# LTI System freq. domain HELP Please

Discussion in 'Homework Help' started by ohwcomp, Oct 14, 2008.

1. ### ohwcomp Thread Starter New Member

Sep 16, 2008
9
0
Howdy Everyone!

Im having trouble with the following interesting question especially with part (c) (a) For this part I realised the steady state principles and applied open and short circuits etc and used the voltage divider rule to get 2V and 1 A, is this correct?

(b) Redrawing this I got

6V turned into 6/s
4Ω stayed the same
1H turned into s
0.05F turned into 1/0.05s

Also for the initial conditions i added a current source under each inductor i(0-) and capacitor vc(0-).

(c) How do I find this Vout? I can see that its the voltage at the node connecting Ir,Il,Ic. So at this node Ir + Il + Ic = 0? From KVL/KCL?

How do I take into the account the -ve sign of the current source from the inductor.

Would it be Ir - Il + Ic = 0?

Ir = (Vo - 6/s)*1/4 ?????

Il = ?

Ic = ?

Then how would I get an expression for Vout? Hence the complete solution?

Cheers for the help in advance!

2. ### mik3 Senior Member

Feb 4, 2008
4,846
70
Maybe tomorrow or in a few days i will be able to answer you because we have just started to study Laplace transforms at university. 3. ### guitarguy12387 Active Member

Apr 10, 2008
359
12
I get 2V and 1A for part A.

Im not sure exactly what you're doing with the voltage in part B.

For part b, i'd get: capacitor --> 1/(.05s) and inductor --> s

For part c, just use nodal analysis. I get the following equation:

(V_0 - 6)/4 + V_0/(s+2)+V_0/(1/(sc)) = 0

From there, you need to get it in the right form. Take the inverse laplace transform and find a solution that satisfies the equation...

Hopefully that gets you started.

4. ### ohwcomp Thread Starter New Member

Sep 16, 2008
9
0
Hi,

(V_0 - 6)/4 + V_0/(s+2)+V_0/(1/(sc)) = 0

took into account the 2Ω which is not there as it is short circuited,

Ir + Il + Ic = 0

V=IR
I = V/R

Il = V_0/s

Ic = V_0*s0.05

ok so rearranging I get some nasty

V_0 = 2s+?/s^2+5s+20 which I cant simplify and get the inverse. Im doing this from memory,so please check my V_0.

But Im sure the bottom bit s^2+5s+20 is correct, how do I simplify this so I can take the LT inverse?

Cheers

5. ### ohwcomp Thread Starter New Member

Sep 16, 2008
9
0
Correcting my mistakes,

You Got,
(V_0 - 6)/4 + V_0/(s+2)+V_0/(1/(sc)) = 0

Shouldnt it be

Ic = 0.05s ( V_0 - (2/s) )

Il = V_0 - (-1) / s

Ir = V_0 - (6/s) /4

Leaving V_0 = 2s+10/s^2+5s+20?

Is that correct? Then how would I go on from here?

Thanks

6. ### guitarguy12387 Active Member

Apr 10, 2008
359
12
Oh good call heh.

That denominator looks good. I'm not sure where the numerator comes from, but going back to the equation i had:

(V_0 - 6)/4 + V_0/(s)+V_0/(1/(sc)) = 0

=> V_0(1/4+1/s+sc) = 6/4

=> V_0(s/4+1+s^2c) = 3s/2

=>V_0(s^2+s(1/4c)+1/c) = 3s/2c

The inverse laplace of that is fairly straightforward, i think. It has been quite a while since i have actually done them, so correct me if i'm wrong haha.

hope that helps a little.

7. ### guitarguy12387 Active Member

Apr 10, 2008
359
12
Oops. I just saw your latest post.

Here's what i get for the currents:

Il = V_0/sL

Ic = V_0*sC

Ir = (6-V_0)/4

I am assuming you're calling the bottom node ground...

8. ### guitarguy12387 Active Member

Apr 10, 2008
359
12
wowza, thats a lie...

Ir = (V_0 - 6)/4

9. ### ohwcomp Thread Starter New Member

Sep 16, 2008
9
0
Heya Mate,
Yours:
Il = V_0/sL
Ic = V_0*sC
Ir = (V_0 - 6)/4

Mine:
Il = V_0 - (-1) / s
Ic = 0.05s ( V_0 - (2/s) )
Ir = V_0 - (6/s) /4

Yours is different to mine, L = 1, C = 0.05 hence yours turns into
Il = V_0/s
Ic = V_0*s0.05
Ir = (V_0 - 6)/4

I believe mine takes into account the initial conditions where as yours assumes initial conditions = 0. I wish I could do that, but unfortunately the complete soltution is taking account these conditions so =((((

Your input is greatly appreciated! Any more ideas?

10. ### guitarguy12387 Active Member

Apr 10, 2008
359
12
I see... Hopefully i'm not causing more problems then i'm helping hah.

Anyway, there very well may be other methods, but the way i was taught this method, the initial conditions are applied to the equations after they are transformed back into the time domain.

Once you get your solution back into a nonhomogeneous equation, you can use elementry diff eqn methods to find the forced and natural solutions.

Once again, thats how i was taught... you may have been taught other methods.