MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems-Fall 2003 PROBLEM SET 6 SOLUTIONS Issued: October 21. 2003 Due: October 29 2003 Exercise for home study O&W6.49 Solution: (a)To determine the time constants of the differential equation P649-1 in O&w,we perform a Fourier transform of the equation. By linearity, the Fourier transform of the quation is the Fourier transform of each of the individual equation terms. This gives The time constants are the zeros of the denominator(1=-1 and c2=-10) (b)Equation 1 can be rewritten as 9 To make this a parallel interconnection of two first order systems, we do a partial fraction expansion on Equation 2 to find u+1 By linearity, the inverse Fourier transform, h(t), is a parallel interconnection of two first order systems. h(t) is the sum of the inverse Fourier transform of each term in Equation 3. Each term can be quickly determined using Table 4.2 of O&W. Thus h(t)=eu(t)-eu(t (c)The dominant time constant in a system with multiple decaying exponentials is the time constant that takes the longest to die out. For this problem, the dominant time constant is C1=-1. We can approximate this to Equation 1 as shown: HGu) (ju)2+110ju)+10
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 6.003: Signals and Systems—Fall 2003 Problem Set 6 Solutions Issued: October 21, 2003 Due: October 29, 2003 Exercise for home study: O&W 6.49 Solution: (a) To determine the time constants of the differential equation P6.49-1 in O&W, we perform a Fourier transform of the equation. By linearity, the Fourier transform of the equation is the Fourier transform of each of the individual equation terms. This gives Y (j�) X(j�) = 9 (j�)2 + 11(j�) + 10 = H(j�). (1) The time constants are the zeros of the denominator (c1 = −1 and c2 = −10). (b) Equation 1 can be rewritten as 9 H(j�) = . (2) (j� + 1)(j� + 10) To make this a parallel interconnection of two first order systems, we do a partial fraction expansion on Equation 2 to find 1 H(j�) = + −1 . (3) j� + 1 j� + 10 By linearity, the inverse Fourier transform, h(t), is a parallel interconnection of two first order systems. h(t) is the sum of the inverse Fourier transform of each term in Equation 3. Each term can be quickly determined using Table 4.2 of O&W. Thus, u(t) − e−10t h(t) = e−t u(t). (4) (c) The dominant time constant in a system with multiple decaying exponentials is the time constant that takes the longest to die out. For this problem, the dominant time constant is c1 = −1. We can approximate this to Equation 1 as shown: 9 1 H(j�) = . (5) (j�)2 + 11(j�) + 10 � j� + 1 1
This approximation has a maximum difference of 10% at w=0 and the difference (d)We want to approximate the faster component of Equation 4 as an impulse with a height equal to its final value. We will then check to see how this approximation affects the overall h(t)of the entire second order system The faster component, h(t)=-eotu(t). Let h,(t)denote the approximation. For h,(t)=s(oo)8(t), we need to determine sf(oo). The step response is defined as s(t)=_o h(t)dt Thus SfIO) h, (t)dt= h(t)=-oo(t) and the approximation to the overall system is h(t)=h(t)+hs(t)=-i(t)+e-u(t Equation 6 enables us to determine the frequency response. We determine B(元u)=-01+-1 0.1ju+0.9 jw+l Jw+1 We then do algebraic manipulations and an inverse Fourier transform to Equation 7 to get the differential equation of the approximation. This gives +y()=0.9(t)-0 The original frequency response, H(u) and the approximate frequency response H jw) are shown below
