Experiment 7.1 Objective To verify the effect of input waveform, loop gain, and

Question

Experiment 7.1 Objective To verify the effect of input waveform, loop gain, and system type upon steady-state errors. Minimum Required Software Packages MATLAB. Simulink, and the Control System Toolbox Prelab 1. What system types will yield zero steady-state error for step inputs? 2. What system types will yield zero steady-state error for ramp inputs? 3. What system types will yield infinite steady-state error for ramp inputs? What system types will yield zero steady-state error for parabolic inputs? What system types will yield infinite steady-state error for parabolic inputs? 5. K(s +6) For the negative feedback system of Figure P7.33, whereas- and H(s) = 1, calculate the steady-state error in terms of K for the following inputs. 5u(),5tu(t), and 5Pu(r). -6. R(s) G(s) H(s) 7. Repeat Prelab 6 for G(s)- and H(s)= 1 s(s +4)(s+7s+9%s+12) FIGURE P7.33 and H(s) 1 8. Repeat Prelab 6 for G(s) Lab 1. Using Simulink, set up the negative feedback system of Prelab 6. Plot on one graph the error signal of the system for an input of 5u(r) and K for inputs of 5tu) and 5Pu) 50. 500, 1000, and 5000, Repeat 2. Using Simulink, set up the negative feedback system of Prelab 7. Plot on one graph the error signal of the system for an input of 5u(r) and K for inputs of Stut) and 5Fu(). 50, 500. I 000, and 5000, Repeat 3. Using Simulink, set up the negative feedback system of Prelab 8. Plot on one graph the error signal of the system for an input of 5u(,) and K = 200, 400800, and i000. Repeat for inputs of Stu and Su) Postlab 1. Use your plots from Lab 1 and compare the expected steady-state errors to those 2. Use your plots from Lab 2 and compare the expected steady-state errors to those 3. Use your plots from Lab 3 and compare the expected steady-state errors to those calculated in the Prelab. Explain the reasons for any discrepancies. calculated in the Prelab. Explain the reasons for any discrepancies. calculated in the Prelab. Explain the reasons for any discrepancies.

Explanation / Answer

Answers

(1) type 1 and more than type 1 (ex: type 2, type 3 etc.. ) will yield zero steady state error. Because the type of the system is more than the type of input (step type 0).

(2) type 2 and more than type 2 (ex: type 3, type 4 etc.. ) will yield zero steady state error. Because the type of the system is more than the type of input (ramp type is 1).

(3) type 0 system will yield infinite state error. Because the type of the system is less than the type of input (ramp type is 1).

(4) type 3 and more than type 3 (ex: type 4, type 5 etc.. ) will yield zero steady state error. Because the type of the system is more than the type of input (parabolic type is 2).

(5) type 0 and type 1 system will yield infinite state error. Because the type of the system is less than the type of input (parabolic type is 2).

(6)

ess for 5u(t) : ess = 1/(1+kp) = 1/(1+ G(0)) = 1/(1+6k/(4*7*9*12)) = 3024/(6k+3024)

ess for 5tu(t) : ess = Infinite (type of input,1 is greater than type of the system,0)

ess for 5t^2u(t) : ess = Infinite (type of input,2 is greater than type of the system,0)

(7)

ess for 5u(t) : ess = 1/(1+kp) = 1/(1+ G(0)) = 1/(1+infinity) = 0

ess for 5tu(t) : ess = 1/kv = 1/(Lt s->0 (s*G(s))) = 1/(k*6*8/(4*7*9*12)) = 3024/48k

ess for 5t^2u(t) : ess = Infinite (type of input,2 is greater than type of the system,1)

(8)

ess for 5u(t) : ess = 1/(1+kp) = 1/(1+ G(0)) = 1/(1+infinity) = 0

ess for 5tu(t) : ess = 1/kv = 1/(Lt s->0 (s*G(s))) = 1/(infinite) = 0

ess for 5t^2u(t) : ess = 1/ka = 1/(Lt s->0 (s^2*G(s))) = 1/(k*1*6*8/(4*7*9*12)) = 3024/48k

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