enee204Lecture24_25_Gomez

# enee204Lecture24_25_Gomez - Transfer Functions Steps: 1....

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1 Lecture 23 1 Welcome to ENEE 204 Basic Circuit Theory Lectures 24 and 25 Transfer functions and Transients Lecture 23 2 Transfer Functions Steps: 1. Look carefully at circuit and see if Norton/Thevenin equivalents are going to be useful. 2. Apply usual current or voltage division, remember to set sL Z sC Z R Z L C R = = = 1 3. H(s) is the ratio of the output signal and the input signal. 4. H(s=0) gives the particular solution. 5. Initial conditions (signal and derivative) ALWAYS refer to the initial voltage and current on the device. 6. Poles of H(s) yields characteristic equation. Lecture 23 3 Transient analysis – Example 1 v S (t)=V o u(t) C R Initial condition: v C (0 + ) =0 Thus, particular solution for v c : + Transfer function and pole: Characteristic equation: 10 sRC += Define u(t) - step function > < = + 0 t 1 0 t 0 ) ( - t u Replaces the switch and source. sRC R sC sC V V s H S C + = + = = 1 1 1 / 1 ˆ ˆ ) ( 1 ) 0 ( = s H 0 V v p = Lecture 23 4 Important tricks for initial conditions and particular soln. Given - step function > < = + 0 t 1 0 t 0 ) ( - t u R Z sL Z sC Z R L c = = = + + + ) 0 ( ; ) 0 ( ; 0 1 ) 0 ( During time of transition: = = + ω j s dt du u ; 1 ) 0 ( For particular solution, 0 ; 1 ) ( = = = j s dt du u R Z Z Z R L c = = = ) ( ; 0 ) ( ; ) (

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2 Lecture 23 5 v s (t)=u(t) + + _ R Initial conditions: V L (0 + )=1, V R (0 + )=0, i L (0 + ) = i R (0 + ) =0 Particular solution for v L : v p =0 Transient analysis – Example 2 Transfer function Characteristic equation: L 1+s =0 R sL R R sL sL V V s H S L + = + = = 1 1 ˆ ˆ ) ( 0 ) 0 ( H Steady state: Lecture 23 6 Transient analysis – Example 3 i S (t)=u(t) C R Initial conditions: V C (0 + ) = V R (0 + )= 0, i C (0 + ) =1, i R (0 + ) =0 Particular solution for i C : i p =0 + Transfer function Characteristic equation: 1 sC+ =0 R R sC sC I I s H S C 1 ˆ ˆ ) ( + = = 0 ) 0 ( = = s H Steady state: Particular solution for v C : v p =R Lecture 23 7 Transient analysis - Examples 4 S (t)=u(t) R L Initial conditions: V L (0 + )= V R (0 + )= R, i L =0, i R =1 Particular solution for i L : i p =1 + Transfer function: Characteristic equation: sL+R=0 sL R R I I s H S L + = = ˆ ˆ ) ( 1 ) 0 ( = = s H Steady state: Particular solution for v L : v p =0 Lecture 23 8 Transient analysis - Example 5 C L + _ R v S (t)=u(t) Initial conditions: V C (0 + )=0, V C (0 + )=0, V L (0 + )=1, V L (0 + )= R/L, Particular solution for v C : v p =1 * _ Transfer function Characteristic equation: 2 1+sRC+s LC=0 sCR LC s R sL sC sC V V s H S C + + = + + = = 2 1 1 1 1 ˆ ˆ ) ( 1 ) 0 ( = = s H Steady state:
3 Lecture 23 9 Transient analysis – Example 6 R 2 C 1 C 2 R 1 Initial conditions: V C1 (0 + )=0, V C1 (0 + )=1/C 1 , V C2 (0 + )=0, V C2 (0 + )= 0 , Part. sol. for v C1 : v p = R 1 +R 2 i S (t)=u(t) + + Transfer function Characteristic equation: 1 ) ( ) ( 2 2 2 1 1 1 2 1 2 1 2 2 2 1 2 1 + + + + + + = R C R C R C s R R C C s R R C R sR s H 0 1 ) ( 2 2 2 1 1 1 2 1 2 1 2 = + + + + R C R C R C s R R C C s 2 1 ) 0 ( R R s H + = = Steady state: Lecture 23 10 Transient analysis – Example 7 Initial conditions: i 2 (0 + )=0, V L2 (0 + )= 0, i 2 (0 + )= 0, V L2 (0 + )= R 1 /L 1 Particular solution for i 2 : i p =1/R 2 * L 2 R 2 R 1 L 1 v S (t)=u(t) + _ Transfer function Characteristic equation: 2 1 2 1 2 1 1 1 2 1 2 1 ) ( ) ( R R R L L R R L s L L s R s H + + + + = 0 ) ( 2 1 2 1 2 1 1 1 2 1 2 = + + + + R R R L L R R L s L L s Lecture 23 11 Transient analysis - Example 8 Initial conditions: V C (0 + )=0, V C (0 + )=1/(R 1 +R 2 )C,

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## This note was uploaded on 04/07/2008 for the course ENEE 204 taught by Professor Gomez during the Fall '04 term at Maryland.

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enee204Lecture24_25_Gomez - Transfer Functions Steps: 1....

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