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MODELING OF ELECTRICAL SYSTEMS 1st METHOD Current Law (Kirchoff) Voltage Law 2nd METHOD Current Law Magnetic Energy, Electrical Energy, Virtual Work: Application of Lagrange’s Equation 2nd method is used in this course. FUNDAMENTAL ELEMENTS OF ELECTRICAL SYSTEMS Passive Elements L x Analogy Between Mechanical and Electrical Systems . .. x Velocity (m/s) x Acceleration (m/s2) Generalized Coordinates q Charge (Coulomb) + R C Displacement (m) Active Elements i Current (Amper) di dt m Mass (kg) k c Spring Stiffness Damping Constant (N/m) (Ns/m) f mx 1 E1 mx 2 2 f kx 1 E 2 kx2 2 L 1/C Inductance C:Capacitance (Farad) (Henry) di dt Generalized Charges VL Lagrange’s Equation 1 E1 Lq 2 2 Op-Amp f Force (N) f cx W cx x R Resistance (Ohm) W f x V Voltage (Volt) 1 V Ri q C W Vq 1 2 W Rq q E2 q 2C V x Analogy Between Mechanical and Electrical Systems . .. Displacement (m) x Velocity (m/s) x Acceleration (m/s2) m Mass (kg) k c Spring Constant Damping Constant (N/m) (Ns/m) f Force (N) θ . θ .. θ IG Kr Cr Angular Dispalement Angular Velocity Angular Acceleration Mass moment of inertia Rotational spring constant Rotational damping constant (Nm/rad) Nm/(rad/s) M Moment (Nm) V Voltage (Volt) δW (rad) (rad/s) (rad/s2) (kg-m2) q Charge (Coulomb) i Current (Amper) L 1/C Inductance (Henry) C:Capacitance (Farad) R Resistance (Ohm) E2 E1 E1 E2 δW 1 2 q 2C W Rq q 1 E1 Lq 2 2 E2 W Vq Current through R1= q q 2 Modeling of Electrical Systems: V1: Input q and q2 : Generalized charges 1 2 1 2 E1 Lq E2 q2 2 2C W V1q R 1 (q q 2 )(q q 2 ) R 2q 2 q 2 Example 3.1: L + V1 C q 2 q - R1 R2 ( V1 R 1q R 1q 2 )q ( R 1q R 1q 2 R 2q 2 )q 2 Qq Qq2 d (E1 E 2 ) (E1 E 2 ) Qq dt q q V1 R 1q R 1q 2 Lq d (E1 E 2 ) (E1 E 2 ) Qq 2 dt q 2 q 2 1 q 2 R 1q ( R 1 R 2 )q 2 C R 1 q 0 0 q V1 0 1 R 1 R 2 q 2 q 0 C 2 L=3.4 mH, C=286 µF, R1=3.2 Ω, R2=4.5 Ω R 1 L 0 q 0 0 q 2 R1 Ls 2 R 1s R 1s R 1s 1 0 ( R 1 R 2 )s C D(s)=0.02618s3+26.288s2+11188.81s=0 Eigenvalues: 0, -502.06±418.70i (ξ=0.768) Example 3.2 C1 q 4 C2 R1 q 1 V1 q 3 q 3 q 2 q f No current flow through Op-Amp q 1 q 2 q 3 q 4 q 4 q 1 q 2 q 3 q 4 q 3 q f R3 E1 0 E2 + R2 q f q 1 q 2 q 3 q 3 q 1 q 2 1 1 2 (q 1 q 2 q 3 ) 2 q3 2C1 2C2 W V1q1 V2 (q1 q 2 ) R 1q 1q1 V2 R 2q 2q 2 R 3q 3q 3 Input : V1 Generalized charges: q1, q2, q3 1 (q1 q 2 q 3 ) V1 V2 R 1q 1 C1 1 (q1 q 2 q 3 ) V2 R 2q 2 C1 1 1 (q1 q 2 q 3 ) q 3 R 3q 3 For Op-Amp : V+=V-=0 C1 C2 1 1 1 C C C 1 1 q1 R 1 0 R 3 q 1 1 V1 1 1 0 R q 1 R q 2 0 2 3 2 C1 C1 C1 0 R 3 q 3 1 0 1 1 1 q 3 0 C1 C1 C1 C 2 R 3q 3 V2 R1 0 0 0 R2 0 1 C R 3 q 1 1 1 R 3 q 2 C1 R 3 q 3 1 C1 R 1s 1 C1 1 C1 1 C1 1 C1 R 2s 1 C1 1 C1 1 C1 1 C1 1 C1 1 C1 q V 1 1 1 q 0 2 C1 1 1 q 3 0 C1 C 2 1 C1 1 R 3s 0 C1 1 1 R 3s C1 C 2 R 3s R1=15.9 kΩ, R2=837 Ω, R3=318 kΩ, C1=C2=0.005 µF Eigenvalues: 0, -628.93±12561.76i (ξ=0.05) GAIN CIRCUIT R2 q q R1 V1 + q V1 R 1q 0 0 R 2q V2 V2 V1 R1 R2 V1 V2 R1 V2 V2 R2 V1 R1 R1 V1 q R2 R2 V1 R1 R q R + V2' + V2 R V2 2 V1 R1 V2 R R R R V2 2 V1 2 V1 R R R1 R1 INTEGRAL CIRCUIT C q R q V1 + V2 V2 1 V1dt RC Rq V1 V1 Rq 0 0 1 q V2 C q 1 V1 R q q dt V2 1 V1dt R 1 V1dt RC Negative sign can be eliminated bu adding an extra gain circuit with gain=1 q 3 R R Vc (V1 V2 V3 ) R V3 R V2 V1 q 1 q V1 Rq 1 0 q 1 + Vc V V V R 1 2 3 Vc R R R V2 Vc V1 V2 V3 R R q 1 V V2 , q 3 3 R R q q 1 q 2 q 3 0 Rq Vc q 2 R V1 V1 R Vc V1 V2 R + R q + V1 Vc q 2 + V2 0 q R Vc 0 q R Vc q q 1 q 2 R V1 Rq 1 0 R V q 1 1 R V2 q 2 R 0 q 2 V2 R V V Vc R 1 2 R R Vc V1 V2 Vc R2 V2 K P V1 K I V1dt K D R1 + R dV1 dt R2 1 K D R 4C4 KP KI R1 R 3C 3 C3 R R3 + R + R4 V1 C4 + R V2 PID Control Circuit PID circuit is used frequently in Control Systems.