FE Electrical Control Systems Practice Problems

Answer: D) Curve 4

The Handbook (p. 233) classifies the second-order step response by \(\zeta\) alone: “If the damping ratio \(\zeta\) is less than unity, the system is said to be underdamped; if \(\zeta\) is equal to unity, it is said to be critically damped; and if \(\zeta\) is greater than unity, the system is said to be overdamped.” Take the curves one at a time.

Curve 3 oscillates for ever between 0 and 2 and never settles, so \(\zeta = 0\): undamped.

Curve 1 overshoots. Any overshoot at all means \(\zeta < 1\). Its peak of about \(1.37\) is a \(37\%\) overshoot, and

$$\%\,OS = 100\,e^{-\pi\zeta/\sqrt{1-\zeta^2}} = 37\% \;\Rightarrow\; \zeta \approx 0.3$$

Curve 2 and Curve 4 both rise without overshoot, so both have \(\zeta \ge 1\); no single feature of the drawing separates them, and that is the point of the question. Critical damping is the fastest response with no overshoot, so of the two monotonic curves the \(\zeta = 1\) one is the faster: it is the one lying above the other everywhere.

Check the two against the formulas. For \(\zeta = 1\) and \(\omega_n = 1\), \(y(t) = 1 - e^{-t}(1 + t)\), which reaches \(0.95\) at \(t = 4.7\,\text{s}\). For \(\zeta = 2\) the poles are real at \(-0.268\) and \(-3.732\), the slow one dominates, and \(0.95\) is not reached until \(t = 11.5\,\text{s}\), so Curve 2 is the overdamped one and Curve 4 is the fast one.

Curve 4 is the critically damped response.

Answer: C) \(\dfrac{s + 1}{5}\)

Handbook: the Handbook has no feedforward result. The word “feedforward” appears in the FE Electrical exam specification (p. 487, topic 12A) and in the Chemical exam specification (p. 481), but nowhere in the Control Systems reference section. What p. 231 (PDF viewer: page 237) does print is the feedback answer to the same problem, for its own classical model in which the disturbance \(L(s)\) enters at the plant input:

$$Y(s) = \frac{G_1G_2}{1+G_1G_2H}R(s) + \frac{G_2}{1+G_1G_2H}L(s)$$

Look hard at the second term: it is never zero for any finite loop gain. Feedback can only attenuate a disturbance, by the factor \(1/(1+G_1G_2H)\). Feedforward is the other approach, and it can cancel exactly. Deriving it takes three lines, and the whole difficulty is reading where \(D(s)\) enters.

Step 1 – Write \(Y(s)\) by superposition, following the figure. The command and the feedforward signal meet at the summing junction ahead of the plant, with the feedforward entering through a minus, so the plant sees \(U - G_{ff}D\). The disturbance then adds at the plant output:

$$Y = G\left(U - G_{ff}D\right) + D$$

Step 2 – Collect the \(D\) terms.

$$Y = GU + \left(1 - G\,G_{ff}\right)D$$

Step 3 – Kill the coefficient of \(D\).

$$1 - G(s)G_{ff}(s) = 0 \quad\Longrightarrow\quad G_{ff}(s) = \frac{1}{G(s)} = \frac{s+1}{5}$$

Answer: (C).

Read the summing junction before you invert anything. Suppose \(D\) entered at the plant input instead — which is where the Handbook's own model on p. 231 puts it. Then the plant would see \(U - G_{ff}D + D\) and the output would be \(G\left[U + \left(1-G_{ff}\right)D\right]\), so the compensator would be \(G_{ff} = 1\), a plain unity path, and inverting the plant would be flatly wrong. Same blocks, same numbers, one summing junction moved — completely different answer. That is option (B), and it is the reason this item has a figure.

Why feedforward is exact and feedback never is. Feedforward measures the disturbance and injects the antidote before it reaches the output, so the cancellation does not depend on an error having appeared first. Feedback cannot begin correcting until the disturbance has already moved the output, and it can only divide the effect by \(1+L(s)\). The price is a model: \(G_{ff} = 1/G\) is exactly right only if the plant is exactly \(5/(s+1)\), and every percent of plant error passes straight through. Feedback needs no model. Real systems use both — feedforward for the disturbance you can measure, feedback to clean up everything else.

The ideal compensator is improper, and that matters. \(G_{ff} = (s+1)/5 = s/5 + 1/5\) contains a pure differentiator: its numerator degree exceeds its denominator's, so its gain grows without bound with frequency and it would amplify measurement noise without limit. It cannot be built exactly. In practice you implement \((s+1)/\left[5(\tau s+1)\right]\) with \(\tau\) small enough that the roll-off sits above the disturbance's bandwidth. The exam answer is the ideal inverse; the engineering answer is the ideal inverse rolled off. Knowing both is what the question is really worth.

Step 4 – Eliminate the distractors:

(A) \(5/(s+1)\) is the plant itself, copied instead of inverted. It leaves the disturbance coefficient at \(1 - G^2\), which at dc is \(1 - 25 = -24\): a unit disturbance comes out twenty-four times larger and upside down. Feeding a disturbance forward through a second copy of the plant makes things worse, not better.

(B) \(1\) is the right compensator for the other topology, with \(D\) entering at the plant input; see above.

(D) \(1/5\) inverts only the dc gain \(G(0) = 5\) and throws the dynamics away. The residual is \(1 - \left[5/(s+1)\right]\left(1/5\right) = s/(s+1)\), which is zero at dc and rises to a full unit at high frequency: perfect rejection of a constant disturbance and none at all of a fast one. A static gain cannot cancel a dynamic path.

