Steady State Error In Control Systems
(Step Inputs)
What Is Steady State Errror (SSE)?
Systems With A Single Pole At The Origin
Problems
You are at:  Analysis Techniques - Performance Measures - Steady State Error

Control systems are used to control some physical variable.  That variable may be a temperature somewhere, the attitude of an aircraft or a frequency in a communication system.  Whatever the variable, it is important to control the variable accurately.

If you are designing a control system, how accurately the system performs is important.  If it is desired to have the variable under control take on a particular value, you will want the variable to get as close to the desired value as possible.  Certainly, you will want to measure how accurately you can control the variable.  Beyond that you will want to be able to predict how accurately you can control the variable.

To be able to measure and predict accuracy in a control system, a standard measure of performance is widely used.  That measure of performance is steady state error - SSE - and steady state error is a concept that assumes the following:

• The system under test is stimulated with some standard input.  Typically, the test input is a step function of time, but it can also be a ramp or other polynomial kinds of inputs.
• The system comes to a steady state, and the difference between the input and the output is measured.
• The difference between the input - the desired response - and the output - the actual response is referred to as the error.

Goals For This Lesson

Given our statements above, it should be clear what you are about in this lesson.  Here are your goals. Given a linear feedback control system, Be able to compute the SSE for standard inputs, particularly step input signals. Be able to compute the gain that will produce a prescribed level of SSE in the system. Be able to specify the SSE in a system with integral control.
In this lesson, we will examine steady state error - SSE - in closed loop control systems.  The closed loop system we will examine is shown below.

• The system to be controlled has a transfer function G(s).
• There is a sensor with a transfer function Ks.
• There is a controller with a transfer function Kp(s) - which may be a constant gain.

What  Is SSE?

We need a precise definition of SSE if we are going to be able to predict a value for SSE in a closed loop control system.  Next, we'll look at a closed loop system and determine precisely what is meant by SSE.

In this lesson, we will examine steady state error - SSE - in closed loop control systems.  The closed loop system we will examine is shown below. • The system to be controlled has a transfer function G(s).
• There is a sensor with a transfer function Ks.
• There is a controller with a transfer function Kp(s).
• A controller like this, where the control effort to the plant is proportional to the error, is called a proportional controller.
In our system, we note the following:
• The input is often the desired output.  In other words, the input is what we want the output to be.  If the input is a step, then we want the output to settle out to that value.
• The output is measured with a sensor.  Often the gain of the sensor is one. That is especially true in computer controlled systems where the output value - an analog signal - is converted into a digital representation, and the processing - to generate the error, E(s) - takes place inside a computer.
The signal, E(s), is referred to as the error signal.
• The error signal is the difference between the desired input and the measured input.
• The error signal is a measure of how well the system is performing at any instant.  When the error signal is large, the measured output does not match the desired output very well.
A step input is often used as a test input for several reasons.
• A step input is really a request for the output to change to a new, constant value.
• If the system is well behaved, the output will settle out to a constant, steady state value.
• The difference between the measured constant output and the input constitutes a steady state error, or SSE.
It helps to get a feel for how things go.  So, below we'll examine a system that has a step input and a steady state error.  That system is the same block diagram we considered above. For the example system, the controlled system - often referred to as the plant - is a first order system with a transfer function:

G(s) = Gdc/(st + 1)

We will consider a system with the following parameters.

• Gdc = 1
• t = 1
• Ks = 1.
• Kp  can be set to various values in the range of 0 to 10,
• The input is always 1.
Here is a simulation you can run to check how this works.  In this simulation, the system being controlled (the plant) and the sensor have the parameters shwon above.  You can adjust the gain up or down by 5% using the "arrow" buttons at bottom right.  You can also enter your own gain in the text box, then click the red button to see the response for the gain you enter.
The actual open loop gain is shown in the text box above the red button.

Vary the gain.  You can set the gain in the text box and click the red button, or you can increase or decrease the gain by 5% using the green buttons.  You should see that the system responds faster for higher gain, and that it responds with better accuracy for higher gain. Try several gains and compare results.  The difference between the desired response (1.0 is the input = desired response) and the actual steady state response is the error.

Problem 1  For a proportional gain, Kp = 9, what is the value of the steady state output?

Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

Problem

P1  For a proportional gain, Kp = 9, what is the value of the steady state error?

Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

Problem

P2  For a proportional gain, Kp = 49, what is the value of the steady state output?

Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

Problem

P3  For a proportional gain, Kp = 49, what is the value of the steady state error?

Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

When you do the problems above, you should see that the system responds with better accuracy for higher gain.  Try several gains and compare results using the simulation.  You need to understand how the SSE depends upon gain in a situation like this.  You need to be able to do that analytically.  That's where we are heading next.  Here is our system again. Now, we can get a precise definition of SSE in this system.  We have the following:

• The input is assumed to be a unit step.  If the input is a step, but not a unit step, the system is linear and all results will be proportional.
• We can calculate the output, Y(s), in terms of the input, U(s) and we can determine the error, E(s).  Later we will interpret relations in the frequency (s) domain in terms of time domain behavior.
• We have:
• E(s) = U(s) - Ks Y(s) since the error is the difference between the desired response, U(s),
• The measured response, =  Ks Y(s).
• And we know:  Y(s) = Kp G(s) E(s).
• Combine our two relations:
• E(s) = U(s) - Ks Y(s) and:
• Y(s) = Kp G(s) E(s), to get:
• E(s) = U(s) - Ks  Kp G(s) E(s)
• Since E(s) = U(s) - Ks  Kp G(s) E(s) we can combine all the E(s) terms on the left side of this equation:
• E(s) [1 + Ks Kp G(s)] = U(s)
• or E(s) = U(s) / [1 + Ks Kp G(s)]
• Now, we know that:
• E(s) = U(s) / [1 + Ks Kp G(s)]
• This relation is in the frequency - s - domain, but has implications for time response behavior.
• Assume a unit step input.  The transformed input, U(s), will then be given by:
• U(s) = 1/s
• With U(s) = 1/s, the transform of the error signal is given by:
• E(s) = 1 / s [1 + Ks Kp G(s)]
• The final value theorem can be applied to get the steady state error.
We get the Steady State Error (SSE) by finding the the transform of the error and applying the final value theorem.  To get the transform of the error, we use the expression found above.

Since E(s) = 1 / s (1 + Ks Kp G(s)) applying the final value theorem Multiply E(s) by s, and take the indicated limit to get:

Ess = 1/[(1 + Ks Kp G(0)]

We can draw a few conclusions from this expression.

• The steady state error depends upon the loop gain - Ks Kp G(0).
• The term, G(0), in the loop gain is the DC gain of the plant.
• To make SSE smaller, increase the loop gain.
• And, the only gain you can normally adjust is the gain of the proportional controller, Kp.
You should also note that we have done this for a unit step input.  If we have a step that has another size, we can still use this calculation to determine the error.  If the step has magnitude 2.0, then the error will be twice as large as it would have been for a unit step.  The system is linear, and everything scales.  We can take the error for a unit step as a measure of system accuracy, and we can express that accuracy as a percentage error.
• If the response to a unit step is 0.9 and the error is 0.1, then the system is said to have a 10% SSE.

Problem 5  What loop gain - Ks Kp G(0) - will produce a system with 5% SSE?

Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

Problem

P4  What loop gain - Ks Kp G(0) - will produce a system with 1% SSE?

Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

Some Observations for Systems with Integrators

This derivation has been fairly simple, but we may have overlooked a few items.

• If the system has an integrator - as it would with an integral controller - then G(0) would be infinite.  That would imply that there would be zero SSE for a step input.
• The pole at the origin can be either in the plant - the system being controlled - or it can also be in the controller - something we haven't considered until now.  However, it should be clear that the same analysis applies, and that it doesn't matter where the pole at the origin occurs physically, and all that matters is that there is a pole at the origin in the controller or the plant.
Reflect on the conclusion above and consider what happens as you design a system.
• You may have a requirement that the system exhibit very small SSE.
• You can get SSE of zero if there is a pole at the origin.
• If there is no pole at the origin, then add one in the controller.  You will have reinvented integral control, but that's OK because there is no patent on integral control.  Click here to learn more about integral control.
• If you want to add an integrator, you may need to review op-amp integrators or learn something about digital integration.  Those are the two common ways of implementing integral control.  You can click here to see how to implement integral control.

Problems

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