EE3054 - Homework 2 - Solutions. The transfer function of the system is given to be H(s) = s 2 +3s+3. Decomposing into partial fractions, H(s) = 0.5774j s +.5 0.866j + 0.5774j s +.5 + 0.866j. () (a) The impulse response h(t) can be found by taking the (causal) inverse Laplace transform of H(s) to be h(t) = 0.5774je (.5+0.866j)t u(t) + 0.5774je (.5 0.866j)t u(t) =.548e.5t sin(0.866t)u(t) (2) (b) The differential equation relating the input signal x(t) and the output signal y(t) can be found from the given transfer function to be ÿ(t) + 3ẏ(t) + 3y(t) = x(t). (3) (c) The step response of the system can be found by integrating the impulse response to be s(t) = t ف h(τ)dτ = 0.5774j e(.5+0.866j)t.5 + 0.866j u(t) + 0.5774j e(.5 0.866j)t u(t).5 0.866j = [ 0.3334e.5t cos(0.866t) 0.5774e.5t sin(0.866t) + 0.3334]u(t). (d) Evaluating H(jω) at ω = 0.5 and ω = π, we get H(j0.5) = 0.392e 0.4993j (4) H(jπ) = 0.0857e 2.2006j. (5) Hence, the steady-state output signal for the given input signal is y s (t) = [.5(0.392)cos(0.5t + 0.4993) 2(0.0857)sin(πt + 3 2.2006)]u(t) = [0.4788 cos(0.5t + 0.5007) 0.74 sin(πt + 0.7994)]u(t)
2. The step response of the system is given to be e 2t u(t) e 4t u(t). (a) The impulse response is found by differentiating the step response to be h(t) = 2e 2t u(t) + 4e 4t u(t) + e 2t δ(t) e 4t δ(t) = 2e 2t u(t) + 4e 4t u(t). (6) Hence, the transfer function of the system is H(s) = 2 s + 2 + 4 s + 4 = 2s s 2 + 6s + 8. (7) (b) From H(s), the differential equation relating the input signal x(t) and the output signal y(t) is seen to be ÿ(t) + 6ẏ(t) + 8y(t) = 2ẋ(t). (8) (c) The input signal is given to be e t u(t) and the initial conditions are given as y(0) =, ẏ(0) =. The output signal of the system can be found using either the one-sided Laplace transform or the method of guessing exponentials. Using one-sided Laplace transform: Taking one-sided Laplace transform of both sides of the differential equation (8), we get s 2 Y (s) sy(0) ẏ(0)+6sy (s) 6y(0)+8Y (s) = 2sX(s) 2x(0). (9) Hence, noting that x(0) =, Y (s) = 2sX(s) 2x(0) + sy(0) + ẏ(0) + 6y(0) s 2 + 6s + 8 = 2s s+ + s + 3 s 2 + 6s + 8 = s 2 + 6s + 3 (s 2 + 6s + 8)(s + ) = 2.5 s + 2 + 0.8333 + 0.6667 s + 4 s +. (0) 2
Therefore, for all t 0, y(t) = 2.5e 2t 0.8333e 4t 0.6667e t. () Using the method of guessing exponentials: The poles of the system are 2 and 4. The pole of the input signal is at. Hence, the output signal can be guessed to be of the form y(t) = c e 2t + c 2 e 4t + c 3 e t (2) for all t 0. Differentiating twice, we get ẏ(t) = 2c e 2t 4c 2 e 4t c 3 e t ÿ(t) = 4c e 2t + 6c 2 e 4t + c 3 e t. (3) Using the given initial conditions y(0) =, ẏ(0) =, we obtain the equations c + c 2 + c 3 = 2c 4c 2 c 3 =. (4) Using the differential equation ÿ(t)+6ẏ(t)+8y(t) = 2ẋ(t), noting that for all t > 0, ẋ(t) = e t, and equating the coefficients of e 2t, e 4t, and e t on both sides, we get 4c + 6( 2c ) + 8(c ) = 0 6c 2 + 6( 4c 2 ) + 8(c 2 ) = 0 c 3 + 6( c 3 ) + 8(c 3 ) = 2. (5) Solving for the coefficients c, c 2, and c 3, we get c = 2.5, c 2 = 0.8333, and c 3 = 0.6667. Hence, for all t 0, y(t) = 2.5e 2t 0.8333e 4t 0.6667e t. (6) 3. It is given that the transfer function of the system is of the form H(s) = a s+b with a and b being unknown constants. 3
