Lecture 12 - (1-exp(t 1-t τ v 1 0 t 1 v 1 0 V in =V 1 u(t-t 1 t t V R =V 1 u(t-t 1)exp(t 1-t τ 0 Spring 2011 ECE 2100 Cornell Univ A Molnar 4

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Spring 2011 ECE 2100, Cornell Univ. A. Molnar 1 Lecture 13 Driven RC circuits Step response
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Spring 2011 ECE 2100, Cornell Univ. A. Molnar 2 Driven RC circuit ( ) ( ) ( ) () ( ) = + = RC t V t V t V t V t V out outn outd outn out exp 0 () () ( ) () () () RC t V RC t V C t I dt dV R t V t V t I in out out out in + = = = + V in - RC, but now with an input R C + V out - Solution has two parts: natural + driven Driven: look at a step input: V 1 V in t 0 t 1 ( ) ( ) () () 1 0 0 0 1 1 = > = < = t u t t u t t t u V t V in
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Spring 2011 ECE 2100, Cornell Univ. A. Molnar 3 Step response: RC circuit ( ) ( ) ( ) () () () () () () () () () ( ) = = = = = = = = > = = < = + = τ t t V t t u t V t t V V V t V t V t t V t t t V V RC t V dt dV dt dV dt dV t V V t V t t t V t V t t RC t V RC t t u V RC t V dt dV out R in out R R R out R out R R out R out out 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 exp 1 exp exp exp 0 , 0 + V in - Driven response: Vout(0)=0 R C + V out - First solve V R (t), then V out (t) v 1 t V out =V 1 u(t-t 1
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Unformatted text preview: )(1-exp[(t 1 -t)/ τ ]) v 1 0 t 1 v 1 0 V in =V 1 u(t-t 1 ) t t V R =V 1 u(t-t 1 )exp[(t 1 -t)/ τ ] 0 Spring 2011 ECE 2100, Cornell Univ. A. Molnar 4 Pulse Response RC circuit + V in-R C + V out-0.5 1 1.5-1.5-1-0.5 0.5 0.5 1 1.5 Can model a pulse as superposition of step functions: response is superposition of step responses Step 1 Step 2 Superposition ( ) ( ) ( ) ( ) ( ) ( ) ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − − − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − − − = − − − = τ t t t t u t t t t u t V t t u t t u t V out in 2 2 1 1 2 1 exp 1 exp 1...
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This note was uploaded on 06/10/2011 for the course ECE 2100 taught by Professor Kelley/seyler during the Spring '05 term at Cornell University (Engineering School).

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Lecture 12 - (1-exp(t 1-t τ v 1 0 t 1 v 1 0 V in =V 1 u(t-t 1 t t V R =V 1 u(t-t 1)exp(t 1-t τ 0 Spring 2011 ECE 2100 Cornell Univ A Molnar 4

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