An Introduction To System Dynamics - Second Order Linear Systems

An Introduction To System Dynamics - Second Order Systems

Introduction
Goals
The System
Impulse Response
Step Response
Problems

You are at Basic Concepts - Time Response - 2nd Order Step and Impulse Response
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Why Study Second Order Systems?

Second order systems are important for a number of reasons. They are the simplest systems that exhibit oscillations and overshoot.

Many important systems exhibit second order system behavior.
Second order behavior is part of the behavior of higher order systems and understanding second order systems helps you to understand higher order systems.

In this lesson you will learn about time response and, in a companion lesson, about frequency response of second order systems. Then you will see how laboratory measurements can be used to determine parameters of second order systems.

Goals For This Lesson

There are a number of goals for you in this lesson.

First, if you have a second order system, you need to be able to predict and understand how it responds to an input, so you need to be able to do this.

Given a second order system,

Determine if the system is underdamped - with an oscillatory response - or if it is overdamped with no oscillations in the response.

Given a transfer function for a second order system,

Determine the DC gain of the system, G_dc, and the damping ratio, z, and the undamped natural frequency, w_n.

Further goals for this lesson include these.

Given a second order system,

Determine the unit step response and the unit impulse response of the system.
Determine the DC gain, damping ratio and natural frequency from a plot of the unit step response of the system.

The System

We start with an example of a system that exhibits second order behavior. You may have seen a similar example in the lesson on first order systems. Click here to look at that example.

A Cartoon Biplane.

Above is a movie of an airplane - actually a biplane - in which the pilot suddenly changes the controls so that the altitude of the biplane changes. The new steady state altititude is higher than the previous altitude. This system shows second order system behavior as the airplane changes altitude. You can click the button at the lower right to see the path followed by the biplane. (Click inside the button to show the path, and release the mouse outside the button to permit the path to show continuously.) This is an example of a second order system that you can see.

OK. You have seen the example. What you see there is an example of a second order step response. It has characteristics - like decaying oscillations - that you can have in second order systems. Those characteristic decaying oscillations are not to be seen in first order systems. If you see decaying oscillations, you know you don't have a first order system. On the other hand, not every second order system will exhibit those decaying oscillations. Second order systems are more complex than that.

With that in mind, let's look at some basic ideas about second order systems. The simplest second order system satisfies a differential equation of this form.

where:

x(t) = Response of the System,
u(t) = Input to the System,
z = Damping Ratio,
w_n=Undamped Natural Frequency,
G_dc= The DC Gain of the System.

The parameters you find in a second order system determine aspects of various kinds of responses. Whether we are talking about impulse response, step response or response to other inputs, we will still find the following relations.

z, the damping ratio, will determine how much the system oscillates as the response decays toward steady state.
w_n,the undamped natural frequency, will determine how fast the system oscillates during any transient response.
G_dc, the DC gain of the system, will determine the size of steady state response when the input settles out to a constant value.

In this lesson we will discuss several particular time responses in second order systems,

The impulse response,
The unit step response,
The response with arbitrary initial conditions and no input.

We're going to assume some background on your part in areas ofLaplace transforms, linear system theory (particularly transfer functions) and differential equations.

So far we have looked at a mathematical description of the system.

Systems do not exist in isolation. They usually have input signals, and we are interested in how they respond to those input signals.
There is a family of standard input signals. We use those standard signals and the response of systems to those signals when we want to compare how different systems respond.
Two common signals are impulses and steps. Click on the hotwords to go to the section on that particular signal viewed as an input to a first order system.

Impulse Response

We'll start with the impulse response.

Assume we have a system described by this differential equation.

There are a number of ways that we can compute the impulse response of a system described by this differential equation.

The first way is to solve the differential equation taking into account how the impulse changes the initial values. Click here to see the differential equation solution.
The second way is to use what you know about Laplace transforms to determine the impulse response. Click here to see how to compute the impulse response from the transfer function.

We will examine the differential equation solution first.

Given the differential equation:

If the input, u(t), is a unit impulse, d(t), then we can compute changes in the first derivative above. If we think about what happens when u(t) is an impulse using the differential equation, we come to the following conclusions.

