Steady State Errors Introduction to Steady State Errors Static Error Constants System Types

The target is to define the errors and deriving the ways to control them. Control system design entails trade offs between desired transient responses, steady state errors and the requirement that the system be stable.

Definition steady state errors it is the difference between the input and output for a prescribed test input as t .

Test InputsINTEGRATIONDIFFERENTIATION

Step inputs represent constant position and examine the ability of a control system to position itself w.r.t. a stationary target.Ramp inputs corresponds to constant velocity input to a system and can be used to test the systems ability to follow a linearly rising input or equivalently track a constant velocity target. Example of satellite tracking system. parabolic Inputs represent constant acceleration inputs to position control systems and can be used to represent accelerating targets like missiles.

Sources of Steady State Errors they mainly arise from non linear sources like backlash in gears, non responsiveness of motors unless the Vin exceeds a threshold. but only the sources of steady state errors from the inputs, systems and input types will be covered.

Steady State errors for a step input having output 1 as zero error and output 2 having finite steady state error

For the ramp input output 1 has zero steady state error and output 2 has a finite error

General Transfer Function

Pole plot for underdamped second order system

Prepared by Mrs. Asma Adeel*Laplace Transform Properties

Prepared by Mrs. Asma Adeel

Steady State Errors for Unity Feedback SystemsSteady State Error in terms of G(s) for closed loop systemsE(s) = R(s) C(s) & C(s) = G(s) E(s)Again applying final value theorem ase( ) = lim s R(s) / 1 + G(s) s 0Now substitute different types of test inputs step, ramp & parabola.

System Types The values of the static error constants depend upon the form of G(s) i.e the number of pure integrations in the forward path. System type corresponds to order of n in the denominator. or the number of integrations in the forward path. n=0 type 0 system n=1 type 1 system n=2 type 2 system

Step Input R(s) = 1/s here n1 atleast one pole must be at the origin atleast one integration in the forward path. if there are no integrations then n=0 and the result is the finite steady state error. so for step input to a unity feedback system there will be a zero error if there is one integration in the forward path. for a simple gain, for the same input the error is finite.

Step Input R(s) = 1/s

Results: There will be zero error if and only if there is atleast one integrator in the signal path. Adding an integrator means adding a pole into the system and thus increase its order. Addition of an integrator always improves the steady state performance of the system.

Ramp Input R(s) = 1/s2 here n2 atleast two poles must be at the origin atleast two integration in the forward path. if there is one integration then n=1 and the result is the finite or constant steady state error. if there are no integrations then error would be infinite become a diverging ramp as output 3. so for ramp input to a unity feedback system there will be a zero error if there are two integration in the forward path.

Ramp Input R(s) = 1/s2

Parabolic Input R(s) = 1/s3 here n3 atleast three poles must be at the origin atleast three integration in the forward path. if there is one or less integration in forward path then the result is infinite. if there are two integrations then error would be finite or a constant error. so for parabolic input to a unity feedback system there will be a zero error if there are three integrations in the forward path.Example 7.2 , 7.3 & skill assessment 7.1 Note: Before finding errors first check system stability

Parabolic Input R(s) = 1/s3

Impulse Input

Static Error Constants & System Types

Static Error ConstantsRepresentation for e() for all three inputs determine the steady state errors where as these limits are called as Static Error Constants. Kp = position Constant = lim G(s) s 0 Kv = Velocity Constant = lim s G(s) s 0 Ka = acceleration Constant = lim s2 G(s) s 0As the value of steady state error decreases static error constant increases.Example 7.4

Relation between Static Error Constants and steady state errors Kp = position Constant = lim G(s) s 0 Kv = Velocity Constant = lim s G(s) s 0 Ka = acceleration Constant = lim s2 G(s) s 0As the value of steady state error decreases static error constant increases.Example 7.4 estep() = 1 / 1 + Kp eramp() = 1 / Kv eparabola() = 1 / Ka

Relationships between input, system type, static error constants and steady state errors

Skill assessment 7.2, example 7.5 & 7.6Analysis and Design using steady state errorsStatic error constants can be used to specify the steady state error x teristics exactly like damping ratio, settling time, peak time, and percent overshoot are used to specify the transient response of the systems Position constant Kp, velocity constant Kv, and acceleration constant Ka can be used to specify system steady state errors.Draw the conclusions about the control system if Kv of a system is 1000?

Review

Further Conclusions

Example no. 1The open loop transfer function of a servo system with unity feedback isG(s) = 10 / s ( 0.1s + 1 ) Evaluate the static error constants ( Kp, Kv , Ka) & corresponding steady state errors for the system. Obtain the steady state error of the system when subjected to an input given by r(t) = a0 + a1t + a2/2 t2

Ans: Kp = , ess( ) = 0 Kv = 1 ess() = 1 Ka = 0 ess() = for parabolic inputs

Example 21. Find the steady state errors for inputs of 5 u(t), 5t u(t) and 5t2 u(t) to the system shown in the figure.

Example 3A unity feedback system has the following forward transfer function G(s) = 10( s+20)(s + 30) / s ( s+25) (s+35). Find the steady state errors for inputs of 15 u(t), 15t u(t) and 15t2 u(t).

Find the values of static error constants as well.

Example 4Find the steady state errors via the static error constants for the following systems.

Example 5A unity feedback system has the following transfer function. G(s) = 1000 ( s+8) / (s+7) (s+9). Evaluate:

System type Kp, Kv and Ka.Find the corresponding steady state errors Example 6What information is contained in the specification Kp =1000.Example 7A unity feedback system has the following forward transfer function G(s) = K ( s+12) / ( s+14) ( s+18). Find the value of K to yield a 10% error in the steady state.

Sensitivity

The degree to which the changes in system parameters affect system transfer functions and thus performance is called sensitivity.A system with zero sensitivity i.e changes in system parameters have no effect on the transfer function is said to be ideal.The greater is sensitivity, the less desirable the effect of a parameter change.Feedback in general reduces sensitivity to parameter changes.

Sensitivity is the ratio of fractional change in the function to the fractional change in parameter as the fractional change of parameter approaches zero.S F:P = lim Fractional change in function F P 0 Fractional change in parameter pS F:P = lim F/F P 0 P/P S F:P = P F/ F P

EXAMPLE Find the sensitivity of steady state error to change in K.

EXAMPLE Calculate the sensitivity of the closed loop transfer function to changes in the parameter a? How sensitivity can be reduced?

EXAMPLE For the system find the sensitivity of steady state error to changes in parameter K and parameter a with ramp inputs?

EXAMPLE Find the sensitivity of the steady state error to changes in parameter K and parameter a for the system with a step input.

Example 7.3, 7.8 Skill Assessment: 7.1, 7.2, 7.6 Case Studies: 1