EE102 Homework 5 and 6 Solutions

Transcription

1 EE2 Prof. S. Boyd EE2 Homework 5 and 6 Soluions 35. The verical dynamics of a vehicle suspension sysem, when he vehicle is driving on level ground, are given by (m v + m l )d () + bd () + kd() =. Here is ime (in seconds) d() is he verical displacemen of he vehicle, wih respec o is neural posiion (in meers) m v = 3 kg is he vehicle mass m l (also given in kg) is he mass of he vehicle load (passengers, cargo, ec.) b = N/m/s is he suspension damping k = 5 N/m is he suspension siffness The iniial condiions are d() =.m, d () = m/s. Wha is he smalles load mass m l for which d is oscillaory? (By oscillaory, we mean ha d() passes hrough zero infiniely many imes.) Soluion. Le m = m v + m l. Then, aking he Laplace ransform of he given differenial equaion, we obain ms 2 D(s) msd () md() + bsd(s) bd() + kd(s) = The above equaion can be rewrien as D(s) = msd () + (m + b)d() ms 2 + bs + k The sysem will exhibi oscillaory behavior when he roos of he characerisic polynomial ms 2 + bs + k are complex. The roos are given by he soluion of he quadraic equaion, i.e., λ = b ± b 2 4mk 2m Therefore, he roos will be complex when b 2 4mk <. We can rewrie his condiion as m > b2 4k = 2 Finally, since m = m v + m l and m v =, we obain ha m l > 2 for d o become oscillaory. 4. Four funcions f, f 2, f 3, and f 4, are shown below. Their Laplace ransforms are F, F 2, F 3, and F 4, respecively. You can assume F,..., F 4 are raional funcions, wih no more han hree poles (couning mulipliciies). Please noe carefully he verical scales hey are no he same!

2 2 2 f() f2() f3() PSfrag replacemens f4() Esimae he poles of F,..., F 4, using he smalles number needed o give a reasonable mach. If you can ge a reasonable mach wih one pole, hen give jus one pole; if wo poles suffice hen give jus wo. Give hree poles only if hree poles are required o mach he given f i. We wan specific numbers, no jus qualiaive answers such as one posiive real, one complex pair or σ + jω (wihou specifying σ or ω). Make clear indicaions on he plos how you go he numbers. Give complex poles separaely, as in + j, j ; we will no auomaically supply conjugaes of complex numbers. Give muliple poles by repeaing hem in your answers, e.g., 3,, (meaning, one pole a s = 3, and anoher pole of mulipliciy wo a s = ). Soluion: (a) From he graph of f, we see ha here is no sinusoidal ringing, and somehing like only one exponenial decay owards a nonzero value, 2. Therefore, we have only 2 real poles: one corresponding o he decaying erm and one corresponding o he consan erm. The cosnan erm corresponds o a pole a s =. The exponenially decaying erm sars wih ampliude ( 2) = 3, and i reaches 37% of he ampliude in approximaely sec. Therefore, ha pole is a. So he poles are:,. (b) From he plo of f 2, we noe ha he waveform exhibis some sinusoidal ringing, and he ampliude of he ringing is approximaely consan, neiher growing nor decaying. This 2

