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Page 1 1 Second-Order Circuits • Introduction • Finding Initial and Final Values • The Source-Free Series RLC Circuit • The Source-Free Parallel RLC Circuit • Step Response of a Series RLC Circuit • Step Response of a Parallel RLC Circuit • General Second-Order Circuits • Duality • Applications Introduction • A second-order circuit is characterized by a second-order differential equation. • It consists of resistors and the equivalent of two energy storage elements. Page 2 1 Second-Order Circuits • Introduction • Finding Initial and Final Values • The Source-Free Series RLC Circuit • The Source-Free Parallel RLC Circuit • Step Response of a Series RLC Circuit • Step Response of a Parallel RLC Circuit • General Second-Order Circuits • Duality • Applications Introduction • A second-order circuit is characterized by a second-order differential equation. • It consists of resistors and the equivalent of two energy storage elements. 2 Finding Initial and Final Values • v(0), i(0), dv(0)/dt, di(0)/dt, v( ? ), and i( ? ) • Two key points: – v and i are defined according to the passive sign convention. – Continuity properties: • Capacitor voltage: (V S -like) • Inductor current: (I S -like) ) 0 ( ) 0 ( ) 0 ( ) 0 ( ? ? ? ? ? ? L L C C i i v v v i + _ Example ). ( , ) ( (c) , ) 0 ( , ) 0 ( (b) ), 0 ( , ) 0 ( (a) Find : Q ? ? ? ? ? ? v i dt dv dt di v i ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? V 4 ) 0 ( ) 0 ( A 2 ) 0 ( ) 0 ( V 4 ) 0 ( 2 ) 0 ( A 2 2 4 12 ) 0 ( 0. for t analysis dc Apply (a) : Sol v v i i i v i Page 3 1 Second-Order Circuits • Introduction • Finding Initial and Final Values • The Source-Free Series RLC Circuit • The Source-Free Parallel RLC Circuit • Step Response of a Series RLC Circuit • Step Response of a Parallel RLC Circuit • General Second-Order Circuits • Duality • Applications Introduction • A second-order circuit is characterized by a second-order differential equation. • It consists of resistors and the equivalent of two energy storage elements. 2 Finding Initial and Final Values • v(0), i(0), dv(0)/dt, di(0)/dt, v( ? ), and i( ? ) • Two key points: – v and i are defined according to the passive sign convention. – Continuity properties: • Capacitor voltage: (V S -like) • Inductor current: (I S -like) ) 0 ( ) 0 ( ) 0 ( ) 0 ( ? ? ? ? ? ? L L C C i i v v v i + _ Example ). ( , ) ( (c) , ) 0 ( , ) 0 ( (b) ), 0 ( , ) 0 ( (a) Find : Q ? ? ? ? ? ? v i dt dv dt di v i ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? V 4 ) 0 ( ) 0 ( A 2 ) 0 ( ) 0 ( V 4 ) 0 ( 2 ) 0 ( A 2 2 4 12 ) 0 ( 0. for t analysis dc Apply (a) : Sol v v i i i v i 3 C o n t ’ d V 12 ) ( A 0 ) ( . 0 for analysis dc Apply (c) : Sol ? ? ? ? ? ? v i t C o n t ’ d V/s 20 0 0 A 2 0 0 find easily can We case. in this source current a as treated be can inductor The abruptly. change cannot current inductor the Since : 0 find To (b) : Sol ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? C i dt dv i i C i dt dv i dt dv C dt dv C C C C ) ( ) ( ) ( ) ( ) ( ? t = 0 + 2 A Page 4 1 Second-Order Circuits • Introduction • Finding Initial and Final Values • The Source-Free Series RLC Circuit • The Source-Free Parallel RLC Circuit • Step Response of a Series RLC Circuit • Step Response of a Parallel RLC Circuit • General Second-Order Circuits • Duality • Applications Introduction • A second-order circuit is characterized by a second-order differential equation. • It consists of resistors and the equivalent of two energy storage elements. 2 Finding Initial and Final Values • v(0), i(0), dv(0)/dt, di(0)/dt, v( ? ), and i( ? ) • Two key points: – v and i are defined according to the passive sign convention. – Continuity properties: • Capacitor voltage: (V S -like) • Inductor current: (I S -like) ) 0 ( ) 0 ( ) 0 ( ) 0 ( ? ? ? ? ? ? L L C C i i v v v i + _ Example ). ( , ) ( (c) , ) 0 ( , ) 0 ( (b) ), 0 ( , ) 0 ( (a) Find : Q ? ? ? ? ? ? v i dt dv dt di v i ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? V 4 ) 0 ( ) 0 ( A 2 ) 0 ( ) 0 ( V 4 ) 0 ( 2 ) 0 ( A 2 2 4 12 ) 0 ( 0. for t analysis dc Apply (a) : Sol v v i i i v i 3 C o n t ’ d V 12 ) ( A 0 ) ( . 