AC Analysis Using Thevenin's Theorem and Superposition - PowerPoint PPT Presentation

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AC Analysis Using Thevenin's Theorem and Superposition

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Title: Basic Concepts Author: Dr. Haskell Last modified by: haskell Created Date: 10/27/2005 8:46:53 PM Document presentation format: On-screen Show – PowerPoint PPT presentation

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Title: AC Analysis Using Thevenin's Theorem and Superposition


1
AC Analysis Using Thevenin's Theorem and
Superposition
  • Discussion D11.2
  • Chapter 4

2
AC Thevenin's Theorem
3
AC Thevenin's Theorem
Thevenins theorem states that the two circuits
given below are equivalent as seen from the load
ZL that is the same in both cases.
VTh Thevenins voltage Vab with ZL
disconnected ( ?) the open-circuit voltage
VOC
4
Thevenin's Theorem
ZTh Thevenins impedance the input impedance
with all independent sources turned off (voltage
sources replaced by short circuits and current
sources replaced by open circuits). This is the
impedance seen at the terminals ab when all
independent sources are turned off.
5
Problem 4.57 in text Solve Problem 4.40 using
Thevenin's Thm.
6
AC Superposition
7
Superposition Principle
Because the circuit is linear we can find the
response of the circuit to each source acting
alone, and then add them up to find the response
of the circuit to all sources acting together.
This is known as the superposition principle.
The superposition principle states that the
voltage across (or the current through) an
element in a linear circuit is the algebraic sum
of the voltages across (or currents through) that
element due to each independent source acting
alone.
8
Steps in Applying the Superposition Principle
  1. Turn off all independent sources except one.
    Find the output (voltage or current) due to the
    active source.
  2. Repeat step 1 for each of the other independent
    sources.
  3. Find the total output by adding algebraically all
    of the results found in steps 1 2 above.

In some cases, but certainly not all,
superposition can simplify the analysis.
9
Example
Note that the voltage source and the current
source have two different frequencies. Thus, if
we want to use phasors, the only way we've solved
sinusoidal steady-state problems, we MUST use
superposition to solve this problem. We will
consider each source acting alone, and then find
v0(t) by superposition.
Remember that
10
Example
Consider first the acting
alone. Since,
,we have w 5 and
11
Example
Use voltage division
12
Example
Now consider first the acting
alone. We have w 10 and
13
Example
For a parallel combination of Y's we have
14
Example
By superposition
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