EE292: Fundamentals of ECEb1morris/ee292/docs/slides19.pdfΒ Β· 2012. 10. 30.Β Β·...
Transcript of EE292: Fundamentals of ECEb1morris/ee292/docs/slides19.pdfΒ Β· 2012. 10. 30.Β Β·...
http://www.ee.unlv.edu/~b1morris/ee292/
EE292: Fundamentals of ECE
Fall 2012
TTh 10:00-11:15 SEB 1242
Lecture 19
121030
Outline
β’ Review
β« Phasors
β« Complex Impedance
β’ Circuit Analysis with Complex Impedance
2
Phasors
β’ A representation of sinusoidal signals as vectors in the complex plane
β« Simplifies sinusoidal steady-state analysis
β’ Given
β« π£1 π‘ = π1cos(ππ‘ + π1)
β’ The phasor representation is
β« π½1 = π1β π1
β’ For consistency, use only cosine for the phasor
β« π£2 π‘ = π2 sin ππ‘ + π2 = π2 cos ππ‘ + π2 β 90Β° =
π½2 = π2β (π2β90Β°)
3
Adding Sinusoids with Phasors
1. Get phasor representation of sinusoids
2. Convert phasors to rectangular form and add
3. Simplify result and convert into phasor form
4. Convert phasor into sinusoid
β« Remember that π should be the same for each sinusoid and the result will have the same frequency
4
Example 5.3 β’ π£1 π‘ = 20 cos ππ‘ β 45Β°
β« π½1 = 20β β45Β°
β’ π£2 π‘ = 10 sin ππ‘ + 60Β°
β« π½2 = 10β 60Β° β 90Β° =
10β β30Β°
β’ Calculate
β« π½π = π½1 + π½2
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Phasor as a Rotating Vector β’ π£ π‘ = ππ cos ππ‘ + π =
π π{ππππ ππ‘+π }
β’ ππππ ππ‘+π is a complex vector
that rotates counter clockwise at π rad/s
β’ π£(π‘) is the real part of the vector
β« The projection onto the real axis of the rotating complex vector
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Source: Wikipedia
Phase Relationships β’ Given
β« π£1 π‘ = 3 cos ππ‘ + 40Β°
β« π£2 π‘ = 4 cos ππ‘ β 20Β°
β’ In phasor notation
β« π½1 = 3β 40Β°
β« π½2 = 4β β20Β°
β’ Since phasors rotate counter clockwise
β« π½1 leads π½2 by 60Β°
β« π½2 lags π½1 by 60Β°
β’ Phasor diagram
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Complex Impedance
β’ Previously we saw that resistance was a measure of the opposition to current flow
β« Larger resistance less current allowed to flow
β’ Impedance is the extension of resistance to AC circuits
β« Inductors oppose a change in current
β« Capacitors oppose a change in voltage
β’ Capacitors and inductors have imaginary impedance called reactance
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Resistance
β’ From Ohmβs Law
β« π£ π‘ = π π π‘
β’ Extend to an impedance form for AC signals
β« π½ = ππ°
β’ Converting to phasor notation for resistance
β« π½π = π π°π
β« π =π½π
π°π
β’ Comparison with the impedance form results in
β« ππ = π β« Since π is real, the impedance for a resistor is purely
real
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Inductance β’ I/V relationship
β« π£πΏ π‘ = πΏπππΏ(π‘)
ππ‘
β’ Assume current
β« ππΏ π‘ = πΌπ sin ππ‘ + π
β« π°πΏ = πΌπβ π βπ
2
β’ Using the I/V relationship
β« π£πΏ π‘ = πΏππΌπ cos ππ‘ + π
β« π½πΏ = ππΏπΌπβ π
β« Notice current lags voltage by 90Β°
β’ Using the generalized Ohmβs Law
β« ππΏ =π½πΏ
π°πΏ=
ππΏπΌπβ π
πΌπβ πβπ
2
= ππΏβ π
2= πππΏ
β« Notice the impedance is completely imaginary
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Capacitance β’ I/V relationship
β« ππΆ π‘ = πΆππ£πΆ(π‘)
ππ‘
β’ Assume voltage
β« π£π π‘ = ππ sin ππ‘ + π
β« π½πΆ = ππβ π βπ
2
β’ Using the I/V relationship
β« ππΆ π‘ = πΆπππ cos ππ‘ + π
β« π°πΆ = ππΆππβ π
β« Notice current leads voltage by 90Β°
β’ Using the generalized Ohmβs Law
β« ππΆ =π½πΆ
π°πΆ=
ππβ πβπ
2
ππΆππβ π=
1
ππΆβ β
π
2= βπ
1
ππΆ=
1
πππΆ
β« Notice the impedance is completely imaginary
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Circuit Analysis with Impedance
β’ KVL and KCL remain the same
β« Use phasor notation to setup equations
β’ E.g.
