Lesson+14

# Lesson+14 - 1 Lesson 14 Lesson 14 Impedance Ch 13.1-3...

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1 Lesson 14

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Impedance Ch 13.1-3 Exam #1 Preview Wednesday Practice Exam #1 on-line HKN review location remains unknown at the tiime (email) Lesson 14 2 Lesson 14
Challenge 13 Suppose V o (s) = (I dc /C)/(s 2 + (1/RC)s + (1/LC)) (given) What is v o ( ) and v o (0)? Simple mechanical problem. sV o (s) = (I dc /C)s/(s 2 + (1/RC)s + (1/LC)) lim sV o (s) = 0 or v o ( ) = 0 (final value theorem applies) s 0 lim sV o (s) = 0 or v o (0) = 0 s 3 Lesson 14

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What’s Up Previously the solution method associated with RLC circuit analysis was based on first deriving the system ODE’s and IC’s. This was a tedious process. Something better was needed. V=40 R=5 R=1 R=4 L=2 C=1 S 0 4 2 4 2 40 4 5 2 2 2 2 2 1 1 2 1 2 1 t d y t y dt t dy t y dt t dy t y t y dt t dy dt t dy ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( y 1 (t) y 2 (t) v 0 (t) Laplace Inverse Laplace Hard Work How can this be simplified. Is there a more direct path from circuit to Y(s)? 4 Lesson 14
Lesson 14 Inductor s Think Impedances (a.k.a., complex Ohm’s Law) Capacitors Cs s I s V s Z s s I C s V d i C t v C C C C C L t t C 1 ) ( ) ( ) ( therefore ) ( 1 ) ( ) ( 1 ) ( 0   sL s Il s V s Z s sI L s V dt t di L t v L L L L L L L   ) ( ) ( ) ( therefore ) ( ) ( ) ( ) ( Resistor s R s R ) ( 5 Lesson 14

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Initial condition generator Capacitors Cs s I s V s Z C C C 1 ) ( ) ( ) ( 6 Lesson 14 Initial condition v(0). s v s s I s V d i t v t ) 0 ( ) ( ) ( ) ( ) ( )) 0 ( ) ( ( 1 ) 0 ( ) ( 1 ) ( )] 0 ( ) ( [ ) ( ; / ) ( ) ( Cv s I Cs s v s I Cs s V v s sV C s I dt t Cdv t i
Initial condition generator Capacitors 7 Lesson 14 C + v(0) - C C/s V(0)/s 1/Cs C/s CV(0) 1/Cs v(s) V(s) v(0)/s Cv(0) )) 0 ( ) ( ( 1 ) 0 ( ) ( 1 ) ( )] 0 ( ) ( [ ) ( ; / ) ( ) ( Cv s I Cs s v s I Cs s V v s sV C s I dt t Cdv t i i(t) + v(t) - I(s)

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Initial condition generator Inductors Ls s I s V s Z L L L ) ( ) ( ) ( 8 Lesson 14 initial condition i(0). )) 0 ( ) ( ( ) ( / ) ( ) ( i s sI L s V dt t Ldi t v ) 0 ( ) ( 1 ) ( ) 0 ( ) ( ) ( ) 0 ( ) ( ) ( Li s V Ls s I s i s I Ls s V li s LsI s V
Initial condition generator Inductors 9 Lesson 14 v(t) V(s) L V(s) L i(t) Li(0) Ls I(s) Ls Li(0) I(0)/s Ls I(s) Ls Ls i(0)/s ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 0 1 0 0 Li s V Ls s I s i s I Ls s V Li s LsI s V

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10 Lesson 14 =2 ICs given i L (0)=2 v C (0)=10 . i L (0)=2 V=10 L=1 R=2 C=1/5 IC=2 IC2=10/s Li L (0)=2 Ls=s R 5/s 10/s y(t) V=10 L=1 R C=1/5 KVL: V(s) = Z(s) Y(s) 10/s +2 +10/s = (s +2 +5/s) Y(s) or Y(s)=2/(s+2+5/s) = 2s/(s 2 +2s+5) y(t) y(t) v C (0)=10 i L (0)=2 Express in impedance form.
V=20 V=4 T=0 R=1 R=1/5 L=1/2 C=1 11 Lesson 14 Initial conditions v C (0)=16 i L 0)=4 Given V=20/s R=1 R=1/5 L=s/2 C=1/s IC/s=16/s L(IC)=2 For t> 0 y 2 (t) y 1 (t) t=0 Express in impedance form.

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12 Lesson 14 V=20/s R=1 R=1/5 L=s/2 C=1/s IC/s=16/s L(IC)=2 For t> 0 y 2 (t) y 1 (t) 2 2 5 6 1 2 5 4 5 1 1 2 2 1 2 1 1 ) ( ) ( ) ( : ) ( ) ( ) ( : s Y s s Y s Y s KVL s Y s Y s s Y KVL Solve 2x2 system for Y 1 (s) and Y 2 (s),
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