Problem Set 2: Solution Due on Wed. 25th Sept. Fall 2013

Transcription

1 EE 561: Digital Control Systems Problem Set 2: Solution Due on Wed 25th Sept Fall 2013 Problem 1 Check the following for (internal) stability [Hint: Analyze the characteristic equation] (a) u k = 05u k 1 03u k 2 Taking the z-transform, we get The characteristic equation turns out to be U(z) ( 1 05z z 2) = 0, U(z) ( z 2 05z + 03 ) = 0 z 2 05z + 03 = 0, with roots z 1,2 = 025 ± 049j z 1,2 = , which is less than 1 Both roots lie inside the unit circle, hence the system is stable (b) u k = 16u k 1 u k 2 The characteristic equation is z 2 16z + 1 = 0, with roots z 1,2 = 08 ± 06j z 1,2 = 1 Roots lie on the unit circle, so the system is marginally (neutrally) stable (c) u k = 08u k u k 2 The characteristic equation is z 2 08z 04 = 0, with roots z 1 = 1148, z 2 = 0348 z 1 > 1 One root lie outside the unit circle, so the system is unstable Problem 2 The first-order system z α has a zero at z = α (1 α)z (a) Plot the step response for this system for α = 08, 09, 11, 12, 2 Step responses are shown in Fig 1 (b) Plot the overshoot of the system on the same coordinates as those appearing in Fig 3 for 1 < α < 1 Overshoot is shown in Fig 2 (c) In what way is the step response of this system unusual for α > 1? For α > 1, responses undershoot instead of overshooting, before settling to the stable point Problem 3 Show that the characteristic equation z 2 2r cos(θ)z + r 2 has the roots z 1,2 = re ±jθ Applying the quadratic roots formula, we get z 1,2 = r cos θ ± r cos 2 θ 1 = r cos θ ± r sin 2 θ = r(cos θ ± j sin θ) = re ±jθ

2 Figure 1: Step responses with changing position of zero z 2 Figure 2: Variation of percentage overshoot with changing position of zero z 2 Problem 4 For a second-order system with damping ratio 05 and poles at an angle in the z-plane of θ = 30, what percent overshoot to a step would you expect if the system had a zero at z 2 = 06? [Hint: You can either analyze it geometrically, or you can also estimate it from the graph in Fig 3] In the graph shown in Fig 3, curves for θ = 18 and θ = 45 are shown We can draw an approximate curve (drawn in blue) for θ = 30 between those curves From the new curve, we can estimate the percentage overshoot at z 2 = 06 as roughly 40% Problem 5 Sketch the inverse z-transform, f k, for each of the following transforms, for up to at least three time constants You can also use an inverse z-transform table 1 (a) F (z) =, z > 1; 1 + z 2 Inverse transform f k can be considered as the impulse response of a transfer function H(z), considering H(z) = U(z) = F (z)

3 Figure 3: So, if Now considering u k = δ[k], y k equals f k, For a causal system f k = 0 for k < 0, so U(z) = z 2, y k = u k y k 2 f k = δ[k] f k 2 f 0 = δ[0] f 2 = 1 0 = 1, f 1 = δ[1] f 1 = 0 0 = 0, f 2 = δ[2] f 0 = 0 1 = 1, f 3 = δ[3] f 1 = 0 0 = 0, f 4 = δ[4] f 2 = 0 ( 1) = 1, f 5 = δ[5] f 3 = 0 0 = 0, z(z 1) (b) F (z) = z 2, z > 1; 125z Considering U(z) = z(z 1) z 2 125z + 025, Now considering u k = δ[k], y k equals f k, y k = u k u k y k 1 025y k 2 f k = δ[k] δ[k 1] + 125f k 1 025f k 2

4 For a causal system f k = 0 for k < 0, so f 0 = δ[0] δ[ 1] + 125f 1 025f 2 = 1 0 = 1, f 1 = δ[1] δ[0] + 125f 0 025f 1 = = 025, f 2 = δ[2] δ[1] + 125f 1 025f 0 = (025) 025(1) = 00625, f 3 = δ[3] δ[2] + 125f 2 025f 1 = (00625) 025(025) = 00156, f 4 = δ[4] δ[3] + 125f 3 025f 2 = (00156) 025(00156) = 00039, f 5 = δ[5] δ[4] + 125f 4 025f 3 = (00039) 025(00156) = , z (c) F (z) = z 2, z > 1; 2z + 1 Considering U(z) = Now considering u k = δ[k], y k equals f k, For a causal system f k = 0 for k < 0, so z z 2 2z + 1, y k = u k 1 + 2y k 1 y k 2 f k = δ[k 1] + 2f k 1 f k 2 f 0 = δ[ 1] + 2f 1 f 2 = 0, f 1 = δ[0] + 2f 0 f 1 = = 1, f 2 = δ[1] + 2f 1 f 0 = 0 + 2(1) 0 = 2, f 3 = δ[2] + 2f 2 f 1 = 0 + 2(2) 1 = 3, f 4 = δ[3] + 2f 3 f 2 = 0 + 2(3) 2 = 4, f 5 = δ[4] + 2f 4 f 3 = 0 + 2(4) 3 = 5, z (d) F (z) = (z 1, z > 2 2 )(z 2) Considering U(z) = z (z 1 2 )(z 2), Now considering u k = δ[k], y k equals f k, y k = u k y k 1 y k 2 f k = δ[k 1] f k 1 f k 2