For an instant, at t = 0, the right hand side of the differential equation is infinite since u(t) is an impulse and the impulse is infinite at t = 0.
At that same instant, something on the left hand side has to be infinite.
For the variable, x(t) to be infinite, the derivative, dx(t)/dt would also have to be infinite.
For the derivative to be infinite, the second derivative would also have to be infinite.

Note, you may want to review what happens in first order systems for an impulse input. Click here to see that material.

It is a good assumption that what happens is that we have:

Integrate both sides around the point t = 0.

And, after integrating, this becomes:

So, the net effect of the impulse is that the first derivative has a different value after the impulse has occurred. Other than that, nothing else changes. So, to compute the impulse response we compute the response of a system with no input (so we get the homogeneous response) but with a different initial condition determined by the impulse.

The homogeneous equation is:

Solution of linear, homogeneous equations is started by assuming that the solution is of the form Ae^st. Doing that, we have:

As²e^st + 2zw_nAse^st + w_n²Ae^st= 0

The common factor, Ae^st, can be removed to get:

s² + 2zw_ns + w_n² = 0

Now, at this point you should realize that there are several things that could happen because the roots of this polynomial equation could be any of the three possibilities below.

Two real - and - distinct roots.
Two equal real roots.
A pair of complex roots.

And, those roots - denoted by s₁ and s₂ - are given by:

The Case of Two Real Roots - Impulse Response

Let us first examine the case of two real, distinct roots. In order to have two real, distinct roots, we must have:

z> 1

When this is true, the impulse response will be of the form:

At t = 0 we can assume zero initial conditions, so we must have:

A₁ + A₂ = 0

or A₁ = - A₂

And, we also have the value of the derivative at t = 0.

s₁A₁ + s₂A₂ = G_dc w_n²

We have two equations in two unknowns - A₁ and A₂ - so we can solve for the unknowns. Doing that we have:

A₁ = G_dc w_n²/(s₁ - s₂)

and: A₂ = -G_dc w_n²/(s₁- s₂)

which means that the solution is:

Example

E1 Let us examine a few examples to help us visualize what this kind of response looks like. A typical situation would be a system that satisfies a differential equation:

d²x(t)/dt² + 6 dx(t)/dt + 5 x(t) = 5 d(t)

This system would have two exponentials in the solution:

A₁e^-t

and A₂e^-5t

Carrying through the algebra above, you should find that the solution works out to be:

1.25e^-t - 1.25e^-5t

The response is the sum of two decaying exponentials. Note the following about this response.

The response starts at 0 at t = 0.
The response decays to zero as t becomes large.
There is a point - and only one point - at which the response reaches a peak.

There would not be a peak if one of the poles were at s = 0.

We can determine where the peak is in the response. Take the expression for the response. Now that we know the response will be zero at t = 0, so we know that the response will have this general form:

r(t) = A (e^-at- e^-bt)

Differentiate this expression, and set the result to zero:

dr(t)/dt = A (ae^-at - be^-bt) = 0

and solving this, we must have:

ae^-at = be^-bt

or:

(a/b) = e^-(b-a)t

ln(a/b) = -(b-a)t_peak

so: t_peak = -ln(a/b)/(b-a)

Some Observations And A Question

How can this result be used? If one of the time constants - i.e. 1/a or 1/b - can be determined, then knowing where the peak is located will allow you to compute the other time constant. If you can measure the impulse response of a system you might be able to figure out what the system is - that is compute G_dc, a and b.

Can you actually get a time constant? If one time constant dies out quickly, then the other one is all that you see after a while. Here's that example plot again.

After about one second, the quick time constant behavior has died out and what's left after one second looks pretty much like a single time constant. So, here's a strategy to figure out the time constants here.

Get the slow time constant using the long-term data. That's the data after one second above.
After you have the slow time constant, use the formula above the lets you relate peak time and the two time constants. If you have the peak time and one of the poles you can putter to get the second pole.

t_peak = -ln(a/b)/(b-a)

An Observation

One interesting observation about this system is that the system can be visualized as two first order systems in cascade, where the output of one system feeds into the next. That cascade system is shown below. Note that the product of the two transfer functions is equal to the transfer function of the composite system.

