Homework 7 Solution - AME 30315, Spring s 2 + 2s (s 2 + 2s + 4)(s + 20)

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1 1 Homework 7 Solution - AME 30315, Spring 2015 Problem 1 [10/10 pt] Ue partial fraction expanion to compute x(t) when X 1 () = Ue partial fraction expanion to compute x(t) when X 2 () = ( ) ( )( + 20) ( ) 1 Plot the repone of each on the ame plot o that you can better explain the difference in the two plot. Comment on the effect of fat pole (LHP pole that are far from the imaginary axi, relative to LHP pole coniderably cloer to the imaginary axi) on ytem repone. Firt, X 1 () = ( ) 1 We aume thi form, and A, B, C, are all unknown coefficient to be olved: ( ) 4 1 X 1 () = = A + B C Multiplying and equating numerator: 4 = (A + C) 2 + (B + 2C) + 4C Which yield three equation and three unknown: which yield [1 pt]: A + C = 0 ( 2 ) B + 2C = 0 ( 1 ) 4C = 4 ( 0 ) A = 1 B = 2 C = 1 The expanded fraction i now: X 1 () =

2 2 Which can be manipulated completing the quare in the denominator: X 1 () = 2 ( + 1) 2 + ( 3) Which can again be plit into two fraction to correpond to Table 8.1 (Goodwine pg. 288): X 1 () = y 1 (t) = L 1 {X 1 ()} = L 1 { Yield: y 1 (t) = 1 ( + 1) 2 + ( 3) ( + 1) 2 + ( 3) ( + 1) 2 + ( 3) 2 } { 1 } { } 3 1 L 1 3 ( + 1) 2 + ( +L 1 3) 2 ( e t co( 3t) 1 e t in( ) 3t) + 1 1(t) [2 pt] 3 Next, we will ue partial fraction expanion on X 2 (): X 2 () = 4 20 ( )( + 20) ( ) 1 We aume thi form, and A, B, C, D, are all unknown coefficient to be olved: ( ) = A + B ( )( + 20) ( ) + C ( + 20) + D Multiplying and equating numerator: 80 = (A + C + D) 3 + (20A + B + 2C + 22D) 2 + (20B + 4C + 44D) + (80D) Which yield four equation and four unknown: Solving yield [1 pt]: A + C + D = 0 ( 3 ) 20A + B + 2C + 22D = 0 ( 2 ) 20B + 4C + 44D = 0 ( 1 ) 80D = 80 ( 0 ) A = B =

3 3 So, Or rewritten a: C = 1 91 D = ( ) ( + 20) (91)( ) 1 (91)( + 20) + 1 Which need to be manipulated to match a form in Table 8.1 (Goodwine, pg. 288): Now, = (91) ( + 1) 2 + ( 3) (91) ( + 1) 2 + ( 3) (91) ( + 20) + 1 { } (91) L 1 +1 (+1) 2 +( 3) 2 y 2 (t) = L 1 {X 2 ()} { 110 3(91) L 1 } 3 (+1) 2 +( 1 3) 2 { (91) L 1 1 (+20) } + L { } 1 1 Which yield: ( y 2 (t) = 90 (91) e t co( 3t) 110 e t in( 3t) 1 ) 3(91) (91) e 20t + 1 1(t) [2 pt] Each i plotted in MATLAB [2 pt]: 1.2 y1(t) y2(t) y(t) t (ec) Fig. 1. Plot of y 1(t) and y 2(t) in blue and red repectively. The inet highlight the difference in the graph. The original econd order ytem ha complex conjugate pole with a real part of -1.

4 4 For y 2 (t), we have added a real pole in the LHP at -20, which i far from the imaginary axi. Thi term decay rapidly and o ha little effect on the ytem repone [2 pt for comment]. 1 % AME30315 HW#7, Problem 1 2 clear 3 clc 4 5 t=linpace(0,8,1000); 6 7 a = ((3)ˆ0.5); 8 y1=-exp(-t).*(co(a.*t)+(1/a).*in(a.*t))+1; 9 10 plot(t,y1,'b','linewidth',1.5) y2= - (90/91)*exp(-t).*co(a.*t)-(110/(a*91)).*exp(-t).*in(a.*t)... -(1/91)*exp(-20.*t) + 1; hold on 16 plot(t,y2,'r','linewidth',1.5) 17 ylabel('y(t)','fontize',12) 18 xlabel('t (ec)','fontize',12) 19 legend('y1(t)','y2(t)') 20 vividoc(14,3) We can alo tudy the effect of adding pole near and far from the imaginary axi a hown in the plot below for a more general equation ( ) 4 r 1 X() = ( )( + r)

