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! !

Spring'2013''

' 5'

!! 0 = ! −250! + 11,180! = 0 → ! = 250!11,180 =

75011,180 → ! = !.!"#$!

! ! = !!!"#$ !"#$!!!,!"#$+ !.!"#$%&'!!!,!"#$ [!]!

!! ! = !!" !!" = −!"#!!!"#$ !"#!!,!"#$ ![!]!

!

!!!" = !!" = 173�0°![!]!

!!" = !!" = 173�− 120°![!]!!!" = !!" = 173�120°![!]!

!!