Problem Set 6 Solutions - DSpace@MIT Home
MASSACHUSETTS INSTITUTE OF TECHNOLOGYDepartment of Electrical Engineering and Computer Science6.003: Signals and Systems—Fall 2003Problem Set 6 SolutionsIssued: October 21, 2003Due: October 29, 2003Exercise for home study:O&W 6.49Solution:(a) To determine the time constants of the differential equation P6.49-1 in O&W, weperform a Fourier transform of the equation. By linearity, the Fourier transform of theequation is the Fourier transform of each of the individual equation terms. This givesY (j )9 H(j ).2X(j )(j ) 11(j ) 10(1)The time constants are the zeros of the denominator (c1 1 and c2 10).(b) Equation 1 can be rewritten asH(j ) 9.(j 1)(j 10)(2)To make this a parallel interconnection of two first order systems, we do a partialfraction expansion on Equation 2 to find 11 .(3)j 1 j 10By linearity, the inverse Fourier transform, h(t), is a parallel interconnection of twofirst order systems. h(t) is the sum of the inverse Fourier transform of each term inEquation 3. Each term can be quickly determined using Table 4.2 of O&W. Thus,H(j ) h(t) e t u(t) e 10t u(t).(4)(c) The dominant time constant in a system with multiple decaying exponentials is thetime constant that takes the longest to die out. For this problem, the dominant timeconstant is c1 1. We can approximate this to Equation 1 as shown:H(j ) (j )219 . 11(j ) 10j 11(5)
This approximation has a maximum difference of 10% at 0 and the differencedecreases as increases.(d) We want to approximate the faster component of Equation 4 as an impulse with aheight equal to its final value. We will then check to see how this approximationaffects the overall h(t) of the entire second order system.The faster component, hf (t) e 10t u(t). Let ĥf (t) denote the approximation. Forĥf (t) sf ( ) (t), we need to determine sf ( ). The step response is defined as ts(t) h(t)dt.Thus,sf ( ) hf (t)dt 0 e 10t dt 1.101 (t) and the approximation to the overall system isĥf (t) 10ˆ hˆ f (t) hs (t) 1 (t) e t u(t).h(t)10(6)Equation 6 enables us to determine the frequency response. We determine:Ĥ(j ) 0.1 0.1j 0.91. j 1j 1(7)We then do algebraic manipulations and an inverse Fourier transform to Equation 7to get the differential equation of the approximation. This gives,dydx y(t) 0.9x(t) 0.1 .dtdtThe original frequency response, H(j ) and the approximate frequency responseĤ(j )are shown below:2
0.90.90.80.80.70.70.60.6 Happrox(j ) H(j ) Exercise: Problem 6.49 This is the approximate to H(j ) 1Exercise: Problem 6.49 H(j ) 10.50.50.40.40.30.30.20.20.10.100204060frequency ( )8001000204060frequency ( )80100The two are shown on the same plot below. The plot of H(j ) is solid and the plot ofĤ(j ) is dashed.3
Exercise: Problem 6.49 Both H(j ) and Happrox(j 1Magnification of H(j ) and Happrox(j at low frequencies1 H(j ) Happrox(j ) 0.9 H(j ) Happrox(j ) 0.90.80.8 H(j ) and Happrox(j ) H(j ) and Happrox(j ) 0.70.60.50.40.30.70.60.50.20.40.100102030frequency ( )400.35000.511.5frequency ( )2ˆFrom the graphs above, it is visible that for 1, H(j )approximates H(j ) well.As increases beyond 1, the two diverge from each other until roughly 20. Thisis due to the fact that the approximation to the fast part of the signal reaches its finalvalue instantaneously, hence it must include all fast frequency components.We can also compare the step response of the original signal, s(t) with the step responseof the approximate signal, ŝ(t).4