5 r=2.5 r=1.5 r=1 Step Repone Original Second Order Sytem w/ r=0.5 r=1 r=1.5 r= r=0.5 y(t) t (econd) Fig. 2. Plot of the original econd order ytem a well a with an added pole at r. Variou value for r have been plotted. Notice that a r increae, the plot approache the original econd order ytem repone. Thi plot wa not required for the homework. 1 % AME30315 HW#7, Problem 1 2 clear 3 clc 4 5 y1 = tf([4],[1,2,4]) 6 r = y2 = tf([4*r],[1,2+r,2*r+4,4*r]) 8 tep(y1) 9 hold on 10 tep(y2) hold on 13 r = 1 14 y3 = tf([4*r],[1,2+r,2*r+4,4*r]) 15 tep(y3) hold on 18 r = y3 = tf([4*r],[1,2+r,2*r+4,4*r]) 20 tep(y3) hold on 23 r = y3 = tf([4*r],[1,2+r,2*r+4,4*r]) 25 tep(y3) ylabel('y(t)','fontize',12) 28 xlabel('t','fontize',12) 29 legend('original Second Order Sytem','w/ r=0.5',' r=1',' r=1.5',' r=2.5')

6 6 Problem 2 [6/6 pt: 0.5 pt for each anwer] Temperature Regulation human body = P 98.6 deg F = R body temperature = Y hypothalamu = H human brain = K activating weat gland = U Campaign Management The campaign manager and candidate = K platform, image, talking point, and propaganda = U public = P a high of poll number a poible = R poll = H public entiment = Y

7 7 Problem 3 [9/9 pt] xp x tr (Franklin, Powell, Emani-Naeini, 2015) Fig. 3. Sytem tep repone with peak diplacement, teady-tate diplacement, and rie time indicated. The equation of motion for the ma-pring-damper ytem i: where F = 21(t) N mẍ(t) + bẋ(t) + kx(t) = F (1) And aume that the unit for thi problem are SI unit. From the repone a a decaying ocillation, we can recognize that our ytem i underdamped (correpond to 0 < ζ < 1). Then the equation of motion can be rewritten a where ẍ(t) + 2ζω n ẋ(t) + ω 2 nx(t) = F m k 1 : ω n = m b 2 : ζ = 2 mk which are the natural frequency and damping ratio repectively (alo find thee on p of textbook).

8 8 Find ω n and ζ by looking at the performance characteritic of the repone. From the plot, we can determine the overhoot (O): O = x p x x = = 0.15 Then uing the formula for OS on p. 360 of the textbook which i a function of ζ, we can olve for ζ: ( ) O = exp πζ 1 ζ 2 become [2 pt] ln 2 (O) ζ = π 2 + ln 2 (O) ln 2 (.15) = π 2 + ln 2 (.15) 0.52 We can alo ue the rie time approximation (Eq. 9.18, p. 363 text): t r 1.8 ω n repone plot, t r 1.4 econd. where from the Then, [2 pt] ω n = 1.8 t r = 1.3 rad/ We have 3 unknown (m, b, k), but only 2 equation 1 and 2. To find the third equation, conider the ytem repone at teady tate where ẋ(t) = ẍ(t) = 0 [2 pt]. Then the equation of motion i: So, And from 1, k = F x = kx = F k = 20 N/m [1 pt] m = 12.1 kg [1 pt] And from 2, b = 2ζ mk = 16.1 N /m We can check that thee parameter uing Matlab: [1 pt] Firt, tranform the equation of motion into the frequency domain, auming zero initial

9 9 condition: Then, m 2 + b + k = F () X() F () = 1 m 2 + b + k And rewrite o that we can tudy the tep repone: X() = Which i implemented in the following code: 2 m 2 + b + k ( ) 1 1 % Check the calculated value of m, b, k 2 cloe all 3 clear all 4 5 x_p=0.115; 6 t_r=1.4; 7 x_=0.1; F=2; 11 O=(x_p-x_)/x_; 12 zeta=qrt((log(o))ˆ2/(piˆ2+(log(o))ˆ2)); 13 omega_n=1.8/t_r; k=f/x_; 16 m=k/(omega_n)ˆ2; 17 b=zeta*(2*qrt(m*k)); num = 2; 20 den = [m, b, k]; y=tf(num,den); 23 tep(y) % plot ytem tep repone

10 Step Repone Amplitude Time (econd) Fig. 4. Repone for the approximate value of m, b, and k. Thi matche cloely with the given plot.