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EKT101 Electric Circuit Theory. Chapter 5 First-Order and Second Circuits. First-Order and Second Circuits Chapter 5. 5.1 Natural response of RL and RC Circuit 5.2 Force response of RL and RC Circuit 5.3 Solution of natural response and force response in RL and RC Circuit
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EKT101Electric Circuit Theory Chapter 5 First-Order and Second Circuits
First-Order and Second CircuitsChapter 5 5.1 Natural response of RL and RC Circuit 5.2 Force response of RL and RC Circuit 5.3 Solution of natural response and force response in RL and RC Circuit 5.4 Natural and force response in series RLC Circuit 5.5 Natural and force response in parallel RLC Circuit
By KCL Ohms law Capacitor law 5.1 Natural response of RL and RC circuit (1) • A first-order circuit is characterized by a first-order differential equation. • Apply Kirchhoff’s laws to purely resistive circuit results inalgebraic equations. • Apply the laws to RC and RL circuits produces differential equations.
Decays more slowly Time constant Decays faster 5.1 Natural response of RL and RC circuit (2) • 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. • 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 t and slower for large t.
where 5.1 Natural response of RL and RC circuit (3) DC source disconnected The key to working with a source-free RC circuit is finding: • The initial voltage v(0) = V0 across the capacitor. • The time constant = RC.
5.1 Natural response of RL and RC circuit (4) Example 1 Refer to the circuit below, determine vC, vx, and io for t ≥ 0. Assume that vC(0) = 30 V. • Please refer to lecture or textbook for more detail elaboration. • Answer: vC= 30e–0.25t V ; vx= 10e–0.25t ; io= –2.5e–0.25tA
Solution 1 vC(0) = 30 V. = vC = 30e–0.25t V vx = 10e–0.25t io = –2.5e–0.25t A
5.1 Natural response of RL and RC circuit (5) Example 2 The switch in circuit below is opened at t = 0, find v(t) for t ≥ 0. • Please refer to lecture or textbook for more detail elaboration. • Answer: V(t) = 8e–2t V
Solution 2 1/6 F V(t) = 8e–2t V
By KVL Inductors law Ohms law 5.1 Natural response of RL and RC circuit (6) • A first-order RL circuit consists of a inductor L (or its equivalent) and a resistor (or its equivalent)
A general form representing a RL where 5.1 Natural response of RL and RC circuit (7) • 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 t and slower for large t. • The general form is very similar to a RC source-free circuit.
A RL source-free circuit A RC source-free circuit where where 5.1 Natural response of RL and RC circuit(8) Comparison between a RL and RC circuit
where 5.1 Natural response of RL and RC circuit(9) The key to working with a source-free RL circuit is finding: • The initial voltage i(0) = I0 through the inductor. • The time constant = L/R.
5.1 Natural response of RL and RC circuit(10) Example 3 Find i and vx in the circuit. Assume that i(0) = 5 A. • Answer: i(t) = 5e–53t A
3 Solution 3find rth 1 5 ? ? ?
5.1 Natural response of RL and RC circuit(11) Example 4 For the circuit, find i(t) for t > 0. • Please refer to lecture or textbook for more detail elaboration. • Answer: i(t) = 2e–2t A
t=0, SC i(0) = current division ??
Unit-Step Function (1) • The unit step functionu(t) is 0 for negative values of t and 1 for positive values of t.
Unit-Step Function (2) • voltage source. • for current source: Represent an abrupt change for:
5.2 Force response of RL and RC Circuit (1) • The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source. • Initial condition: • v(0-) = v(0+) = V0 • Applying KCL, • or • Where u(t) is the unit-step function
Final value at t -> ∞ Initial value at t = 0 Source-free Response 5.3 Solution of natural and force response in RL and RC Circuit • Integrating both sides and considering the initial conditions, the solution of the equation is: Complete Response = Natural response + Forced Response (stored energy) (independent source) = V0e–t/τ + Vs(1–e–t/τ)
5.2 Force response of RC Circuit(3) Three steps to find out the step response of an RC circuit: • The initial capacitor voltage v(0). • The final capacitor voltage v() — DC voltage across C. • The time constant. Note: The above method is a short-cut method. You may also determine the solution by setting up the circuit formula directly using KCL, KVL , ohms law, capacitor and inductor VI laws.