In this composite system you should be able to visualize the step response.

A step input will produce the usual exponential that starts to rise from zero at the output of the first system (like a capacitor charge curve in an RC circuit).
The output of the second system will not begin to change immediately because the second system does not have a step input.
Thus, the second system will not start to respond immediately, and the output of the second system will initially be flat. In other words, the step response does not have a quick jump in the first derivative (slope) of the output.

The Case of Two Complex Roots - Impulse Response

So far we have considered the situation in which the damping ratio, z, is greater than 1. In that situation the exponentials in the solution are decaying exponentials. In other words, if the solution is of the form:

the roots, s₁ and s₂ are real and they are negative. We still need to consider the case in which the damping ratio, z, is less than 1.

If the damping ratio is less than one, then the roots, s₁ and s₂, are both complex. When the roots have an imaginary part there will be oscillations in the impulse response. In order to have two real, distinct roots, we must have:

z < 1

Remember, from above, that the two roots are given by:

When the damping ratio is less than one, these roots will be complex, and the impulse response will be of the form.

Initial conditions will determine the constants A and f.

NOTE: You might want to review the material on first order systems, particularly the material on how an impulse changes the conditions in a system immediately after the impulse occurs. Click here to go to that material where you can find a computation of the change in output of a first order linear system when an impulse occurs at t = 0. You'll need to understand that for the next material.

At t = 0, the output is zero, so we must have sin(f) = 0, or f = 0. That means that the response, call it r(t), is really:

Knowing that we can also examine the derivative at t = 0. Taking the derivative we have the derivative and using t = 0 or 0+, we find:

Now, the question is "What does this have to do with the impulse response?" Earlier we noted that the effect of the impulse was to change the value of the derivative of the response immediately after the impulse occurred. We can calculate the value of the derivative at t = 0+, and equate that to the value of the derivative computed above. Earlier, we found:

But, we also know:

This can be solved for the constant, A:

Knowing A we can determine the expression for the impulse response:

Now, we can examine an example of an impulse response. The plot below shows an impulse response for these parameters.

G_dc= 1.0
z= 0.17
w_n = 1.0

There are some points to note about the response.

The constant, A, in the response is given by:

A = 1/(1 - .17²)^.5 = 1/(1 - .0289)^.5 = 1/(1 - .0289)^.5 = 1.015

Even though this is larger than one, the response only reaches a value of around 0.8 at the first peak because the decay factor has reduced it.

The Impulse Response - Putting It All Together

You need to be able to see how the impulse response changes as parameters change. The most interesting parameter is the damping ratio. Here's a video of the impulse response of a second order linear system showing how the response varies as the damping ratio, z, is varied. Play this video to see how the impulse response changes for different damping ratios in the range from zero to one. In this video, the natural frequency,w_n, is held constant at w_n = 1.

There are a number of points to note in this decaying sinusoidal response.

The frequency of oscillation of the sinusoid is given by:

The frequency of oscillation is always less than the frequency of oscillation with no damping - the undamped natural frequency, w_n . It may be only slightly less, but it is less.

For damping ratios sufficiently small, the frequency of oscillation is very close to the undamped natural frequency. For example, if z = 0.2, we have:

So, for a damping ratio of 0.2, the frequency is 2% lower than the natural frequency.

Other Ways of Calculating Responses

There are other ways that the impulse response can be computed, and we can consider other methods for computing the impulse response next. Let us assume that we have the differential equation as before:

Then, Laplace transforming both sides and solving for the transfer function - the ratio of the transform of the output to the transform of the input, we find the transfer function to be.

Once we have the transfer function, taking the inverse transform of the transfer function gives us the impulse response:

If the damping ratio is less than one, then the impulse response is a decaying sinusoid with a frequency of:

The sinusoid is modulated by a decaying exponential:

The decaying exponential has a time constant of decay given by:

Decay time constant = 1/zw_n

To understand the effect of these two items - the decay time constant and the exponential modulating signal, we'll look at how the impulse response varies as the damping ratio and natural frequency vary.

