51-55
56-60
61-65
66
100
51. A first order lag term 1/(1+0.25s) is present in a transfer function. What value of corner frequency is associated with the Bode magnitude plot?
1/(1+0.25s) → [let s = jw] → = 1/(1+0.25[jw]);
Solve for w = omega sub c = -1/0.25(-1) = 4 rad/s
100
56. When phase margin reduces, what effect is felt on the closed-loop system?
Lower stability as phase margin decreases and is unstable at negative PM values
100
61. A lag compensator can be considered as a ---------------- Filter
Low pass
100
66. The inverted pendulum system used for the final lecture demo is a ____________ Input ____________ Output system
Single, multi
Single (Force), double (X and Theta)
200
52. A second order lag term 1/(s^2+2s+8) is present in a transfer function. What value of corner frequency is associated with the Bode magnitude plot?
1/(s^2+2s+8) → [let s = jw] → = 1/([jw]^2+2[jw]+8);
Solve for w = omega sub c:
(-1)w^2 + 2(-1)w + 8 → [factor out -ve] → - (w^2+2w-8); P =-8; S = 2; (4,-2)
Simplifies to: -(w+4)(w-2) → w = 4 and -2 rad/s → corner frequency should be +ve (?) so w = 4 rad/s
*USE EQUATION SHEET*
200
57. For a second order system, if the phase margin is 20 degrees, estimate the level of damping
ζ= PM100=20100=0.2
200
62. Give two drawbacks of a PD controller when compared to a Lead compensator
PD controllers will keep increasing high frequency gain
Increase noise
SSE is not zero
300
53. What are the two major differences between Bode plots for first order systems and for second order systems?
First order system: higher frequency of -90 dB/decade & slope of -40 dB/decade
Second order system: higher frequency of -180 dB/decade & slope of -80 dB/decade
300
58. A lead compensator works similar to a ------------- controller.
PD
300
63. How would you choose the corner frequencies for a lead compensator given the Bode Plot (Frequency Response) of the open loop system?
Allow more positive phase to be introduced into the system
400
54. A system has four poles and one zero. What would be the value of slope at high frequencies in a Bode magnitude plot? What would be the value of phase at high frequencies?
The slope will be -60 dB/dec and the phase will be -270 degrees at high frequencies
400
59. A lag compensator works similar to a ------------- controller.
PI
400
64. How would you choose the corner frequencies for a lag compensator given the Bode Plot (Frequency Response) of the open loop system?
Undesirable phase lag is minimized- close together
500
55. Typically, when the system gain is increased, phase margin will (circle the correct answer) increase/decrease.
Decrease
500
60. A lead compensator can be considered as a ---------------- Filter
High pass
500
65. Name three functions a digital computer can perform for implementation of closed-loop control
Generate reference signal
Compare reference and output to generate error
Implement compensator