FE Electrical Linear Systems Practice Problems

Answer: D) \(1\text{ s}\)

The Handbook's RC transient has the form \(e^{-t/RC}\) and names \(RC\) the circuit time constant:

$$\tau = RC = (10 \times 10^3)(100 \times 10^{-6}) = 1\text{ s}$$

The units check themselves: an ohm is a volt per ampere and a farad a coulomb per volt, so ohms times farads is coulombs per ampere, which is seconds. One second is a long time constant for an RC circuit, the product of a large resistor and a large capacitor; in one \(\tau\) the capacitor voltage covers 63.2% of its remaining change, and after five it is within 1% of its final value, so this circuit takes about five seconds to settle. The prefixes are where the marks are lost, and carrying \(10^3\) and \(10^{-6}\) explicitly is the defence.

0.001 s takes the resistance as 10 Ω, the kilo dropped.

0.01 s takes the capacitance as 1 µF.

0.1 s takes it as 10 µF.

Answer: A) \(159.2\text{ Hz}\)

The Handbook's resonant frequency is \(\omega_0 = 1/\sqrt{LC} = 2\pi f_0\):

$$\omega_0 = \dfrac{1}{\sqrt{(0.1)(10 \times 10^{-6})}} = \dfrac{1}{\sqrt{10^{-6}}} = 1000\text{ rad/s}, \qquad f_0 = \dfrac{\omega_0}{2\pi} = 159.2\text{ Hz}$$

At this frequency the inductive reactance \(\omega_0 L = 100\ \Omega\) equals the capacitive reactance \(1/(\omega_0 C) = 100\ \Omega\), and they cancel, which is what resonance means; the resistance plays no part in where resonance occurs, only in how sharp it is. The \(2\pi\) between radians per second and hertz is the step most often skipped: the circuit resonates at 1000 rad/s, which is 159 Hz.

1000 Hz is \(\omega_0\) in radians per second reported as hertz.

318.3 Hz is \(1/(\pi\sqrt{LC})\), the 2 of the \(2\pi\) dropped.

6283 Hz multiplies \(\omega_0\) by \(2\pi\) instead of dividing.

Answer: C) \(\dfrac{10}{3}\left(e^{-2t} - e^{-5t}\right)u(t)\)

Two distinct real poles split into partial fractions, \(F(s) = A/(s + 2) + B/(s + 5)\), with the residues found by covering up each factor and evaluating the rest at the pole:

$$A = \dfrac{10}{s + 5}\Big|_{s = -2} = \dfrac{10}{3}, \qquad B = \dfrac{10}{s + 2}\Big|_{s = -5} = -\dfrac{10}{3}$$

Each term is the Handbook's pair \(e^{-at} \leftrightarrow 1/(s + a)\):

$$f(t) = \dfrac{10}{3}\left(e^{-2t} - e^{-5t}\right)u(t)$$

The residues are equal and opposite because the numerator has no \(s\): the Handbook's Initial Value Theorem gives \(\lim_{s\to\infty} sF(s) = 0\), so \(f(0^+)\) must be zero, and only a difference of two exponentials starts at zero. The response rises from zero, peaks near \(t = 0.3\) s, and decays with the slower mode \(e^{-2t}\). Check at \(s = 0\): \(F(0) = 1\), and \(\int_0^\infty f\,dt = \frac{10}{3}(\frac12 - \frac15) = 1\).

10(e^{-2t} - e^{-5t}) has the residues' magnitude wrong, the division by \((-2 + 5) = 3\) omitted.

(10/3)(e^{-5t} - e^{-2t}) has the two signs swapped and is negative for all \(t > 0\), though its transform would have a numerator of \(-10\).

(10/3)(e^{-2t} + e^{-5t}) loses the sign of the second residue; it starts at \(20/3\), contradicting the Initial Value Theorem.

Answer: A) \(\omega_n = 10\text{ rad/s}\), \(\zeta = 0.3\)

Match the denominator to the Handbook's standard form \(s^2 + 2\zeta\omega_n s + \omega_n^2\), constant term first:

$$\omega_n^2 = 100 \Rightarrow \omega_n = 10\text{ rad/s}, \qquad 2\zeta\omega_n = 6 \Rightarrow \zeta = \dfrac{6}{2 \times 10} = 0.3$$

The numerator equals \(\omega_n^2\), so the Handbook's gain \(K\) is 1 and the dc gain is unity. With \(\zeta = 0.3 < 1\) the system is underdamped: its poles are at \(-\zeta\omega_n \pm j\omega_n\sqrt{1 - \zeta^2} = -3 \pm j9.54\), the step response overshoots by \(100e^{-\pi\zeta/\sqrt{1 - \zeta^2}} = 37\%\), and it settles to 2% in \(4/(\zeta\omega_n) = 1.33\) s. The order of the two steps matters: \(\omega_n\) must be found from the constant term before the middle coefficient can be divided into \(\zeta\).

ζ = 0.6 takes the middle coefficient as \(\zeta\omega_n\) rather than \(2\zeta\omega_n\), the factor 2 forgotten.

ωn = 100 reads the constant term as \(\omega_n\) instead of \(\omega_n^2\), the square root forgotten.

Answer: D) Underdamped

Kirchhoff's voltage law around the loop, with the capacitor current \(i = C\,dv_C/dt\), gives a second-order equation for the capacitor voltage, \(LC\,v_C'' + RC\,v_C' + v_C = V\), whose characteristic polynomial is

$$s^2 + \dfrac{R}{L}s + \dfrac{1}{LC}$$

Matching it to the Handbook's second-order model \(s^2 + 2\zeta\omega_n s + \omega_n^2\):

$$\omega_n = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{\sqrt{10^{-8}}} = 10{,}000\text{ rad/s}, \qquad 2\zeta\omega_n = \dfrac{R}{L} = 10{,}000 \Rightarrow \zeta = 0.5$$

The Handbook classifies \(\zeta < 1\) as underdamped: the roots are complex, \(-5000 \pm j8660\), and the capacitor voltage overshoots 24 V and rings at \(\omega_d = \omega_n\sqrt{1 - \zeta^2} = 8660\) rad/s before settling. The dividing line is \(\zeta = 1\), which for this \(L\) and \(C\) needs \(R = 2\sqrt{L/C} = 200\ \Omega\); with half that resistance there is too little damping to prevent oscillation. The source amplitude plays no part in the classification.

Overdamped would need \(\zeta > 1\), a resistance above 200 Ω, giving two real roots and no overshoot.

Critically damped is the single case \(\zeta = 1\), \(R = 200\ \Omega\) exactly.

Undamped is \(\zeta = 0\), no resistance at all, with a ringing that never decays.