Step Response of 2nd Order Systems

So far, we have considered the impulse response of second order systems. However, the step response is also important, and in this section we will examine how second order systems respond to steps. The simplest second order system satisfies a differential equation of this form.

where:

x(t) = Response of the System,
u(t) = Input to the System,
z= Damping Ratio,
w_n= Undamped Natural Frequency,
G_dc= The DC Gain of the System.

As in the impulse response, the parameters you find in a second order system determine aspects of the step response. Whether we are talking about impulse response, step response or response to other inputs, we will still find the following relations.

z, the damping ratio, will determine how much the system oscillates as the response decays toward steady state. And for a step that steady state will depend on the input step size and the DC gain.
w_n,the undamped natural frequency ,will determine how fast the system oscillates during any transient response.
G_dc, the DC gain of the system, will determine the size of steady state response when the input settles out to a constant value.

As in the impulse response, there are two special cases for the step response.

If the damping ratio is less than one, then the step response contains a decaying sinusoid with a frequency of:

If the damping ratio is greater than one, then the step response contains two decaying exponentials. And, like in the impulse response there will not be any decaying sinusoids.
The case of two real roots, when z= 1, produces a special situation, and requires a different analytical form.

Here is a movie you can examine to see how step responses for the two different cases look. After the movie we will examine the analytical solution.

If a system has a transfer function:

There are two cases - as usual. There can be two real roots/poles for the system, and there can be a pair of complex poles. If the two poles are complex, then, the step response of this system can be obtained (Here we rely on your Laplace transform and differential equation background):

Here's the video of the step response of a second order linear system showing how the response varies as the damping ratio, z, is varied. Play this video to see how the step response changes for different damping ratios.

Note the following in the video (for cases where damping ratio, z, is less than one!)

If the damping ratio is less than one, then the step response contains a decaying sinusoid with a frequency of:
When the damping ratio is less than one, the step response has a decaying sinusoid, and the exponential envelope of the decay is:
e^-^zwnt
The case of two real roots, when z = 1, produces a special situation - repeated roots - and requires a different analytical form.

There are some points to note about the second order response.

Neither the response or the first derivative changes suddenly when the step is applied.

Since the step is the integral of the impulse, the step response is the integral of the impulse response. That means that the response does not change at t = 0, and neither does the derivative of the response.
This observation may not be true if the transfer function of the system has an s-term in the numerator.

With three parameters - damping ratio, natural frequency and DC gain - it can get confusing. To give you a chance to examine some different systems, there is a simulator below that will let you vary all three parameters.

Problems and Questions

P1 Using the simulator, determine the percent overshoot when the input is a unit step (which is what is assumed in the simulator), the DC gain is 2, the natural frequency is .5 rad/sec and the damping ratio is 0.2. Note, to compute the percent overshoot, you need to figure the percentage amount that the response - at the first peak - exceeds the steady state value.

P2 Now, change the natural frequency to .25 rad/sec and recompute the overshoot.

Q1 Does changing the natural frequency change the percent overshoot?

A Note On System Responses

Knowing how to compute impulse and step response is not an end in itself. Let's face it, any system you build will not have many impulses or steps as inputs. However, it is important that these responses are important for several reasons.

Step response is an important response. Step inputs are often used as test inputs to determine how well a system is performing. The shape of the step response - how fast it occurs, how much it oscillates, etc. help the designer predict how well the system will respond to other inputs.

How the damping ratio influences the response is important. We have seen how the step response changes when the damping ratio is varied.

Pole location is a function of damping ratio. A change in damping ratio means that the pole location has changed. Pole location is important for predicting responses to all kinds of inputs.

Here's a video that shows how the pole position and step response are related.

There's a funny thing that happens in this video.

The damping ratio, z, changes in these plots, varying from 0 to 1
The undamped natural frequency, w_n,doesn't change.
However, note how the pole moves in the s-plane in the plot on the right. Seems strange, doesn't it? The pole moves in a circular motion.

Let's examine the pole location in a little more detail. Here's the transfer function again.

To get the poles of this system observe the following.

