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- 1. Natural and step response Ahsan Khawaja
- 2. Cont. inductor V L di dt • Where V= voltage between inductor in (volt) L= inductor in (henery) = rate of change of current flow in amper di = rate of change of time in second dt
- 3. dv Cont. capacitor • Mathematical relation . i C dv dt • Where i= current that in capacitor in (ampere) C= capacitance in (farad) d v = rated of change of voltage. d t = rate of change of time in second
- 4. First-Ordersources, resistors and an Circuits A circuit that contains only • inductor is called an RL circuit. • A circuit that contains only sources, resistors and a capacitor is called an RC circuit. • RL and RC circuits are called first-order circuits because their voltages and currents are described by first-order differential equations. R i i L vs – + – + vs R C
- 5. 6 Different First-Order Circuits There are six different STC circuits. These are listed below. • An inductor and a resistance (called RL Natural Response). • A capacitor and a resistance (called RC Natural Response). • An inductor and a Thévenin equivalent (called RL Step Response). • An inductor and a Norton equivalent (also called RL Step Response). • A capacitor and a Thévenin equivalent (called RC Step Response). • A capacitor and a Norton equivalent (also called RC Step Response). RX LX CX RX RX + vS LX - iS RX iS RX LX RX + vS CX - These are the simple, first-order cases. Many circuits can be reduced to one of these six cases. They all have solutions which are in similar forms. CX
- 6. 6 Different First-Order Circuits These are the simplest cases, so we handle them first. There are six different STC circuits. These are listed below. • An inductor and a resistance (called RL Natural Response). RX CX RX LX • A capacitor and a resistance (called RC Natural Response). RX • An inductor and a Thévenin equivalent + vS iS RX LX LX (called RL Step Response). • An inductor and a Norton equivalent (also called RL Step Response). • A capacitor and a Thévenin equivalent RX + vS iS RX CX (called RC Step Response). • A capacitor and a Norton equivalent (also called RC Step Response). These are the simple, first-order cases. CX They all have solutions which are in similar forms.
- 7. The Natural Response of a Circuit • The currents and voltages that arise when energy stored in an inductor or capacitor is suddenly released into a resistive circuit. • These “signals” are determined by the circuit itself, not by external sources!
- 8. Step Response • The sudden application of a DC voltage or current source is referred to as a “step”. • The step response consists of the voltages and currents that arise when energy is being absorbed by an inductor or capacitor.
- 9. Circuits for Natural Response • Energy is “stored” in an inductor (a) as an initial current. • Energy is “stored” in a capacitor (b) as an initial voltage.
- 10. General Configurations for RL • If the independent sources are equal to zero, the circuits simplify to
- 11. Natural Response of an RL Circuit • Consider the circuit shown. • Assume that the switch has been closed “for a long time”, and is “opened” at t=0.
- 12. What does “for a long time” Mean? • All of the currents and voltages have reached a constant (dc) value. • What is the voltage across the inductor just before the switch is opened?
- 13. Just before t = 0 • The voltage across the inductor is equal to zero. • There is no current in either resistor. • The current in the inductor is equal to IS.
- 14. Just after t = 0 • The current source and its parallel resistor R0 are disconnected from the rest of the circuit, and the inductor begins to release energy.
- 15. The Source-Free RL Circuit • A first-order RL circuit consists of a inductor L (or its equivalent) and a resistor (or its equivalent) By KVL vL L vR di 0 iR 0 dt Inductors law Ohms law di R i L dt i (t ) I0 e Rt/L
- 16. The Source-Free RC Circuit • A first-order circuit is characterized by a first-order differential equation. By KCL iR iC Ohms law • • 0 v C R dv dt Capacitor law Apply Kirchhoff’s laws to purely resistive circuit results in algebraic equations. Apply the laws to RC and RL circuits produces differential equations. 0
- 17. Natural Response of an RL Circuit • Consider the following circuit, for which the switch is closed for t < 0, and then opened at t = 0: t=0 Io Ro i L + R v – Notation: 0– is used to denote the time just prior to switching 0+ is used to denote the time immediately after switching • The current flowing in the inductor at t = 0– is Io
- 18. Solving for the Current (t 0) • For t > 0, the circuit reduces to i Io Ro L + R v – • Applying KVL to the LR circuit: • Solution: i (t ) i(0)e ( R / L ) t = I e-(R/L)t 0
- 19. Solving for the Voltage (t > 0) i (t ) ( R / L )t Ioe + Io Ro L R v – • Note that the voltage changes abruptly: v( 0 ) 0 for t 0, v(t) v( 0 ) iR I0R I o Re ( R /L )t
- 20. Time Constant • In the example, we found that i (t ) I oe ( R / L )t and v (t ) I o Re L • Define the time constant ( R / L )t (sec) R – At t = , the current has reduced to 1/e (~0.37) of its initial value. – At t = 5 , the current has reduced to less than 1% of its initial value.
