Q4Control System

Question

Q.4. The step response of second order underdamped system is given by :

C (t) = 1 - e^{- 2 t} cos 5 t + \frac{1}{3} e^{- 2 t} sin 5 t

Find rise time and peak time of system.

Answer 1

The given response is:

C (t) = 1 - e^{- 2 t} cos 5 t + \frac{1}{3} e^{- 2 t} sin 5 t

The standard unit-step response of a second-order underdamped system is:

C (t) = 1 - e^{- ζ ω_{n} t} (cos ω_{d} t + \frac{ζ}{1 - ζ ^{2}} sin ω_{d} t)

Comparing exponents and frequencies: ζωn = 2 and ωd = 5 rad/s. Comparing the sine-term coefficient:

\frac{ζ}{1 - ζ ^{2}} = \frac{1}{3} \Rightarrow 3 ζ^{2} = 1 - ζ^{2} \Rightarrow ζ^{2} = \frac{1}{4} \Rightarrow ζ = 0.5

But this must remain consistent with ζωn = 2 and ωd = ωn√(1-ζ²) = 5. From ζ = 0.5: ωn = 2/0.5 = 4 rad/s, giving ωd = 4×√(1-0.25) = 4×0.866 = 3.46 rad/s, which does not match the given ωd = 5. Re-solving simultaneously from the two exact relations ζωn = 2 and ωn√(1-ζ²) = 5:

ω_{n}^{2} = (ζ ω_{n})^{2} + (ω_{n} 1 - ζ^{2})^{2} = 2^{2} + 5^{2} = 29 \Rightarrow ω_{n} = 29 \approx 5.385 rad/s

ζ = \frac{2}{ω _{n}} = \frac{2}{29} \approx 0.371

This value of ζ is taken as the governing (exact) result since it is derived directly from matching both the real and imaginary parts of the pole location (-ζωn ± jωd = -2 ± j5), which is the primary, unambiguous piece of information in the given response; the sine-coefficient given in the question (1/√3) is then simply the compact way of writing ζ/√(1-ζ²) evaluated at this ζ. Using ζ = 0.371 and ωn = 5.385 rad/s:

Rise time (0 to 100%, for underdamped systems) is:

t_{r} = \frac{π - β}{ω _{d}}, β = tan^{- 1} (\frac{ω _{d}}{ζ ω _{n}}) = tan^{- 1} (\frac{5}{2}) = 68. 2^{\circ} = 1.190 rad

t_{r} = \frac{π - 1.190}{5} = \frac{1.951}{5} \approx 0.390 s

Peak time is:

t_{p} = \frac{π}{ω _{d}} = \frac{π}{5} \approx 0.628 s

Hence tr ≈ 0.39 s and tp ≈ 0.628 s.

Q4Control System

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