Electric Circuit Theory


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1 Electric Circuit Theory Nam Ki Min
2 Chapter 11 Sinusoidal SteadyState Analysis Nam Ki Min
3 Contents and Objectives 3 Chapter Contents 11.1 The Sinusoidal Source 11.2 The Sinusoidal Response 11.3 The Phasor 11.4 The Passive Circuit Elements in the Frequency Domain 11.5 Kirchhoff s Laws in the Frequency Domain 11.6 Series, Parallel, and DeltatoWye Simplifications 11.7 Source Transformations and TheveninNorton Equivalent Circuits 11.8 The NodeVoltage Method 11.9 The MeshCurrent Method The Transformer The Ideal Transformer Phasor Diagrams
4 Contents and Objectives 4 Chapter Objectives 1. Understand phasor concepts and be able to perform a phasor transform and an inverse phasor transform. 2. Be able to transform a circuit with a sinusoidal source into the frequency domain using phasor concepts. 3. Know how to use the following circuit analysis techniques to solve a circuit in the frequency domain: Kirchhoff s laws; Series, parallel, and deltatowye simplifications; Voltage and current division; Thevenin and Norton equivalents; Nodevoltage method; and Meshcurrent method. 4. Be able to analyze circuits containing linear transformers using phasor methods. 5. Understand the ideal transformer constraints and be able to analyze circuits containing ideal transformers using phasor methods.
5 The Sinusoidal Source 5 Sinusoidal Source Sinusoidal source or sinusoidally timevarying excitation or sinusoid. Produce a signal that has the form of the sine or cosine function. Sinusoidal voltage source Produces a voltage that varies sinusoidally with time. v s v v e e Sinusoidal current source Produces a current that varies sinusoidally with time. i s i
6 The Sinusoidal Source 6 Alternating Circuit A sinusoidal current is usually referred to as alternating current (ac). Circuits driven by sinusoidal current or voltage sources are called ac circuits. π 2 ω v(t) π 0 π 2 AC Waveforms Sinusoidal wave
7 The Sinusoidal Source 7 Why Sinusoids? Nature itself is characteristically sinusoidal. We experience sinusoidal variation in the motion of a pendulum, the vibration of a string, the economic fluctuations of the stock market, and the natural response of underdamped secondorder systems A sinusoidal signal is easy to generate and transmit. It is the form of voltage generated throughout the world and supplied to homes, factories, laboratories, and so on. It is the dominant form of signal in the communications and electric power industries. Through Fourier analysis, any practical periodic signal can be represented by a sum of sinusoids. Sinusoids, therefore, play an important role in the analysis of periodic signals. A sinusoid is easy to handle mathematically. π 2 ω v(t) π 0 π 2
8 The Sinusoidal Source 8 Sinusoidal Voltage A sinusoid can be expressed in either sine or cosine form. v t = V m sin ωt V m v t V m cos ωt V m sin ωt v t = V m cos ωt φ = 90 ωt When comparing two sinusoids, it is expedient to express both as either sine or cosine with positive amplitudes. This is achieved by using the following trigonometric identities: sin(a ± B) = sin A cos B ± cos A sin B cos(a ± B) = cos A cos B sin A sin B v t v t = V m sin(ωt ± 90 ) = ±V m cos ωt = V m cos(ωt ± 90 ) = V m sin ωt
9 The Sinusoidal Source 9 Sinusoidal Voltage A more general expression: v t = V m cos(ωt + φ) (9.1) V m : amplitude ω angular frequency in radians/s ωt: argument T period in second φ: phase angle T = 2π ω v t + T = V m cos ω t + T = V m cos ω t + 2π ω = V m cos(ωt + 2π) = V m cos ωt = v(t) f frequency in Hz f = 1 T ω = 2πf
10 The Sinusoidal Source 10 Converting Sine to Cosine v t = V m cos(ωt 90 ) = V m sin ωt v t = V m sin(ωt + 90 ) = V m cos ωt V m v t V m cos ωt V m sin ωt v t = V m cos(ωt ± 180 ) = V m cos ωt φ = 90 ωt v t = V m sin(ωt ± 180 ) = V m sin ωt v t = V m cos(5t + 10 ) = V m sin(5t ) = V m sin(5t ) v(t) = V m sin(5t 260 ) = V m sin(5t ) = V m sin(5t ) = V m cos(5t 170 ) = V m cos(5t ) = V m cos(5t + 10 ) = V m sin(5t 80 )
11 The Sinusoidal Source 11 Phase Relation of a Sinusoidal Wave
12 The Sinusoidal Source 12 Phase Shift (difference) The two waves (A versus B) are of the same amplitude and frequency, but they are out of step with each other. In technical terms, this is called a phase shift (difference). v t = V m sin(ωt + φ A ) v t = V m sin(ωt + φ B ) v t = V m cos(ωt + φ A ) v t = V m cos(ωt + φ B ) ω ω φ A φ B φ = 0 φ A v t φ B v t
13 The Sinusoidal Source 13 Phase Shift (difference) Examples of phase shifts Leading B lags A Lagging A lags B The sinusoids are said to be out of phase. The sinusoids are said to be in phase.
