Control Systems Practice Exam — Free

The closed-loop poles are the roots of $1+G(s) = 0$, i.e. $s(s+2)(s+5) + K = 0$. Expanding $s(s^2 + 7s + 10) + K$ gives the characteristic polynomial

Build the Routh array: the first row takes the coefficients of $s^3$ and $s^1$, the second those of $s^2$ and $s^0$, and each later entry is a cross-product divided by the entry above it.

The $s^1$ entry is $(7\cdot 10 - 1\cdot K)/7$. Stability requires no sign changes in the first column, so every entry must be positive: $70-K>0$ and $K>0$. Hence $0<K<70$.

At the boundary $K = 70$ the $s^1$ row vanishes, and the auxiliary polynomial from the $s^2$ row, $7s^2 + 70 = 0$, gives poles at $s = \pm j\sqrt{10}$: the closed loop oscillates at 3.16 rad/s with constant amplitude. Substituting $s = j\sqrt{10}$ into the characteristic polynomial confirms it: $-j10\sqrt{10} - 70 + j10\sqrt{10} + 70 = 0$. For $K = 71$ the complex pair moves into the right half-plane.

Option A is the usual slip: 10 is only the $s^1$ coefficient, but the Routh test compares $K$ with the product of the inner coefficients, $7\times 10$. Option C reverses the inequality, and option B uses the $s^2$ coefficient alone.

Stable for $0<K<70$ (option D).

Match $T(s)$ to the standard second-order form

From the constant term, $\omega_n^2 = 36$, so $\omega_n = 6$ rad/s. From the $s$ coefficient, $2\zeta\omega_n = 4.8$, so $\zeta = 4.8/12 = 0.40$. Since $0<\zeta<1$ the response is underdamped and overshoots.

Percent overshoot depends on $\zeta$ alone:

The 2 % settling time uses the real part of the poles, $\zeta\omega_n = 2.4$:

Check: the poles are $s = -2.4 \pm j\omega_d$ with $\omega_d = \omega_n\sqrt{1-\zeta^2} = 5.50$ rad/s, so the envelope $e^{-2.4t}$ falls to 2 % at $t = \ln(50)/2.4 = 1.63$ s, consistent with the four-time-constant rule. The peak time is $T_p = \pi/\omega_d = 0.571$ s, and the peak value is $1.254$.

$\zeta = 0.40$, $\omega_n = 6$ rad/s, percent overshoot $\approx 25\%$, 2 % settling time $\approx 1.67$ s.

(a) The loop gain is

so the characteristic equation $1+L(s) = 0$ becomes $s(s+4) + 2(K_p s + K_i) = 0$, i.e.

The specification requires $s^2 + 2\zeta\omega_n s + \omega_n^2 = s^2 + 12s + 100$. Matching coefficients: $4+2K_p = 12$ gives $K_p = 4$, and $2K_i = 100$ gives $K_i = 50$.

(b) With unity feedback, $T(s) = L(s)/(1 + L(s))$. Multiplying numerator and denominator by $s(s+4)$:

(c) For a unit step the final-value theorem gives $y(\infty) = T(0) = 100/100 = 1$, so the steady-state error is 0: the integrator in $C(s)$ makes the loop type 1, which tracks steps exactly. For a unit ramp the error is $1/K_v$, with the velocity constant

so $e_{ss} = 1/25 = 0.04$. Check through the error transfer function $E(s)/R(s) = 1/(1 + L(s)) = s(s+4)/(s^2 + 12s + 100)$: with $R(s) = 1/s^2$, $e_{ss} = \lim_{s\to 0} s\cdot\frac{1}{s^2}\cdot\frac{s(s+4)}{s^2 + 12s + 100} = \frac{4}{100} = 0.04$. Note that $T(s)$ has a zero at $s = -12.5$, so the actual step overshoot will exceed the 9.5 % predicted from $\zeta = 0.6$ alone.

(a) $K_p = 4$, $K_i = 50$; (b) $T(s) = \dfrac{8s + 100}{s^2 + 12s + 100}$; (c) $e_{ss} = 0$ for a step, $e_{ss} = 0.04$ for a ramp.