Solution Manual for Signals and Systems, 3rd Edition

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Solutions 2-1Chapter 2-Mathematical Description ofContinuous-Time SignalsSolutionsExercises With Answers in TextSignal Functions1.Ifgt 7e2t3write out and simplify(a)g 3 7e98.6387104(b)g 2t7e2 2t37e72t(c)gt/1047et/511(d)gjt7ej2t3(e)gjtgjt27e3ej2tej2t27e3cos 2t0.3485 cos 2t(f)gjt32 gjt3227ejtejt27 cost 2.Ifgxx24x4write out and simplify(a)gz z24z4(b)guvuv24uv4u2v22uv4u4v4(c)gejtejt24ejt4ej2t4ejt4ejt22(d)g gt gt24t4t24t424t24t44g gt t48t320t216t4(e)g 248403.Findthe magnitudes and phases of these complex quantities.

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Solutions 2-2(a)e3j2.3e3j2.3e3ej2.3e3cos 2.3jsin 2.3e3j2.3e3cos22.3sin22.3e30.0498(b)e2j6e2j6e2ej6e27.3891(c)1008j131008j131008j13100821326.55124.LetGfj4f2j7f/11.(a)What value does the magnitude of this function approach asfapproachespositive infinity?limfj4f2j7f/11j4fj7f/1147 /114476.285

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Solutions 2-3(b)What value (in radians) does the phase of this function approach asfapproaches zero from the positive side?limf0j4f2j7f/11 limf0j4f2/ 20/ 25.LetXfjfjf10(a)Find the magnitudeX 4and the anglein radiansXfjfjf10X 4j4j4104ej/210.77ej0.38050.3714ej1.19X 40.3714(b)What value (in radians) doesapproach asfapproaches zero fromthe positive side?Shifting and Scaling6.Foreach functiongt graphgt,gt ,gt1, andg 2t.(a)(b)tg(t)24tg(t)1-13-3tg(-t)-24tg(-t)1-13-3t-g(t)24t-g(t)1-13-3

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Solutions 2-47.Find the values of thefollowing signals at the indicated times.(a)xt 2rectt/ 4, x12rect1/ 42(b)xt 5rectt/ 2sgn 2t, x 0.55rect 1/ 4sgn 1 5(c)xt 9rectt/10sgn 3t2,x 1 9rect 1/10sgn3 98.For each pair of functions in Figure E-8provide the values of the constantsA,t0andwin the functional transformationg2t Ag1tt0/w.Figure E-8Answers:(a)A2,t01,w1, (b)A 2,t00,w1/ 2,(c)A 1/ 2,t0 1,w29.For each pair of functions in Figure E-9provide the values of the constantsA,tg(t-1)314tg(t-1)123-3tg(2t)14tg(2t)13-3212--4-2024-2-1012tg1(t)(a)-4-2024-2-1012tg2(t)(a)-4-2024-2-1012tg1(t)(b)-4-2024-2-1012tg2(t)(b)-4-2024-2-1012tg1(t)(c)-4-2024-2-1012tg2(t)(c)

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Solutions 2-5t0andain the functional transformationg2t Ag1w tt0.(a)Amplitude comparison yieldsA2. Time scale comparison yieldsw 2.g222g12 2t02g10 42t00t02(b)Amplitude comparison yieldsA3. Time scale comparison yieldsw2.g223g12 2t03g1042t00t02(c)Amplitude comparison yieldsA 3. Time scale comparison yieldsw1/ 3.g20 3g11/ 30t0 3g12 t0/ 32t0 6ORAmplitude comparison yieldsA 3. Time scale comparison yieldsw 1/ 3.g23  3g11/ 33t0 3g10t0/ 310t03-10-50510-8-4048tg1(t)-10-50510-8-4048A= 2,t0= 2,w= -2tg2(t)g1(t)A= 3,t0= 2,w= 2g2(t)-10-50510-8-4048t-10-50510-8-4048tg1(t)A= -3,t= -6,w= 1/30g2(t)-10-50510-8-4048t-10-50510-8-4048t