s(t) t h(t)dt to t(8)e dt te 10t dt (0.9 e t 0.1e 10t )u(t).ŝ(t) t ˆh(t)dt to te dt (10)(11) t( 0.1) (t)dt(12)0 (0.9 e t )u(t).The two step responses are plotted together below:5(9)0(13)
Exercise: Problem 6.49 Both s(t) and sapprox)t)0.40.90.350.80.30.70.250.60.2s(t) and sapprox(t)s(t) and sapprox(t)10.50.4s(t)s(t)Magnification of s(t) and 1 0.0500246time (sec)810 0.100.10.20.3time (sec)0.40.5From the graphs above, it is apparent that for t .5 seconds, the two step responsesare nearly equal. Furthermore, a general rule of thumb is that an exponential nearlyreaches its final value after 5 time constants. Thus, for e 10t u(t), it should nearly reachits final value by t 0.5 seconds.Problems to be turned in:Problem 1 O&W 4.35Solution:6
(a) We need to find the magnitude, the phase and the impulse response of the continuoustime LTI system with frequency responseH(j ) a j .a j Determining the magnitude, a2 2 1. H(j ) a2 2Determining the phase, e j arctan ae j arctan e 2j arctan a .aeDetermining the impulse response, a j 1 1 1h(t) F {H(j )} F F h1 (t) h2 (t).a j a j j H(j )From table 4.2, h1 (t) ae at u(t). To determine h2 (t), we do the following:h2 (t) F 1 { j 11 dh1 (t).} F 1 { (j )H1 (j )} aa dta j Using the product rule,dh1 (t) ae at (t) a2 e at u(t).dtThus, h2 (t) e at (t) ae at u(t). Combining h1 (t) and h2 (t) gives,h(t) 2ae at u(t) e at (t).Alternatively we can do a partial fraction expansion of H(j ):H(j ) 2a 1.j aNow, h(t) can be found by looking up each term above in the tables to give the sameresult h(t) 2ae at u(t) e at (t). (b) We need to determine y(t) when a 1 and x(t) cos(t/ 3) cos(t) cos( 3t). Wecan use the convolution property, Y (j ) X(j ) H(j ). We have H(j ), we needto determine X(j ).11X(j ) [ ( ) ( ) ( 1) ( 1)33 ( 3) ( 3)] [Xa (j ) Xb (j ) Xc (j )].7(14)(15)
Because the system is LTI, we can write the output as the sum of three terms,Y (j ) Ya (j ) Yb (j ) Yc (j ),(16)where the first term is defined asYa (j ) H(j )Xa (j ) a j Xa (j )a j e 2j tan 1 a(17)(18) Xa (j )(19)11 e 2j arctan ( ) e 2j arctan ( ).33 a a(20)We now substitue a 1 into Ya (j ). Also, because Xa (j ) is the sum of two deltafunctions, we can substitute 13 into the arctangent term in the first term ofYa (j ) and 13 into the arctangent term for the second term of Ya (j ). Thisgives, 11j 3 e .Ya (j ) e33This is now recognizable as a cosine function in the time domain with a phase changeof 3 . Hence, 1 .ya (t) cos t 33 j 3 The other terms, Yb (j ) and Yc (j ) can be solved similarly to give cosine functionswith various phase changes. We get, Yb (j ) e j 2 ( 1) ej 2 ( 1). yb (t) cos t sin(t).2 2 2 Yc (j ) e j 3 ( 3) ej 3 ( 3). 2 yc (t) cos3t .3(21)(22)(23)(24)Summing all the terms,y(t) cos 1 t 33 sin(t) cos 2 3t 3 .These two functions, y(t) and x(t) are plotted together below. They are nonperiodicfunctions. Why are they nonperiodic? Because the three periods that sum to make8
