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