0 for analysis dc Apply (c) : Sol ? ? ? ? ? ? v i t C o n t ’ d V/s 20 0 0 A 2 0 0 find easily can We case. in this source current a as treated be can inductor The abruptly. change cannot current inductor the Since : 0 find To (b) : Sol ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? C i dt dv i i C i dt dv i dt dv C dt dv C C C C ) ( ) ( ) ( ) ( ) ( ? t = 0 + 2 A 4 C o n t ’ d t = 0 + A/s 0 25 0 0 0 0 have we Thus 0 4 8 12 0 0 0 0 0 4 12 gives KVL applying , 0 obtain To case. in this source voltage a as treated be can capacitor The abruptly. change cannot voltage capacitor the Since 0 find To (b) : Sol ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? . ) ( ) ( ) ( ) ( ) ( ) ( ) ( : ) ( L v dt di v v v i v L v dt di v dt di L dt di L L C L L L L ? The Source-Free Series RLC Circuit ? ? ? ? ? ? ) 4 ( 1 ) 0 ( 0 ) 0 ( ) 0 ( gives (2) and (1) required. is ) 0 ( (3), solve To (3) 0 (2) 0 1 gives KVL Applying (1b) 1 0 (1a) 0 : conditions initial Assumed 0 0 0 2 2 0 0 0 V RI L dt di V dt di L Ri dt di LC i dt di L R dt i d idt C dt di L Ri V idt C v I i t C ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 0 0 0 2 2 1 ) 0 ( 0 : conditions Initial 0 V RI L dt di I i LC i dt di L R dt i d Page 5 1 Second-Order Circuits • Introduction • Finding Initial and Final Values • The Source-Free Series RLC Circuit • The Source-Free Parallel RLC Circuit • Step Response of a Series RLC Circuit • Step Response of a Parallel RLC Circuit • General Second-Order Circuits • Duality • Applications Introduction • A second-order circuit is characterized by a second-order differential equation. • It consists of resistors and the equivalent of two energy storage elements. 2 Finding Initial and Final Values • v(0), i(0), dv(0)/dt, di(0)/dt, v( ? ), and i( ? ) • Two key points: – v and i are defined according to the passive sign convention. – Continuity properties: • Capacitor voltage: (V S -like) • Inductor current: (I S -like) ) 0 ( ) 0 ( ) 0 ( ) 0 ( ? ? ? ? ? ? L L C C i i v v v i + _ Example ). ( , ) ( (c) , ) 0 ( , ) 0 ( (b) ), 0 ( , ) 0 ( (a) Find : Q ? ? ? ? ? ? v i dt dv dt di v i ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? V 4 ) 0 ( ) 0 ( A 2 ) 0 ( ) 0 ( V 4 ) 0 ( 2 ) 0 ( A 2 2 4 12 ) 0 ( 0. for t analysis dc Apply (a) : Sol v v i i i v i 3 C o n t ’ d V 12 ) ( A 0 ) ( . 0 for analysis dc Apply (c) : Sol ? ? ? ? ? ? v i t C o n t ’ d V/s 20 0 0 A 2 0 0 find easily can We case. in this source current a as treated be can inductor The abruptly. change cannot current inductor the Since : 0 find To (b) : Sol ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? C i dt dv i i C i dt dv i dt dv C dt dv C C C C ) ( ) ( ) ( ) ( ) ( ? t = 0 + 2 A 4 C o n t ’ d t = 0 + A/s 0 25 0 0 0 0 have we Thus 0 4 8 12 0 0 0 0 0 4 12 gives KVL applying , 0 obtain To case. in this source voltage a as treated be can capacitor The abruptly. change cannot voltage capacitor the Since 0 find To (b) : Sol ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? . ) ( ) ( ) ( ) ( ) ( ) ( ) ( : ) ( L v dt di v v v i v L v dt di v dt di L dt di L L C L L L L ? The Source-Free Series RLC Circuit ? ? ? ? ? ? ) 4 ( 1 ) 0 ( 0 ) 0 ( ) 0 ( gives (2) and (1) required. is ) 0 ( (3), solve To (3) 0 (2) 0 1 gives KVL Applying (1b) 1 0 (1a) 0 : conditions initial Assumed 0 0 0 2 2 0 0 0 V RI L dt di V dt di L Ri dt di LC i dt di L R dt i d idt C dt di L Ri V idt C v I i t C ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? 0 0 0 2 2 1 ) 0 ( 0 : conditions Initial 0 V RI L dt di I i LC i dt di L R dt i d 5 C o n t ’ d ? ? ? ? 0 1 0 1 0 constants. are and : Let 1 ) 0 ( 0 : conditions Initial 0 2 2 2 0 0 0 2 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? LC s L R s LC s L R s Ae e LC A se L AR e As s A Ae i V RI L dt di I i LC i dt di L R dt i d st st st st st ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? LC L R s s LC L R L R s LC L R L R s 1 2 where 1 2 2 1 2 2 0 2 0 2 2 2 0 2 1 2 2 2 1 ? ? ? ? ? ? ? ? Characteristic equation Natural frequencies Damping factor Resonant frequency (or undamped natural frequency) Summary . conditions initial the from determined are and where ) ( : solution general A , : ) (if solutions Two 2 1 2 1 2 2 1 1 2 1 2 1 2 1 A A e A e A t i e A i e A i s s t s t s t s t s ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? LC L R s s s s 1 2 where 0 2 : equation stic Characteri 0 2 0 2 2 2 0 2 1 2 0 2 ? ? ? ? ? ? ? ? ? ? • Three cases discussed – Overdamped case (distinct real roots) : ?> ? 0 – Critically damped case (repeated real root) : ?= ? 0 – Underdamped case (complex-conjugate roots): ?< ? 0Read More
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