Time constant of a transfer function

[jaus tail]

Homework Statement

Homework Equations

Express denominator as 1 + sT₁

The Attempt at a Solution

In denominator
s+2 can be written as : 2(1 + 0.5s)
s + 3 can be written as: 3(1 + 0.3s)
So I have two time constants.
Should I multiply them to get final time constant?

There is another formula: Time constant = 1/ξw_n
But these are used for second order system and in question the TF is of 3rd order.

 

[rude man]

So I have two time constants.
Should I multiply them to get final time constant?

No. In fact, the response to a step input increases without limit as t → ∞. Seems like a really weird question.

There is another formula: Time constant = 1/ξw_n
But these are used for second order system and in question the TF is of 3rd order.

Also, the response to a step input is not oscillatory so there is no w_n.

 

[jaus tail]

If in exam such a question comes, how do I solve it?
Book answer is C.

 

[donpacino]

adding a pole at the orgin will not effect the time constant of the system. You should use your second order formula. If you had a non orgin pole, if it is close to the other two, that complicates things and graphically solving the problem might be best. If it is an order of mag larger, it will not significantly effect the problem.

No. In fact, the response to a step input increases without limit as t → ∞. Seems like a really weird question.Also, the response to a step input is not oscillatory so there is no w_n.

I disagree with both of these statements

 

[rude man]

If in exam such a question comes, how do I solve it?
Book answer is C.

You should have spoken to your instructor about this by now, referring to my remarks in post 2. Hopefully he would then avoid asking such a meaningless question.

 

[rude man]

adding a pole at the orgin will not effect the time constant of the system. You should use your second order formula. If you had a non orgin pole, if it is close to the other two, that complicates things and graphically solving the problem might be best. If it is an order of mag larger, it will not significantly effect the problem.
I disagree with both of these statements

The time response to a unit step input is 6(t/6 - e^-3t/9 + e^-2t/4 - 5/36). There is no single time constant, just two, and the first term indicates unlimited expansion in time. And there is no oscillatory component as you can appreciate. To what do you disagree?

 

[FactChecker]

I disagree with both of these statements

It looks like it's just a couple of low pass filters followed by an integrator. Its response to a step function should be to integrate to infinity without oscillations.

 

[rude man]

If in exam such a question comes, how do I solve it?
Book answer is C.

The response to a unit impulse is also unlimited in time, although there is only one time constant in that case, it being 1/3 sec. Which would give (D), not (C), as the answer.
The problem is not well stated.

 

[rude man]

It looks like it's just a couple of low pass filters followed by an integrator. Its response to a step function should be to integrate to infinity without oscillations.

Right.

 

[LvW]

Gentlemen - let`s first ask for the definition!
What is the time constant of an integrator?
I think, in this case, the time constant is the time which is needed for the output to reach the value of the step input !
Because the given transfer function contains an integrating part - which mostly determines the step response - this definition can be applied here.
Hence, I don`t think that the question is meaningless.

 

[jaus tail]

Isn't time constant the time period of oscillations?
I got an exam in 4 days.
I only know:
If equation is in form: 1 + sT. then T is time constant
or if
s² + 2ζw_ns + w_n²
then T = 1/ζw_n

 

[rude man]

Gentlemen - let`s first ask for the definition!
What is the time constant of an integrator?
I think, in this case, the time constant is the time which is needed for the output to reach the value of the step input !
Because the given transfer function contains an integrating part - which mostly determines the step response - this definition can be applied here.
Hence, I don`t think that the question is meaningless.

That definition is meaningless since it depends on the gain of the integrator: k/s with k the gain in sec^-1.
In order for there to be a time constant you have to have the output approach an asymptote. Integrators don't do that.
cf. https://en.wikipedia.org/wiki/Time_constant

 

[rude man]

Isn't time constant the time period of oscillations?
I got an exam in 4 days.
I only know:
If equation is in form: 1 + sT. then T is time constant
or if
s² + 2ζw_ns + w_n²
then T = 1/ζw_n

If you've been taught that T = 1/ζw_n, use it in your exam. Main thing is to pass the exam!
But, referring again to your 1/(T₁s+1)(T₂s+1) transfer function - no oscillations, sorry, not unless T₁ and T₂ are complex conjugates:
T₁ = T_1r + jT_1i, T₂ = T_1r - jT_1i. But I assumed T₁, T₂ are real and positive.

