EE3118 2006-2007 SemA

# EE3118 2006-2007 SemA - CITY UNIVERSITY OF HONG KONG Course...

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Unformatted text preview: CITY UNIVERSITY OF HONG KONG Course code & title : EE31 18 Linear Systems and Signal Analysis Session: : Semester A 2006/07 Time allowed : Two hours This paper has FIVE pages (including this page). 1. This paper consists of 5 questions. 2. Answer ﬂ questions. 3. Start on a new page for each question. 4. Basic Laplace Transform Pairs are given in Appendix I (page 4). 5. Basic properties of Laplace Transform are given in Appendix 11 (page 5). Materials, aids & instruments permitted to be used during examination 1. Approved calculator Answer ALL the questions Question 1 |26 marksl Consider the RLC system given in Fig. 01. Figure 01. A RLC system (a) Derive the relationship of vC2(t) and i(t) in differential equation form. [9 marks] (b) Based on (a) and assuming that all the initial values are zero, determine the values of R, Cl, C 2 and L such that its transfer function is equal to ch (S) _ S 2 I (s) (s + 1)(s2 + 45 + 5) (c) By the use of Laplace Transform, determine the impulse response of the system. [13 marks] [4 marks] Question 2 l7 marksl The transfer function of a linear system is given as below: 2 s + Ks + 1 H (s ) = 4 3 2 3 +55 +Ks +4s+4(K—1) Determine the range of K such that the system is stable. [7 marks] Question 3 |20 marksl The transfer function of a continuous-time system, SYSTEM A, is given as below: Y(s) 1.5s2 - 0.5s — 2.5 X(s) 53 +3.532 +3.53+1 where X(s) and Y(s) are the Laplace Transforms of the input and the output of SYSTEM A, respectively. (a) Draw the direct realization of the SYSTEM A if only unit-sample delay elements, gains and adders are to be used. The number of unit-sample delay elements used must be minimum. [8 marks] (b) Design a transfer function of a discrete-time system so that its impulse response matches with the impulse response of the SYSTEM A. The sampling time is assumed to be 0.1sec. [12 marks] Page 2 Question 4 [33 marks] Figure Q4a shows a signal conditioning system, which consists of an ampliﬁer and a ﬁltering system. . y(t) Filtering x(t) Ampliﬁer System Z“) Figure Q4a. The block diagram of a system The ampliﬁer has a gain of 5, but it saturates when the amplitude of the input exceeds 1.6V. Mathematically, the input and output relationship can be expressed as {5m if |x(t)| s 1.6 y“) = 8 if Ix(t)| >1.6 (a) Sketch the output of the ampliﬁer, y(t), if the sawtooth signal x(t) below is inputted. [3 marks] x(t) -5 0 5 10 15 Figure Q4b. Sawtooth signal input x(t) (b) Determine the trigonometric Fourier series of y(t). [19 marks] (c) Assuming that the Fourier Transform of the impulse response of the ﬁltering system in Fig. Q4a is given as below: (0.5 — E)?” if 0 s a) < 0.5% rad/s 11(0)) = n 0 if a) 2 0.5” rad/s Sketch the Bode Diagram of the ﬁltering system for the range of O S a) < 7r . [3 marks] (d) Using (b) and (c), determine the steady-state output response of 20‘). [8 marks] Question 5 [14 marks] A discrete-time system is described by the following difference equation: y(n) +%y(n—l) +%y(n— 2) = %x(n) + x(n —1) By solving the difference equation directly, obtain the output sequence y(n) if x(n) =5, given that y(0)=0 and y(1)=3. 75. [14 marks] - END- Page 3 Appendix I: Laplace Transform Pairs sin((00 t)u(t) e'a” cos(a)0 t)u(t) (s + a) (s + a)2 + (002 (s + a)2 + (002 [sin(a)0 t) — we I,‘cos(a)0 t)]u(t) 2608 (s2 + my coo tsin(a)o t)u(t) 2w§s (s2 + (002 2 (00 te'm sin(a)O t)u(t) 2a): (5 + a) [(s + at)2 + 0):]2 e‘“‘ [sin(a)O t) — (00 t cos(a)O t)]u(t) 2603 [(s + a)2 + (03]2 Page 4 Apyendix 11: Basic Properties of Laplace Transform Given that the Laplace Transform L[x(t)] = X (s) , 1. Linearity : L[a1x1(t) + azx2 (1)] = a1X1(s) + aZX2 (s) 2. Time domain differentiation : L[ 61:50] = 5"X (S) — S”'1x(0' )- ..._ x(n-l)(0‘) z (-1) - 3. Time domain integration : LU x(l)dl] = X(S) + gO—J -9: s s 4. s—shift: L[e‘°”x(t]l] = X(s w) 5. Delay: L[x(’—to)u(f—to)]= X(S)e-sto 6. Time Scaling : L[x(at)] = a‘1X[£) where a>0 a 7. Frequency Scaling: Ll:—1-x[£)] = X (as) where a>0 a a 8. Convolution : L[x1(t)* x2 (t)] = XI (s)X2 (s) Page 5 ...
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