EECE310_F05_Lecture_Notes__wk_of_10_3_05_

# EECE310_F05_Lecture_Notes__wk_of_10_3_05_ - College of...

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Unformatted text preview: College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis t=0 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Set-up the differential equation (D.EQ.) for a series resistor-inductor (RL) circuit: di (t ) =0 # V S - Ri (t ) - L dt di (t ) L First-Order Differential Equation: VS di (t ) R + i (t ) = dt L L EECE 310 Preston D. Frazier, Ph.D., P.E., PMP dt + Ri (t ) = VS College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The general solution for a first-order D.EQ. is the following: x(t ) = K1 + K 2 e - at - at i (t ) = K1 + K 2 e Amperes EECE 310 Preston D. Frazier, Ph.D., P.E., PMP VS di (t ) R + i (t ) = # dt L L College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The general solution for a first-order D.EQ. is the following: x(t ) = K1 + K 2 e - at i (t ) = K + K e Solve for the variables a, K1, and K2 in the general solution: Rt - L 1 2 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP Amperes # # College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The general solution for a first-order D.EQ. (continued): Plugging ## into # provides ### RK 2 e - L - Rt L RK1 RK 2 e + + L L Rt - L - Rt L VS = ### L Set t = and discover K1 for ### VS i (t ) = + K 2e R Amperes # # EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The general solution for a first-order D.EQ. (continued): Rt - VS L i (t ) = + K 2 e Amperes # # R Set t = 0 and find K2 for ## (i(t = 0) = 0) VS VS i (t ) = - e R R Rt - L Amperes # # EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Important Note about the General Solutions for Transient Circuits: The form of the particular solution for the general solution is based on the type of particular solution (e.g., constant-K1, exponential-K1e-st, sinusoidal- K1cos(zt) + K2sin(zt)) EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Schematic: RC RL Time Constant () = RC = L/R Important Note about Time Constants: Time constants dictate how long it takes a system to settle to steady-state. EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Set-up the differential equation (D.EQ.) for a parallel resistor-inductor-capacitor (RLC) circuit: 1 t dv (t ) v (t ) - [ v (t ) dt + i L (t 0 )] -C = 0 **** I S (t ) - dt R L t 0 1 v (t ) +[ R L dv (t ) t0 v (t ) dt + iL (t 0 )] +C dt = I S (t ) t d v(t ) d 1 t d d dv(t ) d ( ) + ( v(t )dt ) + iL (t0 ) + (C ) = I S (t ) dt R dt L t0 dt dt dt dt EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis D.EQ. for a parallel RLC circuit (continued): dI S (t ) d v(t ) 1 dv(t ) 1 + + v(t ) = C 2 dt R dt L dt 2 d v(t ) 1 dv(t ) 1 1 dI S (t ) v(t ) = + + **** 2 dt RC dt LC C dt 2 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Set-up the differential equation (D.EQ.) for a series resistor-inductor-capacitor (RLC) circuit: di (t ) 1 -[ V S (t ) - Ri (t ) - L dt C t t0 i (t ) dt + vC (t 0 )] = 0 # # # # 1 di (t ) +[ Ri (t ) + L dt C t t0 i (t ) dt + vC (t 0 )] = V S (t ) d d di(t ) d 1 t d d ( Ri(t )) + ( L ) + ( i(t )dt ) + vC (t0 ) = VS (t ) dt dt dt dt C t0 dt dt EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis D.EQ. for a series RLC circuit (continued): dVS (t ) di(t ) 1 d i(t ) +R + i(t ) = L 2 dt C dt dt 2 d i(t ) R di(t ) 1 1 dVS (t ) i(t ) = + + #### 2 dt L dt LC L dt 2 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The solution, to both the second-order differential equations for a parallel and series RLC networks, is stated as: d v(t ) 1 dv(t ) 1 1 dI S (t ) v(t ) = + + **** 2 dt RC dt LC C dt 2 d i(t ) R di(t ) 1 1 dVS (t ) i(t ) = + + #### 2 dt L dt LC L dt 2 d x(t ) dx(t ) 2 + 2 + 0 x(t ) = f (t ) 2 dt dt EECE 310 Preston D. Frazier, Ph.D., P.E., PMP 2 College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis RLC Configurations: Series Parallel R/(2L) (2RC)-1 0 (LC)-0.5 (LC)-0.5 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering To determine the general solution for a secondorder D.EQ.: Chapter 3-Transient Analysis 2 d x(t ) dx(t ) 2 + 2 + 0 x(t ) = f (t ) 2 dt dt 2 d xh (t ) dxh (t ) 2 + 2 + 0 xh (t ) = 0 2 dt dt d 2 x p (t ) dt 2 x g (t ) = xh (t ) + x p (t ) + 2 dx p (t ) dt 2 + 0 x p (t ) = f (t ) EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Ascertain the characteristic equation for the homogeneous solution of a second-order D.EQ.: Chapter 3-Transient Analysis d xh (t ) dxh (t ) 2 + 2 + 0 xh (t ) = 0 2 dt dt 2 ( s + 2s + 0 ) K1e = 0 2 2 st 2 Characteristic 2 Equation: s + 2s + 0 = 0 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Damping Condition > 0 < 0 = 0 Natural Response Equation xh(t) = K1e -t Type of Response Overdamped Underdamped Critically Damped -(- 2 -02 )t + K2e -(+ 2 -02 )t xh (t) = e [K1 cos( - t ) + K2 sin( - t )] 2 0 2 2 0 2 x h ( t ) = K 1 e - t + K 2 te - t EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The partial solution for second-order networks is commonly called the forced response. d 2 x p (t ) 2 dt dt Important Note about Particular Solutions for Transient Circuits: These solutions may be found directly for DC circuits by replacing the capacitors and inductors with open and short circuits, respectively. EECE 310 Preston D. Frazier, Ph.D., P.E., PMP + 2 dx p (t ) + 0 x p (t ) = f (t ) 2 ...
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