up the function are 3 different irrational numbers and hence we will find no rationalnumber, T , where each term will have an integer number of waveforms in that T . Theperiod of the first term is Ta 2 3,, the period of the second term is Tb 2 , andthe period of the last term is Tc 2 13 .Problem 1 Both x(t) and y(t)3x(t)y(t)2x(t) and y(t)10 1 2 3 10 8 6 4 202time (pi*seconds)46810Problem 2 O&W 6.28 (a)-(iii) and (v)Solution:16(a)-(iii) To sketch the frequency response of H(j ) (j 2)4 , we recognize that this is a cascadeof 4 identical first order systems. It’s magnitude, H(j ) can be written as, H(j ) H1 (j )4 ( ( 1111)()()() 0.5j 1 0.5j 1 0.5j 1 0.5j 11111)( )( )( )2221 0.25 1 0.25 1 0.25 21 0.25 Because the system is LTI, the bode plot of the cascade of the 4 identical first ordersystems is equal to the sum of the bode plots for each first order system. The firstorder system, 20 log H1 (j ) 10 log(1 0.25 2 ) has the plot:2 10 log(1 0.25 ) 90for 2 20 log(0.5 ) for 2(25)
Therefore the bode plot of the cascade of the 4 first order systems has the plot: H(j ) 0for 2 80 log(0.5 ) for 2(26)The following graph illustrates the plot. The two straight-line approximations intersectat 2. The solid line is the approximation to H(j ) and the dashed line is the real H(j ) .Problem 2 Bode plot of approximate and real H(j ) H H(j ) 50 H(j ) and Happrox(j ) (dB)(j ) approx0 100 150 101010frequency ( )121010We can see from the graph and by calculation that at 2, the actual magnitude is H(j ) 40 log(1 0.25 22 ) 40 log(2) 12 dB.The phase is next determined and sketched. Again expressing the function as thecascade of 4 identical first order functions we have,ej H(j ) ej tan (0) 1 ej tan 2 1 ej tan (0) 1 ej tan 2 1 ej tan (0) 1 ej tan 2 1 ej tan (0) 1 ej tan 2 1 e 4j tan 1 2.To plot this we can plot each of the first order phase plots and then sum them to getthe final fourth order plot. For the first order phase plot and making straight lineapproximations, we have for 0.20 (log( 2 ) 1) for 0.2 20 H(j ) (27) 4 2for 2010
By summing this plot with itself four times, we get the overall phase plot characteristics, 0for 0.2 (log( 2 ) 1) for 0.2 20(28) H(j ) 2 for 20The approximation is plotted along with the exact function in the graph below. Theexact function, H(j ) is shown as a dashed line and the approximation is shown asa solid line.0.5Problem 2 Bode plot of H(j ) approximation and exact function Happrox(j ) H(j ) H(j ) and Happrox(j ) in units of radians/ 0 0.5 1 1.5 2 2.5 101010frequency ( )110210(a)-(v) To sketch the Bode plots, we again break the function down into its first order systems,find the bode plot of each first order system and then add the two plots together toget the overall plot. We need to determine the magnitude and phase of the function:j 10 11 j The magnitude is found by dividing the magnitude of the numerator with the magni tude of the denominator:H(j ) 0.01 2 1 H(j ) 2 1The logarithm of H(j ) is then:20 log H(j ) 10 log (0.01 2 1) 10 log ( 2 1)11(29)
20 log H1 (j ) 20 log H2 (j ) To plot 20 log H(j ) we determine the bode plot of each term in Equation 29 andthen sum the two plots to get the bode plot for the magnitude of H(j ) . 10 log 1 0for 1020 log H1 (j ) (30)20 log 20 log 10 for 1020 log H2 (j ) 10 log 1 0 for 1 20 log for 1(31)The bode plots for the two terms are shown below separately. Below the two separateplots is a plot of the addition of the two plots to give the bode plot for the magnitudeof H(j ).Prob 2(v) Bode plot of approximate and exact H (j ) Bode plot of approximate and exact H (j ) 12525 H1approx(j ) H1(j ) H2approx(j ) H2(j ) 020 H2(j ) and H2approx(j ) (dB) H1(j ) and H1approx(j ) (dB) 515105 10 15 20 25 300 35 5 110011010frequency ( )21012 40 110011010frequency ( )210
