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📘 Time Domain Specifications – Complete Explanation

Time domain specifications describe how a system responds in time when subjected to a standard input (usually step input). This topic is very important in GATE examinations.

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🔹 1. Standard Second Order System

G(s) = ω_n² / (s² + 2ζω_ns + ω_n²)

Where:

• ω_n = Natural frequency
• ζ = Damping ratio

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🔹 2. Important Time Specifications

• Rise Time (t_r)
• Peak Time (t_p)
• Settling Time (t_s)
• Maximum Overshoot (M_p)

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🔹 3. Rise Time (t_r)

Time taken for response to rise from 0% to 100% (underdamped case).

t_r ≈ (π − θ) / ω_d

Where:

• ω_d = ω_n√(1 − ζ²)
• θ = tan⁻¹(√(1 − ζ²)/ζ)

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🔹 4. Peak Time (t_p)

t_p = π / ω_d

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🔹 5. Maximum Overshoot (M_p)

M_p = e^{(−πζ / √(1 − ζ²))}

Percentage overshoot:

%OS = M_p × 100

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🔹 6. Settling Time (t_s)

For 2% criterion:

t_s ≈ 4 / (ζω_n)

For 5% criterion:

t_s ≈ 3 / (ζω_n)

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🔹 7. Effect of Damping Ratio

• ζ = 0 → Undamped (oscillatory)
• 0 < ζ < 1 → Underdamped
• ζ = 1 → Critically damped
• ζ > 1 → Overdamped

Higher ζ → Less overshoot but slower response.

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🔹 8. Worked Example

Given:

G(s) = 25 / (s² + 4s + 25)

Step 1: Compare with standard form

ω_n² = 25 → ω_n = 5

2ζω_n = 4 → 2ζ(5) = 4 → ζ = 0.4

Step 2: Calculate ω_d

ω_d = 5√(1 − 0.16) = 5√(0.84) = 4.58

Step 3: Peak Time

t_p = π / 4.58 ≈ 0.69 sec

Step 4: Settling Time

t_s = 4 / (0.4 × 5) = 4 / 2 = 2 sec

Step 5: Maximum Overshoot

M_p ≈ 0.254 %OS ≈ 25.4%

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🎯 GATE Important Points

• Most questions from second order systems
• Damping ratio directly affects overshoot
• Settling time inversely proportional to ζω_n
• Memorize formulas properly

Time Domain = Speed + Stability of Response