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18Spring_hw08_soln.pdf - University of Illinois Spring 2018...

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Unformatted text preview: University of Illinois Spring 2018 ECE210 / ECE211 - Homework 08 Solution 1. The function f (t) is periodic with period T = 4s. Between t=0 and 4s, the function is described by: −2, 0 < t < 1s f (t) = −1, 1 < t < 3s 1, 3 < t < 4s (a) Plot f (t) between t = −5s and t = 7s. (b) Determine the exponential Fourier coecients Fn of f (t) for n = 0, n = ±1s, and n = ±2s. (c) Using the result of part(b), determine the compact-form Fourier coecients C0 , C1 and C2 . Solution: (a) (b) Fn = 1 T Z f (t)e−jnw0 t dt T 1 Z Z Z π π 1 1 3 1 4 −j π nt −2e−j 2 nt dt + −1e−j 2 nt dt + 1e 2 dt 4 0 4 1 4 3 π 3 π 3 2j 1 1 1 = [ (−e−j 2 n + 1) − (e−j 2 πn − e−j 2 n ) + (e−j2πn − e−j 2 πn )] nπ 2 4 4 = Then plug in the number for coecient F0 = F1 = F−1 = F2 = F−2 = Z Z Z 1 1 1 3 1 4 3 −2dt + −1dt + 1dt = − 4 0 4 1 4 3 4 2j 1 1 1 1 + 3j [ (j + 1) − (j + j) + (1 − j)] = π 2 4 4 2π 1 1 1 − 3j 2j 1 [ (−j + 1) − (−j − j) + (1 + j)] = −π 2 4 4 2π 2j 1 1 1 3j [ (1 + 1) − (−1 + 1) + (1 + 1)] = 2π 2 4 4 2π 2j 1 1 1 3j [ (1 + 1) − (−1 + 1) + (1 + 1)] = − −2π 2 4 4 2π Page 1 of 6 (c) The formula to compact-form Fourier coecents is Cn = 2 |Fn |, where n 6= 0. √ 10 π C1 = 2 |F1 | = C2 = 2 |F2 | = 3 π When n = 0, C0 = 2F0 = − 3 2 2. Consider an LTI system whose frequency response is Z +∞ h(t) e−jωt dt = H(ω) = −∞ sin(4 ω) πω If the input to this system is a periodic signal ( f (t) = +1, 0 < t < 4s −1, 4 < t < 6s with period T = 6s. Determine the corresponding system output y(t) Solution: T = 6 and w0 = 2π T = π 3 Fn=0 Fn6=0 Since H(0) = π4 , y(t) = Z 4 6  1 1dt + (−1)dt = 3 0 4 Z 4  Z 6 1 j −j 4 πn 1e−jntπ/3 dt + (−1)e−jntπ/3 dt = = (e 3 − 1) 6 nπ 0 4 1 = 6 Z 3 sin( 43 πn) π 4 j −j 4 πn + Σ∞ (e 3 − 1) × × ejn 3 t n=−∞,n6=0 3π nπ nπ 2 3. Determine the Fourier series representations for the following signals: (a) A periodic signal x(t) with period of T = 2s and x(t) = e−t f or0 < t < 2s Solution: 1 Fn = 2 Z 1 = 2 Z =− 2 e−t e−jnπt dt 0 2 e−(jnπ+1)t dt 0 1 (e−2 − 1) 2(1 + jnπ) (b) A periodic signal x(t) with period T = 1s and ( x(t) = cos(π t), 0 ≤ t ≤ 1s 0, else Page 2 of 6 Solution: 1 1 jπt (e + e−jπt )e−jn2πt dt 2 Z 1 Z Fn = 0 = 1 2 (ejπt(1−2n) + e−jπt(1+2n) )dt 0   1 1 1 1 jπ(1−2n)t −jπ(1+2n)t = e + (− )e 2 jπ(1 − 2n) jπ(1 + 2n) 0 ejπ(1−2n) − 1 e−jπ(1+2n) − 1 + 2jπ(1 − 2n) −2jπ(1 + 2n) 4jn = π (1 − 4n2 ) = (c) Solution: From the plot, we get T = 6 and w0 = F0 = 1 6 Z 2π 6 = π 3 6 f (t)dt = 1.5 0 Z Z πn 1 3 −j πn t 1 