Question 13.4: FINDING THE STEP RESPONSE OF UNITY FEEDBACK CONTROL SYSTEM A...

The relations among V_{r}, V_{o}, and  V_{e}, in Laplace’s s domain, are

V_{e}(s) =V_{r}(s) -V_{o}(s)

\frac{V_{0}(s)}{V_{e}(s)} =\frac{K}{s(1+s)(1+0.2s)}

which gives

[s(1+s)(1+0.2s)]V_{0}(s)=KV_{e}(s)

[s+1.2s^{2}+0.2s^{3}]V_{0}(s)=KV_{e}(s)

which can be written in the time domain as

0.2\frac{d^{3}v_{o} }{dt^{3} } +1.2\frac{d^{2} v_{o} }{dt^{2} } +\frac{dv_{o} }{dt} =Kv_{e}=K(v_{r}-v_{o})    (13.1)

Dividing both sides by 0.2 gives

\frac{d^{3}v_{o} }{dt^{3} } +6\frac{d^{2} v_{o} }{dt^{2} } +5\frac{dv_{o} }{dt} =5Kv_{e}

where v_{e} = v_{r} – v_{o}, or

\frac{d^{3}v_{o} }{dt^{3} }= -6\frac{d^{2} v_{o} }{dt^{2} } -5\frac{dv_{o} }{dt} +5Kv_{e}                                (13.2)

which can be denoted by

\overset{…}{v}_{o} =-6\ddot{v}_{0} –5\dot{v}_{o}+5Kv_{e}                            (13.3)

Integrating the third derivative three times should yield the second derivative of v_{o}, the first derivative of v_{o}, and the output v_{o}. The third derivative on the left- hand side of Equation 13.3 must equal the sum of the terms on the right-hand side. The circuit for PSpice simulation of Equation 13.2 is shown in Figure 13.18.

The gain K is defined in PSpice as a variable.

The input voltage v_{in} is obtained from

v_{in} = \overset{…}{v}_{o}=-6\ddot{ v}_{0} –5\dot{v}_{o}+5Kv_{e}

It should be noted that the output signal of an integrator is inverted. This should be taken into account in summing signals v_{in} and v_{e}.

We can use the ABMs to represent the transfer function and the comparator as shown in Figure 13.18(b).

The listing of the circuit file is as follows:

The plot of the transient response for the feedback control system is shown in Figure 13.19. A higher value of K gives more overshoot and the system tends to be unstable. The transient should settle to the input signal level.