hw4-sol

hw4-sol - MAE 143A – Homework 4 Chapter 4 ∑ 12. ′ a)...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MAE 143A – Homework 4 Chapter 4 ∑ 12. ′ a) ′ " (1) When car is at rest, set all derivates in (1) equal to zero which yields 0 . 0.6 · 0.404 m b) Rearrange (1) as ′ " ′ 0 (2) Define new variable as ′ ′ " ′ " " Substitute into (2) which yields " " 0 " " " " (3) Case 2: m=n, add impulse term to general solution Solve for the homogenous equation 0 " 0 . cos 2.537 . 4 , cos 2.357 sin 2.357 sin 2.357 The impulse response will take the form of . cos 2.357 6.667 2 sin 2.357 2.357 Method 1: Solve simplified problem first " 0 0 0 0 0 0 1 0 0 6.667 2.357 1 0 0 0 0 Thus 1 0 0.424 which yields the homogenous solution as 0.424 . sin 2.357 Then define new variable 1 to solve for (3) where " " " Substitute into (3) which yields " " " " to solve for homogenous solution of (3) Set 0.424 . sin 2.357 2.826 50 0.424 . . sin 2.357 . cos 2.357 . sin 2.357 13.333 2.826 13.333 cos 2.357 . sin 2.357 16.479 sin 2.357 . cos 2.357 Therefore the impulse response is . 13.333 cos 2.357 16.479 sin 2.357 Method 2: Let and for (3), then " " 0 Integrate from (i) ′ ′ 0 ′ 0: 0 0 0 6.667 (ii) 0 ′ 0 2.357 0 ′ 0 0 0 0 0 (a) 0 0 ∞ 0 (b) (iii) ∞ ∞ ∞ 1 (c) 13.333, Solving equations (a) (b) (c) we get 16.497, 1 Therefore the impulse response is . c) 13.333 cos 2.357 16.479 sin 2.357 0.15 Apply convolution integral 0.15 0.15 . 0.15 13.33 cos 2.36 16.5 sin 2.36 Use following properties to solve for z(t) sin asin cos acos . 0.15 13.33 50 . 6.67 cos 2.36 2.36 cos . 0.15 0.15 . 2.81 sin 2.36 2.81 sin 2.36 2.36 sin 2.36 16.5 0.15 cos 2.36 cos 2.36 1 0.15 50 6.67 sin 2.36 0.1 0.05 g(t) 0 -0.05 -0.1 -0.15 -0.5 0 0.5 1 1.5 2 t 20. ∞ Time-Invariance: Let . . Then . Then Let yields into Substitute with , thus the system is time-variant. Stability: / is bounded, then If increases without bound therefore the system is unstable. Invertibility: 3 Thus the system is invertible. Chapter 6 " 18c. 3 Rewrite as " 3 5" 3 3 5" Case 2: m=n, add impulse term to general solution Solve for the homogeneous solution " 3 0 (1) 3 0 0,3 , Method 1: Solve simplified problem first " 3 0 0 3 0 0 30 0 0 1 3 0 0 0 1 3 3 1 0 0 Thus, 1 1 3 Define new variable 5 to solve for (1) where 5 5 " " 5" Substitute into (1) which yields " 5" " 3 3 5 3 3 3 3 15 15 1 1 3 15 1 Therefore the impulse response is 16 1 5" 5 16 Method 2: Let and " 3 3 (i) ′ 0 ′ (ii) 5" 0: 0 Integrate from ′ for (1), then 0 3 0 30 0 0 0 0 3 3 3 3 (a) 3 5 ∞ 0 0 3 0 3 (iii) ′0 5 ′0 3 3 0 3 3 ∞ ∞ 0 (b) 5 ∞ 5 5 1, Solving equations (a) (b) (c) we get 16, (c) 5 Therefore the impulse response is 1 5 16 Plot II-type block diagram with integrators: x(t) ‐1 + + + 5 + + ‐ 3 3 22a. 3 cos 10 12 1 4 10 1 10 10 12 cos 12 12cos 10 y(t) 15 10 5 g(t) 0 -5 -10 -15 -0.5 -0.4 -0.3 -0.2 -0.1 0 t 0.1 0.2 0.3 0.4 0.5 2 22d. ∞ 2 ∞ 1.5 g(t) 1 0.5 0 -0.5 -3 -2 -1 0 t 1 Chapter 15 33b. Express in partial fraction expansion. 