Answer: B) One (type 1)

The Handbook's unity-feedback model on p. 232 writes the open-loop transfer function as \(G(s) = K_B\prod(1 + s/\omega_m)/[s^T\prod(1 + s/\omega_n)]\) and says that \(T\) denotes the type of the system: the type is the power of \(s\) standing alone in the denominator, the number of pure integrators in the loop. Put the given function in that form:

$$G(s) = \dfrac{K(s + 2)}{s(s + 5)} = \dfrac{2K}{5}\cdot\dfrac{1 + s/2}{s(1 + s/5)}$$

One factor of \(s\) remains alone, so there is one open-loop pole at the origin and the system is type 1, with \(K_B = 2K/5\). The zero at \(-2\) and the pole at \(-5\) are finite and do not count. The type fixes the steady-state error table: a type-1 system follows a step with zero error, a ramp with error \(1/K_B\) and cannot follow a parabola. Note that the type is read from the open-loop function before the loop is closed; the closed-loop poles, roots of \(s^2 + (5 + K)s + 2K = 0\), are not at the origin for any \(K > 0\).

None (type 0) overlooks the free \(s\), treating only the pole at \(-5\) as a pole.

Two (type 2) counts both denominator factors as poles at the origin; \(s + 5\) vanishes at \(-5\), not at \(0\).

Three (type 3) counts the zero as well, and zeros never contribute to the type.

Answer: D) \(\dfrac{1}{s^2 + 3s + 2}\)

First \(sI - A\) and its determinant (p. 60 of the Handbook):

$$sI - A = \begin{bmatrix} s & -1 \\ 2 & s + 3 \end{bmatrix}, \qquad \det = s(s + 3) + 2 = s^2 + 3s + 2$$

The inverse of a \(2 \times 2\) matrix is its adjugate over its determinant (p. 59), the diagonal entries swapped and the off-diagonal entries negated:

$$(sI - A)^{-1} = \dfrac{1}{s^2 + 3s + 2}\begin{bmatrix} s + 3 & 1 \\ -2 & s \end{bmatrix}$$

$$(sI - A)^{-1}B = \dfrac{1}{s^2 + 3s + 2}\begin{bmatrix} 1 \\ s \end{bmatrix}, \qquad G(s) = C(sI - A)^{-1}B = \dfrac{1}{s^2 + 3s + 2}$$

\(B\) selects the second column of the inverse and \(C\) its first row, so only the \((1, 2)\) element survives. The result is the transfer function of \(\ddot{y} + 3\dot{y} + 2y = u\) with \(x_1 = y\) and \(x_2 = \dot{y}\), which is what these companion-form matrices encode; its poles, the roots of \(s^2 + 3s + 2 = (s + 1)(s + 2)\), are the eigenvalues of \(A\), and the system is stable and overdamped with dc gain \(1/2\).

\(\dfrac{s}{s^2 + 3s + 2}\) is the transfer function to the second state \(x_2 = \dot{y}\), obtained with \(C = [0\;\;1]\).

\(\dfrac{s + 3}{s^2 + 3s + 2}\) picks the \((1, 1)\) element of the inverse, which \(B\) does not select.

\(\dfrac{1}{s^2 + 2s + 3}\) swaps the two coefficients of the bottom row when forming the determinant.

Answer: B) \(\zeta \approx 0.10\)

Two readings come off the plot: the low-frequency magnitude is \(0\,\text{dB}\), and the resonant peak reaches \(14\,\text{dB}\). In this standard form \(G(0) = 1\), so the dc gain really is \(0\,\text{dB}\) and the peak stands \(14\,\text{dB}\) above it. That height is the resonant peak magnitude \(M_r\), as a ratio:

$$M_r = 10^{14/20} = 5.01$$

For the standard second-order form the peak of the magnitude response is

$$M_r = \dfrac{1}{2\zeta\sqrt{1 - \zeta^2}}$$

For a sharp peak \(\zeta\) is small and the radical is within a percent of 1, so

$$\zeta \approx \dfrac{1}{2M_r} = \dfrac{1}{2(5.01)} = 0.0998$$

Solving the relation exactly, \(\zeta^4 - \zeta^2 + \dfrac{1}{4M_r^2} = 0\) gives \(\zeta = 0.100\), so the approximation costs nothing here.

One detail the figure shows and the formula explains: the peak does not sit exactly at \(\omega_n\). It sits at the damped resonant frequency \(\omega_r = \omega_n\sqrt{1 - 2\zeta^2}\), which for \(\zeta = 0.1\) is \(0.99\) of \(\omega_n\) — one percent low, which is why the maximum falls just short of the \(\omega = 10\) gridline.

The wrong answers are the three slips this problem invites. \(0.02\) converts the \(14\,\text{dB}\) with \(10\log_{10}\) instead of \(20\log_{10}\), giving \(M_r = 25.1\) and then \(1/(2M_r) = 0.020\); magnitude ratios take \(20\log_{10}\), and only power ratios take \(10\log_{10}\). \(0.20\) uses \(M_r = 1/\zeta\), dropping the factor of 2, so \(\zeta = 1/5.01 = 0.20\). \(0.35\) reads the peak as the \(3\,\text{dB}\) half-power level rather than its actual \(14\,\text{dB}\): that gives \(M_r = 1.41\) and \(\zeta = 1/(2 \times 1.41) = 0.354\).