Electric CircuitsDiscussion 5

Keyi YuanTeaching assistant

Apr.24 2018

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Keyi Yuan, Electric Circuit (2018 Spring)

Contents

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• Discussion Review Part : Second-Order circuit

• Homework 6

Keyi Yuan, Electric Circuit (2018 Spring)

1. Second-Order Circuit

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Solving second order equation

𝑑𝑑2𝑣𝑣𝑑𝑑2𝑡𝑡

+𝑅𝑅𝐿𝐿𝑑𝑑𝑣𝑣𝑑𝑑𝑡𝑡

+1

LC𝑣𝑣 =

𝑣𝑣𝑠𝑠𝐿𝐿𝐿𝐿

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And there are THREE cases you should know

First solving the Eigen-function of and Eigenvalues of a second order formula

Case 1: Overdamped (α>ω0)

2 2 2 21 2o os sα α ω α α ω= − + − = − − −0

12RL LC

α ω= =

𝑠𝑠 = −𝑅𝑅2𝐿𝐿

±𝑅𝑅2𝐿𝐿

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−1𝐿𝐿𝐿𝐿

𝑣𝑣 𝑡𝑡 = 𝐴𝐴1𝑒𝑒𝑠𝑠1𝑡𝑡 + 𝐴𝐴2𝑒𝑒𝑠𝑠2𝑡𝑡

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Case 2: Critically Damped (α=ω0)

𝑣𝑣(𝑡𝑡) = 𝐴𝐴1𝑡𝑡 + 𝐴𝐴2 𝑒𝑒−𝛼𝛼𝑡𝑡

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Case 3: Underdamped (α<ω0)

𝑠𝑠1 = −𝛼𝛼 + 𝛼𝛼2 − 𝜔𝜔02 = −𝛼𝛼 + − 𝜔𝜔02 − 𝛼𝛼2 = −𝛼𝛼 + 𝑗𝑗𝜔𝜔𝑑𝑑

𝑠𝑠2 = −𝛼𝛼 − 𝛼𝛼2 − 𝜔𝜔02 = −𝛼𝛼 − − 𝜔𝜔02 − 𝛼𝛼2 = −𝛼𝛼 − 𝑗𝑗𝜔𝜔𝑑𝑑

where 𝑗𝑗 = −1 and 𝜔𝜔𝑑𝑑 = 𝜔𝜔02 − 𝛼𝛼2.

• ω0 is often called the undamped natural frequency.• ωd is called the damped natural frequency.

The natural response

𝑣𝑣 𝑡𝑡 = 𝐴𝐴1𝑒𝑒𝑠𝑠1𝑡𝑡 + 𝐴𝐴2𝑒𝑒𝑠𝑠2𝑡𝑡becomes

𝑣𝑣 𝑡𝑡 = 𝑒𝑒−𝛼𝛼𝑡𝑡 𝐵𝐵1cos𝜔𝜔𝑑𝑑𝑡𝑡 + 𝐵𝐵2sin𝜔𝜔𝑑𝑑𝑡𝑡7

𝑠𝑠 = −𝛼𝛼 ± 𝛼𝛼2 − 𝜔𝜔02

Recall Euler’s formula

Case 3: Underdamped (α<ω0)