3 suggess a complex pole pair along he imaginary axis. Furhermore, when we draw a line down he middle of he oscillaions we find ha here is also a real, exponenially decaying erm wih a negaive coefficien. So, he hird pole is real and negaive. Le s find approximae numerical values for he poles. We see around 9.5 periods of he signal in 6 ime unis, so he oscillaion frequency is roughly ω 2π 6/9.5. So we have he poles ±j. As for he hird pole, we see ha i decays o 37% of he ampliude in approximaely sec. Therefore, he hird pole is a. The poles are: j, j,. (c) From he plo of f 3, we see a sinusoidal ringing wih an exponenially growing ampliude, which suggess a complex pole pair in he righ half plane. The oscillaion frequency of he ringing is he same as in f 2, i.e., ω =. A ime zero, he ampliude of he ringing is abou.5. A 4 seconds, he ampliude is, say,. or.2. Thus we have e σ4 =.2, so σ log(.2/.5)/4.2 or so. Therefore, he poles are.2 ± j. Drawing a line down he middle of he oscillaion indicaes ha here is also a consan offse of, which corresponds o a pole a s =. The poles are:.2 + j,.2 j,. (d) In he las plo, we see no sinusoidal ringing; We see only one exponenial growh (muliplied by negaive coefficien). A ime zero, f 4 () is abou 2; a ime equals 6, f 4 () is abou 2. So, he growh rae is approximaely log( 2/ 2)/6.3. So, he pole is a Reducing he rise-ime of a signal. In a cerain digial sysem a volage signal should ideally swich from V o 5V infiniely fas, i.e., wih zero rise-ime. Bu due o he finie bandwidh of he elecronics ha generaes he signal, i has he form ( ) v in () = 5 e /T for where T = µsec. Thus, he signal has a rise-ime around a few µsec. An engineer claims ha he circui shown below can be used o reduce he rise-ime of he signal, provided he componen values R and C are chosen correcly. Specifically, he engineer claims ha by choosing R and C correcly, we can have ( ) v ou () = a e /T for where a is some nonzero consan. Thus, he rise-ime of v ou is a facor of smaller han he rise-ime of v in, i.e., a few hundred nsec. C PSfrag replacemens v in () R kω v ou () 3

4 Here is he problem: deermine wheher he engineer s claim is rue or false. If he claim is rue, find specific, numerical values of R and C ha validae he claim. If he claim is false, briefly explain why he engineer s idea will no work. (You can assume he circui sars in he relaxed sae, i.e., no charge on he capacior. And no, you canno use negaive R or C.) Soluion: Firs we ll find he relaionship beween v in () and v ou (). V ou = V in + R sc = V in ( + src) ( + R) + src Now le s assume ha he engineer s claim is rue, i.e., for some value of R and C we have v ou () = a( e /T ). Then: V ou (s) = a s ( a 5 s + = 6 s 5 s + 6 Simplifying, we ge 7 ( ) ( ) a 5 6 s + /RC s + 7 = s + 6 s + +R. RC We need o choose R and C o make his equaliy hold. ) s + /RC s + +R RC Bu how can his be? The lef hand side has only one pole, while he righ hand side has wo... The only way his can happen is if he zero on he righ hand side, a s = /RC, cancels he pole a 6, i.e., we ake /RC = 6. This yields 7 a s + 7 = 5 6 s + +R. 3 Now we wan he remaining pole o be a s = 7, so we have ( + R)/( 3 ) = 7. This yields R = 9kΩ, so C = pf. So we see he engineer s claim is rue. Noe ha a =.5V. Thus we see one disadvanage of his circui he rise-ime is en imes faser bu he signal level has been cu by a facor of! Le s also menion anoher mehod of solving his problem. Once we have V ou and V in, we divide o find ou wha he ransfer funcion mus be o make he claim rue. Then we simply see wheher we can choose he parameers R and C o make he ransfer funcion of our circui equal o he one we require, i.e., V ou /V in. (Of course you ge he same answer.) 56. The op plo below shows he sep response of a sysem described by a ransfer funcion. Below ha is a plo of an inpu u() ha we apply o his sysem. Skech he response (oupu) y(). 4

5 .5 s() (nsec).5 PSfrag replacemens u() (nsec) Soluion: Perhaps he easies way o solve his problem is o differeniae he sep response o ge he impulse response, hen convolve he impulse response wih he inpu o find he oupu. The impulse response is zero excep beween = nsec and = 2nsec, where i has he value 9 (noe he ime axis; if you jus ingore he unis everyhing works ou OK anyway!). Thus he oupu a ime is jus he inegral of he inpu beween 2 and 3 nanoseconds ago, muliplied by 9. I is zero unil = ; hen i ramps up quadraically (since we are inegraing a ramp) unil = 2. Beween = 2 and = 3 we inegrae par of a ramp and par of he consan ; his yields anoher quadraic. The maximum value is achieved righ a = 3, when we inegrae he consan, so he oupu is. In he nex wo nanoseconds he same hing happens, backwards in ime. The final soluion is shown below: 5