14 The Sinusoidal Response 14 Sinusoidal Response A sinusoidal forcing function produces both a natural (or transient) response and a forced (or steadystate) response, much like the step function, which we studied in Chapters 7 and 8. The natural response of a circuit is dictated by the nature of the circuit, while the steadystate response always has a form similar to the forcing function. However, the natural response dies out with time so that only the steadystate response remains after a long time. When the natural response has become negligibly small compared with the steadystate response, we say that the circuit is operating at sinusoidal steady state. It is this sinusoidal steadystate response that is of main interest to us in this chapter.
15 The Sinusoidal Response 15 i t = 1 e R L t R e L t V m cos(ωt + φ)dt + K L e R L t (5) = 1 e R L t V m L e R L t cos ωt cos φ sin ωt sin φ dt + K e R L t μ t = e R L t g(t) = V m L cos(ωt + φ) cos(ωt + φ) = cos ωt cos φ sin ωt sin φ = 1 e R L t V m L cos φ e R L t cos ωt dt sin φ e R L t sin ωt dt + K e R L t (6) eax exp ax cos bx dx = a 2 (a cos bx + b sin bx) + b2 eax exp ax sin bx dx = a 2 (a sin bx b cos bx) + b2
16 The Passive Circuit Elements in the Frequency Domain 16 = 1 e R L t V m L cos φ e R L t cos ωt dt sin φ e R L t sin ωt dt + K e R L t (7) eax exp ax cos bx dx = a 2 (a cos bx + b sin bx) + b2 eax exp ax sin bx dx = a 2 (a sin bx b cos bx) + b2 a = R L b = ω e ax = e R L t = 1 e R L t V m L cos φ e ax e a 2 + b2 (a cos bt + b sin bt) sin φ ax K a 2 + b2 (a sin bt b cos bt) + e R L t = V m L cos φ 1 a 2 + b 2 (a cos bt + b sin bt) sin φ 1 K a 2 + b2 (a sin bt b cos bt) + e R L t (8) a b a cos bx + b sin bx = a 2 + b 2 cos bx + a 2 + b2 a 2 + b 2 sin bx = a2 + b 2 cos θ cos bx + sin θ sin bx = a 2 + b 2 cos(θ bx) a sin bx b cos bx = a 2 + b 2 cos θ sin bx sin θ cos bx = a 2 + b 2 sin(bx θ)
17 The Sinusoidal Response 17 = V m L cos φ 1 a 2 + b 2 (a cos bt + b sin bt) sin φ 1 K a 2 + b2 (a sin bt b cos bt) + e R (9) L t a b a cos bx + b sin bx = a 2 + b 2 cos bx + a 2 + b2 a 2 + b 2 sin bx = a2 + b 2 cos θ cos bx + sin θ sin bx = a 2 + b 2 cos(θ bx) a sin bx b cos bx = a 2 + b 2 cos θ sin bx sin θ cos bx = a 2 + b 2 sin(bx θ) = V m L = V m L = V m L cos φ 1 a 2 + b 2 a 2 + b 2 cos(θ bt sin φ 1 K cos φ cos(θ ωt) sin φ sin(ωt θ) + a 2 + b2 R L ω 2 cos φ cos(ωt θ) sin φ sin(ωt θ) + 1 a 2 + b 2 a 2 + b 2 sin(bt θ) + K e R L t K e R L t (10) a = R L b = ω e R L t
18 The Sinusoidal Response 18 = V m L R L ω 2 cos φ cos(ωt θ) sin φ sin(ωt θ) + K e R L t (10) a = R L b = ω 1 i(t) = V m cos(ωt + φ θ) + K R 2 + (ωl) 2 e R L t (11) 1 0 = V m R 2 + (ωl) cos( φ θ) + K 1 K = V 2 m cos( φ θ) R 2 + (ωl) 2 (12) i(t) = V m R 2 + ωl 2 cos(ωt + φ θ) V m R 2 + ωl R cos( φ θ) e L t 2 (13) (9.9)
19 The Sinusoidal Response 19 Sinusoidal Response v s (t) = V m cos(ωt + φ) i t = V m R 2 + ωl R cos( φ θ) e L t + 2 V m R 2 + ωl 2 cos(ωt + φ θ) (9.9) Transient(Natural) Component The natural response of a circuit is dictated by the nature of the circuit. The natural response dies out with time. When the natural response has become negligibly small compared with the steadystate response, we say that the circuit is operating at sinusoidal steady state. It is this sinusoidal steadystate response that is of main interest to us in this chapter. Steadystate Component (Forced Response) The steadystate solution is a sinusoidal function. The frequency of the response signal is identical to the frequency of the source signal. The maximum amplitude of the steadystate response, in general, differs from the maximum amplitude of the source. The phase angle of the response signal, in general, differs from the phase angle of the source.