� This approximation has a maximum difference of 10% at � = 0 and the difference decreases as � increases. (d) We want to approximate the faster component of Equation 4 as an impulse with a height equal to its final value. We will then check to see how this approximation affects the overall h(t) of the entire second order system. The faster component, hf (t)= −e−10t u(t). Let hˆf (t) denote the approximation. For hˆf (t) = sf (∗)�(t), we need to determine sf (∗). The step response is defined as ⎩ t s(t) = h(t)dt. −� Thus, � � � 1 −e−10t sf (∗) = hf (t)dt = dt = − . −� 0 10 hˆf (t) = − 1 �(t) and the approximation to the overall system is 10 hˆ(t) = hˆf (t) + hs(t) = − 1 �(t) + e−t u(t). (6) 10 Equation 6 enables us to determine the frequency response. We determine: 1 Hˆ (j�) = −0.1 + = −0.1j� + 0.9 . (7) j� + 1 j� + 1 We then do algebraic manipulations and an inverse Fourier transform to Equation 7 to get the differential equation of the approximation. This gives, dy dx + y(t) = 0.9x(t) − 0.1 . dt dt The original frequency response, H(j�) and the approximate frequency responseHˆ (j�) are shown below: 2
Exercise: Problem 6.49-IH(joo) Exercise: Problem 6.49- This is the approximate to H(jo) requency(o) requency(oo) The two are shown on the same plot below. The plot of H(w) is solid and the plot of HGw)
Exercise: Problem 6.49− |H(j�)| Exercise: Problem 6.49− This is the approximate to |H(j�)| 1 1 0.9 0.9 0.8 0.8 0.7 0.7 0.6 0.6 0.4 0.4 0.3 0.3 0.2 0.2 0.1 0.1 0 0 20 40 60 80 frequency (�) 0 100 0 20 40 60 80 100 frequency (�) (j�)| |H |H(j approx �)| 0.5 0.5 The two are shown on the same plot below. The plot of H(j�) is solid and the plot of Hˆ (j�) is dashed. 3
Exercise: Problem 6.49-Both (Hjo)l and H( jol Magnification of (Hjo)) and h jol at low frequencies Hgo) Hol 09 08 0.7 0.6 3工E3 04 04 03 0.5 frequency (o) From the graphs above, it is visible that for w<1, Hw) approximates hw) well As w increases beyond 1, the two diverge from each other until roughly w= 20. This is due to the fact that the approximation to the fast part of the signal reaches its final value 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 response gnal, s(t)
Exercise: Problem 6.49− Both |H(j�)| and |Happrox(j�| Magnification of |H(j�)| and |Happrox(j�| at low frequencies 1 1 |H(j�)| |Happr (j�)| ox |H( |Ha j�)| pprox(j�)| 0.9 0.9 0.8 0.8 0.7 |H(j �)| and |Happrox(j�)| |H(j �)| and |Happrox(j�)| 0.6 0.5 0.4 0.3 0.7 0.6 0.5 0.2 0.4 0.1 0 0 10 20 30 40 50 frequency (�) 0.3 0 0.5 1 1.5 2 frequency (�) From ˆ 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. This is due to the fact that the approximation to the fast part of the signal reaches its final value 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 response of the approximate signal, sˆ(t). 4
h(t)dt dt (0.9-e-2+0.le-l0)u(t) (10) s(t) (t)dt dt-/(-0.1)6(t)dt (12) (0.9-e-2)a() The two step responses are plotted together below
� t s(t) = h(t)dt (8) �−� t � t = e−t dt − e−10t dt (9) o 0 = (0.9 − e−t + 0.1e−10t )u(t). (10) � t ˆ sˆ(t) = h(t)dt (11) �−� t � t = e−t dt − (−0.1)�(t)dt (12) o 0 = (0.9 − e−t )u(t). (13) The two step responses are plotted together below: 5
EXercise: Problem 6.49-Both s(t)and s)t) Magnification of s(t)and s nnnn(t) 04 0.9 0.35 0.8 03 025 02 0.5 0.15 04 s(t) 0.1 0 10 00.10.20.30405 it is apparent that for t>5 Is. the two ster are nearly equal. Furthermore, a general rule of thumb is that an exponential nearly reaches its final value after 5 time constants. Thus, for e-lou(t), it should nearly reach its final value by t=0.5 seconds. to be ti d i Problem 1 o&W 4.35
Exercise: Problem 6.49− Both s(t) and sapprox)t) Magnification of s(t) and sapprox(t) 1 0.4 0.5 0.6 0.7 0.8 0.9 s(t) and sapprox(t) s(t) s approx(t) 0.4 s(t) and sapprox(t) 0.35 0.3 0.25 0.2 0.15 0.1 0.3 0.05 0.2 0 0.1 −0.05 0 0 2 4 6 8 time (sec) −0.1 10 0 0.1 0.2 0.3 0.4 0.5 time (sec) From the graphs above, it is apparent that for t > .5 seconds, the two step responses are nearly equal. Furthermore, a general rule of thumb is that an exponential nearly reaches its final value after 5 time constants. Thus, for e−10t u(t), it should nearly reach its final value by t = 0.5 seconds. Problems to be turned in: Problem 1 O&W 4.35 s(t) s a (t) pprox Solution: 6