 

[LvW]

That definition is meaningless since it depends on the gain of the integrator: k/s with k the gain in sec^-1.
In order for there to be a time constant you have to have the output approach an asymptote. Integrators don't do that.
cf. https://en.wikipedia.org/wiki/Time_constant

No - the term "time constant" is not restricted to an asymptotic step response.
The general transfer function of an integrator is (using your notation)
H(s)=k/s=1/(s/k).
Now, it is common usage to set
1/k=Ti resulting in H(s)=1/sTi.

The integration constant Ti is the time constant of the integrator. Ti gives the time the integrator output needs to ramp to the input step.
(Note that the term you call „gain of the integrator“ determines the time constant Ti).

For example, see here:
https://www.electronicshub.org/operational-amplifier-as-integrator/
Quote:
"In the above equation, the output is -{1/(R1.Cf)} times the integral of the input voltage, where the term (R1.Cf) is known as the time constant of the integrator."

 

[FactChecker]

Apparently there no consistent definition of the "time constant" that works for these different examples (integrator, low-pass filter, second order )
I see no pattern for a definition using the gain, such as "T=1/f, where f is the frequency where the gain drops below -3dB".
All the definitions are related to the coefficient of the s term of the denominator, but the time constants of the integrator and low pass filter are 1/Ts and 1/(1+Ts), whereas the time constant of the second order is 1/(s² + s/T + 1) .
That makes it hard to see what the definition might be for a third order 1/(s³+a⋅s²+b⋅s+c)

 

[LvW]

FactChecker - with reference to your definitions: Don`t you think that a time constant should have the unit "seconds"?

 

[rude man]

FactChecker - with reference to your definitions: Don`t you think that a time constant should have the unit "seconds"?

No - the term "time constant" is not restricted to an asymptotic step response.
The general transfer function of an integrator is (using your notation)
H(s)=k/s=1/(s/k).
Now, it is common usage to set
1/k=Ti resulting in H(s)=1/sTi.

The integration constant Ti is the time constant of the integrator. Ti gives the time the integrator output needs to ramp to the input step.
(Note that the term you call „gain of the integrator“ determines the time constant Ti).

For example, see here:
https://www.electronicshub.org/operational-amplifier-as-integrator/
Quote:
"In the above equation, the output is -{1/(R1.Cf)} times the integral of the input voltage, where the term (R1.Cf) is known as the time constant of the integrator."

"Physically, the time constant represents the elapsed time required for the system response to decay to zero if the system had continued to decay at the initial rate, because of the progressive change in the rate of decay the response will have actually decreased in value to 1 / e ≈ 36.8 % in this time (say from a step decrease). In an increasing system, the time constant is the time for the system's step response to reach 1 − 1 / e ≈ 63.2 % of its final (asymptotic) value (say from a step increase)."

(from the link I cited in post 12).

In other words,an asymptote is explicitly required.

 

[LvW]

rude man, I am sure you will agree that such a term like "time constant" is a simple matter of definition.
And - of course - you are right that for a damped first order system the time constant is defined as you have mentioned.
But - does this mean that we are not allowed to define a time constant for other responses?
Of course, we can.
And for an integrating process, it is clear that we have such a definition, which naturally deviates from the definition for a 1st order lowpass.
Thats all!

What is the purpose of a "time constant"?
Answer: It gives a rough information about the timely behaviour of a system (step response).
* For a first order system: As given by you
* For a 2nd order system: Given by the time where the asymptotic line at t=0 (step response) crosses the final value of the step response (horizontal line)
* For an integrator: As I have mentioned.

 

[rude man]

Perhaps we can terminate this idle banter by remebering that none of this relates to the OP's post, its transfer function and "its time constant".

 

[FactChecker]

FactChecker - with reference to your definitions: Don`t you think that a time constant should have the unit "seconds"?

That sounds right.

 

[FactChecker]

"Physically, the time constant represents the elapsed time required for the system response to decay to zero if the system had continued to decay at the initial rate, because of the progressive change in the rate of decay the response will have actually decreased in value to 1 / e ≈ 36.8 % in this time (say from a step decrease). In an increasing system, the time constant is the time for the system's step response to reach 1 − 1 / e ≈ 63.2 % of its final (asymptotic) value (say from a step increase)."

(from the link I cited in post 12).

In other words,an asymptote is explicitly required.

That is a definition that I am familiar with. But I think that definition would not work for an integrator. So I don't know if there is a consistent definition for the different transformation types. That makes it hard to guess what it would mean for 1/(S³+aS²+bS+c).

