.(Code on next page.)35x=[1 2 3 4]; h=[-1 4 -6 8 -10]';Nx=length(x); Nh=length(h);y=conv(x,h)'xr=vrev(x);for n=0:Nx+Nh-2 v=xr(max(1,Nx-n):min(Nx-n+Nh-1,Nx)); if n=Nx v=prepad(v,Nh); end y1(n+1)=v*h;endThe vectors v are:n v0 [1 0 0 0 0]1 [2 1 0 0 0]2 [3 2 1 0 0]3 [4 3 2 1 0]4 [0 4 3 2 1]5 [0 0 4 3 2]6 [0 0 0 4 3]7 [0 0 0 0 4]y = [-1 2 -1 4 4 -20 2 -40].Yes, the matrix formed from the above row-vectors is identical to the matrixproduced by the command: convmtx(x',length(h)).For the direct matrix assembly, a simple non-vectorized approach reads:x=[1 2 3 4];h=[-1 4 -6 8 -10]';Nx=length(x);Nh=length(h);y=conv(x,h)'for i=1:Nx+Nh-1 for j=1:Nh ij=i-j; if ij>=0 & ij 1/3 < 1 as n > . Thus the series converges and h3(n) is a BIBO-stable system.2.33) Find the unit sample response, by setting x(n)=o(n):a0h(n) + a1h(n-1) + a2h(n-2) = o(n). (*)According to classical techniques of solution of difference equations, analogousto those of differential equations, we assume for this second-order equation thath(n)=(Aon + Bn)u(n). For all n > 0, the difference equation saysa0(Aon + Bn) + a1([A/o]on + [B/]n) + a2([A/o2]on + [B/]n) = 0.For this to be true for all n, the coefficients of both on and n must be zero.Focussing on those of on (for the two equations are identical),a0 + a1/o + a2/o2 = 0, or o2 + (a1/a0)o + a2/a0 = 0, which has solutiono = -a1/[2a0] {(a1/[2a0])2 - a2/a0}1/2. We choose + for o and - for .The discriminant is positive, and thus the roots are distinct and real, if(a1/[2a0])2 > a2/a0, or a12 > 4a0a2, which is the condition given in the problem statement. So we may assume theroots are real and distinct. We still need to find the A and B parameters. Forthat, we use (*) for n=0 (right-hand side = 1) and n=1 (right-hand side = 0),where we have h(n0 0, o.w.side of (2.25b): 1/A, t- < A/2 xc()oc(t-) d = xc() lim d - - A>0 0, o.w.We now d e t e r m i n e t he condition o n fo r t h eoc te rm t o be nonzero. Thiscondition is t- < A/2, or -A/2 < t- < A/2, or -(t + A/2) < - < -t + A/2, or48 t - A/2 < < t + A/2. On this interval, as A > 0, xc() is essentially constant and equal to xc(t).Thus xc() can be pulled outside the integral as the constant xc(t), leaving thenormalization integral for oc(), which is A/A = 1. Thus the result of theintegration is xc(t), which proves the sampling property.2.47) (a)

fc(t)oc'(t-)dt = fc(t)oc(-(t-)) - fc'(t)oc(t-)dt = 0 - fc'().- - -u=fc(t); dv=oc'(t-)dtdu=fc'(t)dt; v = oc(t-) (b) fc(t)toc(t) dt = [fc(t)-t]-oc(t) dt. - -From (2A.4), the result is fc(0)-0 = 0. The required result follows. (c) fc(t)(t-)oc'(t-)dt = fc(t)oc(t-) - {fc'(t)-(t-)dt + fc(t)}oc(t-)dt. - - -u=fc(t)(t-), dv=oc'(t-)dtdu=fc'(t)-(t-)dt + fc(t)dt, v=oc(t-). = 0 - fc'()-0 - fc(). = -fc().Thus, toc'(t) and -oc(t) have identical effect under an integral sign. (d) fc(t)oc'(-(t-))dt = fc(t)oc(-(t-)) + fc'(t)oc(t-)dt = 0 + fc'(). - - -u=fc(t); dv=oc'(-(t-))dtdu=fc'(t)dt; v = -oc(-(t-)) = -oc(t-).But from part (a), we know that the result of placing oc'(t-) under the sameintegral yields -fc'(). Thus oc'(-t) under an integral has a result that isequal to -1 times the result of oc'(t) under the same integral. Consequentlyoc'(-t) = -oc'(t). (e) For simplicity, let us first determine what the right-hand side oc(a-b)does. That is, if we substitute oc(a-b) into the sampling formula (2A.3), weobtain

oc(a-b)fc(a)da = fc(b). (*)- Therefore, if we replace oc(a-b) in (*) above by oc(a-t)oc(t-b)dt, -we should get the same result, fc(b). If we do, the two items in the precedingsentence are equal in a distributional sense and thus the result is proved. Letus then perform this substitution:49 fc(a) oc(a-t)oc(t-b)dt da = { fc(a) oc(a-t) da } oc(t-b)dt - - - - = fc(t) oc(t-b) dt = fc(b), which agrees with what oc(a-b) does in the -integral, and the result is proved. [Note: That the integral in {-} above isequal to fc(t) is established by simply reversing the parameters "a" and "t" in(2A.3).]2.48) To prove (2A.15), we letgc(t) = t2 - a2. gc'(t) = 2t. The roots of gc(t), t

, are t1 = a and t2 = -a. Substituting these results into(2A.12), we have oc(t-a) oc(t+a)oc(t2-a2) = + , which is easily seen to reduce to (2A.15). 2a -2aFor our electromagnetics application, we write k2-k02 = kz2 + {kx2 + ky2 - k02}. The roots of this function [our "gc(kz)"] are {k02 - kx2 - ky2}1/2. Substitutinginto (2A.15) gives the result in the problem statement that was to be shown.2.49) Convolution is an example of an operator, in that when it operates on afunction, a second function is produced (rather than just one value), as long asthe independent variable t (or n) is left as a variable. The shifted impulseresponse in convolution is a "distribution," which distributes or weights theinput function in different ways, depending on the variable t (or n).2.50) The im pu lse response hc(t) is gc(t) fo r ze ro conditions on the"boundaries," BEFORE the introduction of the impulse. Remember that the boundarycondition at t = 0 is the limit as t > 0 from BELOW t = 0. Therefore, eventhough h(t=0-) = 0, h(t=0+) will in general NOT be zero. The initial conditionsused in calculating the solution via the Laplace transform are t = 0 - conditions,under the usual procedures. For example, if hc(t) is causal, the "boundaryconditions" are gc(0-) = 0 and gc() = 0. If hc(t) is anticausal, gc(0+) = 0 andgc(-) = 0, with the impulse occurring at t = 0- rather than 0+ as for the causalimpulse response. If the impulse response is assumed to be bicausal, we enforcehc() = 0 and use a superposition of the causal and anticausal responses. Note,that because we allow such unrealistic inputs as FOREVER-lasting step inputs,yc(t) is not guaranteed to go to zero as t>; only if xc()>0 does yc()>0.50END-OF-CHAPTER PROBLEMS FOR CHAPTER 3DIGITAL SIGNAL PROCESSINGTHOMAS J. CAVICCHI3.1) (a) x = 2cos(120) = -1 and y = 2sin(120) = 3. Therefore, z = -1 + j3.(b) z = {32 + 42}1/2 = 5 and U = tan-1(4/3) 53.1. Thus z = 5 /_ 53.1.3.2) 1+3j 4+j 2-j 1+13j -1-2j -12 31 + = + = + j -0.1412+j0.3647. 4-j 4+j 5j 17 5 85 853.3) Equating imaginary parts gives y = -7/2; substituting this into the equatingof the real parts gives3x - 21 = 6, or x = 9.3.4) Let z1 = x1 + jy1 and z2 = x2 + jy2. Then z1 z1z2* (x1+jy1)(x2-jy2) x1x2+y1y2 + j(x2y1-x1y2) = = = . z2 z2z2* z2

