Learning Objectives:

• Identify from the given transfer function, zeros and poles.

• Identify from the given transfer function, type and order of the system.

• Transfer Function for a single input, single output system is defined as ’the ratio of Laplace Transform of the output to the Laplace Transform of the input with zero initial condition.’
  The linear time-invariant transfer function model can be expressed as the ratio of two polynomials in the form:

  \[\dfrac{Y(s)}{X(s)} = G(s) = \dfrac{N(s)}{D(s)} = K \dfrac{\displaystyle \prod_{m=1}^{M} (s - z_m)}{ \displaystyle \prod_{m=1}^{N} (s - p_n)}\]

  where the M zeros, z\(_m\), and the N poles, p\(_n\), are the roots of the numerator polynomial, N(s), and the denominator polynomial, D(s), respectively.
  In other words, \[\textrm{Transfer Function}= K \dfrac{(s-s_{a})(s-s_{b})....(s-s_{m})}{(s-s_{1})(s-s_{2})....(s-s_{n})}\] The values of s obtained by equating the denominator of the transfer function to zero are called poles of the transfer function.
  The values of s which makes the transfer function zero are called zeroes of the transfer function.
  In the above equation, $s_a, s_b, s_c$, ... are zeroes while $s_1, s_2, s_3$, ... are poles.

• Order of a system is the highest power of s present in the denominator polynomial of a closed loop transfer function of a system.

• System type is defined as the order (number) of poles of G(s) at s=0.

• Characteristic equation is obtained by equating the denominator of a closed loop transfer function to zero.

Solved Example: 87-1-01

System type is defined as:

A. No. of poles of G(s) at s= 0

B. No. of poles of G(s) at s= $\infty$

C. No. of poles of G(s) left of origin

D. No. of poles of G(s) right of origin

To get the 'type' of a transfer function count the number of free s's in its denominator of open loop transfer function; free s's are s's that are not part of a first or second order component.

Correct Answer: A

Solved Example: 87-1-02

Basically, poles of transfer function are the Laplace transform variable values which causes the transfer function to become:

A. Zero

B. Unity

C. Infinite

D. Average value

Poles and Zeros of a transfer function are the frequencies for which the value of the denominator and numerator of transfer function becomes zero respectively.
At poles, since the denominator becomes zero, the transfer function becomes infinity.

Correct Answer: C

Solved Example: 87-1-03

The output is said to be zero state response because ___________ conditions are made equal to zero.

A. Initial

B. Final

C. Steady state

D. Impulse response

Correct Answer: A

Solved Example: 87-1-04

By equating the denominator of transfer function to zero, which among the following will be obtained? (RSMSSB JE Electrical (Degree) Nov 2020)

A. Poles

B. Zeros

C. Both A and B

D. None of the above

Poles and Zeros of a transfer function are the frequencies for which the value of the denominator and numerator of transfer function becomes zero respectively.

Correct Answer: A

Solved Example: 87-1-05

Consider a linear time-invariant system whose input r(t) and output y(t) are related by the following differential equation: \[\dfrac{{{d^2}y\left(t \right)}}{{d{t^2}}} + 4y\left(t \right) = 6r\left(t \right)\] The poles of this system are at: (GATE EE 2020)

A. +2j, -2j

B. +2, -2

C. +4, -4

D. +4j, -4j

Given the differential equation is, \[\dfrac{{{d^2}y\left( t \right)}}{{d{t^2}}} + 4y\left( t \right) = 6r\left( t \right)\] By applying the Laplace transform, \[s^2Y(s) + 4 Y(s) = 6 R(s)\] \[\dfrac{{Y\left( s \right)}}{{R\left( s \right)}} = \dfrac{6}{{{s^2} + 4}}\] Poles are the roots of the denominator in the transfer function. \[s^2 + 4 =0 \] \[s = \pm2j\]

Correct Answer: A

Learning Objectives:

• Use poles and zeros of transfer functions to determine the time response of a control system.

• Describe quantitatively the transient response of first-order systems.