The poles are at values of s for which the denominator is zero.
Setting the denominator to zero and solving for s, we obtain two poles:

The poles form a complex conjugate pair if z < 1. There are other connections that we need to understand.

Step response is an important response. Step inputs are often used as test inputs to determine how well a system is performing. The form of this response tells us a great deal about the system.
How the damping ratio influences the response is important. We have seen how the step response changes when the damping ratio is varied, and the video shows pole location as the damping ratio is varied.
Pole location is clearly a function of damping ratio. A change in damping ratio means that the pole location has changed.
There's a funny thing that happens in this video. Although damping ratio, z ,changes in these plots, the undamped natural frequency, w_n, doesn't change.

However, note how the pole moves in the s-plane.
Check how the pole varies.

We're going to look at the pole location a little closer and we're going to look at it analytically.
Here's the expression for the poles.

We'll only consider those cases where the poles are complex, and are off the real axis.
We'll examine the angle the poles are off the axis - and remember the video.
We'll examine how far the poles are from the origin.

Here are the poles.

Work on the magnitude of the poles first - i.e. how far the poles are from the origin.

Square the real and the imaginary parts of either root. Then add.

The result is:

w_n

The important result is that, for a second order pole, the distance from the origin of the s-plane is equal to w_n - the undamped natural frequency.

Since the distance is constant, the pole traverses a circular path around the origin when the damping ratio is varied and the undamped natural frequency is held constant.

That information is summarized in the figure below where the red "X" marks the pole, and the real and imaginary parts are shown. The distance from the origin in the s-plane is w_n.

The angle off the horizontal is also a useful parameter for this pole.

The angle off the horizontal is a measure of the damping ratio.
We have cos ( f ) = z.

The angle off the horizontal thus becomes a measure of the damping ratio.
The larger the angle, the smaller the damping ratio, since the cosine function gets smaller as the angle increases to 90^o.

Observations on Second Order Systems

There are some important points to note about the responses of a second order linear system.

When the step is applied, the derivative of the output does not change immediately.

In a second order system, the second derivative changes suddenly in the step response. The first derivative does not change suddenly in the step response.

In a second order system, the first derivative changes suddenly in the impulse response.

The size of the derivative change depends upon the size of the step, but as long as the step is non-zero, the derivative will have a jump.

To get the steady state value, multiply the input step size by the DC Gain.

If the input is not a step but if it does reach a steady state value, the output will be the DC Gain multiplied by the steady state value of the input.

Problem

P2 A second order system has poles at s = -1 + j and s = -1 - j. Determine the damping ratio for the system with these poles.

Overshoot In Second Order Systems

When you looked at the step response of the second order system, you should have noticed that the system's response went way past the final value. Note the following.

Percentage overshoot depends entirely on the damping ratio of the second order system.

Here's a plot of a system's response.

The system settles out to a steady state of 2.0.
The response goes up to 3.5
3.5 is 75% higher than 2 - the steady state - so the overshoot is 75%.

Overshoot is important and is closely related to damping ratio. (It does not depend upon undamped natural frequency!) We could get an analytical expression for the overshoot by:

Differentiating the expression for the response.
Set the derivative to zero.
Solve for the first non-zero time for which the derivative is zero.

Doing all of that, we get the analytical expression for the overshoot which is given by this expression:

While it is not immediately apparent from this expression, the percent overshoot decreases as the damping ratio increases. If we plot the percent overshoot using this function, we get the plot shown below.

Note the following points about overshoot, the formula above and this plot:

In a second order system, percent overshoot depends entirely upon damping ratio.
As the damping ratio increases, the percent overshoot decreases.

When the damping ratio reaches somewhere around 0.8 the overshoot becomes so small that you will not be able to observe it. It's still there, but you just can't see it in typical lab data. You probably can't see it for damping ratios larger than 0.7 for that matter.

If the system is third or higher order this plot is not valid. You may be able to get some useful information from it but don't count on it.
If the system is second order but has an s-term in the numerator of the transfer function this plot also might not give useful information.
The percent overshoot is measured for the first peak only.

That's it for second order systems. We thought you might want to examine the aircraft's path a little more critically now that you have completed the lesson.