2 x22 + y223.5) da/dU = ja, or da/a = jdU, or ln(a) = jU + c. But from the definition of a, a(0) = 1, so ln{a(0)} = 0 = j-0 + c, or c = 0.Therefore, a = ej.3.6) z1 - z2 is a "vector" pointing from z2 to z1.3.7) Using z = {zz*}1/2, we have 1 + j2 1 - j2 1 + 4z = {()-()}1/2 = {}1/2 = 1/5 0.45. 3 + j4 3 - j4 9 + 163.8) Recall that 1/n (1/n) j(U + 2nk)/nFractional Power: z = z - e, 0 < k < n-1.Thus (1/3) (1/3) j(n + 2nk)/3 (-8) = 8 - e, 0 < k < 2, jn where -1 = e was used. These three roots can be written jn/3k = 0: 2-e j(n/3 + 2n/3) jnk = 1: 2-e = 2-e = -2 (the familiar, real-valued root) 51Rez}jImz}60

60

k 0k 1k 22-1.5 -1 -0.5 0 0.5 1 1.5-1.5-1-0.500.511.5Real AxisImag AxisThe sixth roots of z = -3 + j4. j(n/3 + 4n/3) j5n/3k = 2: 2-e = 2-e. j7n/3 jn/3Note that for k = 3, the root is 2-e = 2-e , which equals the k = 0 root.Thus there are only n = 3 distinct roots. The three roots are shown on thecomplex plane in the figure below.3.9) z = 5/_n-tan-1(4/3), so z1/6 = 51/6/_([(2k+1)n-tan-1(4/3)]/6. = 1.3077/_{21.145, 81.145, 141.145, 201.145, 261.145, 321.145}.The plot of these roots is as shown in the figure below.523.10) (a) Rotate z by 180.(b) Reflect z with respect to the real axis.(c) 1/z = z*/(zz*) = z*/z2. Reflect z with respect to the real axis, andmultiply its magnitude by the factor 1/z2 (or equivalently, make its magnitudebe 1/z rather than z).3.11) x4 = 9 /_ -n+2nk. x = 3 /_ (-n + 2nk)/4 = 3 /_ {-n/4, n/4, 3n/4, 5n/4}.Choosing (3/2)1/2(1j), we have [x-1.51/2(1-j)][x-1.51/2(1+j)] = x2-2-1.51/2x+3.Choosing (3/2)1/2(-1j), we have [x-1.51/2(-1-j)][x-1.51/2(-1+j)] = x2+2-1.51/2x+2.Thus x4+9 = (x2-2-1.51/2x+3)(x2+2-1.51/2x+3).3.12) Following the similar text example, z+3 < 2z-1, orx+jy+3 < 2x+jy-1, orx2 + 6x + 9 + y2 < 4x2 - 8x + 4 + 4y2,or -3x2 + 14x + 5 - 3y2 < 0, orx2 - (14/3)x - 5/3 + y2 > 0, or x2 - (14/3)x + 49/9 + y2 - 49/9 - 15/9 > 0, or(x - 7/3)2 + y2 > (8/3)2,which is the exterior of a circle of radius 8/3 2.667, centered on z0 = 7/3.3.13) (a) Circle of radius 6 centered on z0 = 3, and its interior.(b) Circle of radius centered on the origin.(c) Imaginary axis (distance from a = distance from -a). Mathematically, lettingz = x+jy, z+a2 = (x+a)2+y2 = x2+2ax+a2+y2, whilez-a2 = (x-a)2+y2 = x2-2ax+a2+y2.Equating gives 4ax = 0, or x = Re{z} = 0, the imaginary axis.(d) The line y = mx+d where if a = ar+jai and b = br+jbi, m = (ar-br)/(bi-ai) and d=(b2-a2)/(bi-ai).Proof:z-a2 = (x-ar)2 + (y-ai)2 = x2 - 2xar + ar2 + y2 - 2yai + ai2z-b2 = (x-br)2 + (y-bi)2 = x2 - 2xbr + br2 + y2 - 2ybi + bi2.Equating these, we see the x2 and y2 terms cancel, yieldingy-2-(bi-ai) + x-2-(br-ar) = br2 + bi2 - ar2 - ai2,from which the result follows upon solving for y (result verified on Matlab).53(e) By approach similar to that in (c), x < 5/2.For variety, write this time as follows:{z2 = zz*} < {z-52 = zz* - 5-2-Re{z} + 25}, giving Re{z} < 5/2.(f) By inspection, in words the expression tells us that the distance between zand 3j is between 1 and 5. This is just a ring of width 5-1=4 and inner radius1, centered on 3j.(g) Re{x2+y2+j2xy} > 0, or x2 > y2, or x > y. Geometrically, this is all zhaving angle between -n/4 and n/4, union with all z having angle between 3n/4 and5n/4; i.e., all z within 45 of the real axis (angle measured from the origin).(h) Re{(x-jy)/(x2+y2)} < 1/4, or x/(x2+y2) < 1/4, or 4x < x2 + y2, or0 < x2-4x+y2; completing the square, 4 < (x-2)2 + y2, or the exterior of a circle of radius 2 centered on z0 = 2.3.14) Place "vectors" z1 and z2 end-to-end. When so placed, the angle betweenthe two vectors is U = 180 + U - U. z1,z2 z1 z2From the law of cosines, z1+z2