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Solutions 2-6(d)Amplitude comparison yieldsA 2. Time scale comparison yieldsw1/ 3.g24 2g11/ 34t0 3g12 t0/ 34 / 32t0 2(e)Amplitude comparison yieldsA3. Time scale comparison yieldsw1/ 2.g203g11/ 20t03g11  t0/ 21t0 2Figure E-910.In FigureE-10is plotted afunctiong1t which is zero for all time outside therange plotted. Let some other functions be defined byg2t 3g12t,g3t  2g1t/ 4,g4t g1t32Find these values.(a)g21  3(b)g31 3.5(c)g4t g3t t232 1 32(d)g4t dt31The functiong4t is linear between the integration limits and the area under it isa triangle. The base width is 2 and the height is-2. Therefore the area is-2.g4t dt31 2g1(t)0g2(t)-10-50510-8-4048t-10-50510-8-4048tA= -2,t= -2,w= 1/3g1(t)g2(t)-10-50510-8-4048t-10-50510-8-4048t0A= 3,t= -2,w= 1/2

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Solutions 2-7Figure E-1011.A functionGfis defined byGfej2frectf/ 2.GraphthemagnitudeandphaseofGf10Gf10overtherange,20f20.First imagine whatGflooks like. It consists of a rectangle centered atf0ofwidth, 2, multiplied by a complex exponential.Therefore for frequencies greaterthan one in magnitude it is zero.Its magnitude is simply the magnitude of therectangle function because the magnitude of the complex exponential is one foranyf.ej2fcos2fjsin2fcos 2fjsin 2fej2fcos22fsin22f1The phase (angle) ofGfis simply the phase of the complex exponentialbetweenf 1andf1and undefined outside that range because the phase ofthe rectangle function is zero betweenf 1andf1and undefined outside thatrange and the phase of a product is the sum of the phases.The phase of thecomplex exponential isej2fcos 2fjsin 2ftan1sin 2fcos 2f  tan1sin 2fcos 2fej2f tan1tan 2fThe inverse tangent function is multiple-valued.Therefore there are multiplecorrect answers for this phase. The simplest of them is found by choosingej2f 2ftg (t)1-1-2-3-42341234-4-3-2-11

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Solutions 2-8which is simply the coefficient ofjin the original complex exponential expression.A more general solution would beej2f 2f2n,nan integer.Thesolution of the original problem is simply this solution except shifted up and downby 10 infand added.Gf10Gf10ej2f10rectf102 ej2f10rectf10212.Letx1t 3rectt1/ 6andx2t rampt ut ut4.(a)Graph them in the time range10t10. Put scale numbers onthe vertical axis so that actual numerical values could be read fromthegraph.(b)Graphxt x12tx2t/ 2in the time range10t10. Putscale numbers on the vertical axis so that actual numerical valuescould be read from the graph.t-1010x1(t)-66t-1010x2(t)-66

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Solutions 2-9x12t3rect2t1/ 6andx2t/ 2rampt/ 2ut/ 2ut/ 24x12t3rectt1 / 2/ 3andx2t/ 21 / 2rampt ut ut813.Writean expression consisting of a summation of unit step functions to represent asignal which consists of rectangular pulses of width 6 ms and height 3 which occurat a uniform rate of 100 pulses per second with the leading edge of the first pulseoccurring at timet0.xt 3ut0.01nut0.01n0.006n014.Find thestrengthsof thefollowingimpulses.(a)3 4t3 4t 314t Strength is3 / 4(b)53t153t153t1Strength is 5315.Find thestrengthsand spacing betweenthe impulses in the periodic impulse9115t.9115t 95t11kk 95t11k/ 5kThe strengths are all-9/5 and the spacing between them is 11/5.t-1010x1(2t)-x2(t/2)-88

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Solutions 2-10Derivatives and Integrals of Functions16.Graph the derivative ofxt 1etut .This function is constant zero for all time before time,t0, therefore itsderivative during that time is zero. This function is a constant minus a decayingexponential after time,t0, and its derivative in that time is therefore apositive decaying exponential.xt et,t00,t0Strictly speaking, its derivative is not defined at exactlyt0. Since the value of aphysical signal at a single point has no impact on any physical system (as long as itis finite) we can choose any finite value at time,t0, without changing the effectof this signal on any physical system.If we choose 1/2, then we can write thederivative asxt etut .Alternate Solution using the chain rule of differentiation and the fact that theimpulse occurs at timet0.ddtxt 1et0 fort0t etut etut 17.Find thenumerical value of each integral.(a)u 4tdt211dt246(b)t324tdt18t-14x(t)-11t-14dx/dt-11

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Solutions 2-11t324tdt18t3dt1824tdt18t324tdt18021/ 4t dt18 1/ 2(c)23tdt1/25/223tdt1/25/23t2nndt1/25/213t2n/ 3ndt1/25/223tdt1/25/213 1111(d)t4ramp2tdtramp24ramp 88(e)ramp 2t4dt3102t4dt210t24t210100404864(f)3sin 200tt7dt11820(Impulse does not occur between 11 and 82.)(g)sint/ 20dt55=0Odd function integrated oversymmetrical limits.(h)39t24t1dt21039t2t14kkdt21039t24t1dt21039t2t1t5t9dt21039t24t1dt21039 1252924173(i)e18tut 10t2dt.

