# Determine a transfer function in the s domain and the response to a ramp input

#### David J

Gold Member
1. A process can be represented by the first order equation

$4\frac{dy(t)}{dt}+y(t)=3u(t)$

(a) Determine the transfer function of this process in the s domain

(b) If the input is a ramp change in u(t)=4t, determine the value of y(t) when t=10s

2. For part (a) i had to use laplace transform on $4\frac{dy(t)}{dt}+y(t)=3u(t)$ term by term

For part (b) I used $y(t) = 4[-\tau+t+\tau e^-\frac{t}{\tau}$]

3. First of all for part a

$4\frac{dy(t)}{dt}+y(t)=3u(t)$

$4[sy(s)-y(s)]+y(s)=3u(s)$

So $4[sy(s)+y(s)=3u(s)$

or $y(s)[4s+1]=3u(s)$

So the transfer function (output / input) or gain for this process in the s domain would be:-

$G(s)=\frac{y(s)}{3u(s)}=\frac{1}{4s+1}$

The attachment (part A example) is from my notes which I have basically tried to adapt to this solution and follow
I am still working on part b and will post my solution in a couple of days but I was wondering if someone could look at this so far and advise.

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#### Chestermiller

Mentor
The Laplace transform of y' is sy(s)-y(0)

#### David J

Gold Member
Should this equation

$4[sy(s)-y(s)]+y(s)=3u(s)$

$4[sy(s)-y(0)]+y(s)=3u(s)$ ??

I just noticed in the lesson it shows it as

$4[sy(s)-y_{-0}]+y(s)=3u(s)$

Are these 2 equations more or less the same ??

#### Chestermiller

Mentor
Should this equation

$4[sy(s)-y(s)]+y(s)=3u(s)$

$4[sy(s)-y(0)]+y(s)=3u(s)$ ??

I just noticed in the lesson it shows it as

$4[sy(s)-y_{-0}]+y(s)=3u(s)$

Are these 2 equations more or less the same ??
The latter form is better. The second term should represent the value of y at t = 0 (not s = 0).

#### David J

Gold Member
So would the correct solution be

$4\frac{dy(t)}{dt}+y(t)=3u(t)$

$4[sy(s)-y_{-0}]+y(s)=3u(s)$

So $4[sy(s)]+y(s)=3u(s)$

or $y(s)[4s+1]=3u(s)$

So the transfer function (output / input) or gain for this process in the s domain would be:-

$G(s)=\frac{y(s)}{3u(s)}=\frac{1}{4s+1}$

#### scottdave

Homework Helper
Gold Member
So if u(t) is your input function, I'm not sure about putting 3 coefficient multiplied by u(s) in the denominator. Shouldn't it be just Y(s) / U(s) ?
Also, l like capital letters for the functions in the Laplace domain and lower case in the time domain. That's the way I learned it.

#### David J

Gold Member
I used the coefficient of 3 because I am given that in the original question so my thinking is I need to use all of those original values given in $4\frac{dy(t)}{dt}+y(t)=3u(t)$
I notice in my attachment that they use capitals. Still learning this to be honest so any tips are appreciated

#### David J

Gold Member
Regarding the second part of this question

$4\frac{dy(t)}{dt}+y(t)=3u(t)$

(b) If the input is a ramp change in u(t)=4t, determine the value of y(t) when t=10s

they give me an example in the form of $constant multiplier[-\tau+t+\tau e^-\frac{t}{\tau}]$

I am assuming my constant multiplier is 4 so:-

$y(10)=4[-3+10+3e^-\frac{10}{3}]$

$y(10)=4[7+0.10702198]$

$y(10)=4[14.10707]$

$y(10)=56.42808$

Is this the correct approach ??

#### gneill

Mentor
If

$G(s) = \frac{Y(s)}{U(s)}$

$Y(s) = G(s) U(s)$.

Take the inverse Laplace transform of $G(s) U(s)$ to find the time domain function of the response to the input u(t). You will first need to find the Laplace transform of your input function u(t) = 4t.

#### David J

Gold Member
I have gone back through my notes again and found a few errors in my initial posts so I have tried to start again.

