ECE301 Fall, 2006 Exam 1 Soluation October 7, 2006 1 Name: Score: /100 You must show all of your work for full credit. Calculators may NOT be used. 1. Consider the system described by the differential equation with α β + 1. (D 2 + αd + βy(t (D + αx(t (1 (a (5 points Find the eigenvalues and modes. Solution: The eigenvalues are the solutions of the characteristic equation s 2 + αs + β s 2 + (β + 1s + β (s + 1(s + β 0. (2 These are λ 1 and λ β, which have corresponding modes c 1 e t and c 2 e βt. (b (5 points Is the system stable? (Justify your answer. Solution: The system is stable because all eigenvalues are in the left half-plane (because the given value of β was positive. (c (10 points Find the zero input response y 0 (t corresponding to the initial conditions y(0 1, ẏ(0 1. Solution: Solving the system of linear equations y(0 1 c 1 + c 2 (3 ẏ(0 1 c c 2 (4 for the coefficients c 1 and c 2 yields c 2 2/( and c 1 (1 + β/( so the ZIR is ( ( 1 + β 2 y(t e t + e βt (5 (d (10 points Find the impulse response h(t. Solution: The impulse response is given by h(t b 0 δ(t + [P(Dy n (t] u(t. (6 In (1, b 0 0 and P(D D + α. The y n (t is the ZIR corresponding to initial conditions y(0 0 and ẏ(0 1 (see (2.24a, (2.24b on page 167 of the textbook. Solving the system of linear equations y(0 0 c 1 + c 2 (7 ẏ(0 1 c c 2 (8 for the coefficients c 1 and c 2 yields c 2 1/( and c 1 1/( so y n (t ( 1 e t + ( 1 e βt (9
ECE301 Fall, 2006 Exam 1 Soluation October 7, 2006 2 and h(t [( 1 α e t + (e (5 points Find the transfer function H(s. Solution: The transfer function is given by ( ] [( ( ] α β β 1 e βt u(t e t + e βt u(t. (10 H(s P(s Q(s Checking for consistency, we calculate L {h(t} s + α s 2 + αs + β s + β + 1 (s + 1(s + β. (11 β/( 1/( + s + 1 s + β βs β 2 + s + 1 ((s + 1(s + β (s + ((1 + β ((s + 1(s + β s + 1 + β (s + 1(s + β so our answers to parts (c and (e are consistent. (f (5 points Find the ZIR using Laplace Transforms. Solution: Taking the Laplace Transform of (1 with x(t 0 we obtain ( s 2 Y (s sy(0 ẏ(0 + (β + 1(sY (s y(0 + βy (s 0. (16 Moving the initial conditions to the right hand side we have ( s 2 Y (s + (β + 1sY (s + β (s + 1y(0 + ẏ(0. (17 Solving for Y (s and substituting for the initial conditions from (c, we obtain so Y (s y(t which matches the result of part (c. 2. For the system described by the difference equation with α β. s + β + 2 s 2 + (β + 1s + β A s + 1 + B s + β β + 1 (β 1(s + 1 + 2 (β 1(s + β (12 (13 (14 (15 (18 (19 (20 ( ( 1 + β 2 e t + e βt (21 y[n] + 2αy[n 1] + βy[n 2] x[n 1] (22
ECE301 Fall, 2006 Exam 1 Soluation October 7, 2006 3 (a (5 points Find the eigenvalues and modes. Solution: The eigenvalues are the roots of γ 2 + 2αγ + β γ 2 + 2 βγ + β 0 (23 so the eigenvalues are both γ β α. The corresponding modes are c 1 ( α n and c 2 n(α n. (b (5 points Is the system stable? (Justify your answer. Solution: The system is unstable because it has either eigenvalues outside the unit circle or a repeated eigenvalue on the unit circle, depending on the given value of α. (c (10 points Find the unit impulse response h[n]. Solution: The unit impulse response will have the form h[n] b N a N δ[n] + y c [n]u[n]. (24 In this problem, b N 0 and y c [n] c 1 ( α n + c 2 n( α n with coefficients satisfying the solution obtained by iterative calculation using h[n] 0 n < 0. We find so solving the system of linear equations h[0] 2αh[ 1] βh[ 2] + δ[ 1] 0 (25 h[1] 2αh[0] βh[ 1] + δ[0] 1 (26 h[0] 0 c 1 (27 h[1] 1 c 1 ( α + c 2 ( α (28 for the coefficients yields c 1 0 and c 2 1/α. Thus h[n] ( n( αn u[n] n( α n 1 u[n]. (29 α (d (10 points Use convolution to find an expression for the zero state response corresponding to the input x[n] 3 n u[n]. (30 You need not actually evaluate the sum, but should simplify as much as reasonably possible. Solution: The formula for the ZSR is y ZSR x[m]h[n m] (31 so we have y ZSR m 3 m u[m](n m( α n m 1 u[n m] (32 3 m (n m( α n m 1 (33 n n( α n 1 3 m ( α m ( α n 1 n 3 m m( α m (34 ( α (n n 1 ( 3α m m( 3α m (35