 

[jaus tail]

From what I know time constant is 63.2% of final value if it's rising signal, or 36.8 of initial value if it's decaying signal.
Like charged capacitor discharging via R. Then T will be: V = V(initial)e^-t/RC. V/V(initial) = 0.368. R and C are given. So we can find Time constant.

But if signal is AC. Then time constant is 1/frequency.

 

[LvW]

Perhaps we can terminate this idle banter by remebering that none of this relates to the OP's post, its transfer function and "its time constant".

I rather think, I have given my answer already in post#10.

 

[FactChecker]

I rather think, I have given my answer already in post#10.

Gentlemen - let`s first ask for the definition!
What is the time constant of an integrator?
I think, in this case, the time constant is the time which is needed for the output to reach the value of the step input !

That seems as good a guess as any. But the book answer is time_constant=1/2 and the response of this transformation to a unit step input does not reach 1 till about 1.8 seconds. So it doesn't seem like a match. Also, it doesn't reach the 63.2% value till about 1.4 seconds. The book should give a definition of "time constant" that applies to this example. Without that, we are just guessing.

 

[LvW]

Fact checker, your excellent drawing shows the correct result
Jaus tail: The correct answer is T=0.5 sec.
We have nothing to do than to evaluate the slope of the step response (tangent) and to find the point where the tangent crosses the time scale.
This is in full accordance with the definition of a time constant for an I-T2 system as used in control systems.
For an ideal integrator, the "time constant " (integration constant) is as given in my response in post'#10.

As we know, in control systems the timely behaviour is much more important than the properties in the frequency domain.
For this reason, the various controller are described using the step response and the associated time constants.

Visual interpretation of the time constant:
The timely ramp - caused by the integrator part of the given function - is shifted to the right by T=0.5 seconds

 

[FactChecker]

As we know, in control systems the timely behaviour is much more important than the properties in the frequency domain.
For this reason, the various controller are described using the step response and the associated time constants.

In situations where the system can become unstable, maintaining large enough phase and gain margins is one of the top level system requirements. The margins allow for some uncertainty in the modeling and the equipment condition (age, damage, dirt, etc.). In those cases, getting fast response often makes the designs come close to the margin limits. So they become important constraints.

 

[jaus tail]

I got step response as: 1/(s+2) - 1/(s+3)
In time domain it's: e^-2t - e^-3t
I differentiated this and got: -2e^-2t + 3e^-3t
At t = 0.5 this turns out to be: -0.066 ??
Shouldn't this be 0.632?

 

[jaus tail]

Fact checker, your excellent drawing shows the correct result
Jaus tail: The correct answer is T=0.5 sec.
We have nothing to do than to evaluate the slope of the step response (tangent) and to find the point where the tangent crosses the time scale.

What does time scale mean? When T = 0?

 

[LvW]

What does time scale mean? When T = 0?

I don`t understand this question.
The graph shows the step response of the system.
And - of course - it starts at t=0 becaus the step input starts at t=0.

 

[jaus tail]

So the steps are:
find f(t)
differentiate it to get slope
Then what? Should I equation f'(t) = 0, to get the value of time constant at the values where f'(t) = 0?

 

[LvW]

In situations where the system can become unstable, maintaining large enough phase and gain margins is one of the top level requirements. The margins allow for some uncertainty in the modeling and the equipment condition (age, damage, dirt, etc.). In those cases, getting fast response often makes the designs come close to the margin limits. So they become important.

Yes - full agreement.
The classical procedure for designing/tuning a control system is to find a controller that satisfies the requirements in the time domain (response time, overshoot, etc) and - at the same time - in the frequency domain (stability properties, safety margins)

 

[jaus tail]

I didn't understand. Can you please confirm this:
given transfer function.
Find output for step response using laplace, multiply, then inverse laplace.
Differentiate y(t) to find slope. This gives y'(t)
Then what?
How to find time constant from here?

 

[LvW]

Calculating the time constant is possible, but it is rather involved.:

* Find the step response, which should be named g(t)
* Calculate the first derivative g´(t)
* Find the value for g´(t) for very large times (you need the asymptotic line)
* Now you have a slope of a straight line. For finding the equation of this line you need one point on the line.
* Select one point on the g(t) curve for very large times.
* Applying math basics for calculating the equation for the wanted asymptotic line. .
* Find the time t=T where this line crosses the horizontal time axis.
_________________
This is exactly what you can do (much quicker) graphically (crossing of the asymptotic line).

 

[LvW]

I have corrected an error in my above answer; I have forgotten the step for calculating the first derivative g´(t).