Transfer function of a first order system is: \[\dfrac{\mathrm{Output}}{\mathrm{Input}}= \dfrac{Y(s)}{R(s)}= \dfrac{1}{Ts + 1}\]

Unit-step Response of a first-order System:

The input, r(t) = 1 for \(\displaystyle t\geq 0\).
Laplace transform: \(R(s) = \dfrac{1}{s}\)
and therefore the unit-step response is: \[Y(s) = \dfrac{1}{s(Ts+1)}\] Expanding Y(s) into partial fractions: \[Y(s) = \dfrac{1}{s} - \dfrac{T}{Ts+1} = \dfrac{1}{s} - \dfrac{1}{s+\dfrac{1}{T}}\] Take the inverse Laplace transform: \[y(t) = 1 - e^{\dfrac{-t}{T}},\ \mathrm{where} \ t>0\] The solution has two parts: a steady-state response: \(\displaystyle y_{ss}(t) = 1\) and a transient response: \(\displaystyle y_{t}(t) = e^{\dfrac{-t}{T}}\) which decays to zero as \(\displaystyle t\rightarrow \infty\)

Unit-ramp Response of First-Order System:

The input, r(t) = t for \(\displaystyle t\geq 0\).
Laplace transform: \[R(s) = \dfrac{1}{s^{2}}\] The output transform: \[Y(s) = \dfrac{1}{s^{2}}\dfrac{1}{Ts + 1}\] Expanding Y(s) into partial fractions: \[Y(s) = \dfrac{1}{s^{2}} - \dfrac{T}{s} + \dfrac{T^{2}}{Ts + 1}\] Taking the inverse Laplace transform: \[y(t) = t -T + T e^{\dfrac{-t}{T}} , \ t\geq 0\] Steady-state error: \[e_{ss} = e(\infty ) = \lim_{t\rightarrow \infty }r(t)-y(t) = T\]

Solved Example: 87-2-01

First order system is defined by:

A. Number of poles at origin

B. Order of the differential equation

C. Total number of poles of equation

D. Total number of poles and order of equation

First order system is defined by total number of poles and also which is same as the order of differential equation.

Correct Answer: D

Solved Example: 87-2-02

A unit step is applied at t=0 to a first order system without time delay. The response has the value of 1.264 units at t=10 mins, and 2 units at steady state. The transfer function of the system is:

A. $\dfrac{3}{(1+600s)}$

B. $\dfrac{2}{(1+500s)}$

C. $\dfrac{5}{(1+220s)}$

D. $\dfrac{2}{(1+600s)}$

$y(t)= K[1-e^{-t/T}]$
Here K=2, $0.632= 1-e^{-10/T}$, $T=600 \mathrm{\ sec}$
$G(s)=\dfrac{2}{(1+600s)}$

Correct Answer: D

Solved Example: 87-2-03

A system with transfer function $\dfrac{1}{(Ts+1)}$, subjected to a step input takes to seconds to reach 50% of step height. The value of t is : (UPRVUNL AE EE 2014)

A. 6.9s

B. 10s

C. 14.4s

D. 20s

The response of a first order system is: \[y(t)=K[1-e^{-t/T}]\] \[\dfrac{1}{2}= 1-e^{-10/t}\ \mathrm{Or,\ }T= 14.43 \mathrm{\ sec}\]

Correct Answer: C

Learning Objectives:

• Find the damping ratio and natural frequency of a second-order system.

• Find the settling time, peak time, percent overshoot, and rise time for an underdamped second-order system.

A general form of a second order system is: \[\dfrac{d^{2}y(t)}{dt^{2}} + 2 \xi \omega _{n}\dfrac{dy(t)}{dt} + \omega _{n}^{2}y(t) = K \omega _{n}^{2}u(t)\]

And the transfer function is: \[\dfrac{Y(s)}{R(s)} = \dfrac{K \omega _{n}^{2}}{s^{2}+ 2\xi \omega _{n}s + \omega _{n}^{2}}\] where:
K = Gain of the system,
\(\xi\) = Damping ratio of the system,
\(\omega _{n}\) = The undamped natural frequency of the system
Depending upon the value of \(\xi\) ,there are three possibilities:

