By: Nafees Ahamad, AP, EECE, Dept. DIT University, Dehradun Time Response By: Nafees Ahamad, AP, EECE, Dept. DIT University, Dehradun

Time Response System output in time domain It consists of Transient response Steady state response

Time Response … Time response c(t) is given as Steady State Response Overall Response Transient Response

Transient Response Response just after applying/removing the input to the system is known as transient response. It revels the nature of response (i.e. Oscillatory or overdamped). It also indicates about response speed . The transient response will be zero for large values of ‘t’. Ideally, this value of ‘t’ is infinity [Practically 3T (on 5% error basis) or 4T (on 2% error basis), T = Time Constant] Mathematically,

Steady state Response The response just after transient response is known as Steady State Response. It revels the accuracy of a control system. Steady state error is observed if the actual output does not exactly match with the input. Example Steady State Response Transient Response

Standard Test Signals Standard Test signals are Step signal Ramp Signal Impulse Signal Parabolic Signal

Unit Step Signal It is defined as 𝑈 𝑡 = 1 𝑡≥0 0 𝑡<0

Unit Ramp Signal It is defined as r 𝑡 = 𝑡 𝑡≥0 0 𝑡<0

Unit Impulse Signal It is defined as 𝛿 𝑡 = 1 𝑡=0 0 𝑡≠0

Parabolic Signal It is defined as x 𝑡 = 𝑡 2 /2 𝑡≥0 0 𝑡<0

Relationship

Time Response of a First Order System Consider following system with unity feedback With 𝐺 𝑠 = 1 𝑠𝑇 and 𝐻 𝑠 =1

Time Response of a First Order System… So, overall transfer function 𝐶(𝑠) 𝑅 𝑠 = 𝐺(𝑠) 1+𝐺 𝑠 𝐻(𝑠) , Put the values of G(s) & H(s) 𝐶(𝑠) 𝑅 𝑠 = 1 𝑠𝑇 1+ 1 𝑠𝑇 = 1 𝑠𝑇+1 Note: The power of ‘s’ in denominator is one, so it is first order system So, 𝐶 𝑠 = 1 𝑠𝑇+1 𝑅 𝑠 −−−(1)

Impulse Response of First Order System Consider the unit impulse signal as an input 𝑟 𝑡 =𝛿(𝑡) Put 𝑅 𝑠 =1 in Equ. (1) 𝐶 𝑠 = 1 𝑠𝑇+1 = 1 𝑇 × 1 𝑠+ 1 𝑇 Take inverse Laplace transformation 𝑐 𝑡 = 1 𝑇 𝑒 − 𝑡 𝑇 𝑈 𝑡 −−−(2) 𝑅 𝑠 =1 Take Laplace Transformation

Step Response of First Order System Consider the Unit Step signal as an input 𝑟 𝑡 =𝑈(𝑡) Put 𝑅 𝑠 in Equ. (1) and take partial fractions 𝐶 𝑠 = 1 𝑠 × 1 𝑠𝑇+1 = 𝐴 𝑠 + 𝐵 𝑠𝑇+1 = 1 𝑠 − 1 𝑠𝑇+1 Take inverse Laplace transformation 𝑐 𝑡 = 1− 𝑒 − 𝑡 𝑇 𝑈 𝑡 −−−(3) 𝑅 𝑠 = 1 𝑠 Take Laplace Transformation

Ramp Response of First Order System Consider the Unit Ramp Signal as an input 𝑟 𝑡 =𝑟 𝑡 =𝑡 Put 𝑅 𝑠 in Equ. (1) and take partial fractions 𝐶 𝑠 = 1 𝑠 2 × 1 𝑠𝑇+1 = 𝐴 𝑠 2 + 𝐵 𝑠 + 𝐶 𝑠𝑇+1 = 1 𝑠 2 − 𝑇 𝑠 + 𝑇 2 𝑠𝑇+1 Take inverse Laplace transformation 𝑐 𝑡 =(𝑡−𝑇+𝑇 𝑒 − 𝑡 𝑇 ) 𝑈 𝑡 −−−(4) 𝑅 𝑠 = 1 𝑠 2 Take Laplace Transformation

Ramp Response of First Order System… Equ. (4) has transient as well as steady state response Transient Response 𝑐 𝑡𝑟 𝑡 =𝑇 𝑒 − 𝑡 𝑇 𝑈 𝑡 Steady State Response 𝑐 𝑠𝑠 𝑡 =(𝑡−𝑇)𝑈 𝑡 c(t) T