Bode plot of approximate and exact H(j ) 5 H(j ) approx H j ) ( 5( H j ) and Happrox(j ) (dB)0 10 15 20 25 1100110frequency ( )21010To plot the phase of H(j ), we plot the approximate phase of the numerator of H(j )and we plot the approximate phase of the denominator of H(j ) and we add the twoplots together to get the phase plot of H(j ). For the numerator, we have Hnum (j ) 4 (log( 10) 2 1) 2for 1for 1 100for 100(32)The approximation is plotted along with the exact function in the graph below.1.1Problem 2 Bode plot of numerator of H(j ) (exact and approximate) H(j )num approx H (j )num0.90.80.70.6 Hnum(j ) and Hnum approx(j ) in units of radians/ 10.50.4 110011010frequency ( )13210310
For the denominator, we have Hden (j ) 0 4 (log( ) 2for .1 1) for .1 10for 10(33)The approximation is plotted along with the exact function in the graph below.Problem 2 Bode plot of denominator of H(j ) (exact and approximate) H(j )den approx H (j )0.2den Hden(j ) and Hden approx(j ) in units of radians/ 0.10 0.1 0.2 0.3 0.4 0.5 0.6 0.7 210 1100101frequency ( )10210310The sum of the phases for the numerator and the denominator gives the phase of theoverall function H(j ) and is shown below:14
Problem 2 Bode plot of H(j ) (exact and approximate) H(j )approx H(j ) H(j ) and Happrox(j ) in units of radians/ 10.80.60.40.20 210 1100101frequency ( )10210310Problem 3 O&W 6.32 (b)Solution:The system of Figure P6.32 consists of a cascade of a compensator with a a system with1the frequency response, H2 (j ) j 50. We want the log magnitude of the cascade of thecompensator with H2 (j ) to have the following specs:(a) It should have a slope of 20 dB/decade for 0 10.(b) It should be between 10 and 30 dB for 10 100.(c) It should have a slope of -20 dB/decade for 100 1000.(d) It should have a slope of -40 dB/decade for 1000.To get those specs we need to sum the bode plot of H2 (j ) with the bode plot of thecompensator, C(j ).The bode plot of the magnitude of H2 (j ) is as shown below: 20 log 50 for 5020 log H2 (j ) (34) 20 log for 50The approximation is plotted along with the exact function in the graph below.15
Bode plot of approximate and exact H (j ) for Problem 32 30 H(j ) 2approx H (j ) 2 H2(j ) and H2 approx(j ) (dB) 35 40 45 50 55 60 650110102frequency ( )10310The above bode plot needs to be cascaded with some compensator function, C(j ) togive us the specs listed above. We will determine C(j ) to be some cascaded combination1of a constant and one or more of the four frequency responses, Ha (j ) j , Hb (j ) j ,1Hc (j ) 1 j k and Hd (j ) (1 j ωk ) with appropriate ωk ’s selected.Ck. ByTo meet the first and second spec, we need Ha (j ) j and Hc (j ) 1 j kselecting ωk 0.1, there will be 20 dB/decade for 10 due to Ha (j ) and there will be 0dB/decade for 10 due to the cascade of Ha (j ) and Hc (j ). Ha (j ) will meet the firstspec because it is a continuous slope of 20 dB/decade and Hc (j ) will ‘turn off’ the 20dB/decade at an appropriate and make the slope 0 for frequencies greater than that .For the third spec, we don’t need