6 .5 oupu y If you don ge ha fluffy derivaion, or simply mus see formulas o believe, here is anoher derivaion. The sep response evidenly has he form s() = 2 2 where is in nanonseconds. Differeniae o ge he sep response h: h() = 2 2 The inpu u has he form: Hence he oupu y is jus: u() = y() = h( τ)u(τ) dτ = 2 u(τ) dτ. The oupu will have 5 separae segmens, i.e., differen formulas, depending on. We ll jus work ou an ineresing case, and le you check he ohers. We ll find y() for 2 3. From our formula above, y() = u(τ) dτ = τ dτ + dτ = 2 ( ( 2)2 ) which corresponds o he quadraic ha sars a value /2 for = 2 and ends up a value (and slope zero) a =. 6

7 You could also work his problem using Laplace ransforms, bu i s no a prey sigh. The ransfer funcion is H(s) = e s e 2s s and he Laplace ransform of u is Thus we have U(s) = e s e 2s + e 3s s 2. Y (s) = H(s)U(s) = e s e 2s s e s e 2s + e 3s s 2 = e s 2e 2s + 2e 4s e 5s s 3 which you will insanly recognize as he Laplace ransform of our soluion shown in he plo. 65. The circui below is a simple one-pole lowpass filer. C R kω PSfrag replacemens v ou () v in () Find (posiive) R and C such ha: The (magniude of he) DC gain is +2dB. The magniude of he ransfer funcion a he frequency khz is 3dB less han he magniude of he DC gain. You can assume he op-amp is ideal. Give numerical values for R and C. An acurracy of % will suffice. Soluion. Firs, le s find he ransfer funcion from v in o v ou. You could use sandard circui analysis, bu by now you should jus recognize his circui as an invering amplifier and know ha he ransfer funcion is given by H(s) = V ou(s) V in (s) = R sc kω = R/kΩ + src 7

8 This circui has only one pole, a s = /RC. The DC gain is found by seing s = : H() = R/kΩ. (You can also see his by looking a he circui wih he capacior removed.) We are given ha he (magniude of he) DC gain is +2dB, which is roughly 4 (6dB + 6dB; each is a facor of around 2). Thus, H() = R/kΩ = 4, so we mus have R = 4kΩ. (Noe ha he DC gain is acually negaive, i.e., H() = 4.) The gain is 3dB less a khz, which means ha we have a pole a s = khz = 2π. Since he pole is a s = /RC, we have: C = 4kΩ 2π = 4nF. 68. A simple wo-way crossover circui. A ypical high-fideliy speaker has separae drivers for low and high frequencies. (The driver is he physical device ha vibraes o creae he sound you hear. The old erms for he low and high frequency drivers are woofer and weeer, respecively.) The circui shown below, called a speaker crossover nework, is used o divide he audio signal coming from he amplifier ino a low frequency par for he low frequency driver (LFD) and a high frequency par for he high frequency driver (HFD). Since he audio specrum is divided ino wo pars, his is called a wo-way sysem (hree-way are also common). The amplifier is modeled as a volage source (which is a very good model), and he low and high frequency drivers are modeled as 8Ω resisances (which is no a good model of real drivers, bu we will use i for his problem). PSfrag replacemens C L v amp () Z speaker 8Ω HFD 8Ω LFD speaker The crossover nework is designed so ha he ransfer funcion from he amplifier o each driver has magniude 3dB a a frequency ω c called he crossover frequency of he speaker. (a) Choose C and L so ha he crossover frequency is 2kHz. Do his carefully as you will need your answers in par b. (b) Using he values for L and C found in 8a and 8b, find Z speaker (s), he impedance of he wo-way speaker seen by he amplifier (as indicaed in he schemaic). 8

