FE Electrical and Computer Control Systems Practice Problems

The current FE Electrical and Computer specification assigns 6-9 of the 110 exam questions to Control Systems. Form A contains 7 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A unity-feedback antenna position loop uses a PI controller $C (s) = 4.0 + \frac{10}{s}$ driving a plant $G (s) = \frac{2.5}{s + 5}$ . The command is a ramp of constant rate 3.0 rad/s. The steady-state tracking error is most nearly:

(A)0.20 rad
(B)0.33 rad
(C)0.60 rad
(D)1.50 rad

Show worked solution

Answer: (C)

Integral action raises the loop to type 1, so the 3.0 rad/s ramp is tracked with a constant lag of 0.60 rad rather than a growing one.

Integral action inserts a pole at the origin, making the loop type 1, for which a ramp command produces a finite error set by the velocity error constant.

L (s) = (4 + \frac{10}{s}) \frac{2.5}{s + 5} = \frac{2.5 ( 4 s + 10 )}{s ( s + 5 )}

Only the integral gain survives the limit, so the proportional term contributes nothing to $K_{v}$ .

K_{v} = s \to 0 lim s L (s) = \frac{2.5 ( 10 )}{5} = 5.0 s^{- 1}

Error scales with the commanded rate, so the ramp slope must be carried through rather than assumed unity.

e_{ss} = \frac{R}{K _{v}} = \frac{3.0}{5.0} = 0.60 rad

FE Reference Handbook — Control Systems: Steady-State Error and Static Error Constants

Why the other choices appear

(A)Using the unit-ramp result $1/ K_{v} = 1/5 = 0.20$ rad and ignoring the 3.0 rad/s command slope.
(B)Applying the type-0 position-error form $1/ (1 + K_{p} G (0)) = 1/ (1 + 2) = 0.33$ rad, which discards the integrator.
(D)Building $K_{v}$ from the proportional gain, $4 (0.5) = 2.0$ s $^{- 1}$ , giving $3.0/2.0 = 1.50$ rad.