to add anything since H2 (j ) will give us -20 dB/decadefor 50. However we need to cascade another Hc (j ) to get -40 dB/decade for 1000.Choosing ωk 0.001 for Hc (j ) will cause the slope change from -20 dB/decade to -40dB/decade to occur at 1000.Thus far we have the following frequency response for the compensator:C(j ) Ck j (1 j0.1 )(1 j0.001 )The last part is to determine the constant, Ck of the compensator. Because 20 log H2 (j ) 20 log 50 34 dB for 50 and because the second spec requires the overall logmagnitude to be between 10 dB and 30 dB for 10 100 we select the log magnitudeof C(j ) at 10 to add to the -34 dB and give 29.9 dB. We choose to be close to the16
upper bound of the spec (close to 30 dB as opposed to somewhere in the middle of 10dB and 30 dB) because at 50, H2 (j ) begins to decrease at a slope of -20 dB/decade.Thus, we algebraically determine 34 20 log Ck 20 log 10 29.9yields Ck 156.3. The final compensator frequency response isC(j ) 156.3j .(1 j0.1 )(1 j0.001 )The final frequency response for H(j ) isH(j ) 156.3j .(1 j0.1 )(50 j )(1 j0.001 )The final frequency response for H(j ) is plotted as an approximation and with the exactfunction in the graph below.Bode plot of approximate and exact H(j ) for Problem 3 H(j ) approx30 H(j ) H(j ) and Happrox(j ) (dB)20100 10 20 30 40 1100110102frequency ( )31010Problem 4 O&W 6.39 (j)Solution:We need to sketch the log magnitude and phase of the the function:H(ej ) 1(1 0.25e j )(117 0.75e j ).410
We can write this as a cascade of two frequency responses, H1 (ej ) 1H2 (ej ) (1 0.75e j ) . Then1(1 0.25e j )and20 log H(ej ) 20 log H1 (ej ) 20 log H2 (ej ) and H(ej ) H1 (ej ) H2 (ej )We need to determine 20 log H1 (ej ) and 20 log H2 (ej ) . Using complex number manipu lation, we get,20 log H1 (ej ) 10 log ((1 0.25 cos )2 0.252 sin2 )and20 log H2 (ej ) 10 log ((1 0.75 cos )2 0.752 sin2 )The two bode plots were calculated using Matlab and are shown below separately. Note thedifferent ordinate axis scales for each frequency response. The magnitude of H1 (ej ) is muchless than the magnitude of H2 (ej ). Note also that because the coefficent in the denominatoris positive ( 0.25) for H1 (ej ) whereas it is negative ( 0.75) for H2 (ej ), their peak valuesare shifted by from each other. H1 (ej ) has more low frequency components and H2 (ej )has more high frequency components.j j 20log H (e ) Prob. 4 (O&W 6.39(j))320log H (e ) 1142122108dB20log H2(e ) 1j j 20log H (e ) dB10 16420 2 2 4 3 2 101frequency normalized by ( / ) 6 22 101frequency normalized by ( / )2The two functions are shown summed together to give the overall log magnitude of H(ej ):18
j 20log H(e ) for Problem 4 (O&W 6.39(j))1210j 20log H(e ) dB86420 2 4 2 1.5 1 0.500.5frequency normalized by ( / )11.52To determine H(ej ), we find and plot H1 (ej ) and H2 (ej ) and then sum the twofunctions to get H(ej ). 10.25 sin 1j tan H1 (e ) 1 0.25e j 1 0.25 cos andj H2 (e ) 11 0.75e j tan 1 0.75 sin 1 0.75 cos .Each of these phases are plot separately and summed together to give the overall phase ofH(ej ).19
