Optional homework

1. The curve peaks at about 1.36, so Mᵣ ≈ 1.36. This is an amplitude ratio, not a frequency.
2. The frequency directly below that peak is about 3.3 rad/s, so ωᵣ ≈ 3.3 rad/s.
3. Follow the falling curve to the 0.707 line, then read its frequency: BW ≈ 5.5 rad/s.
4. To check the graph, its model has ωₙ = 4 rad/s and ζ = 0.4. The supplied formulas give
   Mᵣ = 1 divided by 2(0.4)√(1 − 0.4²) ≈ 1.364,
   ωᵣ = 4√(1 − 2(0.4)²) ≈ 3.298 rad/s,
   BW = 4√[0.68 + √1.4624] ≈ 5.498 rad/s.
   Small differences from reading the plotted curve are expected.

Why: peak height, peak location and the 0.707 crossing describe three different features of the same curve.

Watch for: BW is a frequency. It is not the gain value 0.707, and it is not the resonant frequency.

1. Match the constant and s coefficients:
   A:   ωₙ² = 9,   2ζωₙ = 3  ⇒  ωₙ = 3 rad/s,   ζ = 0.5.
   B:   ωₙ² = 36,   2ζωₙ = 6  ⇒  ωₙ = 6 rad/s,   ζ = 0.5.
2. Both peak measures depend only on ζ:
   Mₚ = e^{−π(0.5)/√0.75} ≈ 0.163  (16.3%),   Mᵣ = 1 divided by √0.75 ≈ 1.155.
   They are the same for A and B.
3. At fixed ζ, the standard unit-step response depends on ωₙt. Thus y_B(t) = y_A(2t): B reaches each corresponding feature in half the time. Its delay, rise and ±5% settling times are half those of A.
4. The supplied BW formula has ωₙ as an outside multiplier. At ζ = 0.5,
   BW_A ≈ 3.816 rad/s,   BW_B ≈ 7.632 rad/s.
   B has twice the bandwidth; its frequency curve shifts right by a factor of two.

Why: keeping ζ fixed preserves the curve's shape while ωₙ changes its time and frequency scales.

Watch for: a faster response need not have more overshoot. Here the damping ratios are equal.

1. Both constant coefficients give ωₙ = 3 rad/s. Matching the s coefficients gives
   ζ_A = 1.8 divided by 2(3) = 0.3,   ζ_B = 4.8 divided by 2(3) = 0.8.
2. Use the overshoot formula for both systems:
   Mₚ,A = e^{−π(0.3)/√0.91} ≈ 0.3723,   Mₚ,B = e^{−π(0.8)/0.6} ≈ 0.01516.
   That is about 37.2% for A and 1.52% for B.
3. A has ζ < 1/√2, so
   Mᵣ,A = 1 divided by 2(0.3)√0.91 ≈ 1.747.
   B has ζ > 1/√2: Mᵣ,B = 1 at ω = 0. It has no resonant peak at a positive frequency.
4. Substitute each ζ into the supplied BW expression:
   BW_A = 3√[0.82 + √1.6724] ≈ 4.361 rad/s,
   BW_B = 3√[−0.28 + √1.0784] ≈ 2.613 rad/s.
   B has less overshoot and resonance, but a smaller bandwidth.

Why: damping changes both the step-response shape and the frequency-response peak, even with the same ωₙ.

Watch for: underdamped (ζ < 1) does not automatically mean a positive-frequency resonant peak. Its threshold is ζ < 1/√2.

1. The first 50% crossing is 0.25 s, so t_d = 0.25 s.
2. Rise time is the interval between the first 10% and 90% crossings:
   tᵣ = t₉₀ − t₁₀ = 0.50 − 0.10 = 0.40 s.
3. The peak output value is 1.20 and the final value is 1:
   Mₚ = 1.20 − 1 divided by 1 = 0.20 = 20%.
   The peak value alone does not tell us the time at which the peak occurred.
4. The ±5% band is [0.95, 1.05]. The output first enters at 1.10 s but leaves again at 1.25 s. Only after 1.60 s does it remain inside, so t_s = 1.60 s.

Why: these measurements use the observed crossings and levels directly; no fitted second-order model is needed.

Watch for: entering the band once is not enough. Settling time requires the response to stay there.

1. B has the larger ωₙ. A and B have the same ζ: their step overshoot and resonant peak height match. B's step response is faster, and its frequency-response features lie at higher frequencies.
2. A has the smaller ζ. Compared with C at the same ωₙ, A has more step overshoot and a taller resonant peak. Both clues support the same damping comparison.
3. With ζ fixed, a shorter settling time helps identify the larger ωₙ for A and B. It cannot determine both answers: at fixed ωₙ, settling time is not monotonic in ζ. Use overshoot and resonant peak height to compare A and C.

Why: the same ωₙ and ζ shape both plots; matching the two views gives more evidence than one feature alone.

Watch for: peak height and peak frequency are different. A taller peak points to smaller damping; a horizontal shift with unchanged shape points to a changed frequency scale.

1. Multiply each amplitude ratio by the input amplitude, R = 10 cm:
   ω = 2:   Y = 10(0.95) = 9.5 cm,
   ω = 8:   Y = 10(0.707) ≈ 7.07 cm,
   ω = 30:   Y = 10(0.10) = 1.0 cm.
   The output frequency remains 2, 8 or 30 rad/s, respectively.
2. The low-frequency gain is 1, so its 0.707 level is 0.707. The single downward crossing is at BW ≈ 8 rad/s.
3. At 2 rad/s the output amplitude is close to the command amplitude. At 30 rad/s it is strongly attenuated. The supplied data establish amplitude following at low frequency; close waveform following would also require a small phase shift.
4. Bandwidth does not mark a frequency above which the output is zero. At 30 rad/s, above BW, the output amplitude is still 1 cm.

Why: a magnitude plot tells how much of each sinusoidal input amplitude appears at the output after transients have decayed.

Watch for: gain 0.707 means about 70.7% of the input amplitude, not 0.707 cm and not complete rejection.