ee292: fundamentals of eceb1morris/ee292/docs/slides19.pdfย ยท 2012. 10. 30.ย ยท...
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EE292: Fundamentals of ECE
Fall 2012
TTh 10:00-11:15 SEB 1242
Lecture 19
121030
Outline
โข Review
โซ Phasors
โซ Complex Impedance
โข Circuit Analysis with Complex Impedance
2
Phasors
โข A representation of sinusoidal signals as vectors in the complex plane
โซ Simplifies sinusoidal steady-state analysis
โข Given
โซ ๐ฃ1 ๐ก = ๐1cos(๐๐ก + ๐1)
โข The phasor representation is
โซ ๐ฝ1 = ๐1โ ๐1
โข For consistency, use only cosine for the phasor
โซ ๐ฃ2 ๐ก = ๐2 sin ๐๐ก + ๐2 = ๐2 cos ๐๐ก + ๐2 โ 90ยฐ =
๐ฝ2 = ๐2โ (๐2โ90ยฐ)
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
โซ Remember that ๐ should be the same for each sinusoid and the result will have the same frequency
4
Example 5.3 โข ๐ฃ1 ๐ก = 20 cos ๐๐ก โ 45ยฐ
โซ ๐ฝ1 = 20โ โ45ยฐ
โข ๐ฃ2 ๐ก = 10 sin ๐๐ก + 60ยฐ
โซ ๐ฝ2 = 10โ 60ยฐ โ 90ยฐ =
10โ โ30ยฐ
โข Calculate
โซ ๐ฝ๐ = ๐ฝ1 + ๐ฝ2
5
Phasor as a Rotating Vector โข ๐ฃ ๐ก = ๐๐ cos ๐๐ก + ๐ =
๐ ๐{๐๐๐๐ ๐๐ก+๐ }
โข ๐๐๐๐ ๐๐ก+๐ is a complex vector
that rotates counter clockwise at ๐ rad/s
โข ๐ฃ(๐ก) is the real part of the vector
โซ The projection onto the real axis of the rotating complex vector
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Source: Wikipedia
Phase Relationships โข Given
โซ ๐ฃ1 ๐ก = 3 cos ๐๐ก + 40ยฐ
โซ ๐ฃ2 ๐ก = 4 cos ๐๐ก โ 20ยฐ
โข In phasor notation
โซ ๐ฝ1 = 3โ 40ยฐ
โซ ๐ฝ2 = 4โ โ20ยฐ
โข Since phasors rotate counter clockwise
โซ ๐ฝ1 leads ๐ฝ2 by 60ยฐ
โซ ๐ฝ2 lags ๐ฝ1 by 60ยฐ
โข Phasor diagram
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Complex Impedance
โข Previously we saw that resistance was a measure of the opposition to current flow
โซ Larger resistance less current allowed to flow
โข Impedance is the extension of resistance to AC circuits
โซ Inductors oppose a change in current
โซ Capacitors oppose a change in voltage
โข Capacitors and inductors have imaginary impedance called reactance
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Resistance
โข From Ohmโs Law
โซ ๐ฃ ๐ก = ๐ ๐ ๐ก
โข Extend to an impedance form for AC signals
โซ ๐ฝ = ๐๐ฐ
โข Converting to phasor notation for resistance
โซ ๐ฝ๐ = ๐ ๐ฐ๐
โซ ๐ =๐ฝ๐
๐ฐ๐
โข Comparison with the impedance form results in
โซ ๐๐ = ๐ โซ Since ๐ is real, the impedance for a resistor is purely
real
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Inductance โข I/V relationship
โซ ๐ฃ๐ฟ ๐ก = ๐ฟ๐๐๐ฟ(๐ก)
๐๐ก
โข Assume current
โซ ๐๐ฟ ๐ก = ๐ผ๐ sin ๐๐ก + ๐
โซ ๐ฐ๐ฟ = ๐ผ๐โ ๐ โ๐
2
โข Using the I/V relationship
โซ ๐ฃ๐ฟ ๐ก = ๐ฟ๐๐ผ๐ cos ๐๐ก + ๐
โซ ๐ฝ๐ฟ = ๐๐ฟ๐ผ๐โ ๐
โซ Notice current lags voltage by 90ยฐ
โข Using the generalized Ohmโs Law
โซ ๐๐ฟ =๐ฝ๐ฟ
๐ฐ๐ฟ=
๐๐ฟ๐ผ๐โ ๐
๐ผ๐โ ๐โ๐
2
= ๐๐ฟโ ๐
2= ๐๐๐ฟ
โซ Notice the impedance is completely imaginary
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Capacitance โข I/V relationship
โซ ๐๐ถ ๐ก = ๐ถ๐๐ฃ๐ถ(๐ก)
๐๐ก
โข Assume voltage
โซ ๐ฃ๐ ๐ก = ๐๐ sin ๐๐ก + ๐
โซ ๐ฝ๐ถ = ๐๐โ ๐ โ๐
2
โข Using the I/V relationship
โซ ๐๐ถ ๐ก = ๐ถ๐๐๐ cos ๐๐ก + ๐
โซ ๐ฐ๐ถ = ๐๐ถ๐๐โ ๐
โซ Notice current leads voltage by 90ยฐ
โข Using the generalized Ohmโs Law
โซ ๐๐ถ =๐ฝ๐ถ
๐ฐ๐ถ=
๐๐โ ๐โ๐
2
๐๐ถ๐๐โ ๐=
1
๐๐ถโ โ
๐
2= โ๐
1
๐๐ถ=
1
๐๐๐ถ
โซ Notice the impedance is completely imaginary
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Circuit Analysis with Impedance
โข KVL and KCL remain the same
โซ Use phasor notation to setup equations
โข E.g.