• \(\xi\) = 1 means critically damped

• \(\xi\) \(<\) means 1 underdamped

• \(\xi\) \(>\) means overdamped

The maximum overshoot is the maximum peak value of the response curve measured from unity. \[\%M_{p}= e^{\dfrac{-\zeta \pi }{\sqrt{1- \zeta ^{2}}}} * 100\] Settling time \(t_s\) = It is the time required for the response curve to reach and stay within a specified percentage (usually 2% or 5%) of the final value. \[t_{s} = \dfrac{4}{\zeta \omega _{n}}\]

Solved Example: 87-3-01

For the following second order control system model, the damped natural frequency in rad/s is: \[\dfrac{Y(s)}{R(s)} = \dfrac{20}{s^{2}+5s+24}\]

A. 4.89

B. 0.511

C. 4.2

D. 0.277

\[\dfrac{Y(s)}{R(s)} = \dfrac{20}{s^{2}+5s+24}\] Also, \[\dfrac{Y(s)}{R(s)} = \dfrac{20}{24}\left [\dfrac{24}{s^{2}+5s+24}\right]\] Comparing with, \[\dfrac{Y(s)}{R(s)} = K \left [\dfrac{\omega_{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}\right]\] \[\omega _{n}^{2}= 24\] $\mathrm{Undamped\ natural\ frequency\ \ } \omega _{n} = 4.89 \ \mathrm{rad/s}$
$\mathrm{And\ Also\ }\ 2\xi \omega _{n} = 5$
$\mathrm{The\ damping \ ratio\ \ } \xi = 0.511$
$\mathrm{The \ damped \ natural \ frequency\ \ }$
$\omega _{d} = \omega_{n}\sqrt{1-\xi ^{2}} = 4.2 \ \mathrm{rad/s}$

Correct Answer: C

Solved Example: 87-3-02

Considering the unity feedback system G(s)=$\dfrac{9}{s(s+2.4)}$ , the settling time of the resulting second order system for 2% tolerance band will be:

A. 4.5 sec

B. 3.33 sec

C. 2.84 sec

D. 2.25 sec

\[G(s)=\dfrac{9}{s(s+2.4)}\] Unity feedback system means H(s) = 1
Closed loop transfer function \[\dfrac{Y(s)}{R(s)} = \dfrac{G(s)}{1+G(s)H(s)}\] Substituting the values of G(s) and H(s), Closed loop transfer function \[ \dfrac{Y(s)}{R(s)} = \dfrac{9}{s^2+2.4s + 9}\] Comparing with, \[\dfrac{Y(s)}{R(s)} = K \left [\dfrac{\omega_{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}\right]\] \[\dfrac{9}{s^2+2.4s + 9} = K \left [\dfrac{\omega_{n}^{2}}{s^{2}+2\xi \omega _{n}s+\omega _{n}^{2}}\right]\] \[K = 1, \omega_n = 3\ rad/s, \zeta = 0.4\] The settling time is given by, \[t_s = \dfrac{4}{\zeta \omega_n} = \dfrac{4}{(0.4)(3)} = 3.33\ sec\]

Correct Answer: B

Solved Example: 87-3-03

Peak overshoot of step-input response of an underdamped second-order system is explicitly indicative of: (TANGEDCO AE EE 2018)

A. Damping ratio

B. Rise time

C. Natural frequency

D. Settling time

Peak overshoot refers to the damping of the system as if the damping id less than the peak will be more.