ECE301 Fall, 2006 Exam 1 Soluation October 7, 2006 4 (e (Extra Credit 10 points Evaluate the sum. Solution: The first sum above is (using the formula from Chapter B, given on the formula sheet, ( 3α m ( 3α (n+1 1 ( 3α 1 1 ( 3α n ( 3α 1 ( 3α (36 (37 ( 3α n + 3α. (38 1 + 3α The second sum is (using another formula from Chapter B, given on the formula sheet, m( 3α m ( 3α 1 + [n(( 3α 1 1 1]( 3α (n+1 ( 3α 1 2 (39 ( 3α 1 ( 1 + [n(( 3α 1 1 1]( 3α n (3α + 1 2 (40 Substituting these into (35 yields something rather complicated in terms of α. It would not be difficult (only tedious to compute for a particular value of α, but the general solution is rather messy so I will not provide it here. 3. Consider the transfer function H(s β(s + 2 (s 2 + 25s + α. (41 (a (5 points Determine corner frequencies that you would use in sketching the magnitude plot. Solution: To determine the corner frequencies we must first determine whether the denominator has two real roots or a pair of complex conjugate roots. The roots will be complex if 25 2 < 4α. this was not true for any of the given values of α so the corner frequencies are 2 and 25/2 ± 625 4α 2 /2. The exact corner frequencies depended upon the given value of α, but in all cases, rough estimates would place them somewhere between ω 1 rad/s and ω 25 rad/s. (b (5 points Determine the value of the magnitude (in db at ω 1 rad/s. You need not evaluate the square roots. Solution: We calculate as follows. H(j β(j + 2 j 2 + 25j + α β 5 625 + (α 1 2 (42 so in db we have 20log 10 H(j 20log 10 β + 20log 10 5 20log10 625 + (α 1 2 (43 20log 10 β + 10log 10 5 10log 10 (625 + (α 1 2. (44
ECE301 Fall, 2006 Exam 1 Soluation October 7, 2006 5 (c (5 points Determine the approximate slope of the magnitude (in db and the approximate value of the phase (in deg at ω 10 3 rad/s. Solution: ω 10 3 rad/s is much smaller than any of the corner frequencies so, since there are neither zeros nor poles at the origin, the slope of the magnitude is 0 db/dec and the phase is 0 degrees. (d (5 points Determine the approximate slope of the magnitude (in db and the approximate value of the phase (in deg at ω 10 4 rad/s. Solution: ω 10 4 rad/s is much larger than any of the corner frequencies so the slope of the magnitude is approximately 20(1 2 20 db/dec and the phase is approximately 90(1 2 90 degrees. 4. (10 points Indicate whether the following system is causal, invertible, linear, memoryless, and/or time invariant by circling the correct options. (The system may have more than one of these properties. Justify your answer. y(t x(2t + 3 (causal, invertible, linear, memoryless, time invariant Solution: Causality When t 1 we have y(1 x(2 + 3, so the output depends on a future value of the input. Thus we have shown the system is noncausal. Invertibility The system is invertible by taking x(t y(t/2 3. Linearity We show that the system is nonlinear by defining output y 1 (t in terms of input x 1 (t and output y 2 (t in terms of input x 2 (t, calculating the output y(t corresponding to the input x(t αx 1 (t + βx 2 (t and determining whether y(t y 1 (t + y 2 (t. We calculate as follows. y 1 (t x 1 (2t + 3 (45 y 2 (t x 2 (2t + 3 (46 x(t αx 1 (t + βx 2 (t (47 x(t (αx 1 + βx 2 (t (48 y(t (αx 1 + βx 2 (2t + 3 (49 αx 1 (2t + βx 2 (2t + 3 (50 αx 1 (2t + 3 + βx 2 (2t + 3 (51 y 1 (t + y 2 (t (52 Memorylessness When t 1 we have y( 1 x( 2 + 3, so the output depends on a past value of the input. Thus we have shown the system is not memoryless. Time Invariance To show that the system is time invariant, we define a new function x a (t x(t a, then compute showing that the system is time invariant. y a (t x a (2t + 3 (53 x(2(t a + 3 (54 y(t a + 3, (55