โข ๐ฃ1 ๐ก + ๐ฃ2 ๐ก โ ๐ฃ3 ๐ก = 0
โข ๐ฝ1 + ๐ฝ2 โ ๐ฝ3 = 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 ๐๐ฟ = ๐๐ฟโ ๐
2= ๐๐๐ฟ. Replace
capacitances by their complex impedances
๐๐ถ =1
๐๐ถโ โ
๐
2=
1
๐๐๐ถ. 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
โข Find voltage ๐ฃ๐(๐ก) in steady-state
โข Find the phasor current through each element
โข Construct the phasor diagram showing currents and ๐ฃ๐
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Convert Sources and Impedances โข Voltage source
โซ ๐ฝ๐ = 10โ โ90ยฐ
โซ ๐ = 1000
โข Inductance
โซ ๐๐ฟ = ๐๐๐ฟ = j 1000 0.1 =j100ฮฉ
โข Capacitance
โซ ๐๐ถ =1
๐๐๐ถ=
1
๐1000(10๐)=
106
๐104=
100
๐ฮฉ
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Find Voltage ๐ฃ๐(๐ก) โข Find voltage by voltage divider
โข ๐ฝ๐ถ = ๐ฝ๐ ๐๐๐
๐๐๐+๐๐ฟ
โข Equivalent impedance
โข ๐๐๐ = ๐๐ ||๐๐ถ
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Phasor Diagram โข Source current
โข Capacitor current
โข Resistor current
โข Phasor diagram
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More AC Circuit Analysis
โข Use impedance relationships and convert sinusoidal sources to phasors
โข Techniques such as node-voltage and mesh-current analysis remain the same โซ (Hambley Section 5.4)
โข Thevenin and Norton equivalents are extended the same way โซ (Hambley Section 5.6) โซ Instead of a resistor and source, use an impedance
โซ ๐๐ก =๐ฝ๐๐
๐ฐ๐ ๐
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Example 5.6
โข Use node-voltage to find ๐ฃ1(๐ก)
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Convert to Phasors
โข Sources
โซ 2 sin 100๐ก = 2โ โ90ยฐ
โซ โ90ยฐ because of conversion to cosine from sine
โซ ๐ = 100
โซ 1.5 cos 100๐ก = 1.5โ 0ยฐ
โข Inductor
โซ ๐๐ฟ = ๐๐ฟโ ๐
2= ๐๐๐ฟ = ๐100 0.1 = ๐10
โข Capacitor
โซ ๐๐ถ =1
๐๐ถโ โ
๐
2=
1
๐๐๐ถ= โ๐
1
100 2000๐= โ๐5
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Use Node-Voltage Analysis โข KCL @ 1
โซ๐ฝ1
10+
๐ฝ1โ๐ฝ2
โ๐5= 2โ โ90ยฐ
โข KCL @ 2
โซ๐ฝ2
๐10+
๐ฝ2โ๐ฝ1
โ๐5= 1.5โ 0ยฐ
โข In standard form
โข 0.1 + ๐0.2 ๐ฝ1 โ ๐0.2๐ฝ2 = โ๐2
โข โ๐0.2๐ฝ1 + ๐0.1๐ฝ2 = 1.5
โข Solving for ๐ฝ1
โข 0.1 โ ๐0.2 ๐ฝ1 = 3 โ ๐2
โข Converting to phasor โข 0.2236โ โ63.4ยฐ ๐ฝ1 = 3.6โ โ33.7ยฐ
โข ๐ฝ1 = 16.1โ 29.7ยฐ
โข Convert back to sinusoid
โข ๐ฃ1 ๐ก = 16.1cos(100๐ก + 29.7ยฐ)
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