2 = z1

2 + z2

2 - 2z1z2cos(U ). z1,z2If cos(U ) = -1, i.e., U = n, then z1,z2 z1z2z1+z2

2 = {z1 + z2}2, or z1 + z2 = z1 + z2. ButU = n if and only if U = U; that is, only if z1 = az2 where a is a z1,z2 z1 z2real, positive scalar. That is, if the two "vectors" are colinear, as might havebeen guessed at the outset.3.15) (a) - 0 (indep. of the value of c > 0). We conclude that both roots are in the LHP if and only if both b > 0 and c > 0.(c) Consider first (b/2)2 > c, for which we write z = -b/2 {[b/2]2 - c}1/2. Forthis case, we know z is real and must satisfy -1 < z < 1. Substituting,-1 < -b/2 + {[b/2]2 - c}1/2 < 1 AND -1 < -b/2 - {[b/2]2 - c}1/2 < 1 for bothroots to be within the unit circle. Adding -b/2 to all sides and recognizingthat {[b/2]2 - c}1/2 > 0, we have max(0,b/2-1) < {[b/2]2 - c}1/2< max(0,b/2+1) AND max{0,-(b/2+1)} 2 and (ii) b < 2.(i) b/2-1 > 0 (b>2): In this case, clearly b/2+1 is also > 0 (i.e., b > -2), sosquaring all sides (all are > 0), the first inequality becomes(b/2)2-b+1 < (b/2)2-c < (b/2)2+b+1, or -b+1 < -c < b+1, which we will find we donot need later.(ii) b/2-1 < 0 (b 0 (i.e., 2 > b > -2): the inequality becomes0 < {(b/2)2-c}1/2 < b/2+1, or 0 < (b/2)2-c < (b/2)2+b+1. The lower inequality isredundant; we already have assumed that (b/2)2 > c. The other inequality gives0 < 1+b+c.(iib) b/2+1 < 0 (i.e., b < -2): The inequality becomes 0 < {(b/2)2-c}1/2 < 0,which says (b/2)2 = c, contradicting our assumption that (b/2)2>c.Now we consider the other (lower) original inequality that must hold, namelymax{0,-(b/2+1)} < {[b/2]2 - c}1/2 < max(0,1-b/2). We again consider two cases:(i) -(b/2+1) > 0 and (ii) -(b/2+1) < 0; that is, (i) b < -2 and (ii) b > -2.First, case (i):(i) -(b/2+1) > 0 (i.e., b < -2): This was already found to violate the firstoriginal inequality max(0,b/2-1) < {[b/2]2 - c}1/2< max(0,b/2+1). So we ignorethis case. Next, case (ii):(ii) -(b/2+1) < 0 (i.e., b > -2): In this case, we need to consider twopossibilities for the upper limit, 1-b/2: (iia) 1-b/2 > 0 (-2 < b < 2): the inequality becomes0 < {(b/2)2-c}1/2 < 1-b/2. The lower inequality is redundant [we already haveassumed that (b/2)2 > c]. The upper inequality becomes 0 < 1-b+c.(iib) 1-b/2 < 0 (b > 2). The inequality again becomes 0 < {(b/2)2-c}1/2 < 0,which violates our assumption (b/2)2>c. We finally draw our conclusions for this case (b/2)2>c: If b < -2 or b > 2, theroots cannot both be within the unit circle. If b < 2, then we have shown weneed both (A) 0 < 1+b+c AND (B) 0 < 1-b+c. Now notice that if b < 2 and if(b/2)2>c, it follows that (C) c < 1. All conditions (A), (B), and(C) must hold for the case (b/2)2>c.Next, we consider the case (b/2)2c, distinguishing that (b/2)20. So, combining with the fact c < 1, we have 0 < c < 1. Then (b/2)2>0 can be rewritten -2c < b < 2c. Consider the cases(i) b 0 and (ii) b < 0. (i) b 0: 0 < b < 2c. Let us examine, in consideration of our previousresults for (b/2)2>c, the quantities 1 b + c:1+c < 1+b+c < 1+c+2c and 1+c-2c < 1-b+c < 1+c. The smallest value of 1+c-2c is zero for c approaching 1. So all we need inthis case is 0 < 1-b+c, because for 0 < c < 1 the other inequalities are56guaranteed.(ii) b < 0: -2c < b < 0. Again examine 1 b + c:1+c-2c < 1+b+c < 1+cand 1+c < 1-b+c < 1+c+2c. Again, the smallest value of 1+c-2c is zero for c approaching 1. So all we needin this case is 0 < 1+b+c, because for 0 < c < 1 the other inequalities areguaranteed. Putting these three requirements together, we again have (A) 0 < 1+b+c AND (B) 0 < 1-b+c AND (C) c < 1. The above requirements are seen to guarantee roots within the unit circle for ALLcases.3.19) /_w

= 1/N{tan-1([y1 - y2]/[x1 - x2]) + 2nt} = -n/(2N).Thus, tan-1([y1 - y2]/[x1 - x2]) = -n( + 2t). For 0 < t < N-1 an integer, we have, modulo 2n, tan-1([y1 - y2]/[x1 - x2]) = -n/2, which implies that both x1 = x2 and y1 < y2.3.20) First, note that (z1+z2)* = z1*+z2* and (z1z2)* = z1*z2*. Clearly, theseresults are also true for any combinations of sums and divisions--which make upa rational function. Moreover, if the rational function H(z) has only realcoefficients, the complex conjugate of H(z) is H*(z) = H(z*). This is because,e.g., (a

z

)* = a

(z*)

. What about the exponential function? If z = x+jy, then z* (ez)* = (ex[cos(y)+jsin(y)])* = ex[cos(y)-jsin(y)] = ex-jy = e .Now, cos(z) and sin(z) are linear combinations of ez and e-z, and tan(z) is arational function of ez and e-z. So, to conjugate any of these functions, wesimply conjugate the argument [e.g., (cos(z))* = cos(z*)].Because conjugation and integration can be switched, as can conjugation anddifferentiation, we again can just conjugate the operand to conjugate the entireexpression.What about, e.g., cos-1(z)? Let w = cos-1(z). Then z = cos(w), so z* = [cos(w)]* = cos(w*). So cos-1(z*) = cos-1{cos(w*)} = w*, or w* = cos-1(z*), so the result holds again. One could argue similarly in general about allLaurent-expandable functions as well (apply result to each term in seriesexpansion).3.21) The 3-D plots are as in the figures below; discussion follows after them.Note: In these and all subsequent complex plane plots in this solutions manual,for simplicity we omit placing j in front of each imaginary axis numerical label(as done throughout the text), though we know it belongs there.57-2-1012-2-1012-20246x = Re{z}jy = jIm{z}|ln(z)| cos(U) - sin(U)/z = sin(U) + cos(U)/z cx cy cz cU cz cUwhile cu cv cu cu cv cv = > sin(U) + cos(U)/z = cos(U) + sin(U)/z cy cx cz cU cz cUor, collecting terms, cu cv cv cu{ }cos(U) = { + }sin(U) cz zcU cz zcU cu cv cv cu{ + }cos(U) = { - }sin(U). zcU cz zcU cz65Multiplying the first equation by sin(U), the second by cos(U), and adding, weobtain (using cos2 + sin2 = 1) cu cv cv cv0 = + , giving = . zcU cz cz zcUSimilarly, multiplying the first equation by cos(U), the second by sin(U), andsubtracting the first from the second gives cu cv = 0, and thus we obtain the C-R in terms of z and U:cz zcU cv cv cu cv = and = .cz zcU cz zcU3.27) First write M M H (z - zzm) H z - zzm m=1 m=1 M NLet w(z) = = /_ Z /_(z - zzm) - Z /_(z - zpn). N N m=1 n=1 H (z - zpn) H z - zpn