5.2 Force response of RC Circuit(4) Example 5 Find v(t) for t > 0 in the circuit in below. Assume the switch has been open for a long time and is closed at t = 0. Calculate v(t) at t = 0.5. • Please refer to lecture or textbook for more detail elaboration. • Answer: and v(0.5) = 0.5182V
Solution Ex 5 • 1 -Theinitial capacitor voltage v(0). • 2- final capacitor voltage v() • Use KCL at nod to get v(∞) • 3- time constant. • Use Rth=RC
5.2 Force response of RL Circuit(1) • The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source. • Initial current • i(0-) = i(0+) = Io • Final inductor current • i(∞) = Vs/R • Time constant t = L/R
5.2 Force response of RL Circuit(2) Three steps to find out the step response of an RL circuit: • The initial inductor current i(0) at t = 0+. • The final inductor current i(). • The time constant. Note: The above method is a short-cut method. You may also determine the solution by setting up the circuit formula directly using KCL, KVL , ohms law, capacitor and inductor VI laws.
5.2 Force response of RL Circuit(3) Example 6 The switch in the circuit shown below has been closed for a long time. It opens at t = 0. Find i(t) for t > 0. • Please refer to lecture or textbook for more detail elaboration. • Answer:
Apply source transformation Example 6 (solution) 30V 30 2A
5.4 Natural and force response in series RLC Circuit Second order circuit
RL T-config RC Pi-config RLC Series RLC Parallel Examples of Second Order RLC circuits (1) What is a 2nd order circuit? A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements.
The 2nd order of expression Source-Free Series RLC Circuits (1) • The solution of the source-free series RLC circuit is called as the natural response of the circuit. • The circuit is excited by the energy initially stored in the capacitor and inductor. How to derive and how to solve?
Source-Free Series RLC Circuits (2) • At t=0, • So, • Eliminate integral, differentiate to t, rearrange
where General 2nd order Form Source-Free Series RLC Circuits (3) There are three possible solutions for the following 2nd order differential equation: => The types of solutions for i(t) depend on the relative values of a and w.
1. If a > wo, over-damped case where 2. If a = wo, critical damped case where 3.If a < wo, under-damped case where Source-Free Series RLC Circuits (4) There are three possible solutions for the following 2nd order differential equation:
Source-Free Series RLC Circuits (5) Example 1 If R = 10 Ω, L = 5 H, and C = 2 mF in 8.8, find α, ω0, s1 and s2. What type of natural response will the circuit have? • Please refer to lecture or textbook for more detail elaboration. • Answer: underdamped
Source-Free Series RLC Circuits (6) Example 2 The circuit shown below has reached steady state at t = 0-. If the make-before-break switch moves to position b at t = 0, calculate i(t) for t > 0. • Please refer to lecture or textbook for more detail elaboration. • Answer: i(t) = e–2.5t[5cos1.6583t – 7.538sin1.6583t] A
t>0 under-damped case
The 2nd order of expression Source-Free Parallel RLC Circuits (1) Let v(0) = V0 Apply KCL to the top node: Taking the derivative with respect to t and dividing by C
1. If a > wo, over-damped case where 2. If a = wo, critical damped case where 3.If a < wo, under-damped case where Source-Free Parallel RLC Circuits (2) There are three possible solutions for the following 2nd order differential equation:
Source-Free Parallel RLC Circuits (3) Example 3 Refer to the circuit shown below. Find v(t) for t > 0. • Please refer to lecture or textbook for more detail elaboration. • Answer: v(t) = 66.67(e–10t – e–2.5t) V
The 2nd order of expression Step-Response Series RLC Circuits (1) • The step response is obtained by the sudden application of a dc source. • The above equation has the same form as the equation for source-free series RLC circuit. • The same coefficients (important in determining the frequency parameters). • Different circuit variable in the equation.
(over-damped) (critically damped) (under-damped) Step-Response Series RLC Circuits (2) The solution of the equation should have two components: the transient response vt(t) & the steady-state response vss(t): • The transient response vt is the same as that for source-free case • The steady-state response is the final value of v(t). • vss(t) = v(∞) • The values of A1 and A2 are obtained from the initial conditions: • v(0) and dv(0)/dt.
Step-Response Series RLC Circuits (3) Example 4 Having been in position for a long time, the switch in the circuit below is moved to position b at t = 0. Find v(t) and vR(t) for t > 0. • Please refer to lecture or textbook for more detail elaboration. • Answer: v(t) = {10 + [(–2cos3.464t – 1.1547sin3.464t)e–2t]} V • vR(t)= [2.31sin3.464t]e–2t V
Step-Response Parallel RLC Circuits (1) • The step response is obtained by the sudden application of a dc source. The 2nd order of expression • It has the same form as the equation for source-free parallel RLC circuit. • The same coefficients (important in determining the frequency parameters). • Different circuit variable in the equation.