9 Soluion: We can separae he analysis for he high frequency driver (HFD) and low frequency driver (LFD) since he wo circuis are driven in parallel by a volage source. Noice he HFD is jus a simple RC high-pass filer and LFD is jus a simple RL low-pass filer! For he HFD, we have he ransfer funcion: V HFD V amp = sc = 8sC + 8sC Noe ha crossover frequency is he frequency a which he ransfer funcion is 3dB, i.e., i is he (absolue value of he) pole locaion. Wih he crossover frequency a 2kHz, we have So C = 9.95µF. 8C = (2π)(2) Similarly, for he LFD we have he ransfer funcion: V LFD V amp = Wih he crossover frequency a 2kHz, we have So L =.64mH sl = + s L 8 8 L = (2π)(2) Le s denoe he resisance of he HFD and LFD as R. Now since he poles of he HFD and he LFD have already been designed o be 2kHz, we have RC = L/R. Therefore, Z speaker (s) = (R + sc )(R + sl) R + sc + R + sl = (R + R2 sl )(R + sl) = R. 2R + R2 sl + sl So we see ha Z speaker (s) is consan i is simply 8Ω, independen of frequency. To he amplifier, he speaker looks like a simple resisance of 8Ω. In pracice, real drivers do no have consan 8Ω impedance, and also, 2 and 3 pole crossover neworks are more common han he simple pole nework described in his problem. Bu he idea is he same. 72. The uni sep response s() of a sysem described by a ransfer funcion H, which has hree poles, is shown in he wo plos below. The wo plos have differen ranges; he second plo allows you o see deails for small. 9

10 .5 s() s() PSfrag replacemens (a) Esimae he poles. An accuracy of ±2% is accepable. (b) A high frequencies H(jω) becomes small. From he daa given, can you deermine he rae a which i decreases for large frequency (e.g., 2 db/ocave)? Eiher give he rae (in db/ocave) or sae canno deermine if he daa given is no sufficien o deermine he high-frequency rolloff rae. Soluion: We sar by esimaing he sep response. We should see evidence of he hree poles of he ransfer funcion H, along wih a pole a s = (i.e., a consan) associaed wih he sep inpu. There is a sinusoidal erm ha does no decay or grow, wih an ampliude of abou.. The frequency of his erm is easy o ell: wo cycles fi in.sec so he frequency is 4πrad/sec. Thus here are wo poles a s = ±j4π 26rad/sec. Drawing a line hrough he average value of his oscillaion we see ha somehing like e 4 remains. This corresponds o a pole a s = (which comes from he sep inpu) and s = 4. Thus he poles of H mus be abou 4, ± j26. There are several ways o find he rae a which H(jω) decreases for large ω. Perhaps he easies way is o remember he connecion beween he behavior of he sep response for small and he rolloff rae of he frequency response. The sep response has an iniial nonzero slope, so H mus have one more pole han zero, i.e., wo zeros. This means ha he rolloff rae is 6dB/ocave. We can also find he rae by direcly esimaing he ransfer funcion H from he sep response plo. The sep response has he (approximae) form s() e 4 + cos(26 + φ). We can figure ou he phase of he sinusoid by examining he lower plo, and noing ha s() cos(φ). This suggess ha he sep response is s() e 4 + sin(26).

