eecs 70b s13: midterm solutions
DESCRIPTION
Midterm Solutions for EECS 70BTRANSCRIPT
Spring'2013''
' 1'
EECS$70B"MIDTERM"SOLUTIONS!
'Ideal'Op2Amp:'
!!" = ∞ → !!"! = !!"! = 0! ! → !! = !! ''KCL'at'node'A:'
!! − !!3! − !! − !!! − 0 = 0 → !! − !! − 3!! + 3!! = 0'
1.2− !! − 3!! + 3 = 0 → 4.2 = 4!!'!! =
4.24 = 1.05![!]'
KCL'at'node'B:'!!2! +
!! − !!2! = 0 → !! = 2!! '
!! = 2!! = 2!! = !.!![!]'Ohm’s'Law:'
!! = !!"# ∗ ! → ! = 2.1! !2! !" = 1.05! !Ω = !"#"![!]!
!! !
Spring'2013''
' 3'
!!! =
!!" ∗ !2! = !!"
2 !
!! = ! !!!!" = !! !!" 2
!" !!
!! =!2!!!"!" !(∗)!
!
!!"# =!!"2 + !!! !(∗∗)!!
Combine!equations!(*)!and!(**)!
!!"# =!!"! + !"!
!!!"!" !
!!! !
!At!! = 0!,!the!circuit!is!in!steady!state.!Therefore,!inductor!is!short!and!capacitor!is!open!circuit.!!
! 0! = 124 = 3! ! , !! 0! = 0![!]!
At!! = 0!,!the!switch!is!opened.!The!continuity!conditions!require!that!!! 0! = 3! ! , !! 0! = 0![!]!
Apply!KVL!to!find!!! 0! !
12− !! 0! − !! 0! − 4 ∗ ! 0! = 0!!! 0! = 12− 0− 4 ∗ 3 → !! !! = !![!]!
In!steady!state,!capacitor!will!be!open!circuit,!therefore!
! � = !![!]!It!is!a!series!RLC!circuit,!therefore!
! = !2! =
42 ∗ 8
Ω!" = 250,!! =
1!"
= 11,180!!
Since!! < !!,!the!response!is!underdamped.!The!damped!natural!frequency!is!!
!! = !!! − !! ≈ !! = !!,!"#!
The!form!of!the!transient!response!is!!
!! ! = !!!(!"#$!!! + !"#$!!!)!where!A!and!B!are!constants.!The!steadyEstate!response!is!!
!!!(!) = !(�) = 0!so!that!the!complete!response!is!!
! ! = !! ! + !!! ! = !!!"#!(!"#$!11,180! + !"#$!11,180!)!Now,!let’s!find!A!and!B!constants!with!using!initial!conditions.!
! 0 = ! + 0 = 3 → ! = !!
!! ! = ! !" !!"= ! −250!!!"#! !"#$!11,180! + !"#$!11,180! + !!!"#!
∗ 11,180 −!"#$!11,180! + !"#$!11,180! !