n=1 n=1Let zzm = xzm+jyzm, zpn = xpn+jypn. Then M H [(x - xzm)2 + (y-yzm)2] m=1w(z) = { }1/2 N H [(x - xpn)2 + (y-ypn)2] n=1 M y - yzm N y - ypn -[ cos{ Z tan-1( ) - Z tan-1( ) } m=1 x - xzm n=1 x - xpn M y - yzm N y - ypn +jsin{ Z tan-1( ) - Z tan-1( ) } ] m=1 x - xzm n=1 x - xpn = A[cos(U) + jsin(U)] where M H [(x - xzm)2 + (y-yzm)2] m=1A = { }1/2, N H [(x - xpn)2 + (y-ypn)2] n=1 M y - yzm N y - ypnU = Z tan-1( ) - Z tan-1( ) m=1 x - xzm n=1 x - xpn66or w(z) = u(x,y) + jv(x,y) where u = Acos(U) and v = Asin(U).For use below, we computecU M y - yzm N y - ypn = - Z + Z cx m=1 {1 + [(y-yzm)/(x-xzm)]2}(x-xzm)2 n=1 {1 + [(y-ypn)/(x-xpn)]2}(x-xpn)2cU M 1 N 1 = Z - Z .cy m=1 {1 + [(y-yzm)/(x-xzm)]2}(x-xzm)2 n=1 {1 + [(y-ypn)/(x-xpn)]2}(x-xpn)2 NNow note that if g(z) = H {(x - xpn)2 + (y - ypn)2}, then n=1cg N g(z) = 2 Z (x - xpj) cx j=1 (x - xpj)2 + (y - ypj)2and similarly,cg N g(z) = 2 Z (y - ypj) .cy j=1 (x - xpj)2 + (y - ypj)2 MNext note that if h(z) = H {(x - xzm)2 + (y - yzm)2}, then m=1ch M h(z) = 2 Z (x - xzj) cx j=1 (x - xzj)2 + (y - yzj)2and similarly,ch M h(z) = 2 Z (y - yzj) .cy j=1 (x - xzj)2 + (y - yzj)2Thus, A = {h(z)/g(z)}1/2.Now, M (x-xzj)h(z) N (x-xp)g(z) g(z) Z - h(z) Z cu 1 j=1 (x-xzj)2 + (y-yzj)2 t=1 (x-xp)2 + (y-yp)2ux = = 2 cos(U) cx 2A g2(z) M y - yzm N y - ypn + A sin(U) { Z - Z }. m=1 y - yzm n=1 y - ypn {1 + []2}(x-xzm)2 {1 + []2}(x-xpn)2 x - xzm x - xpnNext,67 M (y-yzj)h(z) N (y-yp)g(z) g(z) Z - h(z) Z cv 1 j=1 (x-xzj)2 + (y-yzj)2 t=1 (x-xp)2 + (y-yp)2vy = = sin(U) cy A g2(z) M 1 N 1 + A cos(U) { Z - Z }. m=1 y - yzm n=1 y - ypn {1 + []2}(x-xzm)2 {1 + []2}(x-xpn)2 x - xzm x - xpnIs ux = vy? Compare the coefficients of cos(U):Coefficient of cos(U) in ux: Note that the numerator has g(z)h(z) and the denominatorhas g2(z); canceling for these factors gives h(z)/g(z) = A2, part of which cancels theA in the denominator, leaving: M x - xzj N x - xpA { Z - Z } j=1 (x - xzj)2 + (y - yzj)2 t=1 (x - xp)2 + (y - yp)2while the coefficient of cos(U) in vy is: M 1 N 1A { Z - Z } j=1 y - yzj t=1 y - yp {1 + ( )2}(x--xzj) {1 + ( )2}(x--xp) x - xzj x - xp M x - xzj N x - xp = A { Z - Z } j=1 (x - xzj)2 + (y - yzj)2 t=1 (x - xp)2 + (y - yp)2which agrees with the coefficient of cos(U) in ux.Next, compare the coefficients of sin(U). Coefficient of sin(U) in ux: M y - yzm N y - ypnA { Z - Z }. m=1 (x - xzm)2 + (y - yzm)2 n=1 (x - xpn)2 + (y - ypn)2Coefficient of sin(U) in vy: Again, (1/A)-g(z)h(z)/g2(z) = A, so, changing j to m andt to n give for the coefficient of cos(U) in vy: M y - yzm N y - ypnA { Z - Z } m=1 (x - xzm)2 + (y - yzm)2 n=1 (x - xpn)2 + (y - ypn)2which agrees with the coefficient of sin(U) in ux. Thus, indeed, ux = vy. Now, M (y-yzj)h(z) N (y-yp)g(z) g(z) Z - h(z) Z cu 1 j=1 (x-xzj)2 + (y-yzj)2 t=1 (x-xp)2 + (y-yp)2uy = = cos(U) cy A g2(z) M 1 N 1 - A sin(U) { Z - Z }. m=1 y - yzm n=1 y - ypn {1 + []2}(x-xzm)2 {1 + []2}(x-xpn)2 x - xzm x - xpn68Next, M (x-xzj)h(z) N (x-xp)g(z) g(z) Z - h(z) Z cv 1 j=1 (x-xzj)2 + (y-yzj)2 t=1 (x-xp)2 + (y-yp)2vx = = sin(U) cx A g2(z) M y - yzm N y - ypn + A cos(U) { - Z + Z }. m=1 y - yzm n=1 y - ypn {1 + []2}(x-xzm)2 {1 + []2}(x-xpn)2 x - xzm x - xpnIs uy = -vx?Compare the coefficients of cos(U). In uy, we have, again using (gh)/[Ag2] = A, M y - yzj N y - ypA { Z - Z } j=1 (x - xzj)2 + (y - yzj)2 t=1 (x - xp)2 + (y - yp)2and this matches the coefficient of cos(U) in -vx if in vx we multiply thedenominator factors and change m to j, and n to t.Compare the coefficients of sin(U). In uy, we have M x - xzm N x - xpn-A { Z - Z }. m=1 (x - xzm)2 + (y - yzm)2 n=1 (x - xpn)2 + (y - ypn)2In -vx, we have, once more using (gh)/[Ag2] = A, and changing j to m and t to n,we obtain exactly the above coefficient. Thus, uy = -vx. It would therefore seem that the rational function is analyticeverywhere. However, because u = Acos(U) and v = Asin(U), and A = at each ofthe poles zp, the C-R equations are undefined at zp, and therefore the rationalfunction is not analytic at its poles.3.28) Taking the hint, let A/_U = kx-k0 and B/_o = kx+k0. Then kx/j = (AB)1/2/_{(U+o)+tn}, t = 0, 1. Let ky1 = j(AB)1/2/_(U+o), ky2 = j(AB)1/2/_(U+o)+n = -ky1; i.e., ky1 and ky2 are the two possible values for ky. Let both U and o berestricted to [-n, n]. Then if we can find a region (called a branch) in whichky1 is single-valued, then so will be ky2. Taking the suggestion in the problemstatement, we draw the figure below. As kx approaches the real axis from above,in between -k0 and k0, we have U > n and o > 0. 69-k Rek

