Rise time

In electronics, when describing a voltage or current step function, rise time is the time taken by a signal to change from a specified low value to a specified high value. Typically, in analog electronics, these values are 10% and 90% of the step height: in control theory applications, according to Template:Harvtxt, rise time is defined as "the time required for the response to rise from x% to y% of its final value", with 0%-100% rise time common for underdamped second order systems, 5%-95% for critically damped and 10%-90% for overdamped.[1] The output signal of a system is characterized also by fall time: both parameters depend on rise and fall times of input signal and on the characteristics of the system.

Overview

Rise time is an analog parameter of fundamental importance in high speed electronics, since it is a measure of the ability of a circuit to respond to fast input signals. Many efforts over the years have been made to reduce the rise times of generators, analog and digital circuits, measuring and data transmission equipment, focused on the research of faster electron devices and on techniques of reduction of stray circuit parameters (mainly capacitances and inductances). For applications outside the realm of high speed electronics, long (compared to the attainable state of the art) rise times are sometimes desirable: examples are the dimming of a light, where a longer rise-time results, amongst other things, in a longer life for the bulb, or digital signals apt to the control of analog ones, where a longer rise time means lower capacitive feedthrough, and thus lower coupling noise.

Simple examples of calculation of rise time

The aim of this section is the calculation of rise time of step response for some simple systems: all notations and assumptions required for the following analysis are listed here.

${\displaystyle BW=f_{H}-f_{L}\,}$
and since the low frequency cutoff ${\displaystyle f_{L}}$ is usually several decades lower than the high frequency cutoff ${\displaystyle f_{H}}$,
${\displaystyle BW\cong f_{H}\,}$
• All systems analyzed here have a frequency response which extends to 0 (low-pass systems), thus
${\displaystyle f_{L}=0\,\Leftrightarrow \,f_{H}=BW}$ exactly.

Gaussian response system

A system is said to have a Gaussian response if it is characterized by the following frequency response

${\displaystyle |H(\omega )|=e^{-{\frac {\omega ^{2}}{\sigma ^{2}}}}}$

where ${\displaystyle \sigma >0}$ is a constant, related to the high frequency cutoff by the following relation:

${\displaystyle f_{H}={\frac {\sigma }{2\pi }}{\sqrt {{\frac {3}{20}}ln10}}\cong 0.0935\sigma }$

The corresponding impulse response can be calculated using the inverse Fourier transform of the shown frequency response

${\displaystyle {\mathcal {F}}^{-1}\{H\}(t)=h(t)={\frac {1}{2\pi }}\int \limits _{-\infty }^{+\infty }{e^{-{\frac {\omega ^{2}}{\sigma ^{2}}}}e^{i\omega t}}d\omega ={\frac {\sigma }{2{\sqrt {\pi }}}}e^{-{\frac {1}{4}}\sigma ^{2}t^{2}}}$

Applying directly the definition of step response

${\displaystyle V(t)=V_{0}{H*h}(t)={\frac {V_{0}}{\sqrt {\pi }}}\int \limits _{-\infty }^{\frac {\sigma t}{2}}e^{-\tau ^{2}}d\tau ={\frac {V_{0}}{2}}\left[1+\mathrm {erf} \left({\frac {\sigma t}{2}}\right)\right]\Leftrightarrow {\frac {V(t)}{V_{0}}}={\frac {1}{2}}\left[1+\mathrm {erf} \left({\frac {\sigma t}{2}}\right)\right]}$

Solving for time the two following equations by using known properties of the error function

${\displaystyle 0.1={\frac {1}{2}}\left[1+\mathrm {erf} \left({\frac {\sigma t_{1}}{2}}\right)\right]\qquad 0.9={\frac {1}{2}}\left[1+\mathrm {erf} \left({\frac {\sigma t_{2}}{2}}\right)\right]}$
${\displaystyle t_{r}={\frac {4}{\sigma }}{\mathrm {erf} ^{-1}(0.8)}\cong {\frac {0.3394}{f_{H}}}}$

and then

${\displaystyle t_{r}\cong {\frac {0.34}{BW}}\quad \Longleftrightarrow \quad BW\cdot t_{r}\cong 0.34}$

One-stage low-pass RC network

For a simple one-stage low-pass RC network, also known as a single-pole filter, the 10% to 90% rise time is proportional to the network time constant ${\displaystyle \tau =RC}$:

${\displaystyle t_{r}\cong 2.197\tau \,}$

The proportionality constant can be derived by using the output response of the network to a unit step function input signal of ${\displaystyle V_{0}}$ amplitude, also known as its step response:

${\displaystyle V(t)=V_{0}\left(1-e^{-{\frac {t}{\tau }}}\right)}$

Solving for time

${\displaystyle {\frac {V(t)}{V_{0}}}=\left(1-e^{-{\frac {t}{\tau }}}\right)}$
${\displaystyle {\frac {V(t)}{V_{0}}}-1=-e^{-{\frac {t}{\tau }}}}$
${\displaystyle 1-{\frac {V(t)}{V_{0}}}=e^{-{\frac {t}{\tau }}}}$
${\displaystyle \ln \left(1-{\frac {V(t)}{V_{0}}}\right)=-{\frac {t}{\tau }}}$
${\displaystyle t=-\tau \;\ln \left(1-{\frac {V(t)}{V_{0}}}\right)}$

