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SJTU1 Chapter 9 Sinusoids and Phasors

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SJTU2 Sinusoids A sinusoid is a signal that has the form of the sine or cosine function.

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SJTU3 t radians/second (rad/s) f is in hertz(Hz)

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SJTU4 Phase difference:

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SJTU5 Complex Number

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SJTU6 Phasor a phasor is a complex number representing the amplitude and phase angle of a sinusoidal voltage or current. Eq.(8-1) and Eq. (8-2) Eq.(8-3)

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SJTU7 When Eq.(8-2) is applied to the general sinusoid we obtain E q.(8-4) The phasor V is written as Eq.(8-5)

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SJTU8 Fig. 8-1 shows a graphical representation commonly called a phasor diagram. Fig. 8-1: Phasor diagram Two features of the phasor concept need emphasis: 1.Phasors are written in boldface type like V or I1 to distinguish them from signal waveforms such as v(t) and i1(t). 2.A phasor is determined by amplitude and phase angle and does not contain any information about the frequency of the sinusoid.

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SJTU9 In summary, given a sinusoidal signal, the corresponding phasor representation is. Conversely, given the phasor, the corresponding sinusoid is found by multiplying the phasor by and reversing the steps in Eq.(8-4) as follows: E q.(8-6) Time domain representation Phase-domain representation

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SJTU10 Properties of Phasors additive property Eq.(8-7) Eq.(8-8) Eq.(8-9)

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SJTU11 derivative property Eq.(8-10) Time domain representation Phase-domain representation

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SJTU12 Integral property Time domain representation Phase-domain representation The differences between v(t) and V: 1.V(t) is the instantaneous or time-domain representation, while V is the frequency or phasor-domain representation. 2.V(t) is a real signal which is time dependent, while V is just a supposed value to simplify the analysis

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SJTU13 The complex exponential is sometimes called a rotating phasor, and the phasor V is viewed as a snapshot of the situation at t=0. Fig. 8-2: Complex exponential

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SJTU15 EXAMPLE 8-1 (a)Construct the phasors for the following signals: (b) Use the additive property of phasors and the phasors found in (a) to find v(t)=v1(t)+v2(t). SOLUTION (a) The phasor representations of v(t)=v1(t)+ v2(t) are

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SJTU16 (b) The two sinusoids have the same frequent so the additive property of phasors can be used to obtain their sum: The waveform corresponding to this phasor sum is V1V1 V2V2 1 j V

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SJTU17 EXAMPLE 8-2 (a)Construct the phasors representing the following signals: (b) Use the additive property of phasors and the phasors found in (a) to find the sum of these waveforms. SOLUTION: (a) The phasor representation of the three sinusoidal currents are

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SJTU18 (b) The currents have the same frequency, so the additive property of phasors applies. The phasor representing the sum of these current is Fig. 8-4

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SJTU19 EXAMPLE 8-3 Use the derivative property of phasors to find the time derivative of v(t)=15 cos(200t-30° ). The phasor for the sinusoid is V= °. According to the derivative property, the phasor representing the dv/dt is found by multiplying V by j. SOLUTION: The sinusoid corresponding to the phasor j V is

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SJTU20 Device Constraints in Phasor Form Voltage-current relations for a resistor in the: (a) time domain, (b) frequency domain. Resistor: Re jIm I V 0

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SJTU21 Device Constraints in Phasor Form Inductor: Voltage-current relations for an inductor in the: (a) time domain, (b) frequency domain.

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SJTU22 Device Constraints in Phasor Form Capacitor: Voltage-current relations for a capacitor in the: (a) time domain, (b) frequency domain.

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SJTU23 Connection Constraints in Phasor Form KVL in time domain Kirchhoff's laws in phasor form (in frequency domain) KVL: The algebraic sum of phasor voltages around a loop is zero. KCL: The algebraic sum of phasor currents at a node is zero.

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SJTU24 The Impedance Concept The IV constraints are all of the form V=ZI or Z= V/I Eq.(8-16) where Z is called the impedance of the element The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, measured in ohms( ) The impedance is inductive when X is positive is capacitive when X is negative

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SJTU25 The Impedance Concept

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SJTU26 EXAMPLE 8-5 Fig. 8-5 The circuit in Fig. 8-5 is operating in the sinusoidal steady state with and. Find the impedance of the elements in the rectangular box. SOLUTION:

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SJTU28 The Admittance Concept The admittance Y is the reciprocal of impedance, measured in siemens (S) Y=G+jB Where G=Re Y is called conductance and B=Im Y is called the susceptance How get Y=G+jB from Z=R+jX ?

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SJTU30 Basic Circuit Analysis with Phasors Step 1: The circuit is transformed into the phasor domain by representing the input and response sinusoids as phasor and the passive circuit elements by their impedances. Step 2: Standard algebraic circuit techniques are applied to solve the phasor domain circuit for the desired unknown phasor responses. Step 3: The phasor responses are inverse transformed back into time- domain sinusoids to obtain the response waveforms.

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SJTU31 Series Equivalence And Voltage Division where R is the real part and X is the imaginary part

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SJTU32 EXAMPLE 8-6 Fig. 8-8 The circuit in Fig is operating in the sinusoidal steady state with (a) Transform the circuit into the phasor domain. (b) Solve for the phasor current I. (c) Solve for the phasor voltage across each element. (d) Construct the waveforms corresponding to the phasors found in (b) and (c)

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SJTU33 SOLUTION:

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SJTU34 PARALLEL EQUIVALENCE AND CURRENT DIVISION Rest of the circuit Y1 Y1Y1 Y2Y2 YNYN I V I1I1 I2I2 I3I3 phasor version of the current division principle

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SJTU35 EXAMPLE 8-9 Fig The circuit in Fig is operating in the sinusoidal steady state with i S (t)=50cos2000t mA. (a) Transform the circuit into the phasor domain. (b) Solve for the phasor voltage V. (c) Solve for the phasor current through each element. (d) Construct the waveforms corresponding to the phasors found in (b) and (c).

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SJTU36 SOLUTION: (a) The phasor representing the input source current is Is=0.05 0° A. The impedances of the three passive elements are Fig. 8-14

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SJTU37 And the voltage across the parallel circuit is The sinusoidal steady-state waveforms corresponding to the phasors in (b) and (c) are The current through each parallel branch is

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SJTU38 EXAMPLE 8-10 Fig Find the steady-state currents i(t), and i C (t) in the circuit of Fig (for Vs=100cos2000t V, L=250mH, C=0.5 F, and R=3k ). SOLUTION: Vs=100 0°

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SJTU41 Y TRANSFORMATIONS The equations for the to Y transformation are

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SJTU42 The equations for a Y-to- transformation are when Z 1 =Z 2 =Z 3 =Z Y or Z A =Z B =Z C =Z N. Z Y =Z N /3 and Z N =3Z Y balanced conditions

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SJTU43 EXAMPLE 8-12 Use a to Y transformation to solve for the phasor current I X in Fig Fig SOLUTION: ABC to Y

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