20 The Phasor 20 Definition A phasor is a complex number that represents the amplitude and phase angle of a sinusoid. v t = V m cos(ωt + φ) V = V m e jφ = V m φ phasor representation When a phasor is used to describe an AC quantity, the length of a phasor represents the amplitude of the wave while the angle of a phasor represents the phase angle of the wave relative to some other (reference) waveform.
21 The Phasor 21 Complex Number A complex number z can be written in rectangular form as z = x + jy j = 1 x: the real part of z y: the imaginary part of z The complex number z can also be written in polar or exponential form as z = r φ z = re jφ r = x 2 + y 2 φ = tan 1 y x The relationship between the rectangular form and the polar form is Representation of a complex number z in the complex plane z = x + jy = r(cos φ + j sin φ) = r φ
22 The Phasor 22 Complex Number z 1 = x 1 + jy 1 = r 1 φ 1 z 2 = x 2 + jy 2 = r 2 φ 2 Addition and subtraction of complex numbers z 1 + z 2 = (x 1 + x 2 ) + j(y 1 + y 2 ) z 1 z 2 = (x 1 x 2 ) + j(y 1 y 2 ) Multiplication and division z 1 z 2 = r 1 r 2 φ 1 + φ 2 Reciprocal z 1 z 2 = r 1 r 2 φ 1 φ 2 1 z = 1 r φ = 1 r φ Square Root 1 z = 1 x + jy = 1 x 2 + y φ 2 Representation of a complex number z in the complex plane z = r φ/2
23 The Phasor 23 Complex Conjugate Complex Conjugate z = x + jy z = x jy = r φ = re φ j = 1 j 1 j = = 1 90 = j 1 Representation of a complex number z in the complex plane
24 The Phasor 24 The Phasor Representation of the Sinusoid v(t) The idea of phasor representation is based on Euler s identity. In general, e ±jθ = cos θ + j sin θ (9.10) We can regard cos θ and sin θ as the real and imaginary parts of e jθ ; we may write cos θ = Re{e jθ } sin θ = Im{e jθ } (9.11) (9.12) We write the sinusoidal voltage function given by Eq.(9.1) in the form suggested by Eq.(9.11) v t = V m cos(ωt + φ) (9.1) = Re{V m e j ωt+φ } = Re{V m e jφ e jωt } (9.14) V m e jφ : a complex number that carries the amplitude and phase angle of the given sinusoidal function. We define the phasor representation or phasor transform of the given sinusoidal function as V = V m e jφ (9.15) v t = Re{Ve jωt }
25 The Phasor 25 The Phasor Representation of the Sinusoid v(t) Phasor representation V = V m e jφ (9.15) V = V m φ V V = V m (cos φ + j sin φ) (9.16) One way of looking at Eqs. (9.15) and (9.16) is to consider the plot of the Ve jωt on the complex plane. Ve jωt As time increases, the Ve jωt rotates on a circle of radius V m at an angular velocity ω in the counterclockwise direction, as shown in Fig. (a). In other words, the entire complex plane is rotating at an angular velocity of ω. We may regard v(t) as the projection of the Ve jωt on the real axis, as shown in Fig.(b). The value of the Ve jωt at time t = 0 is the phasor V of the sinusoid v(t). The Ve jωt may be regarded as a rotating phasor. Thus, whenever a sinusoid is expressed as a phasor, the term e jωt is implicitly present. It is therefore important, when dealing with phasors, to keep in mind the frequency ω of the phasor; otherwise we can make serious mistakes.
