.

 

Hi,

i want to sketch a step response for this transfer function and i was given


This is what i did but im not sure how to get a time constant of 2. Also is it ok to assume the unit step has an amplitude of 2V?



Also how would i know from the transfer function if i get an overshoot like the pic bellow




Any help will be much apprciated

 

By definition the unit step has an amplitude of 1.

 

Hi,

I found this picture online and i would like to find out its specific transfer function

View attachment 199558

The system's general transfer function is View attachment 199559

My expected answer is kp =1, t =490ms ,td = 225ms

Would the transfer function for this specific result of the system be [(1)(exp(-0.225s))]/(0.49s+1)

I found this picture from https://pdfs.semanticscholar.org/5b18/ba1da4b75a01e8b93b4453892c6e0ede02f1.pdf

but this website shows a different answer to my expected solution

Hello,

I think you should label your graph/snapshot a little more clear so we can see the voltages and times more clearly of the excitation wave and the response wave. It's unclear what they are right now.

For example, what is the amplitude of the pulse and what is the max amplitude of the response.
What is the start time of the pulse and what is the start time of the response.
What is the time of the response when the voltage reaches a value of 0.632 times the max amplitude.

These questions need to be answered and also, it does not look like Kp=1 but your answers will help clarify this.

 

Hi,

i want to sketch a step response for this transfer function View attachment 199561and i was given View attachment 199560

View attachment 199563
This is what i did but im not sure how to get a time constant of 2. Also is it ok to assume the unit step has an amplitude of 2V?

View attachment 199564

Also how would i know from the transfer function if i get an overshoot like the pic bellow

View attachment 199565


Any help will be much apprciated

Hello again,

The time constant is the time where the amplitude response reaches a value of 0.632 times the max amplitude. This value 0.632 is an approximation for 1-1/e.

You only get overshoot if you have a 2nd order system or better. That more or less means that you have a power of 's' in the denominator of 2 or higher. For example, 1/(s^2+b*s+c) may have overshoot, but 1/(s+c) does not. The amount of overshoot if there is any depends on the coefficient 'b' in that first example.
So the question is, do you have a second (or higher) order system or just a first order system?

 

Hello again,

The time constant is the time where the amplitude response reaches a value of 0.632 times the max amplitude. This value 0.632 is an approximation for 1-1/e.

You only get overshoot if you have a 2nd order system or better. That more or less means that you have a power of 's' in the denominator of 2 or higher. For example, 1/(s^2+b*s+c) may have overshoot, but 1/(s+c) does not. The amount of overshoot if there is any depends on the coefficient 'b' in that first example.
So the question is, do you have a second (or higher) order system or just a first order system?

Hi, I've been told to sketch a first order system.

Hello,

I think you should label your graph/snapshot a little more clear so we can see the voltages and times more clearly of the excitation wave and the response wave. It's unclear what they are right now.

For example, what is the amplitude of the pulse and what is the max amplitude of the response.
What is the start time of the pulse and what is the start time of the response.
What is the time of the response when the voltage reaches a value of 0.632 times the max amplitude.

These questions need to be answered and also, it does not look like Kp=1 but your answers will help clarify this.

Thank you for your reply,

I have mentioned everything the question gave me. These points that you have mentioned are the main reasons why i have asked for help. l'm not sure what I have to assume to draw this graph.

Thank you

 

Hi,

Well for your second question it looks like you have to do a curve fitting.
You have the graph, now fit it to the expected response in the time domain.

Do you know how to convert that expression (with 'e' in the numerator you supplied) to the time domain?
That would be the first step. In other words, convert this:
Kp*e^(-td*s)/(T*s+1)
to the time domain so you get an expression that is a function of time only.
I am assuming that you learned to do that already but if not just yell and we can do it another way.

One of your drawings seems to have disappeared.

 

Hi,

Well for your second question it looks like you have to do a curve fitting.
You have the graph, now fit it to the expected response in the time domain.

Do you know how to convert that expression (with 'e' in the numerator you supplied) to the time domain?
That would be the first step. In other words, convert this:
Kp*e^(-td*s)/(T*s+1)
to the time domain so you get an expression that is a function of time only.

One of your drawings seems to have disappeared.

Hi,

I removed the first question because I didn't want others to mix up the questions.

How can I change the numerator to time domain

I've never seen S as the power of e in the numerator

I have been taught this but I'm still not confident on doing it on my own and I keep forgetting how to do it (I've never understood the process).

Thanks

 

Hi again,

The conversion requires the Inverse Laplace Transform and you can look that up if you like but for these kinds of problems you would just use a table of Laplace Transforms and their inverses.

e^(-T*s) is just a delay by T seconds, what is sometimes called a time shift.
To find the inverse of an expression with that in the numerator you first separate that from the rest then find the inverse of the rest and then apply the time shift.
So in general you have:
F(s)*e^(-T*s)

and separate them so you just have F(s), then find the inverse of that, then apply the time shift given by T.
The function e^-Ts acts like a turn on function, that turns on the part F(s) in the time domain.
Note that if you have T=0 then e^(-T*s)=1 so you end up with:
F(s)*1
which of course is just the Laplace of f(t).
If you had T=2 then note that you dont get a '1' until t=2 seconds, so the function "turns on" at t=2 seconds.
So back to
F(s)*e^(-T*s)

if T=0 then we just find the inverse of F(s) which is f(t). But if T=2 then we still find the inverse of F(s) but now f(t) does not start until t=2 seconds, so the time function is zero until T=2 seconds and then we get the inverse time function starting up normally.
To find the expression for this time delay in the time domain, you can look up the time shift theorem. But graphically it just looks like the function f(t) is delayed by the time T seconds. You will note that the proper expression involves the unit step as well as the constant change in time t in the function f(t), but sometimes the unit step is implied so we just have the function f(t) combined with the time delay T.