11 (Acually his mehod is exremely prone o approximaion error.) I s Laplace ransform is herefore S(s) s s s = 5s2 + 4s s(s + 4)(s ). Therefore H(s) = ss(s) 5s2 + 4s (s + 4)(s ) which has hree poles and wo zeros. Hence for large ω, he frequency response rolls off a 6dB/ocave. 74. Transfer funcion from rainfall o river heigh. The heigh of a cerain river depends on he pas rainfall in he region. Specifically, le u() denoe he rainfall rae, in inches-per-hour, in a region a ime, and le y() denoe he river heigh, in fee, above a reference (dry period) level, a ime. The ime is measured in hours; we ll only consider. Analysis of pas daa shows ha he relaion beween rainfall and river heigh can be accuraely described by a ransfer funcion: Y (s) = H(s)U(s), H(s) = (3s + )(3s + ) (You don need o know any hydrology o do his problem, bu you migh be ineresed in he physical basis of his wo-pole ransfer funcion. The fas pole is due o runoff from surface waer and small ribuaries, which conribue a relaively small amoun of waer relaively quickly. The slow pole is due o flow from larger ribuaries and deeper ground waer, which conribue more waer ino he river, over a much longer ime scale.) A brief bu inense downpour. (Pars a and b.) Suppose ha afer a long dry spell (i.e., no rain) i rains inensely a 2 inches-per-hour, for 5 minues. This causes he river heigh o rise for a while, and hen laer recede. (a) How long does i ake, afer he beginning of he brief downpour, for he river o reach is maximum heigh? We ll denoe his delay as max (in hours). (b) Wha is he maximum heigh of he river? We ll denoe his maximum heigh as y max (in fee). Noe: you can make a reasonable approximaion provided you say wha you are doing. A coninual rain. (Pars c and d.) Suppose ha afer a long dry spell i sars raining coninuously a a rae of inch-per-hour (and doesn sop). This causes he river heigh o rise. (c) Wha is he ulimae heigh of he river, i.e., y ul = lim y()? (d) A flood occurs when he river heigh y() reaches 8 fee. How long will i ake, afer he onse of he seady rain, o reach flood condiion? We ll denoe his ime as flood. If he river never reaches 8 fee, give your answer as never. Noe: you can make a reasonable approximaion provided you say wha you are doing. Soluion:

12 (a) Firs, we noice ha he ime duraion of he downpour (5 min) is small in comparison o he ime consans of he poles (3 hours and 3 hours). Hence we can reasonably approximae he inense downpour as an impulse funcion. The area under he rainfall u() is (2 inches/hour)(/2 hour), i.e., (inch). So u() δ(), and consequenly he river heigh y is (approximaely) given by he impulse response of he ransfer funcion. (If you model he downpour more accuraely as a sep funcion which urns on a = and off and = /2, he mahemaic ges much more involved, and he final answer is nearly he same.) Thus Y (s) = H(s), so we use a parial fracion expansion: Y (s) = (3s + )(3s + ) =.37 s + / s + /3 The inverse ransform is hen y() =.37(e /3 e /3 ). This is a good poin a which o do a reasonableness check. This y() sars a, rises for a while, hen decays which does seem very reasonable! We find he maximum of y() by seing he derivaive o zero:.37( 3 e /3 + 3 e /3 ) = which can be solved o yield = Thus he river heigh peaks max = 7.7 hours laer. This answer seems reasonable; he river peaks abou a double he fas pole s ime consan. (b) The maximum heigh of he river is found by subsiuing = hours ino our expression for y() found above. We find y max =.257 fee = 3.9 inches. Again, his seems o correlae well wih our inuiive feeling for a river s behavior. A brief downpour occurs, and he river rises 3 inches abou 7 hours laer. (c) The coninual rain is jus a uni sep funcion, so he river heigh is jus he sep response of he ransfer funcion: Y (s) = (3s + )(3s + ) s = s + /9 s + /3 /9 s + /3 Hence y() = + 9 e /3 9 e /3. From inspecion, we see ha as, y fee! Anoher way o solve his problem, which is probably easier, is o noice ha we need o find he limi of he sep response, which is jus he DC gain of he ransfer funcion. Plugging in s = yields, again, he answer y ul =. (d) The flood level is 8 fee, less han he final heigh of fee we solved for in par (c). So we do reach flood condiion; i s jus a quesion of when. Using our expression for y found above, he flood occurs for wih 8 = + 9 e /3 9 e /3. In fac, i is no so easy o solve his equaion! Now is a very good ime for a simple approximaion, as our noe suggess. Noe ha he erm 9 e /3 decays quickly, wihin 2

13 say hours, afer which he river heigh will be approximaely given by 9 e /3. Thus we can find an approximae soluion by solving 8 = 9 e /3 which yields = 5.4 hours. (Noe ha our assumpion ha he firs erm is small is cerainly correc when = 5.4 hours; e 5.4/3 is quie small!) Hence we predic disaser crews have abou 2 days o prepare for flooding afer a big sorm his! (Provided he rain coninues a inch/hour.) 3