Response of Second Order System Consider the following 2nd order control system With, Open loop transfer function 𝐺 𝑠 = 𝜔 𝑛 2 𝑠(𝑠+2𝜉 𝜔 𝑛 ) And H(s) = 1 So, 𝐶(𝑠) 𝑅 𝑠 = 𝐺(𝑠) 1+𝐺 𝑠 𝐻(𝑠) 𝐶(𝑠) 𝑅 𝑠 = 𝜔 𝑛 2 𝑠 2 +2 𝜉𝜔 𝑛 𝑠+ 𝜔 𝑛 2 −−−(5) G(s) E(s) C(s) H(s) R(s) 𝜉=𝐷𝑎𝑚𝑝𝑖𝑛𝑔 𝑅𝑎𝑡𝑖𝑜 𝜔 𝑛 =𝑁𝑎𝑡𝑢𝑟𝑎𝑙 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦

Response of Second Order System Its characteristic equation 𝑠 2 +2 𝜉𝜔 𝑛 𝑠+ 𝜔 𝑛 2 =0 The roots of characteristic equ are 𝑠= −2 𝜉𝜔 𝑛 ± (2 𝜉𝜔 𝑛 ) 2 − ( 4𝜔 𝑛 ) 2 2 ⇒𝑠= −𝜉𝜔 𝑛 ± 𝜔 𝑛 𝜉 2 −1 Note: 𝜉=0, Two Imaginary Roots 𝜉=1, Two Real Equal Roots 𝜉>1, Two Real Unequal Roots 0<𝜉<1, Two Complex Roots

Step Response of Second Order System… Consider the Unit Step signal as an input 𝑟 𝑡 =𝑈(𝑡) Put 𝑅 𝑠 in Equ. (5) and take partial fractions 𝐶 𝑠 = 𝜔 𝑛 2 𝑠 2 +2 𝜉𝜔 𝑛 𝑠+ 𝜔 𝑛 2 𝑅 𝑠 = 𝜔 𝑛 2 𝑠 2 +2 𝜉𝜔 𝑛 𝑠+ 𝜔 𝑛 2 1 𝑠 −−−(6) ⇒𝐶(𝑠)= 1 𝑠 × 𝜔 𝑛 2 (𝑠+ 𝜉𝜔 𝑛 ) 2 + 𝜔 𝑛 2 (1− 𝜉 2 ) 𝑅 𝑠 = 1 𝑠 Take Laplace Transformation Let 𝜔 𝑑 2

Step Response of Second Order System… ⇒𝐶 𝑠 = 1 𝑠 × 𝜔 𝑛 2 𝑠+ 𝜉𝜔 𝑛 2 + 𝜔 𝑑 2 ⇒𝐶 𝑠 = 𝐴 𝑠 + 𝐵 (𝑠+ 𝜉𝜔 𝑛 ) 2 + 𝜔 𝑑 2 , Partial Fraction 𝐴= 𝑠𝐶(𝑠) 𝑠=0 = 𝜔 𝑛 2 𝑠 2 +2 𝜉𝜔 𝑛 𝑠+ 𝜔 𝑛 2 𝑠=0 =1 𝐵= (𝑠+ 𝜉𝜔 𝑛 ) 2 + 𝜔 𝑑 2 𝐶(𝑠) 𝑠=− 𝜉𝜔 𝑛 − 𝑗𝜔 𝑑 = 1 𝑠 𝑠=− 𝜉𝜔 𝑛 − 𝑗𝜔 𝑑 =−(𝑠+2 𝜉𝜔 𝑛 ) So, ⇒𝐶 𝑠 = 1 𝑠 − (𝑠+2 𝜉𝜔 𝑛 ) (𝑠+ 𝜉𝜔 𝑛 ) 2 + 𝜔 𝑑 2

