Introduction

Sensitivity

Problems

        When you design a control system you need to account for all eventualities.  The most likely problem you will have is that the system you are controlling - an airplane, a furnace, a motor, etc. - will not stay the same.  There are numerous reasons for change.

• A motor will have wear in the bearings, and it might also drive different loads at different times.
• An airplane will have changed dynamics over a flight as fuel is consumed (and the weight of the aircraft changes) and as altitude and speed change (both of which change the way control surfaces affect the motion of the airplane).
• A furnace may be full at some times and near empty at others, and the thermal time-constant will change.

        In this lesson we begin to examine how changes in systems affect your designs and you begin to learn how to predict some of those effects.

        As a way of beginning to understand the problem, let's examine a simple feedback control system.  Here is the block diagram for the system we will examine.  We will assume the following for the system.

• You know the transfer function G(s).  We will assume that G(s) is a typical first order system described by a DC gain and a time constant.
• The transfer function is assumed to be G(s) = G_DC/(st + 1)
• For purposes of argument, we will take the following values.
• G_DC = 5
• t = .6
• You want to use proportional control, so you will choose a value for the gain, K.

        Now, let's assume that you have some performance specifications for the closed loop system.

• You need to have less than 2% SSE for a step input.
• You need to have a closed loop time constant less than .1 sec.

The requirement on SSE means that the closed loop DC gain has to be .98, so you get:

Then, you check to see if the closed loop time constant meets the spec.  The closed loop time constant is given by:

That meets the specification, so you have a good design choice with K = 0.98,  What could possibly go wrong?

        Although the design seems simple and straight forward, there are any number of things that could go wrong.  Let's look at what we assumed.  We started off with the following assumption.

• You know the transfer function G(s).  We will assume that G(s) is a typical first order system described by a DC gain and a time constant.
• The transfer function is assumed to be G(s) = G_DC/(st + 1)
• For purposes of argument, we will take the following values.
• G_DC = 5

t = .6

• You might have calculated the transfer function by modelling the system, writing basic equations from physics for the system, and solving for the transfer function.  Then you would have had to calculate paramter values or measure part of the system.
• You might have measured the step response of the system and calculated the transfer function from step response data.
• You might have measured the frequency response of the system, plotted a Bode' plot for the system and used the Bode' plot to get a transfer function.

• When you model a system you often make simplifying assumptions.  With more effort you could generate a more accurate - and probably more complex - model.  If this is the transfer function for a motor, maybe you neglected the inductance in the windings, throwing out an electrical time constant in the process, for example.
• When you measure the step response of a system, you assume that you can determine the order of the system from the response, but there could well be some short time constants that get lost in the data.
• When you measure the frequency response of a system, you assume that you took data over a large enough range of frequencies.  Are you sure that you took enough data?

        Besides the possibilities mentioned above, there are other possibilities that could affect your design.

• You had to measure the parameters of your model.  In this case the parameters you measured were the DC gain and the time constant.  How accurately do you know those parameters?  You surely don't have those parameters nailed down to the 17th decimal place.  If you only know the time constant to within 10% you need to take your limited knowledge into account.
• It's been a while (a day, a month, a year ???) since you measured those parameters.  In that time, things could have changed.  How much could they have change?
• When you measured your system parameters you did that for some particular input step size and in some particular situation.  Are you sure that your parameter values are the same under different circumstances?
• If there are any nonlinearities - even slight ones - that will affect your DC gain and time constant.
• Maybe you measured how a motor changes speed from 1000 to 2000 rpm and used that data.  Will you get the same parameters going from 3000 to 4000 rpm?

        You begin to see the problem.  We need to build an understanding of how systems change when things inside them change.  We'll begin that in the next section.

        In our first order closed loop system, we found that the closed loop DC gain was given by:

Now, the first question is "How much does the closed loop DC gain change when the open loop DC gain changes?"  This is really a question about accuracy.  then the open loop DC gain is larger, the closed loop DC gain is approximately 1.0, and the steady state error (SSE) is small.  You will want the SSE to be small, and you will want the closed loop DC gain close to 1.0 in this system configuration.

        Since this is a question about change, we're going to get into taking derivatives.  The first thing you are tempted to look at is the derivative of the closed loop DC gain with respect to the open loop DC gain.  That's not a bad place to start, but we will use a more sophisticated measure of change later, but we will start by computing this derivative.

That derivative is given by:

E1  For our example system, G_DC is 5.  When K = 9.8, the derivative above is given by:

That means that if the open loop DC gain, G_DC , changes fro 5 to 5.5 (a 10% change), we would estimate the change in the closed loop DC gain to be .00392x 0.5 = .00196.  However, remember that the closed loop DC gain is:

So, we would estimate the closed loop DC gain to change to about 0.984 when the open loop DC gain changes by 10%.  We would also estimate the closed loop DC gain to change to about 0.976 when the open loop DC gain changes by -10%.

        Here is a simulator that will let you compare the response for two different systems.

• Buttons and labels colored yellow are pre-loaded with values for the original system.
• Buttons and labels colored green are pre-loaded with values for a system with a 10% higher open loop DC gain.
• You can input other values for either system.
• Clicking the yellow labelled "Start" button will permit you to see the step response of the original system.
• Clicking the green labelled "Start" button will permit you to see the step response system of the altered system along with the response of the original system.

Click here to get the simulator in a separate window.  Then do the following.

• Check the closed loop DC gain for both cases (original system and 10% increase in open loop DC gain) and verify whether or not the above calculations predict the closed loop DC gain correctly.

