# Laplace transforms for transient analysis

agata78

## Homework Statement

A capacitor of 0.1 F and a resistor of 5 Ω are connected in series; the combination is applied to a step voltage of 20V. Determine the expression for the:

(a) current that flows in the circuit and

(b) the voltage across the capacitor in time domain.

## Homework Equations

For question (b)

Would I be correct to use the following equations?

Where:
Vc - is the voltage across the capacitor
V - is the supply voltage
t - is the elapsed time since the application of the supply voltage
RC - is the time constant of the RC charging circuit
e - is the base of the Natural Logarithm = 2.71828

Time Constant
τ = R x C
τ = 5 x 0.5
τ = 2.5 seconds

The voltage formula is given as:
Vc = V(1-e-t/RC)

which equals:
Vc = 20(1-e-t/5)

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## Answers and Replies

Mentor
Are you expected to do the problem using Laplace transforms (it's in the thread title), or just write down the result from previous knowledge?

EDIT: Also, 5 x 0.1 is not 5.

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agata78
Yes. The general title for these question states 'Use Laplace transforms for the transient analysis of networks'.

Appologies for the typing error.

Mentor
So, the equations you've written are really a result you hope to find using Laplace transforms.

As usual, begin with a circuit equation. Use Laplace domain models for the circuit components and write the appropriate equations...

agata78
I cant find any appropriate equation for this, i dont have any good example for this.

Can you help?

Mentor
You write your usual circuit equations just as you would for an AC circuit, but use the Laplace domain version of the component impedances. Where you were using 'jω' before in the expressions for inductor and capacitor impedances, just use the variable 's'.

I know it looks silly at first glance, writing AC-type equations for what is apparently a DC circuit, but the Laplace domain's 's' variable takes in all frequencies and can readily model the transient response that happens when a switch closes. You'll see You should have in your text or notes the form that a unit step function takes in the Laplace domain. You'll need that to model the 20V source that's suddenly switched on at time t=0.

Mentor
If I(t) is the current at time t, what is the voltage drop across the resistor at time t (in terms of I(t) and R)? What is the voltage drop across the capacitor at time t (in terms of the integral of I(t) with respect to t and C)? Write an equation such that the sum of these two voltage drops is equal to the applied step voltage. Do you know how to take the Laplace Transform of the individual terms in this equation? If so, write the Laplace Transform of the equation.

Chet

agata78
Z = R + (1 / sC)

Z = 5 + (1 / s x 0.1)

Using Ohms Law:

I(s) = Vin / Z

I(s) = 20/s / 5+(1 / s x 0.1)

However:

Vout(s) = I x (1 / sC)

Vout(s) = I x (1 / s x 0.1)

So:

Vout(s) = [ 20/s / (0.1 + (1 / s x 0.1))] x [1 / s x 0.1]

Vout(s) = 20 / s(s x 0.1 x 0.1 + 1)

Vout(s) = 20 / s(s x 0.01 + 1)

Am i right? What value is s in this example?

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Mentor
Something's gone wrong in your algebra when you solved for Vout.

Start by simplifying your Z a bit. Note that 1/(s x 0.1) = 10/s. Makes the numbers easy!

You're going to want to solve for I, so you might as well simplify the expression for I(s) before carrying on to the potential across C. Can you reduce the expression for I(s)?

agata78
To solve Z = R + (1 / sC) i need to add (5 + 10/s) = (10+5s/ s)

for Is= (20s/ 10 +5s) = (20s(10-5s) )/ (10+5s) (10-5s) = (200s - 100s2 ) / (100-25s2)

and if s2=(-1) then

Is= (200s+100) / 125
Is it ok so far?

Mentor
Use more parentheses to keep your operations straight.
$$I(s) = \frac{V(s)}{Z(s)} = \frac{20}{s} \left(\frac{s}{5 s + 10}\right) = ~~?$$

agata78
Is= = 20 / 5s+10 = (100s-200) / (25s2 -100)

Is=( 4s-8) /-5

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Mentor
Is= = 20 / 5s+10 = (100s-200) / (25s2 -100)

Egads! Just divide the top and bottom of 20 / (5s+10) by five. Hint: It is desirable to have the denominator have factors of the form (s + n).