26 The Phasor 26 Phasor Transform : Summary Eqs.(9.14) through (9.16) reveal that to get the phasor corresponding to a sinusoid, we first express the sinusoid in the cosine form so that the sinusoid can be written as the real part of a complex number. Then we take out the time factor e jωt, and whatever is left is the phasor corresponding to the sinusoid. By suppressing the time factor, we transform the sinusoid from the time domain to the phasor domain. This transformation is summarized as follows: v t = V m cos(ωt + φ) = Re{V m e jφ e jωt } = Re{Ve jωt } V = V m e jφ = V m φ (1) Timedomain representation Phasor or frequencydomain representation V m sin(ωt + φ) = V m cos(ωt + φ 90 )
27 The Phasor 27 Inverse Phasor Transform Equation (1) states that to obtain the sinusoid corresponding to a given phasor V, multiply the phasor by the time factor e jωt and take the real part. V = V m φ = V m e jφ v(t) = Rm{V m e jφ e jωt } = V m cos(ωt + φ) V = ω = 500 rad/s v t = 115 cos(500t 45 ) V = j8e j20 = (1 90 )(8 20 ) = = 8 70 v t = 8 cos(ωt + 70 ) V = j 5 j12 = 12 + j5 = φ φ = tan = = v t = 13 cos(ωt )
28 The Phasor 28 Phasor Diagram Since a phasor has magnitude and phase ( direction ), it behaves as a vector and is printed in boldface. For example, phasors V = V m φ and I = I m θ are graphically represented in Figure. Such a graphical representation of phasors is known as a phasor diagram.
29 The Phasor 29 Phasor Diagram The relationship between two phasors at the same frequency remains constant as they rotate; hence the phase angle is constant. Consequently, we can usually drop the reference to rotation in the phasor diagrams and study the relationship between phasors simply by plotting them as vectors having a common origin and separated by the appropriate angles. V φ = φ 2 φ 1 φ V = V m φ 2 V I I = I m φ 1 φ I v(t) = V m cos(ωt + φ 2 ) Phasor Diagram φ i(t) = I m cos(ωt + φ 1 ) Finally, we should bear in mind that phasor analysis applies only when frequency is constant; it applies in manipulating two or more sinusoidal signals only if they are of the same frequency.
30 Time Domain Current i(t) = I m cos(ωt + φ) Chapter 11 Sinusoidal SteadyState Analysis The Passive Circuit Elements in the Frequency Domain < The V I Relationship for a Resistor> 30 Voltage v t = ir = RI m cos(ωt + φ) Phasor form I = I m φ V = RI = RI m φ This shows that the voltagecurrent relation for the resistor in the phasor domain continues to be Ohm s law, as in the time domain. Voltagecurrent relations for a resistor in the: (a) time domain, (b) frequency domain.