j j H e H e10.120.30.080.2 H2(e normalized by (radians/ )0.040.020 0.020.10 0.1j H1(ej normalized by (radians/ )0.06 0.04 0.2 0.06 0.3 0.08 0.1 2 101frequency normalized by ( / ) 0.4 22 101frequency normalized by ( / )2 Hej 0.250.20.10.050 0.05j H(e normalized by (radians/ )0.15 0.1 0.15 0.2 0.25 2 1.5 1 0.500.5frequency normalized by ( / )11.52Problem 5 O&W 6.43 (a)Solution:We need to transform the low pass filter, Hlp (ej ) shown in Figure P6.43 of O&W into ahigh pass filter, Hlp (ej ). In this problem we accomplish this by modulating hlp [n]. Specifi 20
cally, hhp [n] ( 1)n hlp [n]. This is equivalent to shifting the frequency response by sinceFfrom Table 5.1 of O&W we have ej o n x[n] X(ej( o ) ) and we know that ej n ( 1)n .Thus,Hhp (ej ) Hlp ej( ) .The frequency response, Hhp (ej ) is shown below superimposed onto the original Hlp (ej ):Hlp(ej ), Hhp(ej )Hhp(ej )Hlp(ej ) 2 2 Because a discrete signal is periodic with period 2 , its low frequency componentsoccur periodically at 0, 2 , 4 , etc. Thus, its high frequency components occur halfwaybetween the low frequency components at , 3 , etc. The shifted frequency responsecorresponds to a high pass filter since the original signal had its peak magnitude centeredaround 0, 2 , etc. and the new signal is shifted by from that original signal.Problem 6 O&W 6.58 (a), (b)Solution:21
(a)1. We need to find h1 [n] as a function of h[n]. To accomplish this we investigateH1 (ej ). From Figure P6.58 of O&W, we can writeG(ej ),X(ej )R(ej )H(ej ) G(e j )R(e j ) S(ej ).H(ej ) (35)(36)(37)We can manipulate the equations to write,R(e j ) G(ej )H(e j ) X(ej )H(ej )H(e j ).(38)(39)Thus,S(ej )X(ej ) H(ej )H(e j ).H1 (ej ) (40)(41)From this function, we recognize by the convolution property, h1 [n] h[n] h[ n].We can show that the system has zero phase characteristic by noting that sinceh[n] is real, H(e j ) H (ej ). Thus,H1 (ej ) H(ej )H(e j ) H(ej )H (ej ) H(ej ) 2 .2. Because H1 (ej ) H(ej ) 2 , then the magnitude H1 (ej ) H(ej ) 2 and thephase H1 (ej ) 0.(b)1. We look at the Fourier transforms to determine h2 [n]. Since y[n] g[n] r[ n],Y (ej ) G(ej ) R(e j ). From Figure P6.58 of O&W we determineY (ej ) G(ej ) R(e j ) H(ej )X(ej ) H(e j )X(ej )Thus,H2 (ej ) H(ej ) H(e j ) h2 [n] h[n] h[ n].To determine if H2 (ej ) has zero phase characteristic, we manipulate H2 (ej ) to get it into polar form. We use the property in Table 5.1 that since h[n] is real,then e{H(ej )} e{H(e j )} and m{H(ej )} m{H(e j )}. Hence, H2 (ej ) e{H(ej )} m{H(ej )} e{H(e j )} m{H(e j )} (42)j 2 e{H(e )}(43)j j 2 H(e ) cos ( H(e )).(44)From this, we can see that H2 (ej ) has zero phase characteristic.2. Since we already put H2 (ej ) in polar form, we know that H2 (ej ) 2 H(ej ) cos ( H(ej )).and we know that H2 (ej ) 0.22(45)
Problem 7 O&W 7.22Solution:In this problem we need to figure out a range of values for the sampling period, T, to recovery(t) completely from yp (t). To do this we need to determine the bandwidth of the originalY (j ) and use the sampling theorem. By the convolution property, Y (j ) X1 (j )X2 (j ).The bandwidth of Y (j ) then will be the bandwidth of the smaller of the two bandwidths,X1 (j ) or X2 (j ). Hence, Y (j ) 0 for 1000 . Then, using the sampling theorem,2 2 m 2(1000 ).TThis gives the range of T as 0 T 0.001 seconds. s Problem 8 O&W 7.23Solution:1X(j ) (a) We need to sketch Xp (j ) and Y (j ). In the frequency domain, Xp (j ) 2 P (j ). We need to determine P(j ). Since P(j ) is periodic, we need to use theperiodic Fourier