}iImk

}k

k

A B-k Rek

}iImk

}k

k

A B Contrast that with the situation in the next figure, where kx approaches the sameregion of the real axis from below. Now U > -n (note the minus sign) whileo > 0. Consequently, ky1, whose angle is (U+o) is discontinuous in phase-angle on the real axis for Re{kx} < k0--so ky1 is multi-valued. Specifically,/_ky1 = n/2 just above the real axis, while /_ky1 = -n/2 just below the real axis.Notice that ky2 = -ky1 is continuous with ky1 as the crossover is made. Hence thisportion of the real axis is called the branch cut--where we cut over from branchky1 to ky2.Contrast the above phase discontinuity with the approach of the real axis for Re{kx} > k0, shown in the next two figures for Re{kx} > k0, with kx approachingthe real axis from, respectively, above and below. In both approaches, U > 0and o > 0, and there is no discontinuity. If Re{kx} < -k0, there is adiscontinuity in ky1 of 4n = 2n, which is insignificant, for the plane and allcomplex exponentials are 2n-periodic.We conclude that we have two branches, ky1 and ky2 of the function ky, both ofwhich are by definition single-valued within the domain D consisting of all kxexcluding the real axis portion where kx < k0. The branch cut is that excludedportion of the real axis, as discussed above.70-k Rek

}iImk

}k

k

AB-k Rek

}iImk

}k

k

A B 3.29) (a) The value of the integral, by the residue theorem, is 2nj-(j3) = 2n.(b) Because the only pole is not enclosed and thus the integrand is analyticwithin the contour, the integral is zero for this case. 3.30) Only the pole at z = - is enclosed by C. Thus, by the residuetheorem, I = 2nj-Res{f, -} = 2nj-d/dz{1/(z + 2)} = -2nj/(3/2)2 = -8nj/9. z= -3.31) Simple pole at z = -4, so (a) I = 0 and (b) I = -(2nj/5)exp(a-4)sin(4b).3.32) By (3.13), such a function is ecos(z)/{2nj-(z-5)}.713.33) In order to possibly use a-1{za} to calculate the integral of f(z) around C,we must have z0 = zA in (3.23). If z0 = zA, the assertion in the problemstatement would be correct were it not for the fact that C encloses more than onepole. By the residue theorem, we know that in this case the integral around Cis the sum of the residues for all the poles enclosed by C. Thus (3.25) appliesand we have to add 2nja-1{} where {t} signifies each of the poles enclosed by C.Each residue is a-1{}--calculated for a different Laurent expansion of f(z) whereit is valid--for C closer to z

