Time Domain Specifications

In this chapter, let us discuss the time domain specifications of the second order system. The step response of the second order system for the underdamped case is shown in the following figure.

All the time domain specifications are represented in this figure. The response up to the settpng time is known as transient response and the response after the settpng time is known as steady state response.

Delay Time

It is the time required for the response to reach half of its final value from the zero instant. It is denoted by $t_d$.

Consider the step response of the second order system for t ≥ 0, when ‘δ’ pes between zero and one.

$$c(t)=1-left ( frac{e^{-delta omega_nt}}{sqrt{1-delta^2}} ight )sin(omega_dt+ heta)$$

The final value of the step response is one.

Therefore, at $t=t_d$, the value of the step response will be 0.5. Substitute, these values in the above equation.

$$c(t_d)=0.5=1-left ( frac{e^{-deltaomega_nt_d}}{sqrt{1-delta^2}} ight )sin(omega_dt_d+ heta)$$

$$Rightarrow left ( frac{e^{-deltaomega_nt_d}}{sqrt{1-delta^2}} ight )sin(omega_dt_d+ heta)=0.5$$

By using pnear approximation, you will get the delay time t_d as

$$t_d=frac{1+0.7delta}{omega_n}$$

Rise Time

It is the time required for the response to rise from 0% to 100% of its final value. This is apppcable for the under-damped systems. For the over-damped systems, consider the duration from 10% to 90% of the final value. Rise time is denoted by t_r.

At t = t₁ = 0, c(t) = 0.

We know that the final value of the step response is one.

Therefore, at $t = t_2$, the value of step response is one. Substitute, these values in the following equation.

$$c(t)=1-left ( frac{e^{-delta omega_nt}}{sqrt{1-delta^2}} ight )sin(omega_dt+ heta)$$

$$c(t_2)=1=1-left ( frac{e^{-deltaomega_nt_2}}{sqrt{1-delta^2}} ight )sin(omega_dt_2+ heta)$$

$$Rightarrow left ( frac{e^{-deltaomega_nt_2}}{sqrt{1-delta^2}} ight )sin(omega_dt_2+ heta)=0$$

$$Rightarrow sin(omega_dt_2+ heta)=0$$

$$Rightarrow omega_dt_2+ heta=pi$$

$$Rightarrow t_2=frac{pi- heta}{omega_d}$$

Substitute t₁ and t₂ values in the following equation of rise time,

$$t_r=t_2-t_1$$

$$ herefore : t_r=frac{pi- heta}{omega_d}$$

From above equation, we can conclude that the rise time $t_r$ and the damped frequency $omega_d$ are inversely proportional to each other.

Peak Time

It is the time required for the response to reach the peak value for the first time. It is denoted by $t_p$. At $t = t_p$, the first derivate of the response is zero.

We know the step response of second order system for under-damped case is

$$c(t)=1-left ( frac{e^{-delta omega_nt}}{sqrt{1-delta^2}} ight )sin(omega_dt+ heta)$$

Differentiate $c(t)$ with respect to ‘t’.

$$frac{ ext{d}c(t)}{ ext{d}t}=-left ( frac{e^{-deltaomega_nt}}{sqrt{1-delta^2}} ight )omega_dcos(omega_dt+ heta)-left ( frac{-deltaomega_ne^{-deltaomega_nt}}{sqrt{1-delta^2}} ight )sin(omega_dt+ heta)$$

Substitute, $t=t_p$ and $frac{ ext{d}c(t)}{ ext{d}t}=0$ in the above equation.

$$0=-left ( frac{e^{-deltaomega_nt_p}}{sqrt{1-delta^2}} ight )left [ omega_dcos(omega_dt_p+ heta)-deltaomega_nsin(omega_dt_p+ heta) ight ]$$

$$Rightarrow omega_nsqrt{1-delta^2}cos(omega_dt_p+ heta)-deltaomega_nsin(omega_dt_p+ heta)=0$$

$$Rightarrow sqrt{1-delta^2}cos(omega_dt_p+ heta)-deltasin(omega_dt_p+ heta)=0$$

$$Rightarrow sin( heta)cos(omega_dt_p+ heta)-cos( heta)sin(omega_dt_p+ heta)=0$$

$$Rightarrow sin( heta-omega_dt_p- heta)=0$$

$$Rightarrow sin(-omega_dt_p)=0Rightarrow -sin(omega_dt_p)=0Rightarrow sin(omega_dt_p)=0$$

$$Rightarrow omega_dt_p=pi$$

$$Rightarrow t_p=frac{pi}{omega_d}$$

From the above equation, we can conclude that the peak time $t_p$ and the damped frequency $omega_d$ are inversely proportional to each other.

