Sinusoidal Steady-State
Analysis
Qi Xuan
Zhejiang University of Technology
Nov 2015
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Structure
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The Sinusoidal Source The Sinusoidal Response The Phasor The Passive Circuit Elements in the Frequency Domain Kirchhoff's Laws in the Frequency Domain Series, Parallel, and Delta-­‐to-­‐Wye SimplificaEons Source TransformaEons and Thévenin-­‐Norton Equivalent Circuits The Node-­‐Voltage Method The Mesh-­‐Current Method The Transformer The Ideal Transformer Phasor Diagrams Electric Circuits
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Why Sinusoidal Source? •  The genera4on, transmission, distribu4on, and consump4on of electric energy occur under essen4ally sinusoidal steady-­‐
state condi4ons. •  An understanding of sinusoidal behavior makes it possible to predict the behavior of circuits with nonsinusoidal sources. •  Steady-­‐state sinusoidal behavior oAen simplifies the design of electrical systems. A Household
Distribution Circuit
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The Sinusoidal Source
A sinusoidal voltage/current source (independent or dependent) produces
a voltage/current that varies sinusoidally with time.
Vm: Maximum amplitude
ω: Angular frequency
ϕ: Phase angle (radians/degrees)
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f: Frequency
T: Period
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RMS Value
The rms value of a periodic func4on is defined as the square root
of the mean value of the squared func4on.
Power
We can completely describe a specific sinusoidal signal if we know
its frequency, phase angle, and amplitude (either the maximum
or the rms value).
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Example #1
A sinusoidal current has a maximum amplitude of 20 A . The current passes through one complete cycle in 1 ms. The magnitude of the current at zero 4me is 10 A . a)  What is the frequency of the current in hertz? b)  What is the frequency in radians per second? c)  Write the expression for i(t) using the cosine func4on. Express ϕ in degrees. d)  What is the rms value of the current? Electric Circuits
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Solu:on for Example #1
a)  From the statement of the problem, T = 1 ms; hence f = 1/T
= 1000 Hz. b)  ω = 2πf = 2000π rad/s.
c)  We have i(t) = Imcos(ωt + ϕ) = 20 cos(2000πt +ϕ),but i(0)=
10A.Therefore 10 = 20cos ϕ and ϕ = 60°. Thus the expression for i(t) becomes i(t) = 20cos(2000πt + 60°).
d)  The rms value of a sinusoidal current is Im/√2. Therefore the rms value is 20/√2, or 14.14 A. Electric Circuits
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Example #2
•  Calculate the rms value of the periodic triangular current in the given figure. Express your answer in terms of the peak current Ip. Electric Circuits
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Solu:on for Example #2
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The Sinusoidal Response
Transient component
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Steady-state component
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Steady-­‐state Solu4on
•  The steady-­‐state solu4on is a sinusoidal func4on. •  The frequency of the response signal is iden4cal to the frequency of the source signal. This condi:on is always true in a linear circuit when the circuit parameters, R, L, and C, are constant. •  The maximum amplitude of the steady-­‐state response, in general, differs from the maximum amplitude of the source. For the circuit being discussed, the maximum amplitude of the response signal is Vm/√(R2+ω2L2), and that of the signal source is Vm. •  The phase angle of the response signal, in general, differs from the phase angle of the source. For the circuit being discussed, the phase angle of the current is ϕ − θ and that of the voltage source is ϕ.
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The Phasor
The phasor is a complex number that carries the amplitude and phase angle
informa4on of a sinusoidal func4on (please see the Appendix B). Euler’s idenEty: The real part
The imaginary part
Phasor transform: Angle notation
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Inverse Phasor Transform
v = 100cos(ωt − 60)
Inverse phasor transform:
The phasor transform is useful in circuit analysis because it
reduces the task of finding the maximum amplitude and phase
angle of the steady-­‐state sinusoidal response to the algebra of
complex numbers. Electric Circuits
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•  The transient component vanishes as 4me elapses, so the steady-­‐ state component of the solu:on must also sa:sfy the differen:al equa:on. •  In a linear circuit driven by sinusoidal sources, the steady-­‐state response also is sinusoidal, and the frequency of the sinusoidal response is the same as the frequency of the sinusoidal source. •  Using the nota4on introduced in Eq. 9.11, we can postulate that the steady-­‐state solu4on is of the form R{Aejβejωt}, where A is the maximum amplitude of the response and β is the phase angle of the response. •  When we subs4tute the postulated steady-­‐state solu4on into the differen4al equa4on, the exponen4al term ejωt cancel out, leaving the solu4on for A and β in the domain of complex numbers Electric Circuits
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Illustra:on
Substituted into Eq. 9.8
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The phasor transform, along with the inverse phasor transform,
allows you to go back and forth between the 4me domain and
the frequency domain. Therefore, when you obtain a solu4on,
you are either in the :me domain or the frequency domain. You
cannot be in both domains simultaneously. Any solu4on that
contains a mixture of 4me domain and phasor domain
nomenclature is nonsensical.