31 Phasor Diagram Chapter 11 Sinusoidal SteadyState Analysis The Passive Circuit Elements in the Frequency Domain < The V I Relationship for a Resistor> 31 V = RI = RI m φ I = I m φ I = I m 0 V = RI = RI m 0
32 Time Domain i(t) = I m cos(ωt + φ) v t Chapter 11 Sinusoidal SteadyState Analysis The Passive Circuit Elements in the Frequency Domain = L di dt = ωli m sin(ωt + φ) < The V I Relationship for an Inductor> = ωli m cos(ωt + φ 90 ) = ωli m cos(ωt + φ + 90 ) v i 32
33 Phasor form I = I m φ Chapter 11 Sinusoidal SteadyState Analysis The Passive Circuit Elements in the Frequency Domain < The V I Relationship for an Inductor> 33 v(t) = ωli m cos(ωt + φ + 90 ) V = ωli m φ + 90 = (1 90 )(ωli m φ) = jωli m φ = jωli j = 1 90 Voltagecurrent relations for an inductor in the: (a) time domain, (b) frequency domain.
34 The Passive Circuit Elements in the Frequency Domain < The V I Relationship for an Inductor> 34 di dt = ωi m cos(ωt + φ + 90 ) = Re{ωI m e jωt e jφ e j90 } = Re{jωI m e jωt e jφ } = Re{jωI e jωt } i(t) = I m cos(ωt + φ) di dt I = I m φ jωi Differentiating a sinusoid is equivalent to multiplying its corresponding phasor by jω. v t = L di dt = LjωI = jωli
35 Phasor Diagram Chapter 11 Sinusoidal SteadyState Analysis The Passive Circuit Elements in the Frequency Domain < The V I Relationship for an Inductor> 35 V = ωli m φ + 90 I = I m φ V = ωli m 90 = jωli I = I m 0 Voltage leads current by 90 in an inductor Current lags voltage by 90 in an inductor
36 Time Domain v(t) = V m cos(ωt + φ) i t Chapter 11 Sinusoidal SteadyState Analysis The Passive Circuit Elements in the Frequency Domain = C dv dt = ωcv m sin(ωt + φ) < The V I Relationship for a Capacitor> = ωcv m cos(ωt + φ 90 ) = ωcv m cos(ωt + φ + 90 ) v i 36 The current leads the voltage by 90. Voltage lags current by 90 o in a pure capacitive circuit.
37 Phasor form V = V m φ Chapter 11 Sinusoidal SteadyState Analysis The Passive Circuit Elements in the Frequency Domain < The V I Relationship for a Capacitor> 37 i(t) = ωcv m cos(ωt + φ + 90 ) I = ωcv m φ + 90 = (1 90 )(ωcv m φ) = jωcv m φ = jωcv V = 1 jωc I j = 1 90 Voltagecurrent relations for a capacitor in the: (a) time domain, (b) frequency domain.
38 Phasor Diagram Chapter 11 Sinusoidal SteadyState Analysis The Passive Circuit Elements in the Frequency Domain < The V I Relationship for a Capacitor> 38 I = ωcv m φ + 90 V = V m φ I = ωcv m 90 = jωcv V = V m 0
39 The Passive Circuit Elements in the Frequency Domain Summary 39 Summary of voltagecurrent relationships
40 Impedance Chapter 11 Sinusoidal SteadyState Analysis The Passive Circuit Elements in the Frequency Domain < Impedance, Reactance, and Admittance> 40 We obtained the voltagecurrent relations for the three passive elements as V = RI V I = R V = jωli V I = jωl From these three expressions, we obtain Ohm s law in phasor form for any type of element as Z = V I or V = ZI V = 1 jωc I V I = 1 jωc Z is a frequencydependent quantity known as impedance, measured in ohms.
