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Dynamic System Response

Learning Objectives:

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

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?

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:

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:

Compute steady state errors, for various system types and for various input signals.

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:

The system is just switched ’on’.
After any abnormal conditions that disturb the equilibrium, such as change in load or when an input signal is applied.

System Types:

System type is decided by the power of 's' term in the denominator.

e.g. Consider the following system with transfer function:

\[G(s) = \dfrac{K (s+3)}{s (s+ 9) (s+4)}\]

This control system is of Type I, because the 's' ( = s$^1$) term in the denominator has power = 1.

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. Never

B. Type 1 system

C. Type 2 system

D. Type 0 system

Refer steady state error table on page 227 in handbook. For acceleration input, the steady state error becomes finite starting with Type 2 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