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Solutions 2-12e18tut 10t2dte18tut 10t1/ 5dtUsing the scaling property of the impulse,e18tut 10t2dt110e18tut t1/ 5dtUsing the sampling property of theimpulse,e18tut 10t2dt110e18/50.002732(j)9t4/ 5dt2945t4dt2945(k)53t4dt6353t4dt630(l)ramp 3tt4dtramp 3412(m)3t cos 2t/ 3dt117The impulses in the periodic impulse occur at ...-9,-6,-3,0,3,6,9,12,15,18,...At each of these points the cosine value is the same, one, because itsperiod is 3 seconds also. So the integral value is simple the sum of thestrengths of the impulse that occur in the time range, 1 to 17 seconds3t cos 2t/ 3dt117518.Graph the integral from negative infinity to timetof the functions in Figure E-18which are zero for all timet0.This is the integralgdtwhich, in geometrical terms, is the accumulatedarea under the functiongt from timeto timet.For the case of the twoback-to-back rectangular pulses, there is no accumulated area until after timet0and then in the time interval0t1the area accumulates linearly with time up to

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Solutions 2-13a maximum area of one at timet1. In the second time interval1t2the areais linearly declining at half the rate at which it increased in the first time interval0t1down to a value of 1/2 where it stays because there is no accumulation ofarea fort2.In the second case of the triangular-shaped function, the area does not accumulatelinearly, but rather non-linearly because the integral of a linear function is asecond-degree polynomial.The rate of accumulation of area is increasing up totimet1and then decreasing (but still positive) until timet2at which time itstops completely. The final value of the accumulated area must be the total area ofthe triangle, which, in this case, is one.Figure E-1819.If4ut5ddtxt , what is the functionxt ?xt 4rampt5Generalized Derivative20.The generalized derivative of18 rectt23consists of two impulses.Find their numerical locations andstrengths.Impulse #1:Location ist= 0.5 and strength is 18.Impulse #2:Location ist= 3.5 and strength is-18.Even and Odd Functions21.Classify the following functions as even, odd or neither .(a)cos 2ttrit1Neitherg(t)t112312g(t)t1123g(t) dtg(t) dtt112312t1123

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Solutions 2-14Cosine is even but the shifted triangle is neither even nor odd whichmeans it has a non-zero even part and a non-zero odd part. So the productalso has a non-zero even part and a non-zero odd part.(b)sin 2trectt/ 5OddSine is odd and rectangle is even. Therefore the product is odd.22.Aneven functiongt is described over the time range0t10bygt 2t, 0t3153t, 3t72, 7t10.(a)What is the value ofgt at timet 5?Sincegt is even,gt gtg5g 515350.(b)What is the value of the first derivative of g(t) at timet 6?Sincegt is even,ddtgt  ddtgtddtgt t6 ddtgt t6  33.23.Find the even and odd parts of these functions.(a)gt 2t23t6get 2t23t62t23t624t21222t26got 2t23t62t23t62 6t2 3t(b)gt 20cos 40t/ 4get 20 cos 40t/ 420 cos40t/ 42Usingcosz1z2cosz1cosz2sinz1sinz2,

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Solutions 2-15get 20 cos 40tcos/ 4sin 40tsin/ 420 cos40tcos/ 4sin40tsin/ 42get 20 cos 40tcos/ 4sin 40tsin/ 420 cos 40tcos/ 4sin 40tsin/ 42get 20 cos/ 4cos 40t20 /2cos 40tgot 20 cos 40t/ 420 cos40t/ 42Usingcosz1z2cosz1cosz2sinz1sinz2,got 20 cos 40tcos/ 4sin 40tsin/ 420 cos40tcos/ 4sin40tsin/ 42got 20 cos 40tcos/ 4sin 40tsin/ 420 cos 40tcos/ 4sin 40tsin/ 42got 20sin/ 4sin 40t20 /2sin 40t(c)gt 2t23t61tget 2t23t61t2t23t61t2get 2t23t61t2t23t61t1t1t2

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