- 21. The Source-Free RL Circuit Comparison between a RL and RC circuit A RL source-free circuit i(t ) I0 e t/ where A RC source-free circuit L R v (t ) V0 e t/ where RC
- 22. The Complete Solution R i (t ) Ie 0 L t ,t 0
- 23. The voltage drop across the resistor v iR R v I Re L 0 t ,t v(0 ) 0 v(0 ) I R 0 0 .
- 24. The Power Dissipated in the Resistor p vi v 2 iR 2 R p 2 I Re 0 2 R L t ,t 0
- 25. The Energy Delivered to the Resistor t w t pdx w 2 0 0 2 2 I R (1 e 0 L t ,w R L I Re 0 1 R 2 2 1 2 LI 2 0 x dx R L t ),t 0.
- 26. The Source-Free RL Circuit A general form representing a RL i(t ) where I0 e t/ L R • • • The time constant of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value. i(t) decays faster for small and slower for large . The general form is very similar to a RC source-free circuit.
- 27. The Source-Free RC Circuit • The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excitation. Time constant RC Decays more slowly Decays faster • • The time constant of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value. v decays faster for small and slower for large .
- 28. Natural Response Summary RL Circuit RC Circuit i + L C R v R – • Inductor current cannot change instantaneously i(0 ) i (t ) • i(0 ) i(0)e • time constant Capacitor voltage cannot change instantaneously v (0 ) t/ v (t ) L R • v (0 ) v (0)e time constant t/ RC
- 29. General Solution for Natural and Step Responses of RL and RC Circuits ( t t0 ) x (t ) xf [ x (t 0 ) x f ]e Final Value Time Constant Initial Value Determine the initial and final values of the variable of interest and the time constant of the circuit. Substitute into the given expression.
- 30. Example b R1 400kOhm V1 90 V a R3 20 Ohm J1 Key = Space + vC(t) R2 60 Ohm V2 40 V C 0.5uF - • What is the initial value of vC? • What is the final value of vC? • What is the time constant when the switch is in position b? • What is the expression for vC(t) when t>=0?
- 31. Initial Value of vC b R1 400kOhm a R3 20 Ohm J1 + V1 90 V Key = Space V60 + vC(0) C 0.5uF R2 60 Ohm V2 40 V - - • The capacitor looks like an open circuit, so the voltage @ C is the same as the voltage @ 60Ω. v C (0 ) 60 4 0V 20 3 0V 60
- 32. Final Value of vC b R1 400kOhm V1 90 V a R3 20 Ohm J1 Key = Space + vC(∞) R2 60 Ohm V2 40 V C 0.5uF - • After the switch is in position b for a long time, the capacitor will look like an open circuit again, and the voltage @ C is +90 Volts.
- 33. The time constant of the circuit when the switch is in position b R1 400kOhm V1 90 V b a R3 20 Ohm J1 Key = Space R2 60 Ohm C 0.5uF • The time constant τ = RC = (400kΩ)(0.5μF) • τ = 0.2 s V2 40 V
- 34. The expression for vC(t) for t>=0 t vC (t ) vC ( ) [ v C (0) v C ( )]e t vC (t ) 90 [ 30 vC (t ) 90 120 e 90]e 5t V 0.2
- 35. The expression for i(t) for t>=0 b R1 a 400kOhm R3 20 Ohm J1 Key = Space V1 90 V i(t) 30V R2 60 Ohm V2 40 V C 0.5uF + • Initial value of i is (90 - - 30)V/400kΩ = 300μA • Final value of i is 0 – the capacitor charges to +90 V and acts as an open circuit • The time constant is still τ = 0.2 s
- 36. The expression for i(t) (continued) t i (t ) i( ) [ i (0 ) i ( )]e t i (t ) i (t ) 0 [300 10 300 e 5t A 6 0] e 0.2
- 37. How long after the switch is in position b does the capacitor voltage equal 0? vC (t ) 120e e 5t 90 5t 120e 5t 0 90 90 120 5t ln 90 0 .2 8 7 6 8 120 t 0 .0 5 7 5 4 s 5 7 .5 4 m s
- 38. Plot vC(t)
- 39. Plot i(t)

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