My values are :-
$\tau=4$
$t=10$
$K=3$

$\tau\frac{dy(t)}{dt}+y(t)=Ku(t)$

$\frac{dy(t)}{dt}+\frac{1}{\tau}y(t)=\frac{Ku(t)}{\tau}$

$[sY(s)+y(-0)]+\frac{Y(s)}{\tau}=\frac{Ku(s)}{\tau}$

the question tells me to assume the initial steady state is $y=0$ at $t=-0$

$sY(s)+\frac{Y(s)}{\tau}=\frac{KU(s)}{\tau}$

or $Y(s)(s+\frac{1}{\tau})=\frac{KU(s)}{\tau}$

The transfer function in the s domain $G(s)$, is input, $Y(s)$, over output, $U(s)$

So

$G(s)=\frac{Y(s)}{U(s)}=\frac{\frac{K}{\tau}}{(s+\frac{1}{\tau})}=\frac{K}{\tau(s+\frac{1}{\tau})}$

$=\frac{K}{\tau s +1}$

I feel this answers part a of the question

In order to find the value of $Y(t)$ I have to transform $Y(s)$ to the time domain.

I need to re arrange the transfer function equation when the input change is a ramp

$Y(s)=\frac{{K}{\tau}}{(s+{1}{\tau})}*\frac{1}{s^2}=\frac{{K}{\tau}}{s^2(s+\frac{1}{\tau})}$

So $\frac{a}{s^2(s+a)}$where $a=\frac{1}{\tau}$

So $y(t)=K^2(1-e^-\frac{t}{\tau})$

So $y(10)=3^2(1-e^-\frac{10}{4})$

So $y(10)=3^2(1-e^-\frac{10}{4})=8.261235$

I think this answers part b but I am wondering if this equation $\frac{a}{s^2(s+a)}$where $a=\frac{1}{\tau}$ is correct for the ramp function

I have taken this information from my notes but my notes only tell me about the step function. I have to assume the ramp function and adapt it to this particular problem

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#### gneill

Mentor
The input function is u(t) = 4t. What's it's Laplace transform? You've used $\frac{1}{s^2}$, which is not correct as it's missing the constant "4".

You will likely want to employ partial fractions to carve the transfer*input function into digestible portions that you can look up in a table of inverse Laplace transforms (unless your table is particularly extensive ). The Wikipedia page on the Laplace Transform has a short but useful list.

#### David J

Gold Member
I am getting slightly confused now as I have followed the lesson for the step function exactly as it is explained, I thought I just needed to change for the ramp function. I had a look at the tables online and it shows the ramp function as I have used. Whilst I dig into this again can you advise on my answer to part a in the post 10? Is this correct now or still requiring ??? I feel part a is correct now

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#### gneill

Mentor

You used the correct form for the ramp function but forgot its scaling factor of 4.
$Y(s)=\frac{{K}{\tau}}{(s+{1}{\tau})}*\frac{1}{s^2}=\frac{{K}{\tau}}{s^2(s+\frac{1}{\tau})}$

So $\frac{a}{s^2(s+a)}$where $a=\frac{1}{\tau}$

So $y(t)=K^2(1-e^-\frac{t}{\tau})$
I don't know why you chose that inverse transform, but it doesn't match the Laplace form expression above it. The one you chose is for what's called an "exponential approach". The $s^2$ rather than $s$ in the denominator makes a difference.

Quite often, unless you have a pretty extensive list of transforms you may not find an exact match for a given Laplace domain function. In such cases you'll want to break it down into more "primitive" chunks that will be found in the tables. Usually that means employing partial factions on the expression.

#### David J

Gold Member
Good Morning

I attached a copy of the step response example I was given in my lessons. I thought it was just a case of changing $\frac{1}{s}$ to $\frac{1}{s^2}$ but I am not sure it is so straight forward now. The attachment is good for step response but I am now assuming that as part of the learning curve we are supposed to be able to adapt this to a ramp response. (This is an open learning course with limited tutor assistance from the university)

I think the laplace transform of $u(t)=4t$ should have been $U(s)=\frac{4}{s^2}$

I am now looking at the partial fraction idea and will try to update this thread in the next day or so with my attempt

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#### gneill

Mentor
Good Morning

I attached a copy of the step response example I was given in my lessons. I thought it was just a case of changing $\frac{1}{s}$ to $\frac{1}{s^2}$ but I am not sure it is so straight forward now. The attachment is good for step response but I am now assuming that as part of the learning curve we are supposed to be able to adapt this to a ramp response. (This is an open learning course with limited tutor assistance from the university)

I think the laplace transform of $u(t)=4t$ should have been $U(s)=\frac{4}{s^2}$

I am now looking at the partial fraction idea and will try to update this thread in the next day or so with my attempt
That's all good.