(a)We need to find the magnitude, the phase and the impulse response of the continuous- time LTI system with frequency response H a-Jw + w Determining the magnitude, Determining the phase <Hw e J arctan 4 e-2j arctan a Determining the impulse response h()=x-1{(u)}=x--1+x1/二 la+ ju h1(t)+h2() a+1 From table 4.2, h1(t)=aeau(t). To determine h2(t), we do the following h2(t) }=-{--(ju)H1(ju)}= 1 dhi(t) a+w a dt Using the product rule d(t)-aeu(t Thus, h2(t)=eas(t)+ae -atu(t). Combining h1(t) and h2(t) gives h(t)=2ae-au(t)-ea8(t Alternatively we can do a partial fraction expansion of H(w 2a H +a Now, h(t) can be found by looking up each term above in the tables to give the same result h(t)=2ae-atu(teas(t) (b) We need to determine y(t) when a= l and r(t)=cos(t/v3)+cos(t)+cos(v3t)We can use the convolution property, Y(w)=XGw)x HGu). We have H(w), we need to determine X(w) X(ju)=x[6( +6( )+6(u-1)+6(u+1) u-√3)+6(u+√3 (14) [ Xau)+ Xb(jw)+X(w
� � � � (a) We need to find the magnitude, the phase and the impulse response of the continuoustime LTI system with frequency response H(j�) = a − j�. a + j� Determining the magnitude, �a2 + �2 | | H(j�) = � = 1. a2 + �2 Determining the phase, e−j arctan � a e = e j<H(j�) = e−2j arctan � j arctan � a . a Determining the impulse response, + F −1 −j� h(t) = F −1 {H(j�)} = F −1 a = h1(t) + h2(t). a + j� a + j� From table 4.2, h1(t) = ae−atu(t). To determine h2(t), we do the following: { −j� 1 1 dh1(t) h2(t) = F −1 (j�)H1(j�)} = . a + j�} = F −1 {− a −a dt Using the product rule, dh1(t) = ae−at�(t) − a2 e−atu(t). dt Thus, h2(t) = −e−at�(t) + ae−atu(t). Combining h1(t) and h2(t) gives, u(t) − e−at h(t) = 2ae �(t). −at Alternatively we can do a partial fraction expansion of H(j�): 2a H(j�) = j� + a − 1. Now, h(t) can be found by looking up each term above in the tables to give the same u(t) − e−at result h(t) = 2ae �(t). −at (b) We need to determine y(t) when a = 1 and x(t) = cos(t/�3) + cos(t) + cos(�3t). We can use the convolution property, Y (j�) = X(j�) × H(j�). We have H(j�), we need to determine X(j�). 1 1 X(j�) = �[�(� − � ) + �(� + ) + �(� − 1) + �(� + 1) 3 �3 +�(� − � 3) + �(� + � 3)] (14) = �[Xa(j�) + Xb(j�) + Xc(j�)]. (15) 7
Because the system is Lti, we can write the output as the sum of three terms Y(u)= Ya(ju)+yb(u)+Ylw), (16) where the first term is defined as Yaw)= H(uAw (17) Saw 2tan-=丌Xa(ju) 2j arctan as(w We now substitue a= l into Ya(w). Also, because Xa(w) is the sum of two delta functions. we can substitute a into the arctangent term in the first d term of Yalu) and w=- into the arctangent term for the second 8 term of Ya(ju).This 3 +Ted This is now recognizable as a cosine function in the time domain with a phase change of 3.Hence ya(t)=cos The other terms, Yb Gw) and Y(w) can be solved similarly to give cosine functions with various phase changes. We get Y6(ju)=丌e-6(u-1)+re6(u+1) T os vat ng all the t) sin(t)+cos These two functions, y(t)and r(t) are plotted together below. They are nonperiodic functions. Why are they nonperiodic? Because the three periods that sum to make
� � � � � � � � � � �� � � � � � � 3 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 as Ya(j�) = H(j�)Xa(j�) (17) a − j� = �Xa(j�) (18) a + j� = e−2j tan−1 a �Xa(j�) (19) 1 = �e−2j arctan � � 1 3 ) + �e−2j arctan � a �(� − a �(� + ) � . (20) 3 We now substitue a = 1 into Ya(j�). Also, because Xa(j�) is the sum of two delta functions, we can substitute � = � 1 3 into the arctangent term in the first � term of Ya(j�) and � = 1 � into the arctangent term for the second � term of Ya(j�). This 3 − gives, Ya(j�) = �e−j � � 3 3 � � + 1 . 3 � � − 1 + �ej � � 3 This is now recognizable as a cosine function in the time domain with a phase change of � . Hence, 1 � ya(t) = cos � t − . 3 3 The other terms, Yb(j�) and Yc(j�) can be solved similarly to give cosine functions with various phase changes. We get, j � Yb(j�) = �e−j � 2 �(� − 1) + �e 2 �(� + 1). (21) yb(t) = cos t − = sin(t). (22) 2 j 2� Y 3 3 c(j�) = �e−j 2� �(� − � 3) + �e �(� + � 3). (23) 2� yc(t) = cos � 3t − . (24) 3 Summing all the terms, 1 � 2� y(t) = cos t − + sin(t) + cos � � 3t − . 3 3 3 These two functions, y(t) and x(t) are plotted together below. They are nonperiodic functions. Why are they nonperiodic? Because the three periods that sum to make 8
