# s-domain analysis question

Discussion in 'Homework Help' started by derill03420, Mar 4, 2012.

1. ### derill03420 Thread Starter New Member

Oct 15, 2011
26
0
Okay i need to find expressions for v1 and v2 and im using nodal analysis:

I am good finding expression for v1, (2/s)-(v1/1)-((v1-v2)/(2/s)) = 0

Where i am stuck is with the current source and 2 ohm res. in series

For v2, ((v1-v2)/(2/s))-(v2/2s)-(v2/2)-(5/s) = 0 this is my derivation that im unsure of

the reason i think this is wrong for v2 is in my head i see instead of v2/2 - 5/s i see something like ((v2-v)/2)-(5/s)? where v is voltage between current source and res

Can anyone help me understand where im right or wrong thanks

**BTW see attachment for circuit**

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2. ### mlog Member

Feb 11, 2012
276
36
I think your first equation is wrong. I get the following:

2/s - v1/1 - (v1-v2)s/5 = 0

For the second node I get:

(v1-v2)s/5 - v2/(2s) - 5/s = 0

3. ### derill03420 Thread Starter New Member

Oct 15, 2011
26
0
Where is s/2 coming from? The cap is converted to s domain as 1/0.2s which is 2/s?

4. ### IronMod New Member

Jun 14, 2011
15
3
I have a different answer than both of you...

The capacitor in S domain is 5/s...(1/.2s)
The inductor is just simply 2s. (look up inductor s-domain transformations to see that it is 2s and not 2/s).

For Node 1 I have:
(2/s) + (v1/1) + (v1-v2/(5/s))=0

For Node 2 I have:
((v2-v1)/(5/s)) + (v2/2s) + ((v2+5)/2) + (5/s) = 0

dont forget the current source to the right is going in the downward direction, making it positive from v2, not negative. Hope this helps!

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
I believe mlog is correct.

6. ### mlog Member

Feb 11, 2012
276
36
The 2/s term comes from the independent current source. You, derill03420, and I each had a 2/s term too.

The capacitor impedance is 1/(0.2s) or 5/s. Since we're finding the current across the capacitor, the impedance is in the denominator and hence the s/5 term multiplied times the difference of the node voltages.

Here it is again.

2/s - v1/1 - (v1-v2)s/5 = 0

2/s is for the current source. v1/1 is for the 1-resistor current.

For current through capacitor, (v1-v2)/Zc, where Zc = 1/(sC) = 1/(0.2s) = 5/s. Since the Zc is in the denominator, you can invert it and multiply times (v1-v2), which gives (v1-v2)s/5.

For the second node I get:

(v1-v2)s/5 - v2/(2s) - 5/s = 0

Same approach as the first equation. You just have to watch the negative sign. The inductance ZL=2s, but the current is v2/ZL = v2/(2s). (Don't confuse the v2, the latter number is a subscript.)

The last term is the 5 A curent sourse leaving the node. So it is -5/s.

The little wrinkle is in putting the 2-ohm resistor in series with the current srouce. the prof wanted to see if he could fool anybody. Somehow, somebody is going to try and work that resistor into the calculation somehow. A 2-ohm resistor in series is like a 10-volt source in parallel with a 1-ohm resistor.

Oh, my, I'm not watching the time. It's 1:15 past my bed time. I have to get up in 6 hours. Good night.