# Arranging this expression into a laplace transform

• tranceical
In summary: See your table.In summary, the conversation revolves around manipulating an expression into a Laplace transform and finding the time domain response of the system to a Laplace domain input function. The transfer function has been simplified through algebraic manipulation and can be further simplified to the form of a damped sine or cosine transform. The variables tau and omega represent time constant and natural frequency, respectively, and the problem involves components of an RLC circuit. The conversation also includes a discussion on algebraic attempts to simplify the expression and the potential use of partial fractions.
tranceical
Hi Guys,

I have an expression that i am struggling to manipulate into a laplace transform. This expression should fit one or a combination of the common transform pairs. I believe the transform the expression should be fitting is either a unit step 1/s a unit ramp 1/s^2 an exponential 1/s+a or a combination.

Please find the expression and my attempt attached. I multiplied through to simplify and leave multiplied terms in the denominator so i can split the fraction for partial fractions (if necessary)

Please could you give me a nudge in the right direction from here? Can i split the fraction to:
r(ei(s))/[sL1+(r/1+src)]

and

r(ei(s))/(1+src)

and try to arrange into two laplace transforms from there? Or is there a another way i can get the original expression into the laplace form?

Please note: i have values for Ei, R, L and C which i can input at some point.

I hope this isn't too unclear, i would really appreciate any help/tips you can give me. Many thanks

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• IMG.pdf
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Your last step has an algebra error in it.

It looks like you already have the transfer function in the Laplace domain. So are you looking to find the time domain response of the system to some Laplace domain input function ei(s)? In other words, taking the inverse Laplace transform? If so you'll need to specify what ei(s) is.

Note that your transfer function alone can be simplified via algebraic manipulation to the form:
$$G(s) = A \cdot \frac{1}{s^2 + \frac{1}{\tau}s + \omega_o^2}$$

Plugging in your actual values of R, L1, and C will tell you whether the roots of the expression in the denominator will be real or complex, and that will tell you whether to expect exponential or sinusoidal (oscillatory) terms in the transient response (response to an impulse or step as input).

gneill said:
It looks like you already have the transfer function in the Laplace domain. So are you looking to find the time domain response of the system to some Laplace domain input function ei(s)? In other words, taking the inverse Laplace transform? If so you'll need to specify what ei(s) is.

Absolutely, i have a step input of 0.01v and my other values are R=100ohms L=0.5henrys and C = 0.001 farads. I am trying to understand this function
$$G(s) = A \cdot \frac{1}{s^2 + \frac{1}{\tau}s + \omega_o^2}$$

It looks similar to functions from the table I'm working from (I've attached it) could you show me which time function from my table i need to aim for? Then i should do partial fractions before inverse transforming to find the transient response?

Thanks a lot for your help!

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• IMG_0001.jpg
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So your stimulus, in the Laplace domain, is B/s, where B = 0.01. That makes your response:
$$F(s) = A \cdot \frac{1}{s^2 + \frac{1}{\tau}s + \omega_o^2}\cdot\frac{B}{s}$$

The first thing you'll have to do is hammer your original transfer function expression into the form of G(s), so you'll have the expressions for A, ##\tau##, and ##\omega^2##. That's just algebra Note that this G(s) corresponds to the transfer function of an RLC circuit. What you don't know yet is whether the response will be underdamped, critically damped, or overdamped. That depends upon the component values and thus the roots of the characteristic equation...

So plug in your component values and see if the denominator of the transfer function is going to have real or complex roots (I vote for complex at this point, looking at your component values). Complex roots will imply some amount of oscillation along with exponential decay. Figure on decaying sine and cosine terms. If that's the case, you can rework G(s) into a form fitting either a damped sine or a damped cosine transform or both --- probably both (see your table). That's going to involve some possibly tricky partial fraction work.

So, small steps then. Tell us what you get for G(s) and the roots or the denominator.

Please could you explain what tau in the denominator represents/means? In my notes the inverse transform examples haven't ever got too complicated so I'm hoping it won't be a damped sine and damped cosine!

tranceical said:
Please could you explain what tau in the denominator represents/means? In my notes the inverse transform examples haven't ever got too complicated so I'm hoping it won't be a damped sine and damped cosine!