Peak Overshoot

Peak overshoot M_p is defined as the deviation of the response at peak time from the final value of response. It is also called the maximum overshoot.

Mathematically, we can write it as

$$M_p=c(t_p)-c(infty)$$

Where,

c(t_p) is the peak value of the response.

c(∞) is the final (steady state) value of the response.

At $t = t_p$, the response c(t) is -

$$c(t_p)=1-left ( frac{e^{-deltaomega_nt_p}}{sqrt{1-delta^2}} ight )sin(omega_dt_p+ heta)$$

Substitute, $t_p=frac{pi}{omega_d}$ in the right hand side of the above equation.

$$c(t_P)=1-left ( frac{e^{-deltaomega_nleft ( frac{pi}{omega_d} ight )}}{sqrt{1-delta^2}} ight )sinleft ( omega_dleft ( frac{pi}{omega_d} ight ) + heta ight )$$

$$Rightarrow c(t_p)=1-left ( frac{e^{-left ( frac{deltapi}{sqrt{1-delta^2}} ight )}}{sqrt{1-delta^2}} ight )(-sin( heta))$$

We know that

$$sin( heta)=sqrt{1-delta^2}$$

So, we will get $c(t_p)$ as

$$c(t_p)=1+e^{-left ( frac{deltapi}{sqrt{1-delta^2}} ight )}$$

Substitute the values of $c(t_p)$ and $c(infty)$ in the peak overshoot equation.

$$M_p=1+e^{-left ( frac{deltapi}{sqrt{1-delta^2}} ight )}-1$$

$$Rightarrow M_p=e^{-left ( frac{deltapi}{sqrt{1-delta^2}} ight )}$$

Percentage of peak overshoot % $M_p$ can be calculated by using this formula.

$$\%M_p=frac{M_p}{c(infty )} imes 100\%$$

By substituting the values of $M_p$ and $c(infty)$ in above formula, we will get the Percentage of the peak overshoot $\%M_p$ as

$$\%M_p=left ( e^ {-left ( frac{deltapi}{sqrt{1-delta^2}} ight )} ight ) imes 100\%$$

From the above equation, we can conclude that the percentage of peak overshoot $\% M_p$ will decrease if the damping ratio $delta$ increases.

Settpng time

It is the time required for the response to reach the steady state and stay within the specified tolerance bands around the final value. In general, the tolerance bands are 2% and 5%. The settpng time is denoted by $t_s$.

The settpng time for 5% tolerance band is -

$$t_s=frac{3}{deltaomega_n}=3 au$$

The settpng time for 2% tolerance band is -

$$t_s=frac{4}{deltaomega_n}=4 au$$

Where, $ au$ is the time constant and is equal to $frac{1}{deltaomega_n}$.

Both the settpng time $t_s$ and the time constant $ au$ are inversely proportional to the damping ratio $delta$.

Both the settpng time $t_s$ and the time constant $ au$ are independent of the system gain. That means even the system gain changes, the settpng time $t_s$ and time constant $ au$ will never change.

Example

Let us now find the time domain specifications of a control system having the closed loop transfer function $frac{4}{s^2+2s+4}$ when the unit step signal is appped as an input to this control system.

We know that the standard form of the transfer function of the second order closed loop control system as

$$frac{omega_n^2}{s^2+2deltaomega_ns+omega_n^2}$$

By equating these two transfer functions, we will get the un-damped natural frequency $omega_n$ as 2 rad/sec and the damping ratio $delta$ as 0.5.

We know the formula for damped frequency $omega_d$ as

$$omega_d=omega_nsqrt{1-delta^2}$$

Substitute, $omega_n$ and $delta$ values in the above formula.

$$Rightarrow omega_d=2sqrt{1-(0.5)^2}$$

$$Rightarrow omega_d=1.732 : rad/sec$$

Substitute, $delta$ value in following relation

$$ heta=cos^{-1}delta$$

$$Rightarrow heta=cos^{-1}(0.5)=frac{pi}{3}:rad$$

Substitute the above necessary values in the formula of each time domain specification and simppfy in order to get the values of time domain specifications for given transfer function.

The following table shows the formulae of time domain specifications, substitution of necessary values and the final values.