examII_f99 - EXAMINATION II EE 350 Continuous Time Linear...

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Unformatted text preview: EXAMINATION II EE 350 Continuous Time Linear Systems October 25, 1999 Name (Print): ID. number : Section number : Do not turn this page unit you are told to do so Score —« n— um I.“ Test form A Instructions 1. You have 2 hours (120 min. total) to complete this exam. 2. This is a. closed book exam. You are allowed one sheet of 8.5” x 11” of paper for notes as mentioned in the syllabus. 3. Solve each part of the problem in the space following the question. Place your final answer in the box when provided. If you need more space, continue your solution on the reverse side labeling the page with the question number; for example, “Question 4.2) Continued”. NO credit will be given to solution that does not meet this requirement. 4. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted and a grade of ZERO will be assigned. Problem 1: (25 Points) Question 1.1 (10 Points) The impulse response h(t) of a certain system can be expressed as h(t) = gm: + 1) + :60 — t). Find the output of this system due to the input f (t) = e‘2t sin(31rt)u(t). Question 1.2 (15 Points) The input f (t) = 2cos(1rt)u(t) is applied to a system whose impulse response h(t) is shown in Figure 1. Determine the corresponding output y(t) = f (t) at Mt) for ali values of if using the graphical approach. Figure 1: The impulse response h(t). (continued) Problem 2: (25 Points) Question 2.1 (10 Points) Determine if the set composed of :31 (t) and 32 (t), shown in Figure 2, form an orthonormal set. Justify your answer. Figure 2: The signals (a) x1 (15) and (b) :32 (t) for problem 2.1. Question 2.2 (5 Points)Approximate the signal, f (t) = cos (21rt), on the interval [—%, f (t) z: czarz (t) , where 9:2 (t) is shown in Figure 2. Find the coefficient, 62, that minimizes the energy of the error signal, e (t) = f (t) — c222 (t). rah-t as Question 2.3 (10 Points) The signal f (t) = t was estimated on the interval [—é, {I using the basis functions :51 (t) and 9:2 (t), shown in Figure 2. In one approximation, the weighting factors are 01 2 (:2 = 0, resulting in the estimate f (t) z 0. In a second, difi'erent approximation, (:1 = 0 and C2 = %, resulting in the approximation 1 f0) “ 1562(5)- Determine which approximation, f (t) :5 0 or f (t) 3 i326), minimizes the energy t2 2 2 Ee = [II [no — 2mm] dt 1 i=1 Problem 3: (25 Points) Question 3.] (15 Points) The RLC circuit shown in Figure 3 below has the input-output model described by the ODE, L L f(t) L 2 — — D 1 = .__ [( (1)1) + (R1 + R2) + ]y(t) R1, where the zero-state response y(t) has the initial conditions y(0+) = 0 and 55% = 0. Let R1 = t=0+ 2052,1122 = 809, L = 41-1, and C’ = 0.0025F. Find the impulse response h(t) of this system. Figure 3: The circuit for Problem 3.1. (continued) Question 3.2 (10 Points) If the impulse response h(t) of a certain LTI system is described by an equation W) = (x/E- 6‘3" — e‘°'5tlu(t), determine if this system is bounded-input, bounded-output (BIBO) stable? Explain. Hint: |A 1 3| 5 |A| + |B|. 10 Problem 4: (25 Points) Question 4.1 (5 Points)Let f (t) = 7009 (30t + 18°) + 3 sin (30t + %), find the phasor represen- tation I?1 of the signal f(t). Express your answer in terms of a single complex exponential term. 11 Question 4.2 (10 Points) The circuit shown in Figure 4 has an input voltage f (t) and an output current y(t). Find the frequency response, H (jw), for the circuit. Put your final solution in the form 1 5m (.iw')"l + bm—1(jw)m_ + - - - + b1 (3'01) + bo- cw)" + can—1 um)"—1 + . .. + an (M + an L f(t) o C y(t) R2 Figure 4: The circuit for problem 4.2. H (in!) = 12 Question 4.3 (10 Points) Consider the circuit shown in Figure 5. The circuit has the frequency response function m a) — __R__l (M2 _ 4-5 x 10-302»)? 3 (jar)2 + 716(jw) + is (jw)2 + 97031:: + 1.1 x 106' Find the sinusoidal steady state response of the circuit, y (t), given that f (t) = 10 cos (100i —— 60°)+ 10 sin (90001,I + 20°). Figure 5: The circuit for problem 4.3. 13 (continued) 14 ...
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