The phasor transform is also useful in circuit analysis because it
applies directly to the sum of sinusoidal func:on, if When all the voltages on the right-­‐hand side are sinudoidal
voltages of the same frequency, then Electric Circuits
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Example #3
If y1 = 20cos(ωt - 30°) and y2 = 40cos(ωt + 60°), express y = y1 + y2 as a single sinusoidal func4on. a)  Solve by using trigonometric iden44es. b)  Solve by using the phasor concept. Electric Circuits
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Solu:on for Example #2
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The Passive Circuit Elements in the Frequency Domain
The V-­‐I RelaEonship for a Resistor
At the terminals of a resistor, there is no phase
shiA between the current and voltage.
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The V-­‐I Rela:onship for an Inductor
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The V-­‐I Rela:onship for a Capacitor Electric Circuits
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Impedance and Reactance
V=Z I
Impedance: measured in ohms
Reactance: the imaginary part of impedance
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Kirchhoff’s law in the Frequency Domain
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Series, Parallel, and Delta-­‐to-­‐Wye Simplifica4on
The rules for combining impedances in series or parallel and for making delta-­‐to-­‐wye transforma4ons are the same as those for resistors. The only difference is that combining impedances involves the algebraic manipula:on of complex numbers. Impedances in series:
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Impedance in Parallel
Two impedance
susceptance
conductance
admittance
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Delta-­‐to-­‐Wye Transforma:ons
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Example #4
A 90 Ω resistor, a 32 mH inductor, and a 5 μF capacitor are connected in series across the terminals of a sinusoidal voltage source. The steady-­‐state expression for the source voltage vs is 750 cos (5000t + 30°) V. a)  Construct the frequency-­‐domain equivalent circuit. b)  Calculate the steady-­‐state current i by the phasor method. Electric Circuits
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Solu:on for Example #4
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Source Transforma4ons and Thévenin-­‐Norton Equivalent Circuits
The Node-­‐Voltage Method (Example)
The Mesh-­‐Current Method
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Example #5
Use the node-­‐voltage method to find the branch currents Ia, Ib, and Ic in the circuit. Electric Circuits
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Solu:on for Example #5
V1
1
V2
2
Node 1: Node 2: Electric Circuits
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Check
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The Transformer
A transformer is a device that is based on magne:c coupling. Transformers are used in both communica4on and power circuits. •  linear transformer: is found primarily in communica-­‐ 4on circuits •  Ideal transformer: is used to model the ferromagne4c transformer found in power systems. Electric Circuits
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The Analysis of a Linear Transformer Circuit
A simple transformer is formed when two coils are wound on a single core to ensure magne:c coupling. •  Primary winding: the transformer winding connected to the source; •  Secondary winding: the winding connected to the load as the. R1 = the resistance of the primary winding R2 = the resistance of the secondary winding L1 = the self-­‐inductance of the primary winding L2 = the self-­‐inductance of the secondary winding M = the mutual inductance Electric Circuits
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Electric Circuits
The impedance Zab is independent of the magnetic
polarity of the transformer!
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Reflected Impedance
Reflected impedance Zr
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The Ideal Transformer
An ideal transformer consists of two magne4cally coupled coils having N1 and N2 turns, respec4vely, and exhibi4ng these three proper4es: 1.  The coefficient of coupling is unity (k = 1). 2.  The self-­‐inductance of each coil is infinite (L1 = L2 = ∞). 3.  The coil losses, due to parasi4c resistance, are negligible. Electric Circuits
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Exploring Limi:ng Values
M2 = L1L2
X22 = ωL2 + XL
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Factoring ωL2 out of the numerator and denominator yields:
L1/L2 = (N1/N2)2
Scaling factor
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Determining the Voltage and Current Ra:os
(a)
M2 = L1L2
L1/L2 = (N1/N2)2
(b)
R1 = R2 = 0
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Determining the Polarity of the Voltage and Current Ra:os
a.  If the coil voltages V1 and V2 are both posi:ve or nega:ve at the dot
-­‐marked terminal, use a plus sign, Otherwise, use a nega4ve sign. b.  If the coil currents I1 and I2 are both directed into or out of the dot-­‐
marked terminal, use a minus sign. Otherwise, use a plus sign. Electric Circuits
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Important parameter for
ideal transformer!
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Ideal Transformer Used for Impedance Matching
Thus, the ideal transformer's secondary coil reflects the load
impedance back to the primary coil, with the scaling factor 1/a2.
Note that the ideal transformer changes the magnitude of ZL
but does not affect its phase angle. Whether ZIN is greater or
less than ZL depends on the turns ra4o a.
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Example #6
The load impedance connected to the secondary winding of the ideal transformer consists of a 237.5 mΩ resistor in series with a 125 µH inductor. If the sinusoidal voltage source (vg) is generat-­‐ ing the voltage 2500 cos 400t V , find the steady-­‐ state expressions for: (a) i1; (b) v1; (c) i2; and (d) v2. Electric Circuits
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Solu:on for Example #6
jωL
Phasor domain equivalent circuit
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Phasor Diagrams
Construc4ng phasor diagrams of circuit
quan44es generally involves both currents
and voltages. As a result, two different
magnitude scales are necessary, one for
currents and one for voltages. The ability to visualize a phasor quan4ty
on the complex-­‐number plane can be
useful when you are checking pocket
calculator calcula4ons. Electric Circuits
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Example #7 •  For the given circuit, use a phasor diagram to find the value of R that will cause the current through that resistor, iR, to lag the source current, is, by 45° when ω = 5 krad/s. Electric Circuits
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Solu:on for Example #7
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R = 1/3 Ω
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Summary
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