        In the material above we calculated variations in closed loop DC gain, but those calculations did not really give a clear idea of sensitivity because the open loop DC gain was approximately 5, and the closed loop DC gain was close to 1.  In many cases it may be better to have the answer to this question.

• If the open loop DC gain changes by some percentage, what is the percentage change in the closed loop DC gain.

        First, we have to consider what a percentage change is.  If we have some quantity, X, that changes by an amount DX, the percentage change in X is given by:

However, we will be dealing with small changes in most quantities, so we will look at incremental changes (dX).  For example, if we want to get the sensitivity of the closed loop DC gain to changes in open loop DC gain, we define that sensitivity as:

        If we take our example system, we already have the derivative indicated above.  That derivative is:

And, we also have the other terms in the expression for the sensitivity.

so, the sensitivity becomes:

E2  In our example system, we can compute the sensitivity as:

So, a 1% change in the open loop DC gain causes a .02% change in the closed loop DC gain.  Or, a 5% change in the open loop DC gain would cause a .1% change in the closed loop DC gain.

        Note that the percentage change in the closed loop DC gain is substantially smaller than the percentage change in the open loop DC gain.  That's one reason feedback control is used so often.

        Open loop parameters include the following.

• Open loop DC gain
• Open loop time constant
• Proportional gain

• Closed loop DC gain
• Closed loop time constant

We can get the sensitivity of the closed loop time constant to changes in the open loop DC gain.  We define the sensitivity as:

Take the indicated derivative:

 Insert that result into the expression for the sensitivity and we get:

Now, use the expression above for the closed loop time constant (in the last term) and we get:

E3  For the running example system, we can compute a numerical value for the sensitivity.  Using our previous values (K = 9.8 and G_dc = 5) we get:

What this means is that if the DC gain of the system increases 1%, the time constant will decrease almost 1%.

        Here is a simulator that will let you compare the response for two different systems.

• Buttons and labels colored yellow are pre-loaded with values for the original system.
• Buttons and labels colored green are pre-loaded with values for a system with a 10% higher open loop DC gain.
• You can input other values for either system.
• Clicking the yellow labelled "Start" button will permit you to see the step response of the original system.
• Clicking the green labelled "Start" button will permit you to see the step response system of the altered system along with the response of the original system.

Click here to get the simulator in a separate window.  Then do the following.

• Check the closed loop time constant for both cases (original system and 10% increase in open loop DC gain) and verify whether or not the above calculations predict the closed loop time constant correctly.

        We know the closed loop transfer function is given by:

Now, examine the sensitivity of the closed loop DC gain to changes in the open loop time constant.  We know the closed loop DC gain is given by:

The closed loop DC gain does not depend upon the open loop time constant, so when the open loop time constant changes, the closed loop DC gain does not change.  Therefore, we have:

The sensitivity is zero if the closed loop DC gain does not change when the time constant changes.

        We don't always get such nice neat results, and we may be tempted to draw some incorrect conclusions from this result.

        Let's represent the original system a different way.  Consider this.

In other words, instead of using the DC gain and the time constant, we rewrite the transfer function of the controlled system - the plant - to show the open-loop pole explicitly.  Now consider getting the sensitivity of the closed loop DC gain to changes in the pole location.  We want to compute:

And, we know:

Now, notice that the closed loop DC gain depends upon the pole using this representation and that clearly implies that the sensitivity will not be zero.

        To compute the sensitivity, we first take the derivative of the closed loop DC gain with respect to the pole, a.  That derivative is given by:

E4  Consider a situation with these parameters

• G_a = 10
• a = 2

If we wanted to design for 2% SSE, we would require a total DC gain of 50, so we need

• K = 10.

This expression would allow us to compute the change in closed loop DC gain for a change in a.  For example if a change from 2.0 to 2.01, the closed loop DC gain changes from 0.9804 to .9803.  We would estimate exactly those changes, i.e.

E5  Continuing the example above, with these parameter values:

• G_a = 10
• a = 2
• K = 10

or, in other words, a 1% change in a will produce a -0.00019% change in the closed loop DC gain.  The actual values - from Example E4 - are:

• a changes from 2 to 2.01 a 0.5% change in the open loop pole.
• The closed loop DC gain changes from 0.9804 to 0.9803 and change of -0.0001%.
• The ratio of the percentage changes is:
• -.0001%/.5% = -0.0002 - just about right.

       Now, let's summarize what we've learned in this lesson.

• You were introduced to the concept of changes in closed loop system parameter values.
• Small changes in closed loop system parameter values were related to small changes in open loop parameters and the relationship was through derivatives.
• Percentage changes in closed loop system parameter values were related to percentage changes in open loop parameters and the relationship was through a quantity called the sensitivity.

The general form is 1 divided by 1 added to the open loop gain (DC or a function of s).  Many texts refer to this as the "Sensitivity Function".  It does turn up in many of the sensitivity functions you will calculate and has indeed turned up in all of the sensitivity functions we have calculated thus far.

        In this lesson we have used a first order system as a working example.  However, consider the expression we derived for the sensitivity of the closed loop DC gain to changes in open loop DC gain.  That expression is reproduced below.

This expression depends only upon the expression for the closed loop DC gain, given below.

• Perf3SensitivityP01 - Estimating change in closed loop DC gain
• Perf3SensitivityP02 - Estimating change in closed loop time constant
• Perf3SensitivityP03 - Estimating change in closed loop DC gain when the open loop pole changes.
• Perf3SensitivityP04 - Sensitivity of closed loop time constant to changes in open loop time constant.

• Perf1SSE - Steady State Error
• Perf3Sensitivity