I chose the variable names ##\tau## and ##\omega_o## because they happen to have relevance to the problem at hand. I happen to recognize the form of the transfer function and the type of the differential equation it "belongs" to, so I suppose I had an unfair advantage there However! You can use whatever variable names you wish, or pick them out of a hat if you like. The mathematics will be the same. ##\tau## is generally used to denote a time constant of some form, and can appear in a damping term. ##\omega_o## is usually the natural frequency of a system that is prone to oscillations.

I'm afraid you'll be looking at damped sine and cosine here... sorry about that! If you haven't done any examples or problems with them yet then I think that this problem is a bit of a nasty way to get introduced to them.

Thanks Gneill. Attached is my attempt at the algebra to get the expression into the g(s) form. I'm wondering if you can tell if if I'm close or on the right track?

If the Tau term was somehow made to be 1/Tau rather than 2(tau), the ei(s) extracted, making the remaining numerator A then it would be ok?

The algebras been a struggle.

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• IMG.pdf
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Hmm, I'm not "getting" where you're going with your algebra.

You can make your life a bit easier if you set aside the ei(s) to begin with (since it'll just introduce an E/s multiplier, where E is the magnitude of the step) and concentrate on reworking the form of the transfer function.
$$G(s) = \frac{\frac{R}{1 + s R C}}{s L + \frac{R}{1 + s R C}}$$
May I suggest that you begin by multiplying the top and bottom by ##1 + s R C##? That'll immediately reduce the transfer function to:
$$G(s) = \frac{R}{(1 + s R C) s L + R}$$
Then just expand the denominator and work towards the form I showed previously.

Would negative squared and what I've done be a valid match for the form you've suggested? If not i will keep trying. Apologies for my poor algebra. Thankyou for your continued help and patience.

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• IMG_0001.pdf
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gneill said:
Hmm, I'm not "getting" where you're going with your algebra.

Neither do i, there was no mathematical foresight...I have just been multiplying through by all sorts and seeing what it resulted in. I was trying not simplify too much because then i would have more terms to play with. Oh i don't know

tranceical said:
Would negative squared and what I've done be a valid match for the form you've suggested? If not i will keep trying. Apologies for my poor algebra. Thankyou for your continued help and patience.

You want to get the s2 term to have a coefficient of 1. So take your second line and divide through top and bottom by LRC. Don't worry about making the numerator a tad more complicated, it's just a constant and will turn into something excellent later

After multiplying through by LRC it leaves me (1/LC)/[s^2 + (s(1/RC)) + 1/LC]

Which leaves just the ω term 1/LC that i need to square. So i can write (√1/LC)^2 ? Then on to the partial fractions...

I'm still keen to learn how this problem is solved it's just a shame I've needed so much help with it, so much so that i don't think i can justify submitting it as my own work. Let me be as brainy as you one day

tranceical said:
After multiplying through by LRC it leaves me (1/LC)/[s^2 + (s(1/RC)) + 1/LC]

Which leaves just the ω term 1/LC that i need to square. So i can write (√1/LC)^2 ? Then on to the partial fractions...
Yeah, just go ahead and define the constants ##ω_o = 1/\sqrt{L C}## and ##\tau = R C##. Note that you should use ##ω_o## rather than just ω. The reason for this is that the oscillation frequency of a damped system, ω, will be slightly off of the natural frequency ##ω_o##, mainly because the damping extends the time that it takes for the system to return to the equilibrium point (like friction slowing a sliding object). This will become apparent in a bit.

You can calculate values for these constants. The natural frequency should end up with units of radians per second. ##\tau## will be in seconds.

Write out the characteristic equation (the denominator) with the constant values plugged in (you can drop the units for this). Will the roots be real or complex?

I'm still keen to learn how this problem is solved it's just a shame I've needed so much help with it, so much so that i don't think i can justify submitting it as my own work. Let me be as brainy as you one day

"Will the roots be real or complex?" complex... so in this scenario I should multiply out the denominator of the damped sine function to s^2+2as+a^2+omega^2...find my a and omega values, then partial fraction (remembering to add in the b/s from earlier) to form the damped sine and cosine functions?