(a) It is given that the system is BIBO stable. Hence, b should be positive. For the transfer function H(s) = a s+b, we have H(jω) = a ω 2 + b 2 H(jω) = { tan ( ω b ) if a > 0 π tan ( ω b ) if a < 0. (7) It is given that when the input signal is sin(t)u(t), the output signal at steady state is 0.707 sin(t π 4 )u(t). This means that H(j) = 0.707 and H(j) = π 4. From (7), this implies that a = and b =. Hence, the transfer function of the system is s+. The differential equation relating the input signal and the output signal of the system is ẏ(t) + y(t) = x(t). (b) The signal u(t) = e j0t u(t) is a sinusoidal signal with frequency 0. At frequency ω = 0, we get H(jω) = = e j0. Hence, when the input signal is u(t), the steady-state output signal is also u(t). (c) It is required to find the output signal when the input signal is sin(t)u(t) with the initial condition y(0) =. This can be done using either the one-sided Laplace transform or the method of guessing exponentials. Using one-sided Laplace transform: Taking one-sided Laplace transform of both sides of the differential equation of the system, we get Hence, Y (s) = Therefore, for all t 0, sy (s) y(0) + Y (s) = X(s). (8) X(s) + y(0) s + = s 2 + + s + =.5 0.25 + 0.25j + + s + s + j s 2 + 2 = (s + )(s 2 + ) 0.25 0.25j. (9) s j y(t) =.5e t + ( 0.25 + 0.25j)e jt + ( 0.25 0.25j)e jt 4
=.5e t + 2Re(( 0.25 + 0.25j)e jt ) =.5e t + 2[ 0.25 cos(t) + 0.25 sin(t)] =.5e t + 0.5[ cos(t) + sin(t)]. (20) Using the method of guessing exponentials: The system has only one pole located at. The input signal contributes two poles, one at j and the other at j. Hence, the output signal can be guessed to be of the form y(t) = c e t + c 2 e jt + c 3 e jt (2) for all t 0. Differentiating, we get ẏ(t) = c e t jc 2 e jt + jc 3 e jt. (22) Using the given initial condition y(0) =, we obtain the equation c + c 2 + c 3 =. (23) The differential equation ẏ(t) + y(t) = x(t) = sin(t) gives us the equation c e t jc 2 e jt +jc 3 e jt +c e t +c 2 e jt +c 3 e jt = 2j ejt 2j e jt (24) and equating the coefficients of e t, e jt, and e jt on both sides, we also get the equations c + c = 0 jc 2 + c 2 = 2j jc 3 + c 3 = 2j. (25) Solving for the coefficients c, c 2, and c 3, we get c =.5, c 2 = 0.25+ 0.25j, and c 3 = 0.25 0.25j. Hence, for all t 0, y(t) =.5e t + ( 0.25 + 0.25j)e jt + ( 0.25 0.25j)e jt =.5e t + 0.5[ cos(t) + sin(t)]. (26) 5
Yet another alternative solution using the method of guessing exponentials: In (2), we guessed that y(t) includes the terms e t, e jt, and e jt. Since e jt = cos(t) + j sin(t) and e jt = cos(t) j sin(t), we can equivalently guess that y(t) includes the terms e t, cos(t), and sin(t), i.e., that y(t) is of the form y(t) = c e t + c 2 cos(t) + c 3 sin(t) (27) for all t 0. Differentiating, we get ẏ(t) = c e t c 2 sin(t) + c 3 cos(t). (28) Using the given initial condition y(0) =, this gives the equation c + c 2 =. (29) The differential equation ẏ(t) + y(t) = x(t) = sin(t) gives us the equation c e t c 2 sin(t)+c 3 cos(t)+c e t +c 2 cos(t)+c 3 sin(t) = sin(t). (30) Equating the coefficients of e t, cos(t), and sin(t) on both sides, we also get the equations c + c = 0 c 2 + c 3 = 0 c 3 c 2 =. (3) Solving for the coefficients c, c 2, and c 3, we get c =.5, c 2 = 0.5, and c 3 = 0.5. Hence, for all t 0, y(t) =.5e t 0.5 cos(t) + 0.5 sin(t). (32) At steady state, i.e., as t becomes large, the term.5e t goes to zero. Hence, at steady state, the output signal converges to 0.5 cos(t) + 0.5 sin(t) = 0.707 sin(t π 4 ). This is as expected because part of the original data about the system was that when the input signal is sin(t)u(t), the output signal at steady state is 0.707 sin(t π 4 )u(t). This verifies that as t becomes large, the output signal does converge to the expected steady-state output signal. 6