4 8 3 8 0, 8 3 8 3 8 4 5 3 4 5 8 4 3 4 3 4 , 2 3 From the Laplace transform table on pg. 591, we get Initial-value theorem lim 0 4 3 lim ∞ 0 8 Final-value theorem lim lim ∞ 4 3 lim 0 8 33b. 1 1 1 1 1 1 1 1 From the Laplace transform table on pg. 591, we get cos sin Initial-value theorem lim 0 lim ∞ 2 ∞ 1 2 Final-value theorem lim lim ∞ 39a) " lim 2′ 2 2 0 10 First Laplace transform both sides of the equation. 0 2 Then solve for Y(s). 1 20 0 2 1 5 2 1 10 10 1 5 1 9 0 10 1 Express in partial fraction expansion. 1 9 9 1 1 2 9 1 2 10 5, 2 1 10 11 10 5 51 10 1 51 1 0, 10 1 5 9 11 10 1 1 9 49 3 5 5 1 51 , , 2 1 9 3 1 9 From the Laplace transform table on pg. 591, we get 1 1 10 51 cos 3 49 sin 3 3 Additional Item: % Problem 39 (a): Use matlab to compute and plot the response of the % system. You should use commands like 'tf', and 'step'. % Method 1: use 'tf' and 'step' command - simulate response at initial % conditions and add to step response Y = tf([1],[1 2 10]); % construct transfer function a = [0 1; -10 -2]; b = [0; 1]; c = [1 0]; d = [0]; Y1 = ss(a,b,c,d); % construct state space y0 = [-5 10]; % initial conditions t = 0:0.01:10; y1 = step(Y,t); % step response at zero state y2 = initial(Y1,y0,t); % initial condition response y = y1+y2; % time response with I.C.s plot(t,y); % Method 2: use 'lsim' command - compute time response to zero inputs Y = tf([1],[1 2 10]); % construct transfer function a = [0 1; -10 -2]; b = [0; 1]; c = [1 0]; d = [0]; Y1 = ss(a,b,c,d); % construct state space x = sign(t); % x(t) = u(t) y0 = [-5 10]; % initial conditions t = 0:0.01:10; y = lsim(Y1,x,t,y0); % time response with I.C.s plot(t,y); 3 2 1 0 -1 -2 -3 -4 -5 0 1 2 3 1 43. 4 5 6 7 8 9 10 1 1 1 1 3 2 1 2 2 Find the roots of the denominator 3 9 42 2 3 √1 2 4 The system is stable for all positive real values of K. Additional Item: % % % % Problem 43: For a fixed value of K that makes the system stable, use the matlab commands 'tf' and 'feedback' to compute the transfer function from x to y. Use 'impulse' to plot the impulse response of the resulting system. % Solution: set K to any real positive value, i.e. K = 1 num = [1]; den = [1 3 2]; H = tf(num,den) Hf = feedback(H,1) impulse(Hf) >> H = tf(num,den) Transfer function: 1 ------------s^2 + 3 s + 2 >> Hf = feedback(H,1) Transfer function: 1 ------------s^2 + 3 s + 3 Impulse Response 0.3 0.25 Amplitude 0.2 0.15 0.1 0.05 0 -0.05 0 0.5 1 1.5 2 Time (sec) 2.5 3 3.5 4 ...
View Full Document

Ask a homework question - tutors are online