Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

solving circuit using Laplace Transform

Status
Not open for further replies.

andrewllewop

Newbie level 4
Joined
Nov 24, 2012
Messages
6
Helped
0
Reputation
0
Reaction score
0
Trophy points
1,281
Activity points
1,324
having issues with solving this circuit using Laplace transform. firstly i applied KVL to the ciruit[V=V(L)+V(R)+Vo], then i substituted the voltage of the inductor for V(L) and the current through the capacitor to give V(R)=R*(i*dv/dt). resulting in the final equation:
L*di/dt + R*(i*dv/dt) + Vo. I then laplaced that equation to give: V(s) = sLI(s) + RCSVo(s) + Vo. But I'm basically stuck there, just need Vo/Vin maybe its my math but i don't see it.

If there is another way or if i am doinf something wrong I'd appreciate it if someone could look at it for me
Thnx.

click on the image above.

Untitled.jpg
 

You can proceed in two different ways:

1. Using KVL:

VL+VR+Vo=Vi

we know that the current is the same for all the components (single mesh): VR=R*i, VL=L*di/dt and i=C*dVo/dt. Deriving this last di/dt=C*d²Vo/dt². Let's now substitute in the KVL

L*C*d²Vo/dt²+R*C*dVo/dt+Vo=Vi

Laplacing: (S²*L*C+S*R*C+1)*Vo=Vi

2. Using directly the laplace transform of each single component. We have a voltage divider in which the impedance to ground is Zc=1/(S*C) while the series is ZL+R=SL+R then simply:

Vo/Vi=[1/(S*C)]/[1/(S*C)+S*L+R]=1/[1+S²*L*C+S*R*C]
 
hey thnx I see it now the first method is very clear I feel like I"m blind or something. I also get the 2nd part.
Appreciate it.
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top