# S-Plane Analysis

#### Azez

Joined Jan 16, 2012
13
Hi,

My lecturer decided to set us a coursework question WITHOUT teaching us what s-plane analysis is.

I have attached a copy of the question but to be honest I'd just really like someone to tell me what it is?

#### Attachments

• 35.3 KB Views: 31

#### Zazoo

Joined Jul 27, 2011
114
Have you covered Laplace transforms?
That's basically all you need for this problem.

#### Azez

Joined Jan 16, 2012
13
I have done Laplace transforms but not when dealing with circuits only in relation to signals etc.

So I don't know how I would apply it to this

#### Zazoo

Joined Jul 27, 2011
114
I have done Laplace transforms but not when dealing with circuits only in relation to signals etc.

So I don't know how I would apply it to this
There are a couple of ways. Circuit elements can be replaced by their Laplace equivalents, similar to the way you handle circuits using phasors, you should be able to find a table of these online (edit: pages 7, 8 and 9 of this PDF have the Laplace equivalents: http://web.cecs.pdx.edu/~ece2xx/ECE222/Slides/LaplaceCircuits.pdf)

The other option is to form the differential equation for the circuit and apply the Laplace transform to that.

#### Azez

Joined Jan 16, 2012
13
I have changed them all so the values are now

r = R(s) = 1kΩ

l = sL = jω.10mH

c = 1/sC = 1/jω.25μF

Is there somewhere I can get the equation needed to work it out, I have done this when using transfer functions but somehow I dont think it is the same thing.

Last edited:

#### Zazoo

Joined Jul 27, 2011
114
Once you have the Laplace equivalents you can apply standard network analysis techniques to the circuit (e.g. nodal, mesh, etc.)

Relating Is and I2 using current division would probably be the easiest way to proceed here.

#### Azez

Joined Jan 16, 2012
13
Yes but to use node or mesh analysis you need the voltage but that is not given in the question

#### Zazoo

Joined Jul 27, 2011
114
Either method can be used with a circuit regardless of what information is given.

However, current division will be the easiest since it can be applied directly:
$$I_{2} = I_{S}(\frac{Y_{C}}{Y_{R}_+Y_{L}+Y_{C}})$$

Where Y is admittance (Y = 1/Z)

#### t06afre

Joined May 11, 2009
5,934
Also do not forget the rules of complex math.

#### Azez

Joined Jan 16, 2012
13
This is where I've gotten to with the workings, I'm just wondering if I have to add the whole of the bottom of it then substitute ω with what they want in the question. Then finally find the modulus of the equation on the right to solve lI2l/lIsl?

#### Attachments

• 6.5 KB Views: 20

#### Zazoo

Joined Jul 27, 2011
114
Almost, just three things:
1.) Admittance for capacitor is jωC.
2.) The resistance is 1kΩ, so admittance becomes 1/1000.
3.) The inductor is 10mH (10x10^-3)

So:

$$\frac{j\omega 25\times 10^{-6}}{\frac{1}{1000}+\frac{1}{j\omega 10\times 10^{-3}}+j\omega 25\times 10^{-6}}$$