**Time Domain analysis**which will cover the topics such as

**Different Types of Input Signals, Time Response of Ist Order System, Time Response of Second Order System, Overdamped System Underdamped System, Undamped System & Critically Damped System.**

**Time Domain Analysis**

The Time Domain Analyzes of the system is to be done on basis of time. The analysis is only be applied when the nature of the input plus the mathematical model of the control system is known. Expressing the main input signals is not an easy task and cannot be determined by simple equations. There are two components of any system’s time response, which are: Transient response & Steady-state response.

**Transient Response:**This response is dependent upon the system poles only and not on the type of input & it is sufficient to analyze the transient response using a step input.-
**Steady-State Response:**This response depends on system dynamics and the input quantity. It is then examined using different test signals by the final value theorem.

**Standard Test Input Signals**

**Time-Response of First-Order System**

Here consider the armature-controlled dc motor driving a load, such as a videotape. The objective is to drive the tape at a constant speed. Note that it is an open-loop system. ** **

**ω**is the steady-state final speed. If the desired speed is ω_{ss}(t)_{r}, choosing '**a=ω**the motor will eventually reach the desired speed._{r}/k_{1}k_{m}'

** **

- From the time response
**e**we concluded that for^{-t/τm }**t≥5τ**the value of e_{m}^{-t/τm}is less than 1% of its original value. Hence the speed of the motor will reach and stay within 1% of its final speed at 5-time constants.

**Let us now consider the closed-loop system **

If r(t) = a then Response would be ; w(t) = ak_{1}k_{o}- ak_{1}k_{o}e^{-t/τo}

If a is properly chosen, the tape can reach the desired speed. It will reach the desired speed in 5τ_{o} seconds. Here τ_{o=}τ_{m.}So that we can control the speed of response in the feedback system.

**Ramp response of first-order system**

e_{ss}(t)=** **τ_{o}

- Thus, the first-order system will track the unit ramp input with a steady-state error τ
_{o}, which is equal to the time constant of the system.

**Time-Response of Second-Order System**

- Consider the antenna position control system. Its transfer function from r to y is,

where we can define

**(ω _{n})^{2} = k_{1}k_{2}k_{m}/τ_{m }; & 2ξω_{n} = 1/τ_{m}**

The constant **ξ** is called the damping ratio and **ω _{n }**is called the natural frequency. The system above is, in fact, a standard second-order system.

The transfer function **T(s)** has two poles and no zero. Its poles are,

**Natural frequency (ω _{n}): ** The natural frequency of a second-order system is the frequency of oscillation of the system without damping.

**Damping ratio (ξ): **The damping ratio is defined as the ratio of the damping factor **σ**, to the natural frequency **ω _{n.}**

Here,**σ** is called the damping factor,**ω _{d }**is called damped or actual frequency. The location of poles for different

**ξ**are plotted in the given figure below. For

**ξ=0**, the two poles

**±jω**are purely imaginary. If0<

_{n }**ξ<**1, the two poles are complex conjugate.

**Unit Step Response of Second-Order System**

**Second-Order Systems: General Specification**

The second-order system exhibits a wide range of responses that must be analyzed and described. To become familiar with the wide range of responses before formalizing our discussion, we take a look at numerical examples of the second-order system responses shown in the figure.

**Underdamped Response (0<ξ<1)**

This function has a pole at the origin that comes from the unit step and two complex poles that come from the system. The sinusoidal frequency is given the name of damped frequency

of oscillation, ωd. This response is shown in the figure called underdamped.

**Example:**

**Overdamped System (1<ξ)**

This function has a pole at the origin that comes from the unit step input and two real poles that come from the system. The input pole at the origin generates the constant forced response; each of the two system poles on the real axis generates an exponential natural frequency.

**Example: **

**Undamped Response (ξ= 0)**

** **This function has a pole at the origin and two imaginary poles. The pole at the origin generates the constant forced response, and the two system poles on the imaginary axis at ±j3 generate a Sinusoidal natural response.

**Example:**

**Critically Damped Response (ξ = 1)**

** **This function has a pole at the origin and two multiple real poles. The input pole at the origin generates the constant forced response, and two poles at the real axis at -3 generate a natural exponential response.

