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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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1st Order Ckts: Step-by-Step

- This Approach Relies On The Known Form Of The Solution But Finds The ODE Parameters Using Basic Circuit Analysis Tools
- This Method Eliminates the Need For The Determination Of The Differential Equation Model
- Most Useful When Variable of Interest is NOT vC or iL

Basic Concept

- Recall The form of the ODE Solution for a Ckt w/ One E-Storage Element and a Constant Driving Ckt

- Where
- K1 The final Condition for the Variable of Interest
- Can Be determined by Analyzing the Ciruit in Steady State; i.e., t→
- x(0+) The Initial Condition for the Variable
- Provides the Second Eqn for Calculating K2
- Ckt Time Constant
- Determine By Finding RTH Across the Storage Element

FC

The General Approach- Obtain The Voltage Across The Capacitor or The Current Through The Inductor

Thevenin

- With This Analysis Find
- Time Constant using RTH
- Final Condition using vTH

STEP 1. Assume The Form Of The SolutionThe Steps: 1-4

- STEP 3: Draw The Circuit At t = 0+
- The CAPACITOR Acts As a VOLTAGE SOURCE
- The INDUCTOR Acts As a CURRENT SOURCE
- Determine The VARIABLE of INTEREST At t=0+
- Determine x()
- STEP 4: Draw The Circuit a Loooong Time After Switching to Determine The Variable In Steady State

- Determine x(0+)
- STEP 2: Draw The Circuit In Steady State just PRIOR To Switching And Determine Capacitor-Voltage Or Inductor-Current

STEP 5: determine the time constantThe Steps: 5-6

- With These 3-Parameters Write the Solution For the Variable of interest using The Assumed Solution

- RTH Determined at Cap/Ind Connection Terminals

- Step-By-Step DOWNside
- Do NOT have ODE So Can NOT easily Check Solution
- Can usually chk the FINAL Condition

- STEP 6: Determine The Constants K1 & K2,

Step-By-Step: Inductor Example

- STEP-1: The Form of the Soln

- For the Circuit Below Find vO for t>0

- STEP-2: Initial inductor current (L is Short to DC)

- Note That vO is NOT Directly Related to The Storage Element
- → Use Step-by-Step

STEP 3: Determine output at 0+

By Inductor Physics

Inductor Example cont.- Note That at t=0+
- The 6V Source is DISCONNECTED from the Ckt Elements
- No Connection on Supply Side
- Single Loop Ckt

- At t=0+, Replace The L with a 3A Current Src

STEP 4: Find Output In Steady State After The Switching

By Inductor Physics In Steady State

L is SHORT to DC

Inductor Example cont.2- Recall at t=0+The 6V Source is DISconnected from the Ckt Elements
- The Ckt Has NO Power Source
- Over A long Time All the Energy Stored by The Inductor Will be Dissipated as HEAT by The Resistors, Hence

STEP 5: Find Time Constant After Switch

Find RTH With Respect to the L Terminals

Inductor Example cont.3- Then The Time Constant,

- RTH by Series Calc

WhiteBoard Work

- Let’s Work This 1st Order Cap Problem
- Power Source DISengaged

Pulse Response

- Consider The Response Of Circuits To A Special Class Of SINGULARITY functions

VOLTAGE STEP

CURRENT STEP

TIME SHIFTED STEP

Non-Zero Initial Condition (std ODE)PieceWise Transient Repsonse

- This expression will hold on ANY interval where the sources are CONSTANT
- The values of the constants may be different and must be evaluated for each interval
- The values at the END of one interval will serve as INITIAL conditions for the NEXT interval

- The Response is Shifted From the Time Origin by an Amount t0
- For CONSTANT fTH, The Time-Shifted Exponential Solution

PieceWise Example

- The Switch is Initially At a. At Time t=0 It Moves To b, and At t=0.5 it moves back to a.
- Find vO(t) for t>0
- On Each Interval Where The Source is Constant The Response Will Be of the Form

For 0t<0.5 (Switch at b)

t0 = 0

Assume Solution

PieceWise Example cont- Now Piece-2 (Switch at a)
- t0=0.5S

- Find Parameters And Piece-1 Solution

WhiteBoard Work

- Let’s Work This 1st Order Cap Problem
- R1→4 = 2 kΩ
- Power Source ENgaged
- IF we Have Time

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