Is Speed Of Response Important?

Measures Of Speed Of Response

Working In The Frequency Domain

Closed Loop Systems

        How quickly a system responds to a change in input is an important measure of its' performance.

• In an airplane, if the pilot wants the engines to go full speed, s/he doesn't want to wait fifteen minutes for that to happen.  How quickly the engine speed control system responds affects the safety of passengers and crew.
• In your home, you surely consider it important to have the temperature come up to your setting when you return from a trip having turned the heat down before you left.
• It's almost always true that you want a system to respond as quickly as possible whenever you change the setting or desired state of the system.

 Speed of response is an important measure of how quickly a system responds.  When you evaluate how well a system is performing you need to measure speed of response with some metric.  When you are designing a system you need to be able to predict speed of response.  Your goals for this lesson relate to that.

  Given a system in which you need to predict speed of response,

  Be able to predict speed of response using time domain methods.
  Be able to predict speed of response using frequency domain methods.

 Speed of response can be a little tricky because we have so many intuitive ideas of what we mean by speed of response.  Let's review some of the ideas that often form a foundation for this concept.

• First order systems have time constants.  Clearly, in those systems we can take the time constant as a measure of speed of response.
• For two first order systems, the system with the smaller time constant will respond more quickly to a step input or any other input.
• Knowing the time constant allows us to estimate aspects of a response.  For example, a first order system with a time constant, t, will respond to a step input so that the system is within 5% of the final value in 3t seconds - i.e. three time constants.

• Can you determine a time constant for this system?

• A first order system - with a single time constant - has a response that changes slope immediately after the step is applied.  The slope of this system does not change immediately, and  the response seems to take a long time to get started.
• This response seems to take 2 or 3 seconds to get started, but once it does get started, it's over by 8 seconds.
• Can you guess what the transfer function of this system might be?

• This system does have a time constant of 1 second.
• The problem is that it has 5 time constants, all of 1 second.
• If you do encounter a system like this, you will need to have a numerical way to measure the speed of response.  Next, we'll propose a few different ways.  On the other hand, if you encounter a system like this one, you may want to run away.

• Determine the time at which the response reaches 10%.  Click the button on the left, and you should see that time is about 2.5 seconds.
• Determine the time at which the response reaches 90%.  Click the button on the right, and you should see that time is about 7.8 seconds.

• Subtract the two.  7.8 - 2.5 = 5.3 seconds.
• So the ten-to-ninety rise time is 5.3 sec.

        There are other measures, and one of them is the settling time.  We're going to take settling time as the time it takes to get within 10% of the final value, or to 90% of the final value - and, most importantly - stay within that 10%.  That's going to make it interesting if you measure the settling time of a system that has oscillations.

• Determine the time at which the response reaches 90%.  Click the hotword, and you should see that time is about 7.8 seconds.
• The settling time is 7.8 seconds.
• With a little imagination, you may realize that we need to be somewhat more precise than we have been.  Time of response has some subtleties that we need to take into consideration.

• Clearly this system has a short rise time.  It looks to be just a few seconds.
• The settling time is also just a few seconds, since the response gets to within 90% in that time.
• The problem is that the response doesn't stay within 10% of the final value.  It just passes through that range on its way to oscillation after oscillation.

• Rise time is going to be tough, so we're going to leave that definition as it stands.
• Settling time can be defined as the time it takes to get and stay within 10% of the final value.
• Click the button below to get a picture of the 10% range, then estimate the settling time.

• There's no substitute for knowing where are the poles are zeroes are in a system.  Knowing a system has five poles at s = -1 is more information than knowing rise time because you can plot the response and compute rise time and more.  (Root locus analysis will help you determine where the poles are located in a closed loop system.
• Ten-to-Ninety rise time is the time it takes to go from 10% to 90% of the final value.  It can be misleading if the system oscillates or if there is a delay getting started.

Settling time can be a good way to measure response time as long as care is taken to ensure that the response stays within 10% (or 5%?) of the final, steady-state value.

• There are going to be times when you are working in the frequency domain.  In that situation, you will need to get some measure of speed of response from frequency domain data.  We'll discuss that in the next section.
• There are plenty of systems that have peculiar - maybe perverse - responses.
• Consider each case individually, and use good judgement.  Here's a peculiar response.
• Much of what we have discussed does not apply here.  This is just a response that is an aberration - not included in what we have discussed to this point.  However, you may well encounter a system with this sort of response, or - even worse - you could end up inadvertently designing a system like this one.

• The formula for the unit step response of this system is:
• Response(t) = G_dc * (1 - e^-t/t)
• We can calculate the time it takes to get to 0.9*Gain.
• Doing that we find the settling time = 2.3t.
• Similarly, the rise time (10%-90%) is 2.2t.  (It has to be less since both computations, rise time and settling time, use response to 90%.).
• For later work, note that the bandwidth of this system is 1/2pt, so we have:
• rise time = 2.2t = 2.2/(2pBW) = .35/BW.