5 For a causal system f k = 0 for k < 0, so f 0 = δ[ 1] f 1 f 2 = 0, f 1 = δ[0] f 0 f 1 = = 1, f 2 = δ[1] f 1 f 0 = 0 + 5/2(1) 0 = 5/2, f 3 = δ[2] f 2 f 1 = 0 + 5/2(5/2) 1 = 21/4, f 4 = δ[3] f 3 f 2 = 0 + 5/2(21/4) 5/2 = 85/8, f 5 = δ[4] f 4 f 3 = 0 + 5/2(85/8) 21/4 = 341/16, Plots of f k for parts (a)-(d) are shown in Fig 4 Figure 4: Graphs for Problem 5

6 Problem 6 Use the z-transform to solve the difference equation y k 3y k 1 + 2y k 2 = 2u k 1 2u k 2 { k, k 0, u k = 0, k < 0, y k = 0, k < 0 By taking z transf orm of the difference equation with zero initial conditions, we determine the transfer function as U(z) = z(z 1 z 2 ) 2(z 1) 1 3z 1 = + 2z 2 z 2 3z + 2 = 2(z 1) (z 1)(z 2) = 2 z 2 For the given u k, we look up its z-transform from the table as U(z) = = Splitting it up into partial fractions, = ( ) 2z U(z) = U(z) (z 2)(z 1) 2 2z z 2 and taking the inverse z-transform, we finally get 2z z 1 2z (z 1) 2, y k = 2(2 k 1 k), for k 0 z (z 1) 2 This leads to Problem 7 For this problem, refer to section 661 of Signals and Systems (2e) by Oppenheim et al For the first-order causal LTI system described by the difference equation (warning: this is not the same equation that appears in section 661) y k + ay k 1 = ax k (a) Find out the z-transform H(z) of the system The z-transform X(z) = H(z) = a 1 + az 1, (1) (b) Using a suitable substitution to the z-transform found in (a), find out the frequency response H(e jω ) of the system Substituting z = e jω into (1), H(e jω ) = a 1 + ae jω (c) For a = 3/2, 1/3, 0, 1/3, and 3/2, (i) Plot the step responses of the system See Fig 5 (ii) Plot the magnitude responses of the system Comment on the variation in shape of the graph with the variation in a See Fig 6

7 Figure 5: Step responses for Problem 7(c) Problem 8 For this problem, refer to section 662 of Signals and Systems (2e) by Oppenheim et al For the second-order causal LTI system described by with 0 θ π y k 2r cos(θ)y k 1 + r 2 y k 2 = x k, (a) Find out the z-transform H(z) of the system The z-transform 1 H(z) = 1 2r cos θz 1 + r 2 (2) z 2 (b) Find out the frequency response H(e jω ) of the system Substituting z = e jω into (2), H(e jω ) = 1 1 2r cos θe jω + r 2 e j2ω (c) For (r, θ) = (2/3, 0), (3/2, π/4), (1/3, π/2), (2/3, 3π/4), (2/3, π), (i) Plot the step responses of the system Also classify each response as either undamped, underdamped, critically damped or overdamped See Fig 7 (ii) Plot the magnitude responses of the system Comment on the variation in shape of the graph with the variation in r and θ See Fig 8

8 Figure 6: Magnitude responses for Problem 7(c) For a < 0, the system acts as a low-pass filter Whereas, for a > 0, the system acts as a high-pass filter The gain of the filter increases for larger magnitudes of a Figure 7: Step responses for Problem 8(c) are critically damped, undamped, overdamped, underdamped, and underdamped respectively

9 Figure 8: Magnitude responses for Problem 8(c) As θ varies from 0 to π, the system s function shifts from low-pass filtering to high-pass filtering, with band-pass filtering about θ = π/2