Rise time

For a given system output, its rise time depend both on the rise time of input signal and on the characteristics of the system.[9]

For example, rise time values in a resistive circuit are primarily due to stray capacitance and inductance. Since 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%) is approximately equal to 2.2 RC.[10]

Other definitions of rise time, apart from the one given by the Federal Standard 1037C (1997, p. R-22) and its slight generalization given by Levine (1996, p. 158), are occasionally used:[11] these alternative definitions differ from the standard not only for the reference levels considered. For example, the time interval graphically corresponding to the intercept points of the tangent drawn through the 50% point of the step function response is occasionally used.[12] Another definition, introduced by Elmore (1948, p. 57),[13] uses concepts from statistics and probability theory. Considering a step response V(t), he redefines the delay time t_D as the first moment of its first derivative V′(t), i.e.

${\displaystyle t_{D}={\frac {\int _{0}^{+\infty }tV^{\prime }(t)\mathrm {d} t}{\int _{0}^{+\infty }V^{\prime }(t)\mathrm {d} t}}.}$

Finally, he defines the rise time t_r by using the second moment

${\displaystyle t_{r}^{2}={\frac {\int _{0}^{+\infty }(t-t_{D})^{2}V^{\prime }(t)\mathrm {d} t}{\int _{0}^{+\infty }V^{\prime }(t)\mathrm {d} t}}\quad \Longleftrightarrow \quad t_{r}={\sqrt {\frac {\int _{0}^{+\infty }(t-t_{D})^{2}V^{\prime }(t)\mathrm {d} t}{\int _{0}^{+\infty }V^{\prime }(t)\mathrm {d} t}}}}$

All notations and assumptions required for the analysis are listed here.

• Following Levine (1996, p. 158, 2011, 9-3 (313)), we define x% as the percentage low value and y% the percentage high value respect to a reference value of the signal whose rise time is to be estimated.
• t₁ is the time at which the output of the system under analysis is at the x% of the steady-state value, while t₂ the one at which it is at the y%, both measured in seconds.
• t_r is the rise time of the analysed system, measured in seconds. By definition,

${\displaystyle t_{r}=t_{2}-t_{1}.}$

• f_L is the lower cutoff frequency (-3 dB point) of the analysed system, measured in hertz.
• f_H is higher cutoff frequency (-3 dB point) of the analysed system, measured in hertz.
• h(t) is the impulse response of the analysed system in the time domain.
• H(ω) is the frequency response of the analysed system in the frequency domain.
• The bandwidth is defined as

${\displaystyle BW=f_{H}-f_{L}\,}$

and since the lower cutoff frequency f_L is usually several decades lower than the higher cutoff frequency f_H,

${\displaystyle BW\cong f_{H}\,}$

• All systems analyzed here have a frequency response which extends to 0 (low-pass systems), thus

${\displaystyle f_{L}=0\,\Longleftrightarrow \,f_{H}=BW}$ exactly.

• For the sake of simplicity, all systems analysed in the "Simple examples of calculation of rise time" section are unity gain electrical networks, and all signals are thought as voltages: the input is a step function of V₀ volts, and this implies that

${\displaystyle {\frac {V(t_{1})}{V_{0}}}={\frac {x\%}{100}}\qquad {\frac {V(t_{2})}{V_{0}}}={\frac {y\%}{100}}}$

• ζ is the damping ratio and ω₀ is the natural frequency of a given second order system.