Step Response of Second Order System… ⇒𝐶 𝑠 = 1 𝑠 − (𝑠+ 𝜉𝜔 𝑛 + 𝜉𝜔 𝑛 ) (𝑠+ 𝜉𝜔 𝑛 ) 2 + 𝜔 𝑑 2 ⇒𝐶 𝑠 = 1 𝑠 − (𝑠+ 𝜉𝜔 𝑛 ) (𝑠+ 𝜉𝜔 𝑛 ) 2 + 𝜔 𝑑 2 + 𝜉𝜔 𝑛 (𝑠+ 𝜉𝜔 𝑛 ) 2 + 𝜔 𝑑 2 ⇒𝐶 𝑠 = 1 𝑠 − (𝑠+ 𝜉𝜔 𝑛 ) (𝑠+ 𝜉𝜔 𝑛 ) 2 + 𝜔 𝑑 2 + 𝜉𝜔 𝑛 𝜔 𝑑 𝜔 𝑑 (𝑠+ 𝜉𝜔 𝑛 ) 2 + 𝜔 𝑑 2 Take inverse Laplace transformation 𝑐 𝑡 =1− 𝑒 −𝜉 𝜔 𝑛 𝑡 𝑐𝑜𝑠 𝜔 𝑑 𝑡+ 𝜉𝜔 𝑛 𝜔 𝑑 𝑒 −𝜉 𝜔 𝑛 𝑡 𝑠𝑖𝑛 𝜔 𝑑 𝑡 Put 𝜔 𝑑 = 𝜔 𝑛 1− 𝜉 2

Step Response of Second Order System… ⇒𝑐 𝑡 =1− 𝑒 −𝜉 𝜔 𝑛 𝑡 𝐶𝑜𝑠 𝜔 𝑑 𝑡+ 𝜉 1− 𝜉 2 𝑆𝑖𝑛 𝜔 𝑑 𝑡 Put 1− 𝜉 2 =𝑆𝑖𝑛𝜙 So, Cos𝜙=𝜉 and 𝑡𝑎𝑛𝜙= 1− 𝜉 2 𝜉 ⇒𝑐 𝑡 =1− 𝑒 −𝜉 𝜔 𝑛 𝑡 1− 𝜉 2 𝑆𝑖𝑛𝜙 𝐶𝑜𝑠 𝜔 𝑑 𝑡+𝐶𝑜𝑠𝜙 𝑆𝑖𝑛 𝜔 𝑑 𝑡 ⇒𝒄 𝒕 =𝟏− 𝒆 −𝝃 𝝎 𝒏 𝒕 𝟏− 𝝃 𝟐 𝑺𝒊𝒏 𝝎 𝒅 𝒕+𝝓 −−−(𝟕)

Step Response of Second Order System… Put the values of 𝜔 𝑑 and 𝜙 𝜔 𝑑 = 𝜔 𝑛 1− 𝜉 2 and 𝜙= 𝑡𝑎𝑛 −1 1− 𝜉 2 𝜉 ⇒𝒄 𝒕 =𝟏− 𝒆 −𝝃 𝝎 𝒏 𝒕 𝟏− 𝝃 𝟐 𝑺𝒊𝒏 𝝎 𝒏 𝟏− 𝝃 𝟐 𝒕+ 𝒕𝒂𝒏 −𝟏 𝟏− 𝝃 𝟐 𝝃 −−−(𝟖) The error signal for the system 𝑒 𝑡 =𝑟 𝑡 −𝑐 𝑡 ⇒𝑒 𝑡 = 𝑒 −𝜉 𝜔 𝑛 𝑡 1− 𝜉 2 𝑆𝑖𝑛 𝜔 𝑛 1− 𝜉 2 𝑡+ 𝑡𝑎𝑛 −1 1− 𝜉 2 𝜉 −−−(9)

Step Response of Second Order System… 𝜔 𝑛 = Natural frequency of oscillation or undammed natural frequency 𝜔 𝑑 = Dammed frequency 𝜉𝜔 𝑛 = damping factor or actual damping or damping coefficient

Step Response of Second Order System… Case 1: Underdamped Case 0<𝜉<1 Response is given by equ. (8). It is under damped having overshoot and undershoot as shown below.