**Note: In the above specifications of time domain, don't be confused with the number of Poles in G(s), to Specify for which type of Damping is present for a particular case we consider the total number of poles are of transfer function i.e; C(s)/R(s).**

**Summarization: **Here once again we summarize the second-order damping functions as;

**Time Domain **Characteristics

In specifying the Transient-Response characteristics of a control system to a unit step input, we usually specify the following:

**Delay time (**t_{d}**):**It is the time required for the response to reach 50% of the final value in the first attempt.**Rise time, (**t_{r}**):**It is the time required for the response to rise from 0 to 100% of the final value for the underdamped system.**Peak time, (**tp**):**It is the time required for the response to reach the peak of time response or the peak overshoot.**Settling time, ( t**It is the time required for the response to reach and stay within a specified tolerance band ( 2% or 5%) of its final value._{s}):**Peak overshoot ( M**It is the normalized difference between the time response peak and the steady output and is defined as_{p}):

**Steady-state error (**eIt indicates the error between the actual output and desired output as ‘t’ tends to infinity._{ss }):

**Steady-state error **e** _{ss}:** It is found previously that steady-state error for a step input is zero. Let us now consider

**ramp input, r(t)= tu(t)**.

- Therefore, the steady-state error due to ramp input is
**2ξ/ω**_{n.}

### Effect of Adding a Zero to a System**: **

If we add a zero at s = -z be added to a second-order system. Then we have,

- The multiplication term is adjusted to make the steady-state gain of the system unity.

Manipulation of the above equation gives,

- The effect of the added derivative term is to produce a pronounced early peak to the system response.
- The closer the zero to the origin, the more pronounce the peaking phenomenon.
- Due to this fact, the zeros on the real axis near the origin are generally avoided in design. However, in a sluggish system, the artful introduction of a zero at the proper position can improve the transient response.

**Types of Feedback Control System: **

The open-loop transfer function of a system can be written as

- If n = 0, the system is called type-0 system, if n = 1, the system is called type-1 system, if n = 2, the system is called type-2 system, etc.

**Steady-State Error and Error Constants:**

The steady-state performance of a stable control system is generally judged by its steady-state error to step, ramp, and parabolic inputs. For a unity feedback system,

It is seen that steady-state error depends upon the input R(s) and the forward transfer function G(s).

**Time Domain **Characteristics

In specifying the Transient-Response characteristics of a control system to a unit step input, we usually specify the following:

**Delay time (**t_{d}**):**It is the time required for the response to reach 50% of the final value in the first attempt.

The expression of delay time, t_{d} for second-order system is:

**Rise time, (**t_{r}**):**It is the time required for the response to rise from 0 to 100% of the final value for the under-damped system.

The expression of rise time, t_{r} for second-order system is:

**Peak time, (**t_{p}**):**It is the time required for the response to reach the peak of time response or the peak overshoot.

The expression of peak time, t_{p} for second-order system is:

**Settling time, ( t**It is the time required for the response to reach and stay within a specified tolerance band ( 2% or 5%) of its final value._{s}):

The expression of settling time, t_{s} for second-order system is:

**Peak overshoot ( M**It is the normalized difference between the time response peak and the steady output and is defined as_{p}):

The expression of peak overshoot, M_{p} for second-order system is:

**Steady-state error (**eIt indicates the error between the actual output and desired output as ‘t’ tends to infinity._{ss }):

### Effect of Adding a Zero to a System**: **

If we add a zero at s = -z be added to a second-order system. Then we have,

- The multiplication term is adjusted to make the steady-state gain of the system unity.

Manipulation of the above equation gives,

- The effect of the added derivative term is to produce a pronounced early peak to the system response.
- The closer the zero to the origin, the more pronounced the peaking phenomenon.
- Due to this fact, the zeros on the real axis near the origin are generally avoided in design. However, in a sluggish system, the artful introduction of a zero at the proper position can improve the transient response.

**Types of Feedback Control System: **

The open-loop transfer function of a system can be written as

- If n = 0, the system is called type-0 system, if n = 1, the system is called type-1 system, if n = 2, the system is called type-2 system, etc.

**Steady-State Error and Error Constants:**

The steady-state performance of a stable control system is generally judged by its steady-state error to step, ramp and parabolic inputs. For a unity feedback system,

Where,

E(s) is error signal

R(s) is input signal

G(s) H(s) is the open loop transfer function

It is seen that steady-state error depends upon the input R(s) and the forward transfer function G(s).

**1. If input is unit step i.e R(t) = u(t)**

**2. If input is unit ramp i.e R(t) = tu(t)**

**3. If input is unit parabolic i.e R(t) = 0.5tu(t)**

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