• Exponential 𝑒𝑒−𝛼𝛼𝑡𝑡 * Sine/Cosine term• Exponentially damped, time constant =

1/𝛼𝛼• Oscillatory, period 𝑇𝑇 = 2𝜋𝜋

𝜔𝜔𝑑𝑑

v(𝑡𝑡) = 𝑒𝑒−𝛼𝛼𝑡𝑡 𝐵𝐵1cos𝜔𝜔𝑑𝑑𝑡𝑡 + 𝐵𝐵2sin𝜔𝜔𝑑𝑑𝑡𝑡

Properties of Series RLC Network

• Behavior captured by damping• Gradual loss of the initial stored

energy• 𝛼𝛼 determines the rate of damping

• 𝛼𝛼 > 𝜔𝜔0 (i.e., 𝑅𝑅 > 2 𝐿𝐿𝐶𝐶

), overdamped

• 𝛼𝛼 = 𝜔𝜔0 (i.e., 𝑅𝑅 = 2 𝐿𝐿𝐶𝐶

), critically damped𝑣𝑣(𝑡𝑡) = 𝐴𝐴1𝑡𝑡 + 𝐴𝐴2 𝑒𝑒−𝛼𝛼𝑡𝑡

• 𝛼𝛼 < 𝜔𝜔0 (i.e., 𝑅𝑅 < 2 𝐿𝐿𝐶𝐶

), underdamped

𝑣𝑣 𝑡𝑡 = 𝑒𝑒−𝛼𝛼𝑡𝑡 𝐵𝐵1cos𝜔𝜔𝑑𝑑𝑡𝑡 + 𝐵𝐵2sin𝜔𝜔𝑑𝑑𝑡𝑡

𝑣𝑣 𝑡𝑡 = 𝐴𝐴1𝑒𝑒𝑠𝑠1𝑡𝑡 + 𝐴𝐴2𝑒𝑒𝑠𝑠2𝑡𝑡

Series vs. Parallel (Source-Free RLC Network)

• Series

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• Parallel

𝑣𝑣 𝑡𝑡 = 𝐴𝐴1𝑒𝑒𝑠𝑠1𝑡𝑡 + 𝐴𝐴2𝑒𝑒𝑠𝑠2𝑡𝑡

𝑣𝑣(𝑡𝑡) = 𝐴𝐴1𝑡𝑡 + 𝐴𝐴2 𝑒𝑒−𝛼𝛼𝑡𝑡

v(𝑡𝑡) = 𝑒𝑒−𝛼𝛼𝑡𝑡 𝐵𝐵1cos𝜔𝜔𝑑𝑑𝑡𝑡 + 𝐵𝐵2sin𝜔𝜔𝑑𝑑𝑡𝑡

( ) 1 21 2

s t s tv t A e A e= +

( ) ( )2 1tv t A At e α−= +

( ) ( )1 2cos sintd dv t e A t A tα ω ω−= +

Finding Initial and Final Values

• Working on second order system is harder than first order in terms of finding initial and final conditions.

• You need to know the derivatives, dv/dt and di/dtas well.

• Capacitor voltage and inductor current are always continuous.

• For capacitor, 𝑣𝑣 0+ = 𝑣𝑣 0− ;• For inductor, 𝑖𝑖 0+ = 𝑖𝑖 0− .

General Second-Order Circuits

• The principles of the approach to solving the series and parallel forms of RLC circuits can be applied to general second order circuits, by taking the following four steps:1. First determine the initial conditions, x(0) and dx(0)/dt.2. Turn off the independent sources and find the form of the

transient response by applying KVL and KCL.• Depending on the damping found, the unknown constants will be found.

3. We obtain the steady-state response as:

where x(∞) is the final value of x obtained in step 1.4. The total response = transient response + steady-state response.

( ) ( )ssx t x= ∞

( ) ( ) ( )t ssx t x t x t= +12

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Typo:[(𝐷𝐷1cos 𝜔𝜔𝑑𝑑𝑡𝑡 +𝐷𝐷2 sin 𝜔𝜔𝑑𝑑𝑡𝑡 ) 𝑒𝑒−𝛼𝛼𝑡𝑡 +𝑥𝑥(∞)] 𝑢𝑢(t)

2. Homework 6

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Question 1

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Question 2

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Question 2

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Question 3

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Question 3

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Question 4

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Question 4

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Question 5

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Question 5

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Question 6

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Question 6

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Question 7

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Question 7

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Question 8

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Question 8

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Question 8

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Question 8

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Question 9

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Question 9

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Question 9

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Question 9

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Question 10

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Question 10

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Question 10

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Question 10

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Question 10

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Question 10

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Question 10

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