Rise Time: What is it? (Equation & How To Calculate It)

Contents

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Key learnings:

Rise Time Definition: Rise time is defined as the duration it takes for a signal to increase from 10% to 90% of its steady value, indicating how quickly a signal changes.
Rise Time Formula: The rise time formula varies based on the system type, with a common calculation for a first-order system being 𝑡 𝑟 = 2.2 𝑇 t r =2.2T.
Calculation Method: To calculate rise time, use the transfer function to determine time constants and calculate the times when the signal reaches 10% and 90% of its final value.
Practical Measurement: Rise time is typically measured using an oscilloscope, which helps analyze the responsiveness of electronic systems.
Significance in Electronics: Understanding and calculating rise time is crucial for optimizing the performance of electronic and control systems, ensuring devices operate efficiently and responsively.

What is Rise Time?

Rise time is the time a waveform takes to go from a stated low level to a stated high level. In analog and digital work those levels are usually 10% and 90% of the final (steady) value. Some logic families quote 20% to 80% instead.

In analog circuits it is the edge duration of a step. In digital circuits it is the time spent between invalid logic levels, which sets how much timing margin you have.

Control texts pick different X% and Y% by damping. The pair is not one universal pair.

A common textbook split for second-order systems is 0% to 100% if underdamped (first crossing of the final value), 5% to 95% if critically damped, and 10% to 90% if overdamped. Instrument rise time stays 10% to 90% unless a standard says otherwise.

Rise Time Equation

The next derivations use a first-order lag and a second-order plant.

The formula depends on the model. First-order and second-order cases differ.

Rise Time of a First Order System

Take this closed-loop first-order system transfer function:

$G(s) = \frac{1}{Ts+1} = \frac{b}{s+a}$

T is the time constant. For G(s)=1/(Ts+1) the unit-step response is 1−exp(−t/T). (The stored latex later drops some minus signs; the inverse Laplace is still 1−exp(−t/T).)

Drive it with a unit step. The Laplace transform of that input is

$R(s) = \frac{1}{s}$

The output is

$C(s) = G(s) R(s)$

$C(s) = \frac{1}{Ts+1} \times \frac{1}{s}$

$C(s) = \frac{\frac{1}{T}}{s+\frac{1}{T}} \times \frac{1}{s}$

$C(s) = \frac{\frac{1}{T}}{s(s+\frac{1}{T})}$

Partial fractions:

$C(s) = \frac{A_1}{s} + \frac{A_2}{s+\frac{1}{T}}$

Find A₁ and A₂:

$\frac{A_1}{s} + \frac{A_2}{s+\frac{1}{T}} = \frac{\frac{1}{T}}{s(s+\frac{1}{T})}$

$A_1 (s+\frac{1}{T}) + A_2 s = \frac{1}{T}$

At s = 0:

$A_1(0+\frac{1}{T}) + A_2 (0) = \frac{1}{T}$

$A_1 \frac{1}{T} = \frac{1}{T}$

$A_1 = 1$

At s = −1/T:

$A_1(\frac{-1}{T} + \frac{1}{T}) + A_2 (\frac{-1}{T}) = \frac{1}{T}$

$A_1 (0) - A_2 \frac{1}{T} = \frac{1}{T}$

$A_2 = -1$

$C(s) = \frac{1}{s} + \frac{-1}{s+\frac{1}{T}}$

Inverse Laplace:

$C(t) = L^{-1} \left[ \frac{1}{s} -\frac{1}{s+\frac{1}{T}} \right]$

$C(t) = 1-e^{\frac{-t}{T}}$

10% to 90% of the final value (here 1):

$C(t_{10}) = 0.10 \quad and \quad C(t_{90}) = 0.90$

$0.10 = 1 - e^{\frac{t_{10}}{T}}$

$e^{\frac{t_{10}}{T}} = 1-0.10$

$e^{\frac{t_{10}}{T}} = 0.9$

$\frac{-t_{10}}{T} = ln(0.9)$

$t_{10} = -T ln(0.9)$

$t_{10} = -T (-0.1053)$

$t_{10} = 0.1053T$

At 90%:

$0.90 = 1 - e^{\frac{t_{90}}{T}}$

$e^{\frac{t_{90}}{T}} = 1 - 0.9$

$e^{\frac{t_{90}}{T}} = 0.1$

$\frac{-t_{90}}{T} = ln(0.1)$

$t_{90} = -T (-2.3025)$

$t_{90} = 2.3025T$

Then t_r is

$t_r = t_{90} - t_{10}$

$t_r = 2.3025T - 0.1053T$

$t_r = 2.197 T$

$t_r \approx 2.2T = \frac{2.2}{a}$

Rise Time of a Second-Order System

For a second-order step, the 0–100% formula below is for the underdamped case (first time the output hits 1). Overdamped 10–90% and critically damped 5–95% need a different calculation. Do not use (π−φ)/ωd if ζ ≥ 1.

Underdamped unit-step form (ζ < 1):

$C(t) = 1-\frac{e^{-\zeta \omega_n t}}{\sqrt{1-\zeta^2}} sin(\omega_d t_r + \phi)$