So see if you can figure out the inverse transform now and then we can apply it to the problem at hand.

 

Hi again,

The conversion requires the Inverse Laplace Transform and you can look that up if you like but for these kinds of problems you would just use a table of Laplace Transforms and their inverses.

e^(-T*s) is just a delay by T seconds, what is sometimes called a time shift.
To find the inverse of an expression with that in the numerator you first separate that from the rest then find the inverse of the rest and then apply the time shift.
So in general you have:
F(s)*e^(-T*s)

and separate them so you just have F(s), then find the inverse of that, then apply the time shift given by T.
The function e^-Ts acts like a turn on function, that turns on the part F(s) in the time domain.
Note that if you have T=0 then e^(-T*s)=1 so you end up with:
F(s)*1
which of course is just the Laplace of f(t).
If you had T=2 then note that you dont get a '1' until t=2 seconds, so the function "turns on" at t=2 seconds.
So back to
F(s)*e^(-T*s)

if T=0 then we just find the inverse of F(s) which is f(t). But if T=2 then we still find the inverse of F(s) but now f(t) does not start until t=2 seconds, so the time function is zero until T=2 seconds and then we get the inverse time function starting up normally.
To find the expression for this time delay in the time domain, you can look up the time shift theorem. But graphically it just looks like the function f(t) is delayed by the time T seconds. You will note that the proper expression involves the unit step as well as the constant change in time t in the function f(t), but sometimes the unit step is implied so we just have the function f(t) combined with the time delay T.

So see if you can figure out the inverse transform now and then we can apply it to the problem at hand.

i multiplied the transfer function by 1/s

then

i got 0.6/s - 0.6/(s+0.5) = 0.6u(t) - (0.6e^(-0.5t))u(t)

then i used this x(t- τ) u(t- τ)

to give me

x(t- 0.1) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)

then

[0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)]u(t- 0.1)

as you can see ive got three u(t- 0.1) in above is this correct?

 

i multiplied the transfer function by 1/s

then

i got 0.6/s - 0.6/(s+0.5) = 0.6u(t) - (0.6e^(-0.5t))u(t)

then i used this x(t- τ) u(t- τ)

to give me

x(t- 0.1) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)

then

[0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)]u(t- 0.1)

as you can see ive got three u(t- 0.1) in above is this correct?

Hi,

Sorry something does not look right there.

Look up the inverse transform for:
F(s)=1/(s+1)

with a unit step applied:
1/(s*(s+1))

You should get a fairly simple exponential with one u(t-d) multiplication.
Note that is just 1/(s+1) multiplied by 1/s as you know about.

 

Hi,

Sorry something does not look right there.

Look up the inverse transform for:
F(s)=1/(s+1)

with a unit step applied:
1/(s*(s+1))

You should get a fairly simple exponential with one u(t-d) multiplication.
Note that is just 1/(s+1) multiplied by 1/s as you know about.

I've checked my solution many times and I don't know what is wrong with it.

i multiplied the transfer function by 1/s ignoring the e expression for time shift

then
I simplifed it and did partial fractions and I got the following
X(s)= 0.6/s - 0.6/(s+0.5)

Then I did the laplace transform of each term and I got

X(t)=0.6u(t) - (0.6e^(-0.5t))u(t)

then i used this x(t- τ) u(t- τ) which is

to give me

x(t- 0.1) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)

then multiplied the above which is x(t- 0.1) with u(t- 0.1) to get x(t- τ) u(t- τ)

Answer:

[0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)]u(t- 0.1)

 

I've checked my solution many times and I don't know what is wrong with it.

i multiplied the transfer function by 1/s ignoring the e expression for time shift

then
I simplifed it and did partial fractions and I got the following
X(s)= 0.6/s - 0.6/(s+0.5)

Then I did the laplace transform of each term and I got

X(t)=0.6u(t) - (0.6e^(-0.5t))u(t)

then i used this x(t- τ) u(t- τ) which is

to give me

x(t- 0.1) = 0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)

then multiplied the above which is x(t- 0.1) with u(t- 0.1) to get x(t- τ) u(t- τ)

Answer:

[0.6u(t-0.1) - (0.6e^(-0.5(t-0.1)))u(t-0.1)]u(t- 0.1)

Well that is so close, but consider this in the time domain:
5*(1-e^-(t-0.2))*u(t-0.2)

and that is the inverse transform with an amplitude of 5 and delay of 0.2 seconds.
See how we only need to show the u(t) part one time?
Not only that, but if you multiply your answer out (expand it) you would end up with one factor that comes out to:
u(t-0.1)*u(t-0.1)

which looks like a unit step function being driven by another unit step.
So i would say one of them is redundant.
A side point is that the amplitude 5 only needs to be shown once also but that's not strictly required.

Does this make sense now? Only a small adjustment and i think you got it

Another side point is, would you happen to know of a circuit that would have that kind of response using only possibly some of these components:
resistor, capacitor, inductor, voltage source.