FE Electrical Control Systems Practice Problems

Answer: D) \(G_1(s)\,G_2(s)\)

The Handbook defines a transfer function on p. 225 as the ratio of output to input transforms, \(G(s) = Y(s)/X(s)\). Apply that definition twice. The first block gives \(Y_1(s) = G_1(s)X(s)\); the second takes \(Y_1\) as its input and gives \(Y(s) = G_2(s)Y_1(s)\). Substituting,

$$Y(s) = G_2(s)G_1(s)X(s) \quad\Rightarrow\quad \dfrac{Y(s)}{X(s)} = G_1(s)\,G_2(s)$$

Cascaded blocks multiply, which is the same statement as convolution in the time domain becoming a product in the \(s\) domain. For scalar transfer functions the order does not matter, \(G_1G_2 = G_2G_1\), and the poles of the combination are the poles of both blocks together, the zeros likewise. This is also how the Handbook's own diagram is read: the controller \(G_1\) and the plant \(G_2\) in series form the forward path \(G_1G_2\), and with the sensor \(H\) the loop gain is the product \(G_1G_2H\) of the three. The rule assumes that the second block does not load the first, which is built into the block-diagram idealisation.

\(G_1(s) + G_2(s)\) is the parallel connection, two blocks fed by the same input with their outputs summed.

\(\dfrac{G_1(s)}{1 + G_1(s)G_2(s)}\) is the closed-loop transfer function of \(G_1\) with \(G_2\) in a negative feedback path, not a cascade.

\(\dfrac{G_1(s)}{G_2(s)}\) would require the second block to be inverted; nothing in a cascade divides.

Answer: B) \(-90°\)

The Handbook's Bode table on p. 374 lists the term \((j\omega)^{\pm 1}\) with magnitude \(\pm 20\log_{10}(\omega)\) and phase \(\pm 90°\). For the exponent \(-1\), a pure integrator,

$$G(j\omega) = \dfrac{1}{j\omega} = \dfrac{-j}{\omega} \quad\Rightarrow\quad \angle G = 0° - \angle(j\omega) = -90°$$

The phase is constant because \(j\) carries all of it: \(j\omega\) lies on the positive imaginary axis at \(+90°\) for every \(\omega > 0\), and its reciprocal lies on the negative imaginary axis at \(-90°\). Physically the integrator's output lags a sinusoidal input by a quarter cycle, \(\int\sin\omega t\,dt = -\cos(\omega t)/\omega = \sin(\omega t - 90°)/\omega\), while its magnitude \(1/\omega\) falls at 20 dB per decade through 0 dB at \(\omega = 1\). In a loop the integrator's fixed \(-90°\) consumes half the phase budget before any pole is counted, which is why type-1 systems have less phase margin to spare than type-0 systems of the same bandwidth.

\(+90°\) is the phase of \(j\omega\) itself, the differentiator; the sign is reversed by the reciprocal.

\(-180°\) is the phase of a double integrator, \(1/(j\omega)^2 = -1/\omega^2\).

\(-45°\) is the phase of a first-order pole \(1/(1 + j\omega/\omega_c)\) at its corner, a frequency-dependent value; the integrator has no corner.

Answer: B) \(\omega_n = 5\) rad/s, \(\zeta = 0.6\)

Match the denominator to the Handbook's second-order model on p. 227, \(s^2 + 2\zeta\omega_n s + \omega_n^2\):

$$\omega_n^2 = 25 \quad\Rightarrow\quad \omega_n = 5\text{ rad/s}$$

$$2\zeta\omega_n = 6 \quad\Rightarrow\quad \zeta = \dfrac{6}{2 \times 5} = 0.6$$

The numerator 25 equals \(\omega_n^2\), so the steady-state gain \(K\) is 1 and a unit step settles at 1. With \(\zeta = 0.6 < 1\) the system is underdamped: the poles are \(-\zeta\omega_n \pm j\omega_n\sqrt{1 - \zeta^2} = -3 \pm j4\), the damped frequency is \(\omega_d = 4\) rad/s, the percent overshoot \(100e^{-\pi\zeta/\sqrt{1 - \zeta^2}} = 9.5\%\), the peak time \(\pi/\omega_d = 0.785\) s and the 2% settling time \(4/(\zeta\omega_n) = 1.33\) s (p. 228). The pole coordinates are a useful check: the real part \(3\) is \(\zeta\omega_n\) and the distance from the origin \(\sqrt{9 + 16} = 5\) is \(\omega_n\).

