MEGR 3171  |  UNC Charlotte Mechatronics 2
Dr. Roger Tipton
Mechanical Engineering & Engineering Science
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Midterm Exam 2 — Week 14  |  Module 3

Midterm 2 Study Guide

Comprehensive review of Module 3: mathematical modeling, Laplace transforms, block diagrams, first- and second-order response, stability, and Bode plots (Weeks 11-13).

Exam Coverage

Section 1: Transfer Functions and Laplace

Q1 A parallel RLC circuit has R = 1000 Ω, L = 0.1 H, C = 10 µF. Derive the transfer function V_out(s) / I_in(s) where V_out is the voltage across the parallel combination.

Parallel impedances: 1/Z_total = 1/R + 1/(sL) + sC

Z_total = V/I = 1/(1/R + 1/(sL) + sC) = sL/(s²LC + sL/R + 1)

= (L/C) · s / (s² + s/(RC) + 1/(LC))

= (0.1/10×10&sup-⁶) · s / (s² + 100s + 10&sup6;) = 10000s / (s² + 100s + 10&sup6;)

G(s) = 10000s / (s² + 100s + 10&sup6;)
Q2 What is the final value (steady-state) of the step response of G(s) = 6 / (s² + 5s + 6)?

Y(s) = G(s)/s. Final value = lim_{s→0} s·G(s)/s = G(0) = 6/(0+0+6) = 1.

Steady-state value = 1 (DC gain = 1)

Section 2: First and Second-Order Response

Q3 A first-order system has time constant τ = 3 s and DC gain K = 4. A unit step is applied at t = 0. At what time does the output reach 2.0?

y(t) = K(1-e^(-t/τ)) = 4(1-e^(-t/3)). Set equal to 2.0:

2.0 = 4(1-e^(-t/3)) → 0.5 = 1 - e^(-t/3) → e^(-t/3) = 0.5 → -t/3 = ln(0.5) = -0.693

t = 3×0.693 = 2.079 s

t = 2.08 s (note: 50% of final value at t = τ·ln2)
Q4 A second-order system exhibits 40% overshoot in its step response. Calculate the damping ratio ζ and the phase margin estimate.

ζ = -ln(OS/100) / √(π² + [ln(OS/100)]²) = -ln(0.40) / √(π² + [ln(0.40)]²)

= 0.9163 / √(9.870 + 0.8396) = 0.9163 / √10.710 = 0.9163 / 3.273 = 0.280

PM ≈ 100ζ = 28°

ζ ≈ 0.28; PM ≈ 28° (lightly damped, close to instability)
Q5 A system has ωn = 5 rad/s and ζ = 0.7. Find: (a) settling time (2%), (b) peak time, (c) percent overshoot.

(a) t_s = 4/(ζω_n) = 4/(0.7×5) = 4/3.5 = 1.14 s

(b) ω_d = ω_n√(1-ζ²) = 5√(1-0.49) = 5√0.51 = 5×0.714 = 3.57 rad/s. t_p = π/ω_d = π/3.57 = 0.880 s

(c) %OS = e^(-π×0.7/√(1-0.49)) × 100 = e^(-2.199/0.714) × 100 = e^(-3.081) × 100 = 4.6%

t_s = 1.14 s, t_p = 0.88 s, %OS = 4.6%

Section 3: Stability and Steady-State Error

Q6 — Routh-Hurwitz Determine stability for characteristic equation: s&sup4; + 2s³ + 3s² + 4s + 5 = 0.

Row 1: 1, 3, 5    Row 2: 2, 4, 0

Row 3, c1: (2×3 - 1×4)/2 = (6-4)/2 = 1.0    c2: (2×5 - 1×0)/2 = 5

Row 4, c1: (1.0×4 - 2×5)/1.0 = (4-10)/1 = -6

Row 5, c1: (-6×5 - 1.0×0)/(-6) = 5

First column: 1, 2, 1, -6, 5. Sign changes: + to + to + to - (1 change) to + (1 change) = 2 sign changes = 2 RHP poles.

UNSTABLE. Two sign changes → two right-half-plane poles.
Q7 — Steady-State Error A unity-feedback system has open-loop transfer function G(s) = 10 / (s(s+5)). Find: (a) system type, (b) steady-state error for a unit step input, (c) steady-state error for a unit ramp input.

(a) One pole at s=0 → Type 1 system.

(b) Type 1: zero steady-state error for step input. e_ss(step) = 0.

(c) K_v = lim_{s→0} s·G(s) = lim_{s→0} s·10/(s(s+5)) = 10/5 = 2. e_ss(ramp) = 1/K_v = 1/2 = 0.5

Type 1; e_ss(step) = 0; e_ss(ramp) = 0.5

Section 4: Bode Plots and System ID

Q8 — Bode Plot For G(s) = 50 / (s(s/10 + 1)(s/100 + 1)), find the low-frequency magnitude in dB at ω = 1 rad/s using asymptotic approximation.

At low frequency (ω < 10), the two poles at 10 and 100 are not yet active. Only gain 50 and integrator 1/s contribute.

|G(j1)| ≈ 50/1 = 50. Magnitude in dB = 20·log(50) = 20×1.699 = 33.98 dB ≈ 34 dB

|G(j1)| ≈ 34 dB
Q9 — System Identification from Step Response A measured step response has 18% overshoot and reaches its first peak at 0.8 s. Find ζ, ωd, and ωn.

ζ = -ln(0.18) / √(π² + [ln(0.18)]²) = 1.715 / √(9.870 + 2.941) = 1.715 / √12.811 = 1.715 / 3.579 = 0.479

ω_d = π / t_p = π / 0.8 = 3.927 rad/s

ω_n = ω_d / √(1-ζ²) = 3.927 / √(1-0.230) = 3.927 / 0.877 = 4.48 rad/s

ζ = 0.479, ω_d = 3.93 rad/s, ω_n = 4.48 rad/s