The aim of this section is the calculation of rise time of step response for some simple systems:

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 σ > 0 is a constant,[14] related to the high cutoff frequency by the following relation:

${\displaystyle f_{H}={\frac {\sigma }{2\pi }}{\sqrt {{\frac {3}{20}}\ln 10}}\cong 0.0935\sigma .}$

Even if this kind frequency response is not realizable by a causal filter,[15] its usefulness lies in the fact that behaviour of a cascade connection of first order low pass filters approaches the behaviour of this system more closely as the number of cascaded stages asymptotically rises to infinity.[16] 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,

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

To determine the 10% to 90% rise time of the system it is necessary to solve for time the two following equations:

${\displaystyle {\frac {V(t_{1})}{V_{0}}}=0.1={\frac {1}{2}}\left[1+\mathrm {erf} \left({\frac {\sigma t_{1}}{2}}\right)\right]\qquad {\frac {V(t_{2})}{V_{0}}}=0.9={\frac {1}{2}}\left[1+\mathrm {erf} \left({\frac {\sigma t_{2}}{2}}\right)\right],}$

By using known properties of the error function, the value t =  - t₁ = t₂ is found: since t_r = t₂ - t₁ = 2t,

${\displaystyle t_{r}={\frac {4}{\sigma }}{\mathrm {erf} ^{-1}(0.8)}\cong {\frac {0.3394}{f_{H}}},}$

and finally

${\displaystyle t_{r}\cong {\frac {0.34}{BW}}\quad \Longleftrightarrow \quad BW\cdot t_{r}\cong 0.34.}$[17]

For a simple one-stage low-pass RC network,[18] the 10% to 90% rise time is proportional to the network time constant τ = RC:

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

The proportionality constant can be derived from the knowledge of the step response of the network to a unit step function input signal of V₀ amplitude:

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

Solving for time

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

and finally,

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

Since t₁ and t₂ are such that

${\displaystyle {\frac {V(t_{1})}{V_{0}}}=0.1\qquad {\frac {V(t_{2})}{V_{0}}}=0.9,}$

solving these equations we find the analytical expression for t₁ and t₂:

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

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

The rise time is therefore proportional to the time constant:[19]

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

Now, noting that

${\displaystyle \tau =RC={\frac {1}{2\pi f_{H}}},}$[20]

then

${\displaystyle t_{r}={\frac {2\ln 3}{2\pi f_{H}}}={\frac {\ln 3}{\pi f_{H}}}\cong {\frac {0.349}{f_{H}}},}$

and since the high frequency cutoff is equal to the bandwidth,

${\displaystyle t_{r}\cong {\frac {0.35}{BW}}\quad \Longleftrightarrow \quad BW\cdot t_{r}\cong 0.35.}$[17]

Finally note that, if the 20% to 80% rise time is considered instead, t_r becomes:

${\displaystyle t_{r}=\tau \cdot \ln {\frac {8}{2}}=(2\ln 2)\tau \cong 1.386\tau \quad \Longleftrightarrow \quad t_{r}={\frac {\ln 2}{\pi BW}}\cong {\frac {0.22}{BW}}}$

Even for a simple one-stage low-pass RL network, the 10% to 90% rise time is proportional to the network time constant τ = ​^L⁄_R. The formal proof of this assertion proceed exactly as shown in the previous section: the only difference between the final expressions for the rise time is due to the difference in the expressions for the time constant τ of the two different circuits, leading in the present case to the following result

${\displaystyle t_{r}=\tau \cdot \ln 9={\frac {L}{R}}\cdot \ln 9\cong {\frac {L}{R}}\cdot 2.197}$

According to Levine (1996, p. 158), for underdamped systems used in control theory rise time is commonly defined as the time for a waveform to go from 0% to 100% of its final value:[6] accordingly, the rise time from 0 to 100% of an underdamped 2nd-order system has the following form:[21]

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

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.

Consider a system composed by n cascaded non interacting blocks, each having a rise time t_ri, i = 1,...,n, and no overshoot in their step response: suppose also that the input signal of the first block has a rise time whose value is t_rS.[22] Afterwards, its output signal has a rise time t_r0 equal to

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

According to Valley & Wallman (1948, pp. 77–78), this result is a consequence of the central limit theorem and was proved by Wallman (1950):[23][24] however, a detailed analysis of the problem is presented by Petitt & McWhorter (1961, §4–9, pp. 107–115),[25] who also credit Elmore (1948) as the first one to prove the previous formula on a somewhat rigorous basis.[26]