\(\omega_n = 25\) rad/s reads the constant term as \(\omega_n\) instead of \(\omega_n^2\).

\(\zeta = 1.2\) forgets the factor 2 in \(2\zeta\omega_n\), dividing 6 by 5 alone; it would put the system overdamped, contradicting the complex poles.

Answer: A) 25.4%

The Handbook gives the percent overshoot of the underdamped second-order model on p. 227 as \(\%OS = 100e^{-\pi\zeta/\sqrt{1 - \zeta^2}}\). The exponent first:

$$\dfrac{\pi\zeta}{\sqrt{1 - \zeta^2}} = \dfrac{\pi(0.4)}{\sqrt{1 - 0.16}} = \dfrac{1.257}{0.917} = 1.371$$

$$\%OS = 100e^{-1.371} = 25.4\%$$

The overshoot depends on \(\zeta\) alone; \(\omega_n\) sets how fast the peak arrives, \(t_p = \pi/(\omega_n\sqrt{1 - \zeta^2}) = \pi/7.33 = 0.43\) s, but not how high it is, which is why the natural frequency is not needed here. The value sits in the usual design band: \(\zeta = 0.4\) is lightly damped, and the response rings visibly; \(\zeta = 0.5\) would give 16.3%, \(\zeta = 0.7\) about 4.6%, and \(\zeta = 1\) none. The Handbook's peak value \(M_p = 1 + e^{-\pi\zeta/\sqrt{1 - \zeta^2}} = 1.254\) is the same result expressed as the peak of a unit step response, 25.4% above the final value of 1.

28.5% is \(100e^{-\pi\zeta}\), the square root in the denominator of the exponent omitted.

40% reports the damping ratio itself as a percentage.

74.6% is \(100(1 - e^{-1.371})\), the complement, the fraction of the way to the peak rather than the overshoot beyond the final value.

Answer: A) \(s^3 + 8s^2 + 15s + K = 0\)

The Handbook gives the closed-loop characteristic equation on p. 226 as \(1 + G_1(s)G_2(s)H(s) = 0\); with unity feedback the loop gain is \(G(s)\):

$$1 + \dfrac{K}{s(s + 3)(s + 5)} = 0 \quad\Rightarrow\quad s(s + 3)(s + 5) + K = 0$$

$$s(s^2 + 8s + 15) + K = 0 \quad\Rightarrow\quad s^3 + 8s^2 + 15s + K = 0$$

The roots of this cubic are the closed-loop poles. At \(K = 0\) they are the open-loop poles \(0\), \(-3\), \(-5\); as \(K\) grows they move, and the equation shows that \(K\) enters only the constant term, so the sum of the poles stays at \(-8\) whatever the gain. That is why increasing \(K\) eventually drives a pair of poles into the right half-plane: two of them move right while the third moves left to keep the sum fixed. The boundary is where a pair sits on the imaginary axis, at \(s = \pm j\sqrt{15}\) with \(K = 120\), found by substituting \(s = j\omega\) and separating real and imaginary parts.

\(s^3 + 8s^2 + 15s - K = 0\) corresponds to \(1 - G = 0\), positive feedback.

\(s^3 + 3s^2 + 5s + K = 0\) copies the numbers 3 and 5 from the factors as if they were the expanded coefficients; \((s + 3)(s + 5) = s^2 + 8s + 15\).

\(s^2 + 8s + K = 0\) drops the pole at the origin, expanding \((s + 3)(s + 5)\) without the leading \(s\).