# Homework Help: Simple Op-amp/Laplace question

1. May 6, 2007

### trickae

1. The problem statement, all variables and given/known data

http://img504.imageshack.us/img504/7714/assn2dn1.jpg [Broken]
Find H(s) = V2(S) / V1(s)

2. Relevant equations

Laplace transformed ckt is topologically equivalent to the standard circuit - except all dynamic elements are replaced by either s (L) or 1/s (C) and then treated as resistances. inverse laplce of the solution will give the solution

3. The attempt at a solution
Code (Text):

Laplace transform of the circuit shown above
find H(s) = V2(S) / V1(s)

[B][U]i) KCL @ Va[/U][/B]

(Va - V1 )/ R + VaSC1 + (Va - Vb)/R = 0

or  Va-V1 + RVaSC1 + Va-Vb = 0
or  2Va-V1-Vb +RSC1Va = 0
or  (2+RSC1)Va - Vb = V1 -------------------------(1)

[B][U]ii)KCL @ Vb[/U][/B]

(Vb-Va)/R + (Vb-V2)SC3 + (Vb-V2)SC2 = 0
or   Vb - Va + RSC3(Vb-V2) + R(Vb-V2)SC2 = 0
or   (1 + RSC3 + RSC2)Vb - Va - (RSC3+RSC2)V2 = 0
or   [(1 + RSC3 + RSC2)Vb - Va] / (RSC3+RSC2) = V2 -------------------(2)

divide V2/v1 = H(S)

H(S) = V2(s)/V1(s)
=       [(1 + RSC3 + RSC2)Vb - Va]
----------------------------
(RSC3+RSC2)
---------------------------------------
[(2+RSC1)Va - Vb]

=         [(1 + RSC3 + RSC2)Vb - Va]
----------------------------
[(RSC3+RSC2)[(2+RSC1)Va - Vb]

Problem is that I need Va and Vb to cancel for the next three parts of the question where we determine plots etc.

Any help? Should i have substituted Va and Vb with each other and get them in terms of v2/v1 then?

Last edited by a moderator: May 2, 2017
2. May 6, 2007

### trickae

i found my mistake in the second KCL equation :

it should be Vb - V2/R
and i'll get another KCL equation at the third node v2 directly above the C2 capacitor.

Last edited: May 7, 2007
3. May 6, 2007

### trickae

Code (Text):

I think i messed up again - when trying to cancel Va and Vb
I end up getting a cubic in the denominator and that can't
be simplified that easily with inv laplace transforms.

So I got
KCL @ Va

(2 + RSC1)Va - Vb = V1 -----------------------------(1)

KCL @ Vb

(Vb-Va)/R  +  (Vb-V2)SC3 + (Vb-V2)/R = 0
or ....
(2 + RSC3)Vb - Va - (RSC3+1)V2 = 0 ---------------(2)

KCL @ V+

(V2-Vb ) / R = V2SC2
or ..
(1-RSC2)V2 = Vb --------------------------------------(3)

Sub (3) into (2)

(2 + RSC3)(1-RSC2)V2 - Va - (RSC3 + 1)V2 = 0
or ..
[-(RS)^2 . C3C2 - 2(RS)C2 + 1 ] V2 = Va ------------------------(4)

Sub (4),(3) into (1)

[(2 + RSC1)(-(RS)^2 . C3C2   - 2RSC2 + 1)  - (1-RSC2)]V2 = V1
V2/V1 = 1 / [(2 + RSC1)(-(RS)^2 . C3C2   - 2RSC2 + 1)  - (1-RSC2)]

or better visually

V2                              1
----   = --------------------------------------------------------------
V1           (2 + RSC1)(-(RS)^2 . C3C2   - 2RSC2 + 1)  - (1-RSC2)

If i expand the last term i get a cubic. Help :'(