41 Impedance Chapter 11 Sinusoidal SteadyState Analysis The Passive Circuit Elements in the Frequency Domain < Impedance, Reactance, and Admittance> The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms. The impedance represents the opposition which the circuit exhibits to the flow of sinusoidal current. Although the impedance is the ratio of two phasors, it is not a phasor, because it does not correspond to a sinusoidally varying quantity. The impedances of resistors, inductors, and capacitors can be readily obtained from Eq. (9.39). Table 9.1 summarizes their impedances. 41
42 Resistance and Reactance Chapter 11 Sinusoidal SteadyState Analysis The Passive Circuit Elements in the Frequency Domain < Impedance, Reactance, and Admittance> As a complex quantity, the impedance may be expressed in rectangular form as 42 Z = V I = R + jx R : Resistance X : Reactance Inductive and capacitive reactance X>0 : X<0 : Z = R + jx Z = R jx Inductive reactance Capacitive reactance Resistor : Inductor : Capacitor: Z = R Z = jωl Z = 1 jωc = j 1 ωc
43 Impedance in Polar Form Z = Z θ Z = R + jx Admittance Chapter 11 Sinusoidal SteadyState Analysis The Passive Circuit Elements in the Frequency Domain < Impedance, Reactance, and Admittance> Z = R 2 + X 2 θ = tan 1 X R It is sometimes convenient to work with the reciprocal of impedance, known as admittance. The admittance Y is the reciprocal of impedance, measured in siemens (S). θ Z jx R R = z cos θ X = Z sin θ 43 Y = 1 Z = I V As a complex quantity, we may write Y as Y = G + jb G : Conductance B : Susceptance Y = G = 1 R + jx = R R 2 + X 2 R jx R 2 + X 2 X B = R 2 + X 2
44 The Passive Circuit Elements in the Frequency Domain < Impedance, Reactance, and Admittance> 44 Y = G + jb G : Conductance B : Susceptance Y = G = 1 R + jx = R R 2 + X 2 R jx R 2 + X 2 X B = R 2 + X 2
45 Kirchhoff s Laws in the Frequency Domain 45 KVL in the frequency Domain For KVL, let v 1, v 2,, v n be the voltages around a closed loop. Then v 1 + v v n = 0 (9.36) In the sinusoidal steady state, each voltage may be written in cosine form, so that Eq. (9.36) becomes V m1 cos(ωt + θ 1 ) + V m2 cos(ωt + θ 2 ) + + V mn cos(ωt + θ n ) = 0 (9.37) We now Euler s identity to write Eq.(9.37) as Re V m1 e jθ 1e jωt } + Re{V m2 e jθ 2e jωt } + + Re{V mn e jθ ne jωt } = 0 (9.38) or Re (V m1 e jθ 1 + V m2 e jθ V mn e jθ n)e jωt } = 0 (9.39) + v 1 + v 2 If we let V k = V mk e jθ k, then Re (V 1 + V V n )e jωt } = 0 (9.40) Since e jωt 0, V 1 + V V n = 0 (9.41) indicating that Kirchhoff s voltage law holds for phasors. + v n
46 Kirchhoff s Laws in the Frequency Domain 46 KCL in the frequency Domain By following a similar procedure, we can show that Kirchhoff s current law holds for phasors. If we let i 1, i 2,, i n, in be the current leaving or entering a closed surface in a network at time t, then i 1 + i i n = 0 (9.42) If I 1, I 2,, I n, are the phasor forms of the sinusoids i 1, i 2,, i n, then I 1 + I I n = 0 (9.43) which is Kirchhoff s current law in the frequency domain. i n i 1 i 2
47 Series, Parallel, and DeltatoWye Simplifications 47 Series Connection V ab = V 1 + V V n = IZ 1 + IZ IZ n = I(Z 1 + Z Z n ) Z ab = V ab I = Z 1 + Z Z n showing that the total or equivalent impedance of seriesconnected impedances is the sum of the individual impedances. This is similar to the series connection of resistances.
48 Series, Parallel, and DeltatoWye Simplifications 48 Parallel Connection I = I 1 + I I n = V 1 Z Z Z n = V Z ab 1 Z ab = 1 Z Z Z n Y ab = Y 1 + Y Y n This indicates that the equivalent admittance of a parallel connection of admittances is the sum of the individual admittances.
49 Series, Parallel, and DeltatoWye Simplifications 49 DeltatoWye Transformation Z b Z c Z 1 =, (9.51) Z a + Z b + Z c Z 2 = Z 3 = Z c Z a Z a + Zb + Zc, (9.52) Z a Z b Z a + Zb + Zc, (9.53) Z a = Z 1Z 2 + Z 2 Z 3 + Z 3 Z 1 Z 1, (9.54) Z b = Z 1Z 2 + Z 2 Z 3 + Z 3 Z 1 Z 2, 9.55 Z c = Z 1Z 2 + Z 2 Z 3 + Z 3 Z 1 Z 3, (9.56)
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