Mentor
This is the same answer I got, but I used a different method. My starting equation was
$$\frac{\int_0^t{I(t')dt'}}{C}+RI(t)=Vu(t)$$
where u is the unit step function. Taking the Laplace Transform gives:
$$\frac{I(s)}{sC}+RI(s)=\frac{V}{s}$$
Solving for I(s):
$$I(s)=\frac{V}{R}\frac{1}{\left(s+\frac{1}{RC}\right)}$$

agata78
Then Is= 4/ (s+2)

Should i calculate now Vout?

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Mentor
Then Is= 4/ (s+2)

Should i calculate now Vout?

You could, or you could find I(t) first. You need to provide it as one of the answers...

agata78
It= ( V/R ) ( 1-e (-Rt/L) )

But i dont have L for this equation. Should i find a different one for It?

Mentor
It= ( V/R ) ( 1-e (-Rt/L) )

But i dont have L for this equation. Should i find a different one for It?

There's no inductor in this circuit.

Find I(t) by finding the inverse Laplace transform of the I(s) that you found above. You should have a table of Laplace transforms to work with.

Mentor
What is the Laplace Transform of the function eat?

agata78
ok then,

It= 4e -2t

am i right?

Mentor
ok then,

It= 4e -2t

am i right?

Right. And that should be I(t); it's current as a function of time.

agata78
Trying to calculate now Vout:

(4+/ (s+2) ) x (10/s) = 40 / ( s2 +2s)

how to deal with s2

Mentor
Trying to calculate now Vout:

(4+/ (s+2) ) x (10/s) = 40 / ( s2 +2s)

how to deal with s2

Factor the denominator.

In fact, when you're deriving these things it's a good idea to keep a lookout for factors like s, (s + n), etc., and not combine them. They'll be needed later!

agata78
Im stuck now, I out of ideas what to do next. Help!

Mentor
After factoring the denominator, use partial fractions to split the expression into two terms. Both terms should have forms that you can recognize in your Transform tables (If they're good tables you will probably also be able to spot an entry for your expression before the partial fraction procedure is done).

agata78
40 / s2 +2s = -8+16s

am i right so far? What you mean by split the expression? which expression ?

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Mentor
40 / s2 +2s = -8+16s

am i right?

Nope. You want to use partial fractions on the expression:
$$\frac{40}{s(s + 2)}$$
so
$$\frac{40}{s(s + 2)} = \frac{A}{s} + \frac{B}{s + 2}$$

...carry on...

agata78
I dont know if im doing that right,

after some calculations i have if s= (-2) then B=-20
and if s = 0 then A= 20

is it ok?

Mentor
You've found the correct values for A and B, but I don't recognize your method. Usually one multiplies out the expression on the right and then equates like terms of the numerators on the LHS and RHS to create a pair of equations to find A and B (look up the method of partial fractions).

Nevertheless, since you have values for A and B you now have two expressions in the Laplace domain that you should be able to find in your table of Laplace transforms. Write down their time domain translations; they comprise the terms of the current I(t) that you want.

agata78
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Mentor
I used this website to educate myself a bit more then i remembered from school www.mash.dept.shef.ac.uk/Resources/web-partialfractions.pdf

But does it matter which way i used to calculate A and B.

i(t)= 20e 1 + (-20) -2t

But what next?

Your transformation of the Laplace terms does not look right! You have found the that in the Laplace domain:
$$I(s) = \frac{20}{s} - \frac{20}{s + 2}$$
Find the forms of those two terms in your Laplace Transform tables and convert them to time domain terms. The result should look familiar!

agata78
the only way it could be is:

20-20e(-2t)

Mentor
the only way it could be is:

20-20e(-2t)

Right. Technically each term is multiplied by a unit step function to indicate the the voltage is "switched on" at time t = 0. Mathematically it means that this solution applies only for t ≥ 0.

So, writing it in proper form for a function:

[STRIKE]I(t)[/STRIKE]Vc(t) = 20-20e(-2t) = 20(1 - e(-2t))
[STRIKE]
Now you can move on to finding Vc(t). Use the same methods.[/STRIKE]

EDIT: Sorry about that. I misremembered where we were at in the problem sequence! I think you're done with this one!

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agata78
Yes i was actually trying to ask you why would i have to calculate it again.

Hurrayyyyy!

Thank you so much for help all the way! One to go!