than any other pole of f(z) is to z

. All theresidues are summed to obtain the value of the integral of f(z) around C. dz3.34) The integrand is even, so I = where C is as described C z4 + 1in the problem statement. By the residue formula, we know that I = -2nj-[sum of residues of poles enclosed by C].z0 = 1 /_ 45, z1 = 1 /_ 135, z2 = 1 /_ -135, z3 = 1 /_ -45. If the semicircle of "D" contour is in the upper half-plane, then only z0 and z1are enclosed. Thus 1 1I = nj[ + . (z0 - z1)(z0 - z2)(z0 - z3) (z1 - z0)(z1 - z2)(z1 - z3)By either manual evaluation, or by numerical calculation, we obtain the exactresult I = n/(22) 1.107.For example, note that (z - z2)(z - z3) = z2 - z[1 /_ -45 + 1 /_ -135] + 1 /_ 180 = z2 - z[-j2/2] - 1 = z2 + j2-z - 1. In particular,(z0 - z2)(z0 - z3) = 1/_90 + j2-(1 /_ 45) - 1 = j + (j2/2)-(1 + j) - 1 = -2 + j2. Also, z0 - z1 = 1 /_ 45 - 1 /_ 135 = 2/2 = 2. By using this and similar reasoning for the second term, we have I = (nj/2)[1/(-2 + j2) + 1/(2 + j2)] = n/(22).3.35) Again, use the D-shaped contour lying on its side (flat part on real axis).The poles enclosed by C are at z = j and z = 2j. Thus j2 (2j)2I = 2nj[ + ] = n/3. (j+j)(j2+4) ([2j]2+1)(2j+2j)723.36) Consider first the case n -3, where the integrand has no pole at theorigin. (i) C encloses only the pole at z = -0.4, so I = [(2nj)/(2nj)]{(-0.4)n+3/(-1.4)2} = -0.0327(-0.4)n. (ii) C encloses both z = -0.4 and z = 1. Residue @ z = 1:d zn+3 (z+0.4)(n+3)zn+2 - zn+3-1 = , which evaluated at z = 1dz z + 0.4 (z+0.4)2gives {1.4(n+3) - 1}/(1.4)2 = (1.4n + 3.2)/(1.4)2 = n/1.4 + 1.6327.The other residue is the same as above, so I = -0.0327(-0.4)n + n/1.4 + 1.6327.Next, consider the case n = -4, where we also have a pole now at the origin.(i) C encloses the pole at the origin, and the pole at z = -0.4, soI = 1/0.4 + 1/[(-0.4)(-1.4)2] = 1.2245.(ii) C encloses all poles. For z = 1, we haved 1 -(2z+0.4) = which at z = 1 gives -2.4/1.42 = -1.2245.dz z2 + 0.4 (z2+0.4z)2We thus obtainI = -1.2245 + 1.2245 = 0.All of the above formulas have been numerically checked by quadrature on Matlab. 3.37) clg% For all but Ex. 3.6:xmin=-0.3; xmax=0.3; ycent=0;% For Ex. 3.6:xmin=-1.3; xmax=1.3; ycent=0;Nx=3000; xcent=0.5*(xmin+xmax); xcent2=xcent^2;B=xmax; A=0.1;x1=linspace(xmin,xmax,Nx);x2=vrev(x1);a=xcent-xmin;a2=a^2;for i=1:Nx% circle:% y1(i)=ycent+sqrt(a2-(x1(i)-xcent)^2); y2(i)=ycent-sqrt(a2-(x2(i)-xcent)^2);% ellipse: y1(i)=ycent+A*sqrt(1-((x1(i)-xcent)/B)^2); y2(i)=ycent-A*sqrt(1-((x1(i)-xcent)/B)^2);endzc=[x1+j.*y1 x2+j.*y2];zc=zc(1:length(zc)-1);[Code continues on next page.]73% want CCW contour, not CW contour, so reverse array zc.zc=vrev(zc);% (view contour)%plot(zc); pause%clgintegral=h(zc(1))*(zc(1)-zc(2*Nx-1));for i=2:2*Nx-1 integral=integral+h(zc(i))*(zc(i)-zc(i-1));enddisp('Integral of F(z) around contour, divided by 2pij = ')integral/(2*pi*j)where in file h.m we have:function H=h(z)%H=1;%H=cos(z);%H=exp(2*z)/(z-0.2);H=(z^5-3*sin(z))/(z-pi/6)^4;The results are as the theory says, within numerical approximation: The firsttwo integrals are equal to zero, while the third integral is equal to exp(2-0.2).Note that it takes a large number of terms for the results to be quite accurate.In addition, the numerical result derived in Example 3.6 is numerically verified.3.38) (a) Solving for the sum in (3.42) immediately gives (3.41a). Then N N M-1 Z an = Z an - Z an n=M n=0 n=0 1 - aN+1 1 - aM aM - aN+1 = - = , which is (3.40). 1 - a 1 - a 1 - a(b) a=0.9; N=400; M=100;n=[M:N];s=sum(a.^n)s1=(a^M-a^(N+1))/(1-a)The result in either case is obtained in Matlab as 2.6561e-004.3.39) In (3.14), let n = Mk-1. Then [Mk-1] (Mk-1)! f(z')f (z) = O dz', 2nj C (z' - z)MkNow let g(z) = f(z)/(z - zk)Mk; thus, g(z) is analytic within C except for a poleof order Mk at z = zk. Substituting this into the equation above and dividingboth sides by (Mk-1)!, 1 cMk-1 1 {(z - zk)Mkg(z)} = O g(z')dz' Res{g,zk}.(Mk-1)! czMk-1 2nj CFinally, the "f" in (3.29) is of exactly the same type as g(z) above, becausewithin C both "f" and g(z) are analytic except for a pole of order Mk at z = zk.This proves (3.29).743.40) h(z) - h(zm) h(z)lim = lim = h'(zm) 0,z>zm z - zm z>zm z - zm by the definition of the derivative when we look at the leftmost expression on this line of expressions.so z - zmlim = 1/h'(zm)z>zm h(z)and g(z) g(zm)Res{f(z) at zm} = lim (z - zm)- = . z>zm h(z) h'(zm)3.41) The problem is that the two series have different regions of convergence,which do not overlap. z/(2-z) = (1/2)z + (1/4)z + (1/8)z3 + ... converges for z < 2, while z/(z-2) = 1 + 2/z + 4/z + 8/z3 +... converges for z > 2. As there is no overlapping ROC, the statement ...+ 8/z3 + 4/z + 1/z + 1 + (1/2)z + (1/4)z + (1/8)z3 +... = 0is not true for any value of z.3.42) With the hint, this problem is not difficult.(a) z0 = 0, and h(n) = (-1)n. R = lim h(n)1/n = 1. Therefore, n>the sum converges for z < 1. Of course, this should be no surprise when weconsider that this is the ROC of the z-transform of the sequence (-1)nu(-n).(b) Now z0 = 2+j and h(n) = 1/n2 for n 0. R = lim h(n)1/n = lim n2/n = 1. Therefore, n> n>the sum converges for z-2-j < 1, that is, within a circle of radius 1 centeredon z0 = 2+j. (c) Now z0 = 4j and h(n) = 1/2n for n 0. R = lim h(n)1/n = lim 2-n/n = 1/2. Therefore, n> n>the sum converges for z-4j < 1/2, that is, within a circle of radius centeredon z0 = 4j. 3.43) 1 (a) f(z) = 1, nf,m = 0: a-n-1{m} = O (z' - zm)n dz' = o(n+1), (1) 2nj C [C encloses zm]where by defining -tnew = nold + 1 so that -(nold + 1) = nnew and nold = -nnew - 1, weare able to write (3.33a) in a form allowing immediate use of (3.11). Noticethat (1) is merely a statement that the Laurent series of 1 is just 1; thecoefficients are an{m} = o(n) for any zm so that (3.32) becomes 75f(z) = 1 = 1-(z - zm)0 + 0.(b) In (3.33) we set n = 0 and evaluate at z = zm, where we assume nf,m 0, sothat f(z) is analytic within C. Then evaluating (3.32) at z = zm we obtain f(zm)= a0{m}(zm-zm)0 = a0{m} [the only nonzero term in the entire sum, because nf,m 0],or, using (3.33) with n = 0, 1 f(z')f(zm) = O dz' = a0{m}, (2) 2nj C z' - zmwhich is just the Cauchy integral formula (3.13), and which if differentiated ntimes gives n!-an{m} in (3.33c). We recognize (2) as the coefficient of the firstterm in the Taylor expansion: f(z) = f(zm)(z - zm)0 + ... = f(zm) + ...(c) n = -1, nf,m < 0 and C = Cm where Cm is a contour in a deleted neighborhood ofzm (e.g., C3 for z3 in Fig. 3.25), where f(z) is a rational function having polesof multiplicity Mm at zm (previously notated as zk):a-1{m} = a~-1{m} = Res{f, zm}. (3a)From (3.33a) with n = -1 and by the definition of the residue when C = Cm =closed contour of infinitesimal size around zm, (3a) can be written [see (3.29)] 1 (3b)a~-1{m} = Res{f, zm} = O f(z')dz' 2nj Cm Mm [Mm-1] {(z - zm) f(z)} = , (3c) (Mm - 1)! z=zmwhere again [Mm - 1] symbolizes the (Mm - 1)st derivative. In the special casethat Mm = 1 (simple pole), (3c) reduces to [see (3.30)]a~-1{m} = Res{f, zm} = {(z-zm)f(z)} , only if Mm = 1. (3d) z=zmIt cannot be overly stressed that, in addition to depending on f(z) and zm, thean{m} depend on the location of the point of evaluation z in (3.32); for differentregions of analyticity of f(z), there will be different Laurent series about zm.That is, there are different Laurent expansions of the same function f(z) indifferent regions of the z-plane, even for expanding about the same point zm.In particular, if z is in a deleted neighborhood of zm, then a-1{m} becomes theresidue of f at zm. In that case, we may make use of a-1{m} as the sole residuein calculating the closed contour integral of f(z) in the residue theorem,(3.25a). 3.44) jIm{z} z-plane(a) The pole-zero diagram is: Oxx->. -2 -0.9 0 z0=1 Re{z} We will write z = z0 + rej, so z - z0 = r.From the text, 1 (z+2)dzan = O . 2nj C z(z+0.9)(z-1)n+176For n 0, use the p-form, due to the increasingly higher-order pole at z = z0:p = 1/(z-z0), so dz = -dp/p2, and z = 1/p + z0.In this case, p = 1/(z-1), so dp = -dz/(z-1)2 = -dz-p2 or dz = -dp/p2. 1 (1/p + 3) pn-1dp 3 (p + 1/3)pndpan = O = O . 2nj C' (1/p + 1)(1/p + 1.9) (1.9)2nj C' (p + 1)(p + 1/1.9)C' is a circle of radius 1/r centered on the origin of the p-plane, and the pole-zero diagram in the p-plane is as follows: jIm{p} p-planexxO>Re{p} -1 -1/1.9 -1/3 For r < 1, both poles are enclosed. Thus, by the residue theorem, 3 -2/3 -1/1.9 + 1/3 -1an = { (-1)n + ( )n} 1.9 -1 + 1/1.9 -1/1.9 + 1 1.9 2 1.1 = (-1)n - (-1/1.9)n. 0.9 1.71For 1 < r < 1.9, only the pole at -1/1.9 is enclosed. Thus 1.1an = (-1/1.9)n. 1.71 For r > 1.9, no poles are enclosed and an = 0.Next, we consider n < 0. In this case, we use the usual z-form.For r < 1, no poles are enclosed, so an = 0.For 1 < r < 1.9, only the pole at the origin is enclosed:an = (2/0.9)(-1)n+1 = -(2/0.9)(-1)n.For r > 1.9, both poles are enclosed, so an = -(2/0.9)(-1)n + (1.1/[-0.9])(1/[-1.9])n+1 2 1.1 = (-1)n - (-1/1.9)n. 0.9 1.71We now put all our results together: 2 1.1 Z (1-z)n - ((1-z)/1.9)n, z-1 < 1 n=0 0.9 1.71 -2 -1 1.1 1-zf(z) = Z (1-z)n Z ()