6 3 Fn = te dt + (6 − t) e−j 3 t dt 6 0 6 3 (−1)n (9jπn − 9(−1)n + 9) −9 + (−1)n (9 − 9jπn) = + 6π 2 n2 6π 2 n2 3 = − 2 2 ((−1)n − 1)2 2π n 4. Let t + 2, 1, x(t) = −t + 2, 0, −2 < t ≤ −1, −1 ≤ t ≤ 1, 1 < t ≤ 2, 2 < t ≤ 4. be a periodic signal with fundamental period T = 6s and Fourier coecients ak . (a) Determine the value of a0 . (b) Determine the Fourier series representation of dx(t) dt Page 3 of 6 (c) Use the result of part (b) and the dierential property of the Fourier series to help determine the Fourier series coecients of x(t). Solution: (a) F0 = 1 6 Z −1 (t + 2)dt + −2 1 6 Z 1 1dt + −1 1 6 2 Z (2 − t) dt + 1 1 6 Z 4 0 dt = 2 1 2 a0 = 2F0 = 1 (b) Let Gn be the Fourier coecients of Gn6=0 = 1 6 dx dt , −1 Z π e−jn 3 t dt − −2 1 6 Z 2 π 1e−jn 3 t dt 1 2 π π π je−j 3 πn jej 3 n = (−ejn 3 + 1) − (−ej 3 n + 1) 2nπ 2nπ j πn 2πn = (cos( ) − cos( )) nπ 3 3 Z Z 1 2 1 −1 1dt − 1dt = 0 Gn=0 = 6 −2 6 1 (c) Using the dierential property Gn jnωo 3 Gn = πnj j πn 2πn 3 × (cos( ) − cos( )) = πnj nπ 3 3 3 πn 2πn = 2 2 (cos( ) − cos( )) π n 3 3 Fn6=0 = When n = 0, F0 = 1 2 5. Let the signal f (t) = cos3 (t) be the input of an LTI system with frequency response H(ω) = 2 e−jωπ/2 for ω[−2, 2] rad/s and zero elsewhere. Obtain the steady-state response y(t) of the system to the input f (t). Solution : f (t) = cos3 (t) = ( = ejt + e−jt 3 ) 2 1 j3t (e + e−j3t + 3ejt + 3e−jt ) 8 Evaluate the H(w) at w = 1, −1, we get H(1) = −2j H(−1) = 2j 3 jt 3 e × (−2j) + e−jt × (2j) 8 8 3 = sin(t) 2 y(t) = 6. Below is the circuit you will build and test in Lab 3. Please Page 4 of 6 (a) using phasor method to calculate the phasor of Vo as a function of ω , assuming your input Vi (t) = cos(ωt) (b) Please plot Vo (ω) in log-log scale, from 100Hz to 100KHz. Hint: Treating ω as a constant, calculate the phasor of Vo in terms of ω . Then, you may get your Vo as a complex function of ω , plot the amplitute of phase of your phasor of Vo . Solution: (a) By applying KCL at node V2 , we have: V1 − V2 1 jωC = V2 , 1.7kΩ By applying KCL at node V1 , we have: V1 V1 − V2 vi − V1 V 1 − vo = 1 + + , 1 5kΩ 1kΩ jωC jωC By applying the KCL at the node between two right 3.6kΩ resistor: vo − V 2 V2 = . 3.6kΩ 3.6kΩ We combine the three equations to get: vi = 10 5vo 5 vo 1 vo + 2500vo jωC + − vo + + vo , 3.4 3400jωC 2 3400jωC 2 after the simplication, we have: vi = ( 16 3 + 2500jωC + )vo . 17 1700jωC Since the phasor of vi = 1, we have: vo = 16 17 1 + 2500jωC + 3 1700jωC = 16 17 + 10−8 1 × 2500jω + 3×108 1700jω Page 5 of 6 (b) The log-log plot is as following: Page 6 of 6 ...
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