Note that in the Laplace domain, dividing by s is equivalent to integration. So if you used your step response version to find the time domain response to the step, you could then integrate that time domain response (in the time domain) to find the ramp response, but you'd also have to find the constant of integration. Easily done since you do have the value of y(0).

#### David J

Gold Member
Ok here goes for attempt number ???

This is taken from my lesson where it explains the partial fraction method. I have posted this to work out if this is the correct way to work out this problem. I think this is what you are relating to.

$u(t)=4t$

$u(s)=4*\frac{1}{s^2} or \frac{4}{s^2}$

$y(s)=[\frac{3}{4s+1}]u(s)=[\frac{3}{4s+1}][\frac{4}{s^2}]$

$y(s)=\frac{12}{s^2(4s+1)}=\frac{3}{s^2(s+\frac{1}{4})}$ or $\frac{a}{s^2(s+a)}$

$a=\frac{1}{\tau}$
$\frac{a}{s^2(s+a)}$
is not in the list of transforms so we have to use partial fractions to find available transforms

$\frac{a}{s^2(s+a)}$ = $\frac{(As+B)}{s^2}$ + $\frac{C}{(s+a)}$

A, B and C are constants

Multiply both sides by
$s^2(s+a)$

$a=As^2 +Bs+Aas+Ba+C^2$

$S=0$
$a =Ba$
$B=1$

$s^1$
0=$B+Aa$

Substituting for $B$ from above we get:-
$0=1+aA$

and

$A=\frac{-1}{a}$

$s^2$
0=$A+C$
$A+C=0$

Substituting for $A, C=\frac{1}{a}$

Putting these values back into the standard partial fraction equation, we get:-

$\frac{a}{s^2(s+a)}$ = $\frac{(-\frac{s}{a^2}+\frac{1}{a})}{s^2}$ + $\frac{\frac{1}{a^2}}{{(s+a)}}$

This equals $\frac{(-\frac{s}{a}+1)}{s^2}$ + $\frac{\frac{1}{a}}{{(s+a)}}$

This can be simplified by separating off the $\frac{(-\frac{s}{a}+1)}{s^2}$ term into fractions

$-\frac{\frac{s}{a}}{s^2}$ + $\frac{1}{s^2}$=$-\frac{\frac{1}{a}}{s}$+$\frac{1}{s^2}$

Substitute this back in the equation to give the final result

$\frac{a}{s^2(s+a)}$ = $-\frac{\frac{1}{a}}{s}$ + $\frac{1}{s^2}$ + $\frac{\frac{1}{a}}{(s+a)}$

These 3 terms each have a transform:-

$-\frac{\frac{1}{a}}{s}$ transforms to $-\frac{1}{a}$

$\frac{1}{s^2}$ transforms to $t$

$\frac{\frac{1}{a}}{(s+a)}$ transforms to $\frac{1}{a}e^-at$

at this point the lesson explains that equation can now be transformed for the ramped input.

I am given a constant multiplier of 0.5

i.e. $G(s)=\frac{0.5a}{s^2(s+a)}$

$a=\frac{1}{\tau}$

$G(s)=0.5[-\frac{{1}}{\frac{1}{\tau}}+t+\frac{{1}}{\frac{1}{\tau}}e^-\frac{t}{\tau}]$

$=0.5[-\tau+t+\tau e^-\frac{t}{\tau}]$

This is the equation I posted in post 8 of this thread but I used the following values

$\tau=4$
$t=10$
$K=3$ (Constant multiplier)

I appreciate the above is a little long winded but all I am trying to work out at the moment is, is the above method the correct way to solve this problem?

#### gneill

Mentor
You wrote:

How did both $3$ and $\frac{1}{4}$ both come $a$?

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#### David J

Gold Member
I see that mistake, yes.
I have removed that first section from this post as I was getting confused with past attempts.

$u(t)=4t$ so $u(s)=\frac{4}{s^2}$
The transfer function for this ramped output is $\frac{4}{s^2}$

$G(s)=\frac{u(s)}{\tau s+1}$ = $\frac{\frac{4}{s^2}}{\tau s +1}$ = $\frac{4}{s^2(\tau s +1)}$

=$\frac{4}{\tau s^2(s+\frac{1}{\tau})}$ = $\frac{4}{s^2}$ x $\frac{\frac{1}{\tau}}{(s+\frac{1}{\tau})}$

So I have $G(s)=\frac{4a}{s^2(s+a)}$ where $a=\frac{1}{\tau}$

The general form $\frac{a}{s^2(s+a)}$ is not in the list of transforms so we have to use partial fractions to find available transforms