Many thanks

Last edited:
tranceical said:
"Will the roots be real or complex?" complex... so in this scenario I should multiply out the denominator of the damped sine function to s^2+2as+a^2+omega^2...find my a and omega values, then partial fraction (remembering to add in the b/s from earlier) to form the damped sine and cosine functions?

Many thanks

Yup. That's the plan.

I found a = 5 and omega = √2000-a^2 = 44.44

Then inputting step input, my partial fractions were:

A/s +(Bs+C)/(s+5)^2+(44.44^2)

The result:

[-78.2/s + (78.2)s+782] / (s+5)^2 + 1975

Doesn't seem correct to me...

tranceical said:
I found a = 5 and omega = √2000-a^2 = 44.44

Then inputting step input, my partial fractions were:

A/s +(Bs+C)/(s+5)^2+(44.44^2)

The result:

[-78.2/s + (78.2)s+782] / (s+5)^2 + 1975

Doesn't seem correct to me...

Yeah, your a and ω look fine, but something went awry with your partial fraction expansion. Can you elaborate a bit on your work there?

Heres my working (sorry its rough)

Many thanks

#### Attachments

• IMG.pdf
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Okay, looks like you might have not multiplied the "s" through the whole denominator in your first line.

Symbolically, your Laplace domain response looks like:
$$R(s) = ω_o^2 b \left(\frac{1}{s}\right) \cdot \left( \frac{1}{(s - a)^2 + ω^2} \right)$$
Leaving off the leading constants for now, the partial fraction expansion becomes:
$$\frac{A}{s} + \frac{B s + C}{(s - a)^2 + ω^2} = \left(\frac{1}{s}\right) \cdot \left( \frac{1}{(s - a)^2 + ω^2} \right)$$
Now, if you combine the terms on the LHS by putting them over a common denominator, both sides will then have the same denominator. Equate like terms of the numerators. Bonus! The right hand side's numerator is just 1.

1 person
gneill said:
Okay, looks like you might have not multiplied the "s" through the whole denominator in your first line.

Symbolically, your Laplace domain response looks like:
$$R(s) = ω_o^2 b \left(\frac{1}{s}\right) \cdot \left( \frac{1}{(s - a)^2 + ω^2} \right)$$
Leaving off the leading constants for now, the partial fraction expansion becomes:
$$\frac{A}{s} + \frac{B s + C}{(s - a)^2 + ω^2} = \left(\frac{1}{s}\right) \cdot \left( \frac{1}{(s - a)^2 + ω^2} \right)$$
Now, if you combine the terms on the LHS by putting them over a common denominator, both sides will then have the same denominator. Equate like terms of the numerators. Bonus! The right hand side's numerator is just 1.

Thanks a lot for help with this problem, it's much appreciated.

I see an error in your first line in the first attachment. Looking through the thread it is not clear that it has been caught.

you need to have A*1975 after clearing the denominator.

## What is a Laplace transform?

A Laplace transform is a mathematical operation that converts a function from the time domain to the frequency domain. It is used to solve differential equations and simplify complex mathematical expressions.

## How do you arrange an expression into a Laplace transform?

To arrange an expression into a Laplace transform, you need to first identify the function and its variable. Then, use the Laplace transform table to find the corresponding transformed function. Finally, apply any necessary algebraic manipulations to simplify the expression.

## What is the purpose of arranging an expression into a Laplace transform?

The purpose of arranging an expression into a Laplace transform is to make it easier to solve complex mathematical problems, such as differential equations. The Laplace transform allows us to convert a function into a simpler form in the frequency domain, where it is easier to analyze and manipulate.

## What are some common properties of Laplace transforms?

Some common properties of Laplace transforms include linearity, time-shifting, differentiation, and integration. These properties allow us to manipulate and solve Laplace transforms more easily.

## What are some applications of Laplace transforms in science?

Laplace transforms have many applications in science, including in physics, engineering, and mathematics. They are commonly used to solve differential equations in fields such as mechanics, electromagnetics, and control systems. They are also used in signal processing, circuit analysis, and data analysis.

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