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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 settling time is known as transient response and the response after the settling time is known as steady state response.

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 âδâ lies between zero and one.

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

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.

$$c(t_d)=0.5=1-\left ( \frac{e^{-\delta\omega_nt_d}}{\sqrt{1-\delta^2}} \right )\sin(\omega_dt_d+\theta)$$

$$\Rightarrow \left ( \frac{e^{-\delta\omega_nt_d}}{\sqrt{1-\delta^2}} \right )\sin(\omega_dt_d+\theta)=0.5$$

By using linear approximation, you will get the **delay time t _{d}** as

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

It is the time required for the response to rise from **0% to 100% of its final value**. This is applicable 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 = t_{1} = 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.

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

$$c(t_2)=1=1-\left ( \frac{e^{-\delta\omega_nt_2}}{\sqrt{1-\delta^2}} \right )\sin(\omega_dt_2+\theta)$$

$$\Rightarrow \left ( \frac{e^{-\delta\omega_nt_2}}{\sqrt{1-\delta^2}} \right )\sin(\omega_dt_2+\theta)=0$$

$$\Rightarrow \sin(\omega_dt_2+\theta)=0$$

$$\Rightarrow \omega_dt_2+\theta=\pi$$

$$\Rightarrow t_2=\frac{\pi-\theta}{\omega_d}$$

Substitute t_{1} and t_{2} values in the following equation of **rise time**,

$$t_r=t_2-t_1$$

$$\therefore \: t_r=\frac{\pi-\theta}{\omega_d}$$

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

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

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

Differentiate $c(t)$ with respect to âtâ.

$$\frac{\text{d}c(t)}{\text{d}t}=-\left ( \frac{e^{-\delta\omega_nt}}{\sqrt{1-\delta^2}} \right )\omega_d\cos(\omega_dt+\theta)-\left ( \frac{-\delta\omega_ne^{-\delta\omega_nt}}{\sqrt{1-\delta^2}} \right )\sin(\omega_dt+\theta)$$

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

$$0=-\left ( \frac{e^{-\delta\omega_nt_p}}{\sqrt{1-\delta^2}} \right )\left [ \omega_d\cos(\omega_dt_p+\theta)-\delta\omega_n\sin(\omega_dt_p+\theta) \right ]$$

$$\Rightarrow \omega_n\sqrt{1-\delta^2}\cos(\omega_dt_p+\theta)-\delta\omega_n\sin(\omega_dt_p+\theta)=0$$

$$\Rightarrow \sqrt{1-\delta^2}\cos(\omega_dt_p+\theta)-\delta\sin(\omega_dt_p+\theta)=0$$

$$\Rightarrow \sin(\theta)\cos(\omega_dt_p+\theta)-\cos(\theta)\sin(\omega_dt_p+\theta)=0$$

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

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

$$\Rightarrow \omega_dt_p=\pi$$

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

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 **M _{p}** is defined as the deviation of the response at peak time from the final value of response. It is also called the

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 -

$$c(t_p)=1-\left ( \frac{e^{-\delta\omega_nt_p}}{\sqrt{1-\delta^2}} \right )\sin(\omega_dt_p+\theta)$$

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

$$c(t_P)=1-\left ( \frac{e^{-\delta\omega_n\left ( \frac{\pi}{\omega_d} \right )}}{\sqrt{1-\delta^2}} \right )\sin\left ( \omega_d\left ( \frac{\pi}{\omega_d} \right ) +\theta\right )$$

$$\Rightarrow c(t_p)=1-\left ( \frac{e^{-\left ( \frac{\delta\pi}{\sqrt{1-\delta^2}} \right )}}{\sqrt{1-\delta^2}} \right )(-\sin(\theta))$$

We know that

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

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

$$c(t_p)=1+e^{-\left ( \frac{\delta\pi}{\sqrt{1-\delta^2}} \right )}$$

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

$$M_p=1+e^{-\left ( \frac{\delta\pi}{\sqrt{1-\delta^2}} \right )}-1$$

$$\Rightarrow M_p=e^{-\left ( \frac{\delta\pi}{\sqrt{1-\delta^2}} \right )}$$

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

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

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

$$\%M_p=\left ( e^ {-\left ( \frac{\delta\pi}{\sqrt{1-\delta^2}} \right )} \right )\times 100\%$$

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

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 settling time is denoted by $t_s$.

The settling time for 5% tolerance band is -

$$t_s=\frac{3}{\delta\omega_n}=3\tau$$

The settling time for 2% tolerance band is -

$$t_s=\frac{4}{\delta\omega_n}=4\tau$$

Where, $\tau$ is the time constant and is equal to $\frac{1}{\delta\omega_n}$.

Both the settling time $t_s$ and the time constant $\tau$ are inversely proportional to the damping ratio $\delta$.

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

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 applied 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

$$\frac{\omega_n^2}{s^2+2\delta\omega_ns+\omega_n^2}$$

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

$$\omega_d=\omega_n\sqrt{1-\delta^2}$$

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

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

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

Substitute, $\delta$ value in following relation

$$\theta=\cos^{-1}\delta$$

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

Substitute the above necessary values in the formula of each time domain specification and simplify 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.

Time domain specification | Formula | Substitution of values in Formula | Final value |
---|---|---|---|

Delay time |
$t_d=\frac{1+0.7\delta}{\omega_n}$ |
$t_d=\frac{1+0.7(0.5)}{2}$ |
$t_d$=0.675 sec |

Rise time |
$t_r=\frac{\pi-\theta}{\omega_d}$ |
$t_r=\frac{\pi-(\frac{\pi}{3})}{1.732}$ |
$t_r$=1.207 sec |

Peak time |
$t_p=\frac{\pi}{\omega_d}$ |
$t_p=\frac{\pi}{1.732}$ |
$t_p$=1.813 sec |

% Peak overshoot |
$\%M_p=\left( e^{-\left (\frac{\delta\pi}{\sqrt{1-\delta^2}} \right ) }\right )\times 100\%$ |
$\%M_p=\left( e^{-\left (\frac{0.5\pi}{\sqrt{1-(0.5)^2}} \right ) }\right )\times 100\%$ |
$\% \: M_p$=16.32% |

Settling time for 2% tolerance band |
$t_s=\frac{4}{\delta\omega_n}$ |
$t_S=\frac{4}{(0.5)(2)}$ |
$t_s$=4 sec |

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