$\frac{a}{s^2(s+a)}$ = $\frac{(As+B)}{s^2}$ + $\frac{C}{(s+a)}$

A, B and C are constants

Multiply both sides by
$s^2(s+a)$

$a=As^2 +Bs+Aas+Ba+C^2$

$S=0$
$a =Ba$
$B=1$

$s^1$
0=$B+Aa$

Substituting for $B$ from above we get:-
$0=1+aA$

and

$A=\frac{-1}{a}$

$s^2$
0=$A+C$
$A+C=0$

Substituting for $A, C=\frac{1}{a}$

Putting these values back into the standard partial fraction equation, we get:-

$\frac{a}{s^2(s+a)}$ = $\frac{(-\frac{s}{a^2}+\frac{1}{a})}{s^2}$ + $\frac{\frac{1}{a^2}}{{(s+a)}}$

This equals $\frac{(-\frac{s}{a}+1)}{s^2}$ + $\frac{\frac{1}{a}}{{(s+a)}}$

This can be simplified by separating off the $\frac{(-\frac{s}{a}+1)}{s^2}$ term into fractions

$-\frac{\frac{s}{a}}{s^2}$ + $\frac{1}{s^2}$=$-\frac{\frac{1}{a}}{s}$+$\frac{1}{s^2}$

Substitute this back in the equation to give the final result

$\frac{a}{s^2(s+a)}$ = $-\frac{\frac{1}{a}}{s}$ + $\frac{1}{s^2}$ + $\frac{\frac{1}{a}}{(s+a)}$

These 3 terms each have a transform:-

$-\frac{\frac{1}{a}}{s}$ transforms to $-\frac{1}{a}$

$\frac{1}{s^2}$ transforms to $t$

$\frac{\frac{1}{a}}{(s+a)}$ transforms to $\frac{1}{a}e^-at$

at this point the lesson explains that equation can now be transformed for the ramped input.

my constant multiplier is 4

i.e. $G(s)=\frac{4a}{s^2(s+a)}$

$a=\frac{1}{\tau}$

$G(s)=4[-\frac{{1}}{\frac{1}{\tau}}+t+\frac{{1}}{\frac{1}{\tau}}e^-\frac{t}{\tau}]$

$=4[-\tau+t+\tau e^-\frac{t}{\tau}]$

This is the equation I posted in post 8 of this thread but I used the following values

$\tau=4$
$t=10$
$K=3$
I think these values were wrong.
I think $\tau = 3$ and $K = 4$
If that is the correct assumption then:-
$y(10)=4[-3+10+3 e^-\frac{10}{3}]$
$y(10)=4(7+0.10702198)$
$y(10)=28.428 units$

????

#### gneill

Mentor
Let's go back to your first post where you wrote: $y(s)[4s+1]=3u(s)$. I'll use Y for for Y(s) and U for U(s). So,

$Y(4s + 1) = 3U$
$G(s) = \frac{Y}{U} = \frac{3}{4s + 1}$

Using our given ramp input, $U(s) = \frac{4}{s^2}$ we find:

$Y(s) = \frac{12}{(4s + 1) s^2}$

Now apply partial fractions:

$\frac{a}{4s + 1} + \frac{b}{s^2} + \frac{c}{s} = \frac{12}{(4s + 1) s^2}$

multiplying out and equating the numerators:

$as^2 + 4bs + b +4cs^2 +cs = 12$

Collecting:

$(4c+a)s^2 +(c+4b)s + b = 12$

Equating like terms on either side:

1) $b = 12$
2) $4c + a = 0$
3) $c + 4b = 0$

Eq (1) gives the value for $b$ directly. This can be plugged into eq (3) to give us the value of $c$, and the that in turn can be plugged into eq (2) yielding a value for $a$. In the end we have:

$a = 192$, $b = 12$, $c = -48$ so that:

$Y = \frac{192}{4s + 1} + \frac{12}{s^2} - \frac{48}{s}$

Adjusting the first term to in "standard form",

$Y = \frac{48}{s + \frac{1}{4}} + \frac{12}{s^2} - \frac{48}{s}$

#### David J

Gold Member
$u(t)=4t$

$u(s)=4\frac{4}{s^2}$

$y(s)=[\frac{3}{4s+1}]u(s)=[\frac{3}{4s+1}][\frac{4}{s^2}]$

$y(s)=\frac{12}{s^2(4s+1)}$

I didn't quite understand how you applied the partial fractions but with the help of a software package I use I managed to get an idea of it. I wont post it on here as the step by step is quite long and drawn out

