# Laplace Transforms and non ideal inductors

1. Mar 14, 2005

### brendan_foo

Hi there,

I'm trying to derive an expression for the transient response (for a step input of magnitude V), for a non-ideal inductor modelled in the schematic I have drawn. This non-ideal inductor includes its inductance ( ), a parasitic parallel resistance and a parasitic capacitance.

The schematic is here: http://homepage.ntlworld.com/b.preece/RLC.JPG

So my initial thoughts are:

$$Z_{c}(S) = \frac{1}{cS}$$

$$Z_{l}(S) = lS$$

$$Z_{r}(S) = R$$

Right... so the parallel combination of all of these three operational impedances will be defined as:

$$Z_{eff} = \left[\frac{1}{Z_{r}} + \frac{1}{Z_{c}} + \frac{1}{Z_{l}}\right]^{-1}$$

So using those values quoted up above, we should have :

$$Z_{eff} = \left[\frac{1}{R} + Cs + \frac{1}{sL}\right]^{-1}$$

Which if I'm not mistaked is evaluated to:

$$Z_{eff} = \frac{sLR}{(LCR)s^2 + sL + R_p}$$

$R_p$ Any subscript 'p' indictating that its a parasitic value.

Ok... Now based upon the Voltage divider equation, we can say that the transfer function is as below:

$$H(s) = \frac{V_{out}}{V_{in}} = \frac{Z_{eff}}{Z_{eff} + Z_{fix}}$$

Now we have:

$${Z_{eff} + Z_{fix} = \frac{sLR + R_{fix}((LCR)s^2 + sL + R_{p})}{(LCR)s^2 + sL + R_p}$$

And from there we have the Transfer function $H(s)$ to be:

$$H(s) = \frac{sLR}{sLR + R_{fix}((LCR)s^2 + sL + R_{p}}$$

Knowing that the Laplace transform of the unit step function at t = 0, multiplied by a scaling value will be:

$$L(u(t)) = \frac{1}{s} \cdot V_{ip}$$

And the fact that :

$$V_{out} = H(s) \cdot V_{ip}$$

So that the $S$ in the numerator will disappear to form :

$$V_{out} = \frac{V_{ip} \times LR}{sLR + R_{fix}((LCR)s^2 + sL + R_{p})}$$

This is one of the expressions I've derived, but I cant seem to hammer it into some useful form to take inverse laplace transforms of the second-shift nature, to form expressions which show a damped harmonic oscillations (which i know happens).

In a word...HELP!!

If you can spot any obvious mistakes then please point them out, but I would be very grateful for anyone who can get me to a final solution for $V(t)$

Cheers people
Brendan

Last edited by a moderator: Apr 21, 2017
2. Mar 14, 2005

### Davorak

3. Mar 14, 2005

### egsmith

Solution via Mathematica

I was able to confirm your math in mathematica and find an inverse laplace transform. Didn't attempt it by hand. I confirmed mathematica's solution in spice by picking some dummy values.

One idea if your frequency domain solution does not match any table entry which may result in a simpler expression (didn't try it) would be to solve for the frequency domain solution of the upper resistor, inverse it to get the time domain solution, then subtract that from the source to get the voltage across the lower net.

4. Mar 14, 2005

### egsmith

2nd try

I guess the filetypes .nb and .sp are invalid so I bundled them in a zip...

#### Attached Files:

• ###### inductance.zip
File size:
7.1 KB
Views:
64
5. Mar 15, 2005

### brendan_foo

Cheers mate!!! Much appreciated...I got something very similar to that by hand, although I didnt get a hyperbolic sine... but i'll investigate into it.. Very pleasing results with the oscillatory harmonic motion.

Thankyou very much... can I purchase mathematica direct from wolfram and download it or something?! I have matlab at the moment and I'm not a big fan of it, although I have no choice but to learn it..

Thanks guys
Rock on!!!

6. Mar 15, 2005

### egsmith

My pleasure.

Your answer may be right just in a different form. The mathematica FullSimplify is pretty good, especially at applying trig identities. My math has improved greatly just trying to figure out how it got from A to B!

I use Mathmatica all the time when analyzing simple systems (1st, 2nd order) because I can do it symbolically (much more intuitive for me) and it's wonderfull for quickly checking the math in homework assignments. However I use matlab almost exclusively when doing numerical partial differential equations and for analyzing hspice output files (I highly recommend the MIT matlab toolbox [1]). So in a nutshell, use both. Right tool for the job kinda thing.

Mathematica can be purchased through the wolfram website however I got mine through the campus bookstore so I could get the substantial student discount. I believe you can get a student version online by faxing a schedule or something...

There is also an opensource alternative to Mathematica called Maxima [2]. I haven't really used it seriously though.

[1]
http://ocw.mit.edu/OcwWeb/Electrical-Engineering-and-Computer-Science/6-012Fall2003/Assignments/ [Broken]
Scroll down to problem set 8.
[2]
http://maxima.sourceforge.net/

Last edited by a moderator: May 1, 2017
7. Mar 15, 2005

### brendan_foo

I'm a newbie with Matlab really...and just trying to plot my equation is a nightmare... If I post you my equation could you plot it with mathematica using the same conditions you did for the one you zipped up and send it to me?!

If you've got msn im brendan_online@hotmail.com... I'll post my equation to see if its right when you give me the go ahead.

Cheers buddy
Brendan

8. Mar 15, 2005

### egsmith

Sure. What are your component values?
You could always just spice it too ;)

9. Mar 15, 2005

### brendan_foo

I've not got any Spice tools etc... Everything thats been done here is a complete pen and paper job :D...