Correct Answer: A

Solved Example: 87-3-04

Match the transfer functions of the second order systems with the nature of the systems given below: (GATE EE 2018)

Transfer Function
P.$\dfrac{15}{s^2+5s+15}$
Q.$\dfrac{25}{s^2+10s+25}$
R.$\dfrac{35}{s^2+18s+35}$

Nature of System
I. Overdamped
II. Critically damped
III. Under damped

A. P-I, Q-II, R-III

B. P-II, Q-I, R-III

C. P-III, Q-II, R-I

D. P-III, Q-I, R-II

Compare each equation with the standard second order system equation: \[\dfrac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2}\]
For system P.$\dfrac{15}{s^2+5s+15}$
$\omega_n^2 = 15$ and $2\zeta \omega_n = 5$
$\zeta = \dfrac{5}{2\sqrt{15}}< 1$ (Underdamped)

For system Q.$\dfrac{25}{s^2+10s+25}$
$\omega_n^2 = 25$ and $2\zeta \omega_n = 10$
$\zeta = 1$ (Critically Damped)

For system R.$\dfrac{35}{s^2+18s+35}$
$\omega_n^2 = 35$ and $2\zeta \omega_n = 18$
$\zeta = \dfrac{9}{\sqrt{35}}> 1$ (Overdamped)

Correct Answer: C

Learning Objectives:

• Compute steady state errors, the influence of open-loop system type on steady state error, hints on the trade-off between closed-loop errors and stability.

Transient response represents the response of the system in situations such as just after switching on the system or just after sudden changes in load or after short circuiting.
Transient response is important when:

1. The system is just switched ’on’.

2. After any abnormal conditions that disturb the equilibrium, such as change in load or when an input signal is applied.

Steady State Error \(e_{ss}\), is the difference between the steady state value (desired output) and the actual output as time tends to infinity. \[e_{ss} = \lim_{t\rightarrow \infty } e(t) = \lim_{t\rightarrow \infty } r(t)-y(t)\] where y(t) is the output signal which is tracking reference input r(t).

Solved Example: 87-4-01

The steady-state error of a feedback control system with an acceleration input becomes finite in a:

A. Type 3 system

B. Type 1 system

C. Type 2 system

D. Type 0 system

Correct Answer: C

Solved Example: 87-4-02

Consider a unity feedback system with forward transfer function given by $G\left( s\right) =\dfrac {1}{\left( s+1\right) \left( s+2\right) }$. The steady-state error in the output of the system for a unit-step input is:

A. 0.25

B. 0.50

C. 0.67

D. 0.75

This is a unity feedback system. H(s) = 1, so $G\left( s\right) H\left( s\right) =\dfrac {1}{\left( s+1\right) \left( s+2\right) }$ \[K_{p} = \lim_{s\rightarrow 0} G\left( s\right) \cdot H\left( s\right) =\lim_{s\rightarrow0}\dfrac {1}{\left( s+1\right) \left( s+2\right) } =0.5\] Steady state error to unit step input is, \[e_{ss}=\dfrac {1}{1+K_p} =\dfrac {1}{1.5} =0.67\]

Correct Answer: C

Solved Example: 87-4-03

The initial response when the output is not equal to input is called:

A. Transient response

B. Error response

C. Dynamic response

D. Either of the above

Let us see the options individually.
The initial response is called transient response. After ’sufficient’ time has passed, the response is referred as steady state response. Error is the difference between input and output. A dynamic response is the response of a structure to a dynamic load (such as an explosion, or earthquake shaking) whereas a static response is the response of a structure to static loads (such as the self weight of a structure).

Correct Answer: A

Solved Example: 87-4-04

Closed-loop transfer function of a unity-feedback system is given by: $\dfrac{Y(s)}{R(s)} = \dfrac{1}{\tau s+1}$ Steady-state error to unit-ramp input is:

A. $\tau$

B. $\infty$

C. 1

D. $\dfrac{1}{\tau}$

The input, r(t) = t for $ t\geq 0$.
Laplace transform: $R(s) = \dfrac{1}{s^{2}}$
The output transform: $Y(s) = \dfrac{1}{s^{2}}\dfrac{1}{\tau s + 1}$
Expanding Y(s) into partial fractions: $Y(s) = \dfrac{1}{s^{2}} - \dfrac{\tau}{s} + \dfrac{\tau^{2}}{\tau s + 1}$
Taking the inverse Laplace transform: $y(t) = t -\tau + \tau e^{\frac{-t}{\tau}} , \ t\geq 0$
Steady-state error: $e_{ss} = e(\infty ) = \lim_{t\rightarrow \infty }r(t)-y(t) = \tau$

Correct Answer: A