$Y=\frac{48}{s+\frac{1}{4}}+\frac{12}{s^2}-\frac{48}{s}$

So i have got $Y=\frac{48}{(s+0.25)}+\frac{12}{s^2}-\frac{48}{s}$

Using inverse laplace transform, first of all, $\frac{48}{(s+0.25)}$

$^{L^{-1}}\left\{ \frac{48}{s+0.25}\right\}$

So $^{L^{-1}}\left\{ 48(\frac{1}{s+0.25)}\right\}$

So $48^{L^{-1}}\left\{ \frac{1}{s+0.25}\right\}$

From the inverse transform tables

$^{L^{-1}}\left\{ \frac{1}{s-a}\right\}=e^{at}$

$^{L^{-1}}\left\{ \frac{1}{s+0.25}\right\}=e^{-\frac{t}{4}}$

$48e^{-\frac{t}{4}}$

Using inverse laplace transform, $^{L^{-1}}\left\{ \frac{12}{s^2}\right\}$

$^{L^{-1}}\left\{ 12(\frac{1}{s^2})\right\}$

$12{L^{-1}}\left\{ \frac{1}{s^2}\right\}$

From the inverse transform tables

$^{L^{-1}}\left\{ \frac{1}{s^2}\right\}=t$= $12t$

Using inverse laplace transform, $^{L^{-1}}\left\{ \frac{-48}{s}\right\}$

From the inverse transform tables

$^{L^{-1}}\left\{ \frac{a}{s} \right\}=a$

so $^{L^{-1}}\left\{ \frac{-48}{s} \right\}=-48$

So I think $Y(t)=-48+12(t)+48e^{-\frac{t}{4}}$

$Y(10)=-48+12(10)+48e^{-\frac{10}{4}}$
$Y(10)=72+48e^{-\frac{10}{4}}$
$Y(10)=72+3.94$
$Y(10)=75.94$units

#### gneill

Mentor
Yup. That looks better.

#### David J

Gold Member
Thanks for your help with this one, its most appreciated.

I just have a quick question about this, I have spent this afternoon going through the partial fraction part again, trying to get an understanding of it

When I run $\frac{12}{s^2(4s+1)}$ through the partial fraction software it gives me $\frac{12}{s^2}+\frac{192}{4s+1}-\frac{48}{s}$

$\frac{192}{4s+1}$ can be reduced to $\frac{48}{s+1}$

In your post you adjusted the first term to "in standard form"

$\frac{48}{s+\frac{1}{4}}$

If I give s a dummy value, for example, 13, then these 2 equations both give a very similar answer.

$\frac{48}{s+1}$so$\frac{48}{13+1}=3.42857$

$\frac{48}{s+\frac{1}{4}}$so$\frac{48}{13+0.25}=3.6226$

The inverse laplace of $\frac{48}{s+1}=48e^{-t}$

The inverse laplace of $\frac{48}{s+\frac{1}{4}}=48e^-{\frac{t}{4}}$

These 2 different equations give different values of the final answer

$\frac{48}{s+1}=48e^{-t}$ gives an answer of 72.002 units

$\frac{48}{s+\frac{1}{4}}=48e^-{\frac{t}{4}}$ gives an answer of 75.94 units

what is the correct method to follow or are they both considered acceptable?

#### gneill

Mentor
Thanks for your help with this one, its most appreciated.

I just have a quick question about this, I have spent this afternoon going through the partial fraction part again, trying to get an understanding of it

When I run $\frac{12}{s^2(4s+1)}$ through the partial fraction software it gives me $\frac{12}{s^2}+\frac{192}{4s+1}-\frac{48}{s}$

$\frac{192}{4s+1}$ can be reduced to $\frac{48}{s+1}$
No. If you divide numerator and denominator by some value you must divide every term in each by that value. The "1" in the denominator cannot remain untouched.
In your post you adjusted the first term to "in standard form"

$\frac{48}{s+\frac{1}{4}}$
Right. Everything was divided by 4, top and bottom.
If I give s a dummy value, for example, 13, then these 2 equations both give a very similar answer.
Close only counts in horseshoes and hand grenades
what is the correct method to follow or are they both considered acceptable?
You simply have to get the algebra correct. It's the only way.

#### David J

Gold Member
Fully understood now, thanks again for your help. I will mark this as solved now

"Determine a transfer function in the s domain and the response to a ramp input"

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