Step Response of Second Order System… Case 2: Undamped Case 𝜉=0 Response of equ (8) becomes ⇒𝑐 𝑡 =1−𝑆𝑖𝑛 𝜔 𝑛 𝑡+ 90 0 ⇒𝑐 𝑡 =1−𝐶𝑜𝑠 𝜔 𝑛 𝑡 −−−(10)

Step Response of Second Order System… Case 3: Critical damped Case 𝜉=1 Put 𝜉=1 in original equation 𝐶(𝑠)= 𝜔 𝑛 2 𝑠 2 +2 𝜔 𝑛 𝑠+ 𝜔 𝑛 2 1 𝑠 𝐶(𝑠)= 𝜔 𝑛 2 𝑠+ 𝜔 𝑛 2 1 𝑠 𝐶 𝑠 = 𝐴 𝑠 + 𝐵 𝑠+ 𝜔 𝑛 + 𝐶 𝑠+ 𝜔 𝑛 2 𝐶 𝑠 = 1 𝑠 − 1 𝑠+ 𝜔 𝑛 − 𝜔 𝑛 𝑠+ 𝜔 𝑛 2 𝐴= 𝑠𝐶(𝑠) 𝑠=0 =1 𝐶= 𝑠+ 𝜔 𝑛 2 𝐶(𝑠) 𝑠=− 𝜔 𝑛 =− 𝜔 𝑛 𝐵= 𝑑 𝑑𝑠 𝑠+ 𝜔 𝑛 2 𝐶(𝑠) 𝑠=− 𝜔 𝑛 = 𝑑 𝑑𝑠 𝜔 𝑛 2 𝑠 𝑠=− 𝜔 𝑛 =−1

Step Response of Second Order System… Take inverse Laplace transform 𝑐 𝑡 =1− 𝑒 − 𝜔 𝑛 𝑡 − 𝑡 𝜔 𝑛 𝑒 − 𝜔 𝑛 𝑡 𝑐 𝑡 =1− 𝑒 − 𝜔 𝑛 𝑡 1+𝑡 𝜔 𝑛 −−−(11)

Step Response of Second Order System… Case 4: Over damped Case 𝜉>1 When 𝜉>1 Equ (6) can be written as ⇒𝐶(𝑠)= 1 𝑠 × 𝜔 𝑛 2 (𝑠+ 𝜉𝜔 𝑛 ) 2 − 𝜔 𝑛 2 ( 𝜉 2 −1) ⇒𝐶(𝑠)= 1 𝑠 × 𝜔 𝑛 2 (𝑠+ 𝜉𝜔 𝑛 ) 2 − 𝜔 𝑑 2 ⇒𝐶(𝑠)= 1 𝑠 × 𝜔 𝑛 2 (𝑠+ 𝜉𝜔 𝑛 + 𝜔 𝑑 )(𝑠+ 𝜉𝜔 𝑛 − 𝜔 𝑑 ) Let 𝜔 𝑑 2

Step Response of Second Order System… ⇒𝐶 𝑠 = 𝐴 𝑠 + 𝐵 (𝑠+ 𝜉𝜔 𝑛 + 𝜔 𝑑 ) + 𝐶 (𝑠+ 𝜉𝜔 𝑛 − 𝜔 𝑑 ) Put values of A, B, & C and take inverse Laplace transformation 𝑐 𝑡 =1+ 𝑒 𝜉+ 𝜉 2 −1 𝜔 𝑛 𝑡 2 𝜉 2 −1 𝜉+ 𝜉 2 −1 − 𝑒 𝜉− 𝜉 2 −1 𝜔 𝑛 𝑡 2 𝜉 2 −1 𝜉− 𝜉 2 −1 −−−(12) 𝐴= 𝑠𝐶(𝑠) 𝑠=0 =1 𝐵= (𝑠+ 𝜉𝜔 𝑛 + 𝜔 𝑑 ) 𝐶(𝑠) 𝑠=−( 𝜉𝜔 𝑛 + 𝜔 𝑑 ) = 1 2 𝜉 2 −1 𝜉+ 𝜉 2 −1 𝐶= (𝑠+ 𝜉𝜔 𝑛 − 𝜔 𝑑 ) 𝐶(𝑠) 𝑠=−( 𝜉𝜔 𝑛 − 𝜔 𝑑 ) = −1 2 𝜉 2 −1 𝜉− 𝜉 2 −1

Step Response of Second Order System… Above equ (12) has two time constant 𝑇 1 = 1 𝜉+ 𝜉 2 −1 𝜔 𝑛 and 𝑇 2 = 1 𝜉− 𝜉 2 −1 𝜔 𝑛 with 𝑇 1 < 𝑇 2 So, we can neglect terms with time constant 𝑇 1 , Hence response can be approximated as 𝑐 𝑡 =1− 𝑒 𝜉− 𝜉 2 −1 𝜔 𝑛 𝑡 2 𝜉 2 −1 𝜉− 𝜉 2 −1 =1− 𝑒 − 𝑡 𝑇 2 2 𝜉 2 −1 𝜉− 𝜉 2 −1