up the function are 3 different irrational numbers and hence we will find no rational number, T, where each term will have an integer number of waveforms in that T. The period of the first term is Ta =2T 3, the period of the second term is T=2T,and the period of the last term is Tc= 2o oblem 1- Both x(t) and y(t) Problem 2 O&W 6.28(a)-(ii)and(v) Solution (a)-(ii) To sketch the frequency response of H (w)=ttgr, we recognize that this is a cascade of 4 identical first order systems. It's magnitude, H(w)l can be written as JHGu)|=|( I 0.5ju+10.5ju+10.5 0.5 √1+0.252√1+0.2521+0.25 -0. Because the system is lti, the bode plot of the cascade of the 4 identical first order systems is equal to the sum of the bode plots for each first order system. The first order system, 20 log H1()=-10 log(1 +0.25w 2)has the plot for 10log(1+0.2542)≈ 20 log(0.5w) for
� up the function are 3 different irrational numbers and hence we will find no rational number, T, where each term will have an integer number of waveforms in that T. The period of the first term is Ta = 2� �3,, the period of the second term is Tb = 2�, and the period of the last term is Tc = 2� 1 � . 3 Problem 1− Both x(t) and y(t) x(t) and y(t) 3 2 1 0 −1 −2 −3 x(t) y(t) −10 −8 −6 −4 −2 0 2 4 6 8 10 time (pi*seconds) Problem 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 cascade of 4 identical first order systems. It’s magnitude, | | H(j�) can be written as, 1 1 1 1 |H(j�) = H1(j�) 4 | | | | = ( )( )( )( ) 0.5j� + 1 0.5j� + 1 0.5j� + 1 0.5j� + 1 | 1 1 1 1 = (� )( )( )( ) 1 + 0.25�2 � 1 + 0.25�2 � 1 + 0.25�2 � 1 + 0.25�2 Because the system is LTI, the bode plot of the cascade of the 4 identical first order systems is equal to the sum of the bode plots for each first order system. The first order system, 20 log H1(j�) = −10 log(1 + 0.25�2 | | ) has the plot: 0 for � � 2 −10 log(1 + 0.25� (25) 2 ) � −20 log(0.5�) for � ≤ 2 9
Therefore the bode plot of the cascade of the 4 first order systems has the plot for w2 (26) The following graph illustrates the plot. The two straight-line approximations intersect at w=2. The solid line is the approximation to H(w) and the dashed line is the real Problem 2- Bode plot of approximate and real [H(jo) We can see from the graph and by calculation that at w= 2, the actual magnitude is log(1 2)=-40log(2)=-12dB. The phase is next determined and sketched. Again expressing the function as the cascade of 4 identical first order functions we have ∠H(ju) (=)()(-)(= To plot th can plot each of the first order phase plots and then sum them to get the final plot. For the first order phase plot and making straight line for w 20
� 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�) and the dashed line is the real | | H(j�) . Problem 2− Bode plot of approximate and real |H(j�)| |H(j�)| and |Happrox(j�)| (dB) 0 −50 −100 −150 |Happrox(j�)| |H(j�)| 10−1 100 101 102 frequency (�) We 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 the cascade of 4 identical first order functions we have, ⎧ ⎨ ⎧ ⎨ ⎧ ⎨ ⎧ ⎨ j�H(j�) e = ej tan−1 (0) ej tan−1 (0) ej tan−1 (0) ej tan−1 (0) ej tan−1 � 2 ej tan−1 � 2 ej tan−1 � 2 ej tan−1 � 2 = e−4j tan−1 � 2 . To plot this we can plot each of the first order phase plots and then sum them to get the final fourth order plot. For the first order phase plot and making straight line approximations, we have � �H(j�) � � � 0 −� 4 (log(� 2 ) + 1) −� 2 for � √ 0.2 for 0.2 < � < 20 for � → 20 (27) 10