Test: Step Response of Second Order Circuits - Electrical Engineering (EE) MCQ

Concept:

Under steady-state:

• When a capacitor, is present with a D.C supply, it behaves as an open circuit.
• When an inductor is present with D.C supply, it behaves as a short circuit

At t = 0⁺ :

Inductor replace with current source I_L (0⁺) = I_L (0^-)

The inductor does NOT allow the sudden change in current”

The capacitor is replaced with the voltage source

V_c (0⁺) = V_c (0^-)

The capacitor does NOT allow the sudden change in voltage.

Calculation:

The circuit at t = 0^-

V_c (0^-) = 5 V

Circuit at t = 0⁺

I = 5/4 = 1.25mA

For ζ>1\zeta > 1ζ>1, the system is overdamped, which means no oscillations occur.

Maximum overshoot, rise time and delay time are the major factor of the transient behaviour of the system and determines the transient characteristics.

Thus, substituting the values will give the required peak time in seconds.

• The peak time (tp) for a second-order system with a damping ratio of ζ = 0.6 is determined by the following formula:

• tp is calculated using the formula: tp = π / (ωn * √(1 - ζ²)).
• In this formula:
  • ωn represents the natural frequency of the system.
  • ζ is the damping ratio.
• To find the peak time, you will need to know the value of ωn.

• Using this approach allows you to assess how quickly the system responds to changes.

The output is given as a function of time. The final value of the output is lim_n->∞c(t) = 1; .

Hence Td (at 50% of the final value) is the solution of 0.5 = 1-e^-Td, and is equal to ln 2 or 0.693 sec.

For a system with a damping ratio of 0.2, the time to reach 90% is approximated by:
t ≈ 2.2 / ωn
If ωn = 5 rad/s,
t ≈ 2.2 sec

C(s)/R(s) = 1/s²+s+1
C(s)/R(s) = w/ws² + 2Gws + w²
Compare both the equations,
w = 1 rad/sec
2Gw = 1
Mp = 16.3 %

The time to reach 98% of the final value is approximately:
t = 4 / (ζ * ωn)
For ζ = 0.5 and ωn = 10,
t = 4 seconds

The rise time tr for a second-order system is approximately:
tr ≈ (π - θ) / (ωn * √(1 - ζ²))
where θ = cos⁻¹(ζ).
For ζ = 0.7 and ωn = 15,
tr ≈ 0.2 sec

C(s) = 1/s-1/s+5-5/(s+5)^2
C(s) = 25/s(s²+10s+25)
R(s) = 1/s
G(s) = 25/(s²+10s+25 )
w= √25
w = 5 rad/sec
G = 1.

The maximum overshoot Mp is given by:
Mp = 100 * exp( - (ζ * π) / √(1 - ζ²))
For ζ = 0.2,
Mp ≈ 20%

C(s)/R(s) = 2/s+1
R(s) = 1/s (step input)
C(s) = 2/s(s+1)
c(t) = 2[1-e^-t]
1.96 = 2[1-e^-T]
T= 4sec.

The time constant, τ, for a critically damped second-order system can be determined using the natural frequency, ωn. The relationship is as follows:

• The time constant is defined as τ = 1/(2ωn).
• This indicates that the system responds quickly without oscillating.
• Understanding this relationship is crucial for analysing system behaviour.

 s + 6/K[s² + s/K + 6/K]
Comparing with s² + 2Gw + w²
w = √6/K
2Gw = 1/K
2 x 0.5 x √6/K = 1/K
K = 1/6.

For a second-order system, the percentage overshoot Mp is given by the formula:
Mp = 100 * exp( - (ζ * π) / √(1 - ζ²))
Substituting ζ = 0.5 and ωn = 10 rad/s,
Mp ≈ 16.3%