Location of roots of characteristic equation and Time response We know closed loop response of 2nd order system is given as 𝐶(𝑠) 𝑅 𝑠 = 𝜔 𝑛 2 𝑠 2 +2 𝜉𝜔 𝑛 𝑠+ 𝜔 𝑛 2 Its characteristic equation 𝑠 2 +2 𝜉𝜔 𝑛 𝑠+ 𝜔 𝑛 2 =0 Its roots are ⇒ 𝑠 1 = −𝜉𝜔 𝑛 + 𝜔 𝑛 𝜉 2 −1 ⇒ 𝑠 2 = −𝜉𝜔 𝑛 − 𝜔 𝑛 𝜉 2 −1 𝜔 𝑛 𝜉 2 −1 )Damped Frequency ( −𝜉𝜔 𝑛 )Damping

Location of roots of characteristic equation and Time response … 𝜉=0 Underdamped Undamped 𝜔 𝑛 𝜉 2 −1 𝜉<1 𝜔 𝑛 𝜃 Critical damped 𝜉=1 Re 𝜉>1 −𝜉𝜔 𝑛 𝐶𝑜𝑠𝜃=𝜉 Over damped

Transient Response Specifications of 2nd Order System For 2nd order system with unit step input following are the transient response specifications Delay Time (td) Rise Time (tr) Peak Time (tp) Maximum Overshoot (Mp) Settling Time (ts) Steady-State Error (ess)

Transient Response Specifications of 2nd Order System … c(t) c(t)max T=1/ 𝜉 𝜔 𝑛 Mp 1.0 ±2% 𝑜𝑟 ±5% 0.5 Time td tr tp ts

Delay time Delay time (td) is the time required to reach at 50% of its final value by a time response signal during its first cycle of oscillation.

Rise time Rise time (tr) is the time required to reach at final value by a under damped time response signal during its first cycle of oscillation. If the output is over damped, then rise time is counted as the time required by the response to rise from 10% to 90% of its final value. We know the output of 2nd order system with step input is given as 𝑐 𝑡 =1− 𝑒 −𝜉 𝜔 𝑛 𝑡 1− 𝜉 2 𝑆𝑖𝑛 𝜔 𝑛 1− 𝜉 2 𝑡+𝛷 , Equ (7) put c(t)=1 and solve for t 𝑡 𝑟 = 𝜋−𝜙 𝜔 𝑛 1− 𝜉 2

Peak time Peak time (tp) is simply the time required by response to reach its first peak i.e. the peak of first cycle of oscillation, or first overshoot. For maxi value of c(t), put 𝑑𝑐(𝑡) 𝑑𝑡 =0 𝑡= 𝑛𝜋 𝜔 𝑛 1− 𝜉 2 For first overshoot, n=1 𝑡 𝑝 = 𝜋 𝜔 𝑛 1− 𝜉 2

Maximum overshoot Maximum overshoot (Mp) is straight way difference between the magnitude of the highest peak of time response and magnitude of its steady state. Put 𝑡=𝑡 𝑝 = 𝜋 𝜔 𝑛 1− 𝜉 2 in c(t), (equ. (7)) 𝑐(𝑡) 𝑚𝑎𝑥 =1+ 𝑒 − 𝜋𝜉 1− 𝜉 2 𝑡 % 𝑀 𝑝 = 𝑐(𝑡) 𝑚𝑎𝑥 −1 1 ×100 ⇒% 𝑀 𝑝 = 𝑒 − 𝜋𝜉 1− 𝜉 2 𝑡 ×100

Settling time Settling time (ts) is the time required for a response to become steady. It is defined as the time required by the response to reach and steady within specified range of 2 % to 5 % of its final value. Time period of the system 𝑇= 1 𝜉 𝜔 𝑛 𝑡 𝑠 =4𝑇 on the basis of ±2% error 𝑡 𝑠 =3𝑇 on the basis of ±5% error

Steady-state error  Steady-state error (ess ) is the difference between actual output and desired output at the infinite range of time. 𝑒 𝑠𝑠 = lim 𝑡→∞ 𝑟 𝑡 −𝑐(𝑡)

Assignment 1. Find out time response of 2nd order system subjected to unit impulse input. 2. Find out time response of 2nd order system subjected to unit ramp input. 3. Explain how time response of higher order system is determined?

Thanks