β’ π£1 π‘ + π£2 π‘ β π£3 π‘ = 0
β’ π½1 + π½2 β π½3 = 0
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Steps for Sinusoidal Steady-State Analysis
1. Replace time descriptions of voltage and current sources with corresponding phasors. (All sources must have the same frequency)
2. Replace inductances by their complex
impedances ππΏ = ππΏβ π
2= πππΏ. Replace
capacitances by their complex impedances
ππΆ =1
ππΆβ β
π
2=
1
πππΆ. Resistances have
impedances equal to their resistances. 3. Analyze the circuit by using any of the
techniques studied in Chapter 2, and perform the calculations with complex arithmetic
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Example 5.5
β’ Find voltage π£π(π‘) in steady-state
β’ Find the phasor current through each element
β’ Construct the phasor diagram showing currents and π£π
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Convert Sources and Impedances β’ Voltage source
β« π½π = 10β β90Β°
β« π = 1000
β’ Inductance
β« ππΏ = πππΏ = j 1000 0.1 =j100Ξ©
β’ Capacitance
β« ππΆ =1
πππΆ=
1
π1000(10π)=
106
π104=
100
πΞ©
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Find Voltage π£π(π‘) β’ Find voltage by voltage divider
β’ π½πΆ = π½π πππ
πππ+ππΏ
β’ Equivalent impedance
β’ πππ = ππ ||ππΆ
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Phasor Diagram β’ Source current
β’ Capacitor current
β’ Resistor current
β’ Phasor diagram
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More AC Circuit Analysis
β’ Use impedance relationships and convert sinusoidal sources to phasors
β’ Techniques such as node-voltage and mesh-current analysis remain the same β« (Hambley Section 5.4)
β’ Thevenin and Norton equivalents are extended the same way β« (Hambley Section 5.6) β« Instead of a resistor and source, use an impedance
β« ππ‘ =π½ππ
π°π π
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Example 5.6
β’ Use node-voltage to find π£1(π‘)
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Convert to Phasors
β’ Sources
β« 2 sin 100π‘ = 2β β90Β°
β« β90Β° because of conversion to cosine from sine
β« π = 100
β« 1.5 cos 100π‘ = 1.5β 0Β°
β’ Inductor
β« ππΏ = ππΏβ π
2= πππΏ = π100 0.1 = π10
β’ Capacitor
β« ππΆ =1
ππΆβ β
π
2=
1
πππΆ= βπ
1
100 2000π= βπ5
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Use Node-Voltage Analysis β’ KCL @ 1
β«π½1
10+
π½1βπ½2
βπ5= 2β β90Β°
β’ KCL @ 2
β«π½2
π10+
π½2βπ½1
βπ5= 1.5β 0Β°
β’ In standard form
β’ 0.1 + π0.2 π½1 β π0.2π½2 = βπ2
β’ βπ0.2π½1 + π0.1π½2 = 1.5
β’ Solving for π½1
β’ 0.1 β π0.2 π½1 = 3 β π2
β’ Converting to phasor β’ 0.2236β β63.4Β° π½1 = 3.6β β33.7Β°
β’ π½1 = 16.1β 29.7Β°
β’ Convert back to sinusoid
β’ π£1 π‘ = 16.1cos(100π‘ + 29.7Β°)
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