Chapter 6(b) Sinusoidal Steady State Analysis
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Transcript of Chapter 6(b) Sinusoidal Steady State Analysis
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Chapter 6(b)Chapter 6(b)Sinusoidal Steady State AnalysisSinusoidal Steady State Analysis
Chapter Objectives: Apply previously learn circuit techniques to sinusoidal steady-
state analysis. Learn how to apply nodal and mesh analysis in the frequency
domain. Learn how to apply:
superposition, Source Transformation Thevenin’s and Norton’s theorems
in the frequency domain.
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Steps to Analyze AC Circuits
Transform the circuit to the Phasor Domain.
Solve the problem using circuit techniques listed below
1) *Nodal Analysis2) *Mesh Analysis3) Superposition4) Source transformation5) *Thevenin or Norton Equivalents
Transform the resulting circuit back to time domain.
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Steps to Analyze AC Circuits Transform the circuit to the phasor or
frequency domain. Solve the problem using circuit techniques
(nodal analysis, mesh analysis, superposition, etc.).
Transform the resulting phasor to the time domain.
Time to Freq Solve Variables in Freq
Freq to Time
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Nodal Analysis Since KCL is valid for phasors, we can analyze AC circuits by
NODAL analysis. Determine the number of nodes within the network. Pick a reference node and label each remaining node with a
subscripted value of voltage: V1, V2 and so on. Apply Kirchhoff’s current law at each node except the reference.
Assume that all unknown currents leave the node for each application of Kirhhoff’s current law.
Solve the resulting equations for the nodal voltages. For dependent current sources: Treat each dependent current
source like an independent source when Kirchhoff’s current law is applied to each defined node. However, once the equations are established, substitute the equation for the controlling quantity to ensure that the unknowns are limited solely to the chosen nodal voltages.
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Nodal Analysis
Practice Problem 10.1: Find v1 and v2 using nodal analysis
Since KCL is valid for phasors, we can analyze AC circuits by NODAL analysis.
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Nodal Analysis Practice Problem 10.1
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Nodal Analysis Practice Problem 10.1
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Mesh Analysis
Practice Problem 10.4: Calculate the current Io
Meshes 2 and 3 form a supermesh as shown in the circuit below.
Since KVL is valid for phasors, we can analyze AC circuits by MESH analysis.
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Mesh Analysis Practice Problem 10.4: Calculate the current Io
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Mesh Analysis Practice Problem 10.4: Calculate the current Io
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Superposition TheoremThe superposition theorem eliminates the need for solving simultaneous linear equations by considering the effect on each source independently. To consider the effects of each source we remove the remaining sources; by setting the voltage sources to zero (short-circuit representation) and current sources to zero (open-circuit representation). The current through, or voltage across, a portion of the network produced by each source is then added algebraically to find the total solution for current or voltage. The only variation in applying the superposition theorem to AC networks with independent sources is that we will be working with impedances and phasors instead of just resistors and real numbers. The superposition theorem is not applicable to power effects in AC networks since we are still dealing with a nonlinear relationship. It can be applied to networks with sources of different frequencies only if the total response for each frequency is found independently and the results are expanded in a nonsinusoidal expression . One of the most frequent applications of the superposition theorem is to electronic systems in which the DC and AC analyses are treated separately and the total solution is the sum of the two.
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Superposition Theorem
When a circuit has sources operating at different frequencies, • The separate phasor circuit for each frequency
must be solved independently, and • The total response is the sum of time-domain
responses of all the individual phasor circuits.
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Superposition Theorem
a) All sources except DC 5-V set a) All sources except DC 5-V set to zeroto zero
b) All sources except 10cos(10t) set to zero
Exp. 10.6 Superposition Technique for sources having different frequencies
Superposition Theorem applies to AC circuits as well.
For sources having different frequencies, the total response must be obtained by adding individual responses in time domain.
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c) All sources except 2 sin 5t set to c) All sources except 2 sin 5t set to zerozero
Superposition Theorem Exp. 10.6 Superposition Technique for sources having different frequencies
vo= v1+ v2+ v3
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Superposition Theorem
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Superposition Theorem
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Superposition TheoremP.P.10.6 Superposition Technique for sources having different Frequencies
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Superposition Theorem
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Superposition Theorem
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20Eeng 224
Transform a voltage source in series with an impedance to a Transform a voltage source in series with an impedance to a current source in parallel with an impedance for simplification or current source in parallel with an impedance for simplification or vice versa.vice versa.
Source Transformation
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Source Transformation
If we transform the current source to a voltage source, we obtain the circuit shown in Fig. (a).
Practice Problem 10.4: Calculate the current Io
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Source Transformation Practice Problem 10.4: Calculate the current Io
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23Eeng 224
Thevenin Equivalent Circuit Thévenin’s theorem, as stated for sinusoidal AC circuits, is changed only to include the term impedance instead of resistance. Any two-terminal linear ac network can be replaced with an equivalent circuit consisting of a voltage source and an impedance in series. VTh is the Open circuit voltage between the terminals a-b. ZTh is the impedance seen from the terminals when the independent sources are set to zero.
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Norton Equivalent Circuit The linear circuit is replaced by a current source in parallel with an impedance. IN is the Short circuit current flowing between the terminals a-b when the terminals are short circuited.
Thevenin and Norton equivalents are related by:
Th N N Th NV Z I Z Z
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Thevenin Equivalent CircuitP.P.10.8 Thevenin Equivalent At terminals a-b
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Thevenin Equivalent CircuitP.P.10.9 Thevenin and Norton Equivalent for Circuits with Dependent Sources
To find Vth , consider the circuit in Fig. (a).
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Thevenin Equivalent CircuitP.P.10.9 Thevenin and Norton Equivalent for Circuits with Dependent Sources
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28Eeng 224
Thevenin Equivalent CircuitP.P.10.9 Thevenin and Norton Equivalent for Circuits with Dependent Sources
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29Eeng 224
Thevenin Equivalent CircuitP.P.10.9 Thevenin and Norton Equivalent for Circuits with Dependent SourcesSince there is a dependent source, we can find the impedance by inserting a voltage source and calculating the current supplied by the source from the terminals a-b.