FE Electrical Linear Systems Practice Problems

Answer: A) \(\dfrac{1}{s + 3}\)

The Handbook's table has the pair \(e^{-at} \leftrightarrow 1/(s + a)\), and with \(a = 3\):

$$\mathcal{L}\{e^{-3t}u(t)\} = \dfrac{1}{s + 3}$$

The Handbook's definition \(F(s) = \int_0^\infty f(t)e^{-st}dt\) derives it in one line, \(\int_0^\infty e^{-(s + 3)t}dt = 1/(s + 3)\) for \(\text{Re}(s) > -3\). A decaying exponential has a single real pole at \(s = -a\), in the left half-plane, and the sign in the denominator is opposite to the sign in the exponent: \(e^{-3t}\) decays and its pole sits at \(-3\). The Handbook's Initial Value Theorem checks the amplitude, \(\lim_{s\to\infty} s/(s + 3) = 1 = f(0^+)\).

1/(s - 3) has the pole at \(+3\), in the right half-plane; it is the transform of the growing \(e^{+3t}\).

3/(s + 3) copies the 3 into the numerator; that transform has initial value 3, not 1.

s/(s + 3) is not proper and has a nonzero value as \(s \to \infty\), which no ordinary function's transform has; it is \(1 - 3/(s + 3)\), the transform of \(\delta(t) - 3e^{-3t}\).

Answer: D) \(\frac{s^2LC + 1}{s^2LC + sRC + 1}\)

Handbook: FE Reference Handbook 10.6, Electrical and Computer Engineering → Analog Filter Circuits – Band-Reject Filters, p. 385 (PDF viewer: page 391). The second circuit shown there is this one: a series resistor feeding an \(L\)–\(C\) series branch, with the output across that branch. The Handbook gives

$$H(s) = \frac{R_P}{R_1}\cdot\frac{s^2 + 1/LC}{s^2 + sR_P/L + 1/LC}, \qquad R_P = \frac{R_1R_2}{R_1 + R_2}$$

With no load resistor across the output (\(R_2 \to \infty\)), \(R_P \to R_1 = R\) and the leading factor becomes 1. Multiplying numerator and denominator by \(LC\) turns it into the form the options use. Deriving it directly takes two lines:

Step 1 – The impedance of the output branch. \(L\) and \(C\) are in series, so their impedances add (p. 366):

$$Z_{LC} = sL + \frac{1}{sC} = \frac{s^2LC + 1}{sC}$$

Step 2 – Voltage division. No current leaves the output terminals, so \(R\) and \(Z_{LC}\) carry the same current:

$$H(s) = \frac{Z_{LC}}{R + Z_{LC}} = \frac{\dfrac{s^2LC+1}{sC}}{R + \dfrac{s^2LC+1}{sC}}$$

Step 3 – Multiply through by \(sC\).

$$H(s) = \frac{s^2LC + 1}{sRC + s^2LC + 1} = \frac{s^2LC + 1}{s^2LC + sRC + 1}$$

Answer: (D).

Why this is a notch. The numerator vanishes at \(s^2 = -1/LC\), that is at \(\omega_0 = 1/\sqrt{LC}\) — the resonant frequency of the branch. There \(Z_{LC} = 0\): the inductor's reactance and the capacitor's cancel exactly, the branch is a short circuit, and the output drops to zero. Away from \(\omega_0\) the branch impedance grows in both directions, so low and high frequencies pass. That is a band-reject, or notch, response, which is exactly what the Handbook's sketch for these circuits shows.

The three limits worth checking. At DC, \(s \to 0\) gives \(H \to 1\): the capacitor blocks, no current flows, and the input appears undivided at the output. At high frequency, \(s \to \infty\) gives \(H \to 1\) as well: the inductor dominates and again takes the whole voltage. In between, \(H = 0\) at \(\omega_0\). Only one option does all three.

Step 4 – Eliminate the distractors:

(A) \(\frac{sRC}{s^2LC+sRC+1}\) is the output taken across the resistor instead. It is the band-pass response of the same circuit: zero at DC, zero at high frequency, and a peak of 1 at \(\omega_0\) — the exact complement of the notch.

(B) \(\frac{s^2LC}{s^2LC+sRC+1}\) is the output across the inductor alone: a high-pass, zero at DC and unity far above resonance.

(C) \(\frac{1}{s^2LC+sRC+1}\) is the output across the capacitor alone: the low-pass response. All four options share the same denominator because they all belong to the same circuit; only the numerator says where the voltage is measured.

Answer: D) \(\frac{5}{s - 7}\)

Handbook: FE Reference Handbook 10.6, Instrumentation, Measurement, and Control → Control Systems, p. 231 (PDF viewer: page 237). The Handbook gives the classical negative-feedback result,

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

so the \(+\) in that denominator belongs to a summing junction that subtracts the feedback. This diagram adds it, and one sign is the whole problem. Derive it from the junction rather than recalling a formula:

Step 1 – Write what the summing junction does. Both of its signs are plus, so the signal entering \(G\) is

$$E(s) = R(s) + H(s)Y(s)$$

Step 2 – Follow the loop. The output is that signal through \(G\):

$$Y = G\left(R + HY\right) \;\Longrightarrow\; Y - GHY = GR \;\Longrightarrow\; T(s) = \frac{Y}{R} = \frac{G}{1 - GH}$$

Step 3 – Substitute. With \(G = 5/(s+3)\) and \(H = 2\), the loop gain is \(GH = 10/(s+3)\):

$$T(s) = \frac{\dfrac{5}{s+3}}{1 - \dfrac{10}{s+3}} = \frac{\dfrac{5}{s+3}}{\dfrac{s+3-10}{s+3}} = \frac{5}{s-7}$$

Answer: (D).

