ME 365 Exam 2 Review 2

ME 365 Exam 2 Review Guide

Transfer Functions • Stability • Bode Plots • Step Response • Frequency Response


1. First Order System Time Constant

For first order systems, the time constant determines how fast the system responds. A larger time constant means a slower response.

The time constant occurs when the response reaches:

63.2% of the final value

From the exam figure, system B decays slower, meaning:

System B has the larger time constant.

2. Transfer Function from Differential Equation

Given:

y'' + 4y' + 6y = 5u'

Take the Laplace transform assuming zero initial conditions:

s²Y(s) + 4sY(s) + 6Y(s) = 5sU(s)

Factor Y(s):

Y(s)(s² + 4s + 6) = 5sU(s)

Transfer function:

G(s)=Y(s)/U(s)=5s/(s²+4s+6)

3. Phase Shift from Sinusoidal Response

Phase difference can be estimated using:

φ = (Δt / T) × 360°

From the graph:

  • Δt = 0.5 s
  • T = 4 s
φ = (0.5 / 4) × 360 = 45°
The output leads the input by 45°.

4. Settling Time of a Second Order System

Given:

y'' + 6y' + 25y = 2u(t)

Compare with standard form:

y'' + 2ζωₙy' + ωₙ²y
  • ωₙ = 5
  • ζ = 0.6

2% settling time approximation:

t₂% ≈ 4/(ζωₙ)
t₂% ≈ 4/(0.6×5)=1.33 s

5. Stability from Pole Locations

Transfer function:

G(s)=(s−2)/(s+3)²

Poles:

s = -3, -3

A system is stable if all poles have negative real parts.

This system is stable.

6. Normal Distribution Probability

Probability problems use the z-score formula:

z=(x−μ)/σ

For:

  • x = 0.8
  • μ = 1.0
  • σ = 0.1
z=(0.8−1.0)/0.1 = -2

From standard normal tables:

P(Z < -2) ≈ 2.3%

7. Confidence Interval

95% confidence interval:

μ = x̄ ± t(S/√n)
  • x̄ = 37.02 mm
  • S = 1.12 mm
  • n = 5
  • t = 2.776
μ = 37.02 ± 1.39 mm

8. Bode Plot Interpretation

Break frequency:

ωb = 1/τ

Smaller break frequency means larger time constant.

System A has the larger time constant.

9. Stability with Parameter K

Closed-loop transfer function:

G(s)=K/(s−1+K)

Pole:

s = 1−K

For stability:

1−K < 0
K > 1

10. Frequency Response Estimation

Low frequency magnitude:

30 dB = 20log(K)
K ≈ 31.62

Natural frequency estimated from peak:

ωₙ ≈ 7 rad/s

Estimated transfer function:

G(s)=1550/(s²+0.221s+49)

11. Steady State Sinusoidal Response

Given:

G(s)=3/(s²+2s+7)
u(t)=4cos(2t)

Magnitude ratio:

M(2)=3/5

Phase:

φ=-tan⁻¹(4/3)

Steady state response:

yss(t)=12/5 cos(2t−tan⁻¹(4/3))

12. Straight-Line Bode Approximation

Given:

G(s)=(s+400)/(s+20)²
  • Zero at 400 rad/s → +20 dB/decade
  • Double pole at 20 rad/s → -40 dB/decade

Combined slope:

-20 dB/decade after 400 rad/s

Final Exam Tips

  • Memorize standard second-order equations
  • Know how to estimate damping ratio from overshoot
  • Understand break frequency relationships
  • Practice transfer function derivations
  • Learn how to read Bode plots quickly
  • Know stability rules using poles