Rise time for a first-order system is approximately 2.2 times the time constant.

Rise time for a first-order system is approximately 2.2 times the time constant.

• A. True
• B. False
• C. Only for underdamped systems
• D. It depends on the gain

Answer: (A)

When a first‑order system responds to a step input, its output follows an exponential approach to its final value. The time constant τ sets how fast that approach happens. If we define rise time as the time it takes to go from 10% to 90% of the final value, you solve for the times when the output equals 0.1 and 0.9 of the final value. For a standard first‑order response y(t) = K(1 − e^(−t/τ)), the times are: - t when y = 0.1K gives e^(−t/τ) = 0.9, so t1 ≈ 0.105τ - t when y = 0.9K gives e^(−t/τ) = 0.1, so t2 ≈ 2.303τ Rise time = t2 − t1 ≈ (2.303 − 0.105)τ ≈ 2.20τ. Thus rise time is about 2.2 times the time constant. The gain only scales the final value and does not affect τ, and first‑order systems aren’t oscillatory, so the underdamped or damping ideas don’t apply here. In short, the statement is true for the common 10%–90% rise-time definition.

When a first‑order system responds to a step input, its output follows an exponential approach to its final value. The time constant τ sets how fast that approach happens. If we define rise time as the time it takes to go from 10% to 90% of the final value, you solve for the times when the output equals 0.1 and 0.9 of the final value.

For a standard first‑order response y(t) = K(1 − e^(−t/τ)), the times are:

• t when y = 0.1K gives e^(−t/τ) = 0.9, so t1 ≈ 0.105τ

• t when y = 0.9K gives e^(−t/τ) = 0.1, so t2 ≈ 2.303τ

Rise time = t2 − t1 ≈ (2.303 − 0.105)τ ≈ 2.20τ.

Thus rise time is about 2.2 times the time constant. The gain only scales the final value and does not affect τ, and first‑order systems aren’t oscillatory, so the underdamped or damping ideas don’t apply here. In short, the statement is true for the common 10%–90% rise-time definition.