, 1 < z-1 < 1.9 0.9 n=- 1.71 t=0 1.9 -1 -2 1.1 1-z Z (1-z)n + ()n, z-1 > 1.9. n=- 0.9 1.71 1.977(b) The code is as follows; agreement was found in all three regions ofconvergence:z0=1;z=input('Enter z for checking Laurent expansion ')zdif=z-z0; r=abs(zdif); Nterm=4000; F=(z+2)/z/(z+0.9)sum=0;if r where X is a double pole. - 0.8 Re{z}We will write z = z0 + rej, so z - z0 = r. zdzan = O . C (z + )2(z - 0.8)(z -1-j)n+1For n 0, use the p-form to avoid having to take successively higher derivativesof 1/[(z + )2(z - 0.8)] for the Laurent coefficients. Thus,p = 1/(z-z0), so dz = -dp/p2, and z = 1/p + z0.The pole-zero diagram in the p-plane is as follows.With p=1/(z-1-j), we have z = 0.8 translating to p = -1/(0.2+j) -0.192+j0.962, and z = -0.5 translates to -1/(1.5+j) -0.46+0.308j. 78The approximate diagram is jIm{p} x j X p-plane-> where X is a double pole. -1 Re{p}The Laurent coefficients integral becomes 1 (1/p +1+j)pn-1dpan = O 2nj C' (1/p + 3/2+j)2(1/p + 0.2+j) 1+j (p + 1/[1+j])pn+1dp = O . 2nj(3/2+j)2(0.2+j) C' (p + 1/[3/2+j])2(p+1/[0.2+j])From our pole-zero diagram in the z-plane, we see that the critical distancesfrom z0 are r1 = {1 + 0.22}1/2 1.019804 and r2 = {1 + 1.52}1/2 1.8027756.For r < r1, both poles are enclosed in the p-form integral. Thus 1+jan = { Z residues} (1.5+j)2(0.2+j) -(0.0819299+j0.4187528){ Z residues} = o{ Z residues}.Consider first the residue for the double pole [at p = -1/(1.5+j)]. It is d (p + 1/[1+j])pn+1 { }

dp p + 1/[0.2+j] p=-1/[1.5+j] (p+1/[0.2+j])((n+2)pn+1 + (n+1)p2/(1+j) - [pn+2 + pn+1/(1+j)] = (p + 1/[0.2+j])2 p=-1/[1.5+j] pn+2(n+2-1) + pn+1{(n+2)/[0.2+j] + (n+1)/[1+j] - 1/[1+j]} + pn(n+1)/[(0.2+j)(1+j)] = , (p + 1/[0.2+j])2evaluated at p = -1/[1.5+j], or n p2(n+1) + p{(n+2)/[0.2+j] + n/[1+j]} + (n+1)/[(0.2+j)(1+j)]Res = p , (p + 1/[0.2+j])2evaluated at p = -1/[1.5+j], or n+1 1 n(1+j+0.2+j)+2(1+j) n+1 - {} + -1 n (1.5+j)2 1.5+j (0.2+j)(1+j) (0.2+j)(1+j)Res = {} , 1.5+j (-1/[1.5+j] + 1/[0.2+j])279or -1 nRes = {} {(0.23076923+j0.1538462)n -0.00591716-j0.414201183}. 1.5+j -1 n = {} {1n + 2}. 1.5+jThe residue at the simple pole is n+1 (p + 1/[1+j])p -1 nRes = = {} {1.00591716 + j0.41420118} 2 0.2+j (p + 1/[1.5+j]) p = -1/[0.2+j] -1 n = {} -3. 0.2+jThese are then substituted into the formula above for an.For r1 < r < r2, only the double pole is enclosed, so only its residue is used.For r > r2, no poles are enclosed and an = 0.Next, consider n < 0. Now the z-form is used.For r < r1, no poles are enclosed and an = 0.For r1 < r < r2, only the simple pole (z=0.8) is enclosed. Its residue is 0.8 -1 nan = = {-0.0910332+j0.455166}{} = 4/[-0.2-j]n. (0.8+0.5)2(0.8-1-j)n+1 0.2+jFor r > r2, both poles are enclosed. We now need the residue for the double pole (forthe z-form integral): d z (z-0.8)(z-1-j)n+1-z[(z-1-j)n+1+(z-0.8)(n+1)(z-1-j)n] Res= { = , dz (z-0.8)(z -1-j)n+1 z=- (z-0.8)(z-1-j)2n+2evaluated at z = -, which under partial simplification becomes -1.3(z-1-j) + [(z-1-j)n+1-1.3(n+1)(z-1-j)n] Res =