Bear with me, I'll sling you that equation in about 5 minutes...let me check it first

10. Mar 15, 2005

### egsmith

11. Mar 15, 2005

### brendan_foo

Equation coming now...its huge

12. Mar 15, 2005

### brendan_foo

DEEP BREATH:

$$\frac{V}{R_{f}C \cdot \sqrt{\frac{4(RR_fC)^2 - R_fLC(R_f+1)^2}{4(R_fLC)(RR_fC)^2}}} \cdot e^{\frac{-(R+1)t}{RR_fC}}\cdot sin\left [{\sqrt{\frac{4(RR_fC)^2 - R_fLC(R_f+1)^2}{4(R_fLC)(RR_fC)^2}}t \right]$$

Where $R$ is the parasitic resistance
$R_f$ is the fixed resistance
and the rest are the other values... phew..

If you could simulate that with the values you entered before for C, Rf, R and L and see what happens...HOPEFULLY something similar..but it'll probably not work at all :D

Cheeeeeeeers...I owe you a beer, if you drink

Last edited: Mar 15, 2005
13. Mar 15, 2005

### egsmith

There is no t
I'll assume the equation is of the form
kV*Exp[-b*t]*Sin[t*c]

14. Mar 15, 2005

### brendan_foo

Hahah yeah sorry, i got caught up in the latex.... yeah there's a -t in the power of e and a t in the argument of sin... Amended above

Last edited: Mar 15, 2005
15. Mar 15, 2005

### egsmith

I believe I entered your equation correctly however when I tried to plot it I got floating point underflow errors. Unfortunately the two equations are probably not equivalent.

PS
It turns out Mathematica can export TeX code.

Here is the solution Mathematica found.

$\frac{2\multsp {{e}^{-\frac{t\multsp (R+{R_p})}{2\multsp R\multsp {C_0}\multsp {R_p}}}}\multsp L\multsp \sinh \big[\frac{t\multsp {\sqrt{L\multsp \big(L\multsp {R^2}+2\multsp L\multsp R\multsp {R_p}+\big(L-4\multsp {R^2}\multsp {C_0}\big)\multsp R_{p}^{2}\big)}}}{2\multsp L\multsp R\multsp {C_0}\multsp {R_p}}\big]\multsp {R_p}}{{\sqrt{L\multsp (L\multsp {R^2}+2\multsp L\multsp R\multsp {R_p}+(L-4\multsp {R^2}\multsp {C_0})\multsp R_{p}^{2})}}}$

Here is how it saw your equation.

$\frac{2\multsp {{e}^{-\frac{(1+R)\multsp t}{{\sqrt{{R^2}\multsp C_{0}^{2}\multsp R_{f}^{2}}}}}}\multsp \sin \big[\frac{1}{2}\multsp t\multsp {\sqrt{\frac{4\multsp {R^2}\multsp C_{0}^{2}\multsp R_{f}^{2}-L\multsp {C_0}\multsp {R_f}\multsp {{(1+{R_f})}^2}}{L\multsp {R^2}\multsp C_{0}^{3}\multsp R_{f}^{3}}}}\big]}{{C_0}\multsp {R_f}\multsp {\sqrt{\frac{4\multsp {R^2}\multsp C_{0}^{2}\multsp R_{f}^{2}-L\multsp {C_0}\multsp {R_f}\multsp {{(1+{R_f})}^2}}{L\multsp {R^2}\multsp C_{0}^{3}\multsp R_{f}^{3}}}}}$

Edit: Mathematica exports 'e' as ExponentialE which the TeX image generator doesn't seem to like.

#### Attached Files:

• ###### inductance2.zip
File size:
4.5 KB
Views:
62
Last edited: Mar 15, 2005
16. Mar 15, 2005

### brendan_foo

In the eq' Mathematica produced...is the argument of the Sinh a t-th root of the expression or is it multiplied by t?

The first expression is correct, guarantee'd...it produced the output I expected, but i cant get to a similar expression... Whenever I hammer it into some form it always comes out as:

$$k \cdot \frac{a}{(s + \beta)^2 + a^2}}$$

which leads to a damped sine function...

ARRRGH :surprised

I really need to solve this by hand, else I will look like a chump!!

Hmmmm where to next eh

17. Mar 15, 2005

### brendan_foo

Ok so now i've come up with this...some of it seems to have symmetry with the first, but I dont know how to use this Opensource Mathematica style software...

$$\frac{V}{ZC \cdot \sqrt{\frac{4(RZC)^2 - LC(R+Z)^2}{4LC)(RZC)^2}}} \cdot e^{\frac{-(R+Z)t}{2RZC}}\cdot sin\left [{\sqrt{\frac{4(RZC)^2 -LC(R+Z)^2}{4LC(RZC)^2}}t \right]$$

Where for simplicity's sake I used Z for the fixed resistance and R for the parasitic resistance....ARRRRRRRRGH...seems to be symmetry in the exponent though, to mathematicas!!

18. Mar 15, 2005

### brendan_foo

Gah I'm not getting luck with any of this stuff at all.... Tried it all out on a trial version of Mathematica and its not working at all... I got near enough the same damping factor for my solution as the one Mathematica produced {in the example you gave to me}... and yet trying to plot that on its own {ke^(-at)} for that particular time range wont plot...

GRRR

19. Mar 15, 2005

### egsmith

When ploting look at the order of the factor of exponential decay (e^-at, look at 'a'). If you choose to plot a range that is much larger than this value you will cause a floating point underflow and get no graph.