transform formula. That isP (j ) 2 k ak ( k o ). Here, o 2 . We need to determine the ak ’s using the formula ak T1 T p(t)e jk o t .TA few are shown below: 2 1a0 ( (t) (t ))dt 02 0 2 111( (t) (t ))e j t dt (1 1 · e j ) a1 2 02 2 11a2 (1 1 · e j2 ) 0( (t) (t ))e j 2 t dt 2 02 2 111a3 (1 1 · e j3 ) ( (t) (t ))e j 3 t dt 2 02 1Thus, ak 0 for k even and ak for k odd andP (j ) 2 ( k ) (2k 1) k 2 koddFrom this Fourier transform for P (j ), we can sketch Xp (j ) as copies of X(j ) scaled . for all k. This can be seen in theby 1 and replicated at intervals of (2k 1) figure below:23
Xp(j )1/ 4 / 3 / 2 / / / 2 / 3 / 4 / H(j ) is a sum of two ideal unity gain bandpass filters. Thus, Y (j ) is the part ofXp (j ) that is passed through H(j ). This is shown below:Y(j )1/ 4 / 3 / 2 / / / 2 / 3 / 4 / (b) To recover x(t) from xp (t) we need to do two things. First, we need to multiply xp (t) with a cosine function,cos t. This will shift Xp (j ) such that one of the copies ofX(j ) is centered around 0. Second, we send the shifted signal through a lowpassfilter, R(j ), to eliminate the extra copies of X(j ). To achieve this we have a filter,and centered around 0. This is shown below:R(j ) with gain , bandwidth 2 R(j ) / / 2 / The overall system is shown below:cos( / t)xp(t)x24R(j )x(t)
(c) To recover x(t) from y(t) we need to run Y (j ) through two parallel filter systems. The top parallel path will multiply y (t) by cos t which will shift the demi-replicate of X(j ) that is centered at over to 0. The shifted signal then passes throughthe lowpass filter, R(j ) described above in part (b) to eliminate the extra copies.The bottom parallel path will multiply y (t) by cos 3 t which will shift the demi-replicate 3 of X(j ) that is centered at over to 0. The shifted signal then passesthrough the lowpass filter, R(j ) described above in part (b) to eliminate the extracopies. Thus, the two halves combine together to form a complete X(j ) and x(t) isrecovered. The overall system is shown below: cos( t/ )R(j )xy(t) x(t)R(j )xcos(3 t/ )(d) To recover x(t) from xp (t) and y(t), Xp (j ) cannot have any overlap in the copies of for allX(j ). Because of this particular p(t), the copies of X(j ) are at (2k 1) k. Thus, just looking at one interval to make sure the copies of X(j ) don’t overlap, we have one copy of X(j ) centered at and one copy of X(j ) centered at.(SeeFigureofX(j )above).Fornooverlapbetween these copies, 3 p 3 m m , which gives or max . m m25
Therefore the bode plot of the cascade of the 4 first order systems has the plot: H(j ) 0 for 2 (26) 80 log(0.5 ) for 2 The following graph illustrates the plot. The two straight-line approximations intersect at 2. The solid line is the approximation to H(j ) an
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SOLUTIONS TO PROBLEM SET 4 MAT 141 Abstract. These are the solutions to Problem Set 4 for the Euclidean and Non-Euclidean Geometry Course in the Winter Quarter 2020. The problems were posted online on Saturday Feb 15 and due
Accessing a solution in CS50 Vault to some problem prior to (re-)submitting your own. Asking a classmate to see his or her solution to a problem set’s problem before (re-)submitting your own. Decompiling, deobfuscating, or disassembling the staff’s solutions to problem sets.
