Modelling, Analysis and Control

Question 1

A DC gain equal to one in the closed loop transfer function 𝑇(𝑠) means what in terms of control system performance?

Answer 1

A DC gain of one means the steady-state error for a step response will be 0. The ratio of the output amplitude Y(s) and reference amplitude R(s) are the same.

DC gain = |Y(w=0)|/|R(w=0)|=1

Question 2

If there is a peak in gain at a frequency of 2 Hz in T(s), then what is this frequency known as?

Answer 2

A peak in gain is known as a Resonant peak, and 2Hz will therefore be the resonant frequency corresponding to to the peak.

Question 3

Why are resonant frequencies a problem in control systems?

Answer 3

Resonant frequencies can correspond to a region of high gain, which means the
output will be higher amplitude than the reference, which is undesirable. In excessive regions of high gain caused by a resonance, a system can sustain damage if operated at or near this frequency.

Question 4

The closed loop bandwidth of a control system is an important quantity – why?

Answer 4

The closed loop bandwidth of a control system indicates the maximum frequency a control system can track. Frequencies above the closed loop bandwidth are relatively low gain, meaning the output cannot match the reference in terms of amplitude above the closed loop bandwidth.

Question 5

How would you find the DC gain from a Step Response graph?

Answer 5

You would read the final settling value of the step response of the graph - For example, if it settled on 0.23 then that would be the DC gain.

Question 6

What is the equation for a standard first order system with an input u(t) and an output y(t)?

Answer 6

a dy/dt + b y(t) = c u(t)

Question 7

This is the Laplace transform of a standard first order system:
Y(s)/U(s) = x/z s + 1
What are ‘x’ and ‘z’?

Answer 7

‘x’ is the DC gain of the system, K_DC
‘z’ is the time constant of the system, T

Question 8

Knowing the standard first order system with an input u(t) and out y(t), what are the equations to find the K_DC and T?

Answer 8

K_DC = c / b
T = a / b

Question 9

At what percentages on a standard unit step input would you find T, 2T, 3T and 4T respectively

Answer 9

T = 63%
2T = 87%
3T = 95%
4T = 98%

Question 10

A system G(s) = 0.09/(s+0.24) is connected with proportional compensator K in unity negative feedback. Determine the value of K which will cause the closed-loop to converge to within 5% of its steady-state in 16 seconds. Give the answer to 3 decimal places. Also explain it to yourself like you are trying to teach it to someone.

Hint: It is a standard feedback loop except instead of C(s) its K :)

Answer 10

1. Y(s)/U(s)=G(s)K/1+G(s)K
2. Y(s)/U(s) = 0.09K/s+0.24+0.09K
3. Realize that it needs to be in the standard first order system form.
   So, T = 1/0.24+0.09K
4. 3T = 16
5. sub in an solve
   Answer:
   K = -0.583

Question 11

In MATLAB, what would the code be find the partial fractions of a transfer function G(s).
3 s + 5
G(s) = ————-
s^2 + 4 s + 3

Answer 11

[R,P,K] = residue([3 5],[1 4 3])
3,5 are the numerators
1,4,3 are the denominators

Question 12

Obtain the steady-state error of the following proportional control system to a unit step reference input, where the plant and controller as decribed as follows:

Plant: Y(s)/U(s) = 1/(1.5s2 + 1.0s + 3.0)

PD controller: C(s) = 8.1

Answer 12

SSE = 0.27
SSE equation = c/c+K
K = C(s)
c is from the transfer function, last coefficent is left because it doesnt have an an ‘s’ which tends towards zero

Question 13

what is the purpose of the integral action in the PID?

Answer 13

Remove the steady-state error