We call t1 the time needed to go from 0% to 10% of the steady-state value, and t2 the one to 90%. Thus t1 is such that ${\displaystyle {\frac {V(t)}{V_{0}}}=0.1}$ and t2 is such that ${\displaystyle {\frac {V(t)}{V_{0}}}=0.9}$. Solving the previous equation for these two values we find the analytical expression for t1 and t2:

${\displaystyle t_{1}=-\tau \;\ln \left(1-0.1\right)=-\tau \;\ln \left(0.9\right)=-\tau \;\ln \left({\frac {9}{10}}\right)=\tau \;\ln \left({\frac {10}{9}}\right)=\tau ({\ln 10}-{\ln 9})}$

We obtain t2 in the same way, resulting in

${\displaystyle t_{2}=\tau \ln {10}\,}$

Subtracting ${\displaystyle t_{1}}$ from ${\displaystyle t_{2}}$ we obtain the rise time, which is therefore proportional to the time constant:

${\displaystyle t_{r}=t_{2}-t_{1}=\tau \cdot \ln 9\cong \tau \cdot 2.197}$

Now, noting that

${\displaystyle \tau =RC={\frac {1}{2\pi f_{H}}}}$

(see here for the proof of the previous equation) then

${\displaystyle t_{r}\cong {\frac {2.197}{2\pi f_{H}}}\cong {\frac {0.349}{f_{H}}}}$

and since the high frequency cutoff is equal to the bandwidth (signal processing)

${\displaystyle t_{r}\cong {\frac {0.35}{BW}}\quad \Longleftrightarrow \quad BW\cdot t_{r}\cong 0.35}$

Note that, if the rise time is 20% to 80% instead of 10% to 90%, ${\displaystyle t_{r}}$ becomes:

${\displaystyle t_{r}\cong 1.386\tau \quad \Longleftrightarrow \quad t_{r}\cong {\frac {0.22}{BW}}}$

Consider a system composed by ${\displaystyle n}$ cascaded non interacting blocks, each having a rise time ${\displaystyle \scriptstyle {t_{r_{i}}}}$ and no overshoot in their step response: suppose also that the input signal of the first block has a rise time whose value is ${\displaystyle \scriptstyle {t_{r_{S}}}}$. Then its output signal has a rise time ${\displaystyle \scriptstyle {t_{r_{O}}}}$ equal to

${\displaystyle t_{r_{O}}={\sqrt {t_{r_{S}}^{2}+t_{r_{1}}^{2}+\dots +t_{r_{n}}^{2}}}}$

This result is a consequence of the central limit theorem, as reported in Template:Harvnb and proved by Henry Wallman in Template:Harvnb.[2]

Factors affecting rise time

Rise time values in a resistive circuit are primarily due to stray capacitance and inductance in the circuit. Because every circuit has not only resistance, but also capacitance and inductance, a delay in voltage and/or current at the load is apparent until the steady state is reached. In a pure RC circuit, the output risetime (10% to 90%), as shown above, is approximately equal to ${\displaystyle 2.2RC}$.

Rise time in control applications

In control theory, for overdamped systems, rise time is commonly defined as the time for a waveform to go from 10% to 90% of its final value.[1]

The quadratic approximation for normalized rise time for a 2nd-order system, step response, no zeros is:

${\displaystyle t_{r}\cdot \omega _{0}=2.230\zeta ^{2}-0.078\zeta +1.12\,}$

where ζ is the damping ratio and ω0 is the natural frequency of the network.

However, the proper calculation for rise time from 0 to 100% of an under-damped 2nd-order system is:

${\displaystyle t_{r}\cdot \omega _{0}={\frac {1}{\sqrt {1-\zeta ^{2}}}}\left(\pi -\tan ^{-1}\left({\frac {\sqrt {1-\zeta ^{2}}}{\zeta }}\right)\right)}$

where ζ is the damping ratio and ω0 is the natural frequency of the network.

Notes

1. Precisely, Template:Harvtxt states: "The rise time is the time required for the response to rise from x% to y% of its final value. For overdamped second order systems, the 0% to 100% rise time is normally used, and for underdamped systems... the 10% to 90% rise time is commonly used". See also the textbook Template:Harvnb.
2. This beautiful one-page paper does not contain any calculation. Henry Wallman simply sets up a table he calls dictionary paralleling concepts from electronics engineering and probability theory: the key of the process is the use of Laplace transform. Then he notes that, following the correspondence of concepts established by the dictionary, that the step response of a cascade of blocks corresponds to the central limit theorem and states that: "This has important practical consequences, among them the fact that if a network is free of overshoot its time-of-response inevitably increases rapidly upon cascading, namely as the square-root of the number of cascaded network".Template:Harv

References

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