• As the gain is increased, the closedloop pole moves to the left.
• We know that settling time = 2.3t, so we can say that the smaller the time constant, the smaller the settling time.
• A settling time of 1 second would require t of 1/2.3 or 0.435 seconds.
• That would require a pole to the left of s = - 1/.435 or s = -2.3.
• That can be seen by clicking the button at the right of the root locus in the plot below.

• How far into the left half plane the pole is defines how quickly the system will respond.
• Here are two responses.  The oscillatory system has two complex poles (only one of two shown), and the exponential system has one real pole.
• On the figure below there are hot spots around the pole locations.  Move the mouse over those hotspots for more information.

What do we note?

• The real part of the two poles are the same for the two systems.
• The settling times are the same for the two systems.

• The speed of response of a system is determined by how far into the left half plane the poles of the system are.
• The distance into the left half plane is -1/t since the pole is located there.

       Working in the frequency domain is often easier than working in the time domain.  For many system designs, working with Bode' plots is easier than working with a root locus.  The flip side of that is that you may often have to design to time domain specifications like rise time or settling time while you are doing your design in the frequency domain.

        In this section we're going to look at how you can estimate time response parameters from frequency response parameters.  It's not hard, and the connections are more obvious than you think.

        Here are the Bode' plots for two systems.  The transfer functions are shown below.  G₁(s) is the red line and G₂(s) is the blue line.

• G₁(s) = 1/(s + 1)
• G₂(s) = 1/(.1s + 1)

• G₁(s) has a one second time constant.
• G₂(s) has a .1 second time constant - ten times smaller.

• G₁(s) has a bandwidth of .159 Hz.
• GG₂(s) has a bandwidth of 1.59 Hz, ten times larger.
• Bandwidth is calculated as the frequency at which the Bode' plot is 3 db down from the DC gain.  That's the 3 db bandwidth.

• In general, the wider the system bandwidth, the faster the system responds!
• We may be able to develop a rule of thumb that will allow you to make reasonably accurate estimates of response time from system bandwidth.
• To get a rule of thumb, we should examine a few more systems.

• The 3 db point is at approximately f = .07Hz.
• We have examined the time response of this system earlier in this lesson, and we concluded that rise time = 5.3s and settling time = 7.8s.
• Now, let's examine the frequency response data for the system.  Here's the Bode' plot for the system.

        Now we can multiply the 3 db frequency times the rise time and settling time.  Doing that, we obtain the results in the next table.
 
 

        Are there any conclusions here?  Well, the product of 3 db frequency, f3db, and the rise time seems to work halfway well.  We might be justified placing confidence in an expression like:

or:

        We previously found rise time to be .35/BW for a first order system, and that's exactly what we have here - for another system.

        It's time to try a few more systems.

        Response time of closed loop systems presents an interesting problem.

• Many times it is much easier to design in the frequency domain.
• There is often some type of specification on response time for the closed loop system.
• It is valuable to have some measure of closed loop response time that can be estimated from open loop frequency response data - the kind of data that you would use for the design.

• Here's the closed loop transfer function.

• From the closed loop transfer function, we can compute the closed loop frequency response by substituting jw for s.

• It is valuable to have some measure of closed loop response time that can be estimated from open loop frequency response data - the kind of data that you would use for the design.

• There are frequencies - usually low frequencies - for which the magnitude of the frequency response function, |KG(jw)|, is large.
• When |KG(jw)| is large, the closed loop frequency response, |CLTF(jw)|, is close to one in value.
• There are frequencies - usually high frequencies - for which the magnitude of the frequency response function, |KG(jw)|, is small.
• When |KG(jw)| is small, the closed loop frequency response, |CLTF(jw)|, is close to |KG(jw)| because the denominator in |CLTF(jw)| is essentially 1.

• At low frequencies, the closed loop frequency response, |CLTF(jw)|, is close to one - i.e. zero db - in value.
• Click the top green button to see the closed loop frequency response at low frequencies.
• At high frequencies, the closed loop frequency response, |CLTF(jw)|, is close to |KG(jw)|.
• Click the lower green button to see the closed loop frequency response at high frequencies.
• At the frequencies in between, we find the zero db crossing of the open loop response, and we know the 3db closed loop bandwidth is in the same region.

• The closed loop bandwidth is close to the open loop zero db crossing.
• We also know that:
• rise time = .35/f3db ~ .35/fopen loop zero db crossing.
• We conclude that we can make a closed loop system respond faster if we can make the open loop zero db crossing higher.
• Remember, all of these conclusions are approximate, but still good estimates to use in design.

        That's it for this lesson.  Obviously, there's a lot of folklore and estimation in this area.  Still, one main point of the lesson is that time response and bandwidth are somehow reciprocally related.  Don't forget that point because it's important in system design.  We'll need to use that idea later.

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