The pole moved the wrong way, and that is the point. The open-loop pole sits at \(s = -3\); closing a positive loop around it has dragged it to \(s = +7\), into the right half-plane. The system is unstable: its natural response contains \(e^{+7t}\), which grows without bound. Negative feedback would have pushed the pole to \(s = -13\), further from the imaginary axis and faster to settle. Same components, opposite behaviour, one sign apart.

A check that costs nothing. The dc gain is \(T(0) = 5/(-7) = -0.71\): negative, even though every block in the diagram has a positive gain. That sign inversion is a signature of \(1 - GH\) having gone negative, which happens whenever the loop gain exceeds 1 — here \(GH(0) = 10/3 = 3.3\).

Step 4 – Eliminate the distractors:

(A) \(\frac{5}{s+3}\) is \(G\) alone, the open-loop transfer function. It is what the system would be with the feedback path cut; the diagram plainly closes the loop.

(C) \(\frac{5}{s+13}\) is the negative-feedback result, \(G/(1 + GH)\) — the Handbook formula applied without checking the signs at the summing junction. This is the single most common error on this item, and the diagram is the only place the sign is written.

(B) \(\frac{5}{s+8}\) is negative feedback with \(H = 1\): the loop gain has lost its factor of 2 as well as its sign. Two errors, and it corresponds to no path through this diagram.

Answer: B) \(25000\)

The Handbook's resonant frequency:

$$\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{(0.5)(32 \times 10^{-12})}} = \dfrac{1}{\sqrt{1.6 \times 10^{-11}}} = \dfrac{1}{4 \times 10^{-6}} = 250{,}000\text{ rad/s}$$

and its series quality factor:

$$Q = \dfrac{\omega_0 L}{R} = \dfrac{250{,}000 \times 0.5}{5} = 25{,}000$$

A \(Q\) of 25 000 is impossible for a coil and capacitor, whose losses limit them to a few hundred, but it is typical of a quartz crystal, whose mechanical resonance is modelled by this absurdly large inductance and tiny capacitance. The bandwidth is \(\omega_0/Q = 10\) rad/s, about 1.6 Hz at a 40 kHz centre, which is why crystals hold an oscillator's frequency to parts per million. The reactance at resonance, \(\omega_0 L = 125\ \text{k}\Omega\), dwarfs the 5 Ω of loss.

250000 is \(\omega_0\) itself, in radians per second, reported as \(Q\).

125000 is \(\omega_0 L\), the inductive reactance in ohms, not yet divided by \(R\).

2500 is the answer with \(L\) a decade too small.

Answer: A) \(H(s) = \frac{10{,}000\,s}{(s + 10)(s + 100)}\)

Handbook: FE Reference Handbook 10.6, Electrical and Computer Engineering → Decibels and Bode Plots, p. 378 (PDF viewer: page 384), which states that Bode magnitude plots use the straight-line asymptotic approximation and tabulates what each term contributes:

• a constant \(K_0\): a horizontal line at \(20\log_{10}(K_0)\);

• \((j\omega)^{+1}\): a straight \(+20\) dB/decade;

• \((1 + j\omega/\omega_c)^{-1}\): flat below \(\omega_c\), then \(-20\) dB/decade above it.

A plot made of straight segments is read by matching each change of slope to one of these terms.

Step 1 – Read the slopes and the corners. Below 10 rad/s the trace climbs 20 dB per decade — from 20 dB at \(\omega = 1\) to 40 dB at \(\omega = 10\). A rising \(+20\) dB/decade at the low-frequency end is a zero at the origin, a factor \(s\) in the numerator. The slope then changes by \(-20\) dB/decade at 10 rad/s (from \(+20\) to flat) and again by \(-20\) dB/decade at 100 rad/s (from flat to \(-20\)): each break of that size is one pole. So

$$H(s) = \frac{Ks}{(s+10)(s+100)}$$

Step 2 – Use the flat band to find \(K\). Between the two corners, \(\omega\) is well above 10 and well below 100, so \(|s + 10| \approx \omega\) and \(|s + 100| \approx 100\):

$$|H| \approx \frac{K\omega}{\omega(100)} = \frac{K}{100}$$

The \(\omega\) cancels, which is exactly why that band is flat. The plot puts it on the 40 dB line, and 40 dB is a factor of 100:

$$\frac{K}{100} = 100 \qquad\Longrightarrow\qquad K = 10{,}000$$

$$H(s) = \frac{10{,}000\,s}{(s+10)(s+100)}$$

Answer: (A).

Check at a second frequency. Take \(\omega = 1\) rad/s, where the plot reads 20 dB, a gain of 10:

$$|H(j1)| = \frac{10{,}000(1)}{|1 + j10|\,|1 + j100|} \approx \frac{10{,}000}{(10)(100)} = 10 \;\checkmark$$

Two points on the trace agreeing is what separates the correct constant from the three that differ only by powers of ten.

What the shape is. A zero at the origin and two poles make a band-pass: the response climbs at low frequency because of the zero, flattens once the first pole cancels that rise, and falls once the second pole takes over. The passband here is one decade wide, from 10 to 100 rad/s.

Step 3 – Eliminate the distractors:

All three share the correct pole-zero structure and differ only in the constant, so each simply puts the flat band at the wrong height:

(D) \(K = 100\) gives a flat level of \(K/100 = 1\), that is 0 dB — the trace would run along the bottom axis.

(B) \(K = 10\) gives 0.1, or \(-20\) dB, off the bottom of the plot entirely.

(C) \(K = 1\) gives 0.01, \(-40\) dB. Reading the flat level is the whole of Step 2; getting the shape right and the level wrong leaves a system with the correct bandwidth and the wrong gain.