1.69(z-1-j)2n+2 z=- -1 n+1 (n+1) -1 n+1 = -0.4733728{} - 0.3846154 {} 1.5+j (-1.5-j) 1.5+j -0.4733728 + (0.177514793-j0.118343195)(n+1) = (-1.5-j)n+1 (-0.04551661+j0.109239873)n+(0.1729631-j0.0364133) = (-1.5-j)n = (5n+6)/(-1.5-j)n.80Putting all the results together, we have Z o{(1n+2)[(z-z0)/(-1.5-j)]n + 3[(z-z0)/(-0.2-j)]n}, z-z01/4, while the second converges for z < 2. ThusH(z) converges for 1/4 < z < 2. 1 H(z) = + 3{ Z (z/2)m - 1} 1 - 1/(4z) m=0 1 1 -z 3z = + 3{ 1} = 1 - 1/(4z) 1 - z/2 z - 1/4 z - 2 -z2 + 2z - 3z2 + (3/4)z = (z - 1/4)(z - 2) -4z(z - 11/16) = ; ROC is 1/4 < z < 2. (z - 1/4)(z - 2)(b) Purely causal sequence (n 0): use z-form. -4 z(z - 11/16)h1(n) = O zn-1dz. 2nj C (z - 1/4)(z - 2)We take z > 2, so both poles are enclosed. Thus, noting that for n = 0 thereis no pole at z = 0 due to the z factor in the numerator of H(z), we can use thefollowing expression for all n 0: 87 z(z - 11/16)h1(n) = -4 - Res{ zn-1dz (z - 1/4)(z - 2) (1/4)(1/4 - 11/16) 2(2 - 11/16) = -4-{ (1/4)n-4 + 2n(1/2)} 1/4 - 2 2 - 1/4 -7/16 21/16 = -4-{(1/4)n + 2n} -7/4 7/4 = -[(1/4)n + 3-(2n)]u(n).Purely anticausal sequence (n < 0): use p-form. 1 -4p-1(p-1 - 11/16)h2(n) = O p-n-1dp 2nj C' (p-1 - 1/4)(p-1 - 2) 1 (11/4) (p - 16/11) = O p-n-1dp 2nj (-1/4)(-2) C' (p - 4)(p - 1/2) 11 p - 16/11 = - Z res{ p-n-1, enclosed poles}. 2 (p - 4)(p - 1/2)We take z < 1/4 or p > 4, so both poles are enclosed. Thus for n > 0, 11 4 - 16/11 1/2 - 16/11h2(n) = { 4-n-1 + (1/2)-n-1} 2 4 - 1/2 1/2 - 4 = (1/4)n + 3-2n, n > 0.The only difference from the above expression for n = 0 is that at n = 0 thereis a simple pole at the origin of the p-plane, which is enclosed by C'.Therefore, to the above expression we must add the residue at p = 0:For n = 0, p - 16/11h2(n) = (11/2)- Z res{ } p(p - 4)(p - 1/2) = [above expression evaluated at n = 0] - (11/2)[(16/11)/2] = 1 + 3 - 4 = 0.Thus, h2(n) = ((1/4)n + 3-2n)u(-n-1).Interestingly, the only differences between h(n), h1(n), and h2(n), aside fromthe step functions, are the signs of the coefficients!88(c) The code is as shown below.z=input('Enter z for checking Laurent expansion ')r=abs(z); Nterm=4000;F=z*(-4*z+11/4)/(z-0.25)/(z-2);p1=0.25; p2=2;r1=abs(p1); r2=abs(p2);sum=0;if r Re{z} -1/2 j/2 X (c) Let region A be 0 < z < 1, and region B be z > 1. The Laurent expansionswill differ in A and B.(d) Region A (0 < z < 1):In the series expansion formula, let a = e-j/4 and b = z. Then (e-j/4 + z)-1 = ej/4 Z (-1)kejk/4zk k=0 (-e-j/4 + z)-1 = -ej/4 Z ejk/4zk. k=0The sum of these two is (e-j/4 + z)-1 + (-e-j/4 + z)-1 = ej/4 Z ((-1)k - 1)ejk/4zk k=0 but ((-1)k - 1) = -2 for k odd and 0 otherwise. Thus(e-j/4 + z)-1 + (-e-j/4 + z)-1 = -2{jz - z3 - jz5 + z7 + jz9 + ...}soH(z) = -j/z + z + jz3 - z5 - jz7 + z9 + ... This sequence of coefficients repeats.Region B (z > 1):Now let a = z, b = e-j/4. Then (z + e-j/4)-1 = (1/z) Z (-1)ke-jk/4z-k k=0 (z - e-j/4)-1 = (1/z) Z e-jk/4z-k. k=0The sum of these two is (z + e-j/4)-1 + (z - e-j/4)-1 = (1/z) Z ((-1)k + 1)e-jk/4z-k k=0 but ((-1)k + 1) = 2 for k even and 0 otherwise. Thus(z + e-j/4)-1 + (z - e-j/4)-1 = 2{1/z - j/z3 - 1/z5 + j/z7 + 1/z9 - j/z11 + ...}90soH(z) = -j/z + j/z + z-3 - jz-5 - z-7 + jz-9 + ... This sequence of coefficients repeats. Notice that, however, for this first repetition the -j/z and the j/z cancel.3.57) (a) 0.7-2 -2 H(z) z - 2 (0.72-1.41(0.7)+1.21)0.7 1.21(-0.7) = = + z (z2 - 1.41z + 1.21)(z - 0.7) z - 0.7 zplus anticausal pole terms. Thus H(z) = -2.6047z/(z - 0.7) + 2.3613 plus anticausal pole terms. Thus h(n)n0 = -2.6047(0.7)n + 2.3613-o(n).Note that there is no point in calculating complex pole components!! (z - 2)zn-1(b) h(n) = Z Res{, poles within contour}. (z2 - 1.41z + 1.21)(z - 0.7)For stable h(n), an appropriate choice of C is the unit circle, for it must bein the ROC of H(z) for a BIBO system. Only z = 0.7 is enclosed, except at n =0 there is also a pole at the origin. Thush(n)n0 = Res{(z - 2)zn-1/A, 0.7} + o(n)-Res{(z - 2)/A, 0} = -2.6047(0.7)n+2.3613-o(n). 3.58) z-10H1(z) = H(z)/z = (z-1)2(z+)(z2+0.3126z+0.81) A B C D D* = + + + + z - 1 (z - 1)2 z + z - 0.9/_100 z - 0.9/_-100where d z - 10 A = { } . dz (z+)(z2+0.3126z+0.81) z=1Letting A = (z+)(z2+0.3126z+0.81) = z3+0.8126z2+0.9663z+0.4050, A - (z-10)(3z2+1.6251z+0.9663) 3.1839+50.3226A = = = 5.2784. A2 z=1 10.1369910 5 10 15 20 25 30-80-70-60-50-40-30-20-10010* stems = result from "filter" w/conv to neg z powers.Stairs = dlsim result; O stems = direct calc; -9 -10.5B = = -2.8268, C = = -5.1639, 3.1839 (-1.5)2(0.9037) -10.1563+j0.8863D = = -0.0573+j2.8497 = 2.8502 /_ 91.1513. 0.3825+j3.5563Thus, for n 0,h(n) = A + Bn + C(-)2 + Dz1n + D*z1*n,where the last term can be written 2Dz1

ncos{(/_z1)-n + /_D}.This expression was calculated in Matlab and compared with the result for "dlsim"with a unit sample input, and the results are identical, as shown in the plotbelow. 3.59)(a) From the hint, we already know that Z z-n/n = ln{z/(z-1)}, z > 1 n=1and possibly for z = 1 with z 1.Moreover, using ZT{anx(n)} = X(z/a) with a = e-, we have Z e-nz-n/n = ln{e

z/[e

z-1]} = ln{z/[z - e-]}, z>e-. n=1We can add the above two results for z > max(1,e-), but we also need to addx(0)z-0, where by L'Hopital's rule, x(0) = o. ThusX(z) = o + ln{[z/(z-1)]-[z - e-]/z} = o + ln{(z - e-)/(z - 1)}. The 3-D plot is as shown below for o = 0.8.92-2-1012 -202-20246x = Re{z}jy = jIm{z}|X(z)| where X(z) = ZT{[(1-exp(-sig n)/n]u(n)}; sig = 0.8. M). There is a zero at the origin oforder N - (N1 - M1) = N+K, which in part (a) is equal to 3 - 1 = 2, as seen inthe solution of part (a). Now, concerning the apparent contradiction. In thetext we say, e.g., that the ROC z > o corresponds to a causal sequenceregardless of the a

and b

coefficients. We see that a more refined way ofsaying this is that the ROC z > o corresponds to a right-sided sequence, whichmay possibly begin at a finite value of n, n

, where n

< 0. This will be thecase for K < 0. Thus, there is no contradiction, but rather a refinement ofcharacterization needed to cover this case. We of course may always obtain theanticausal