Chapter 9 Sinusoids and Phasors SJTU

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

t  radians/second (rad/s) f is in hertz(Hz) SJTU

Phase difference: SJTU

Complex Number  SJTU

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

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

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

Phase-domain representation Time domain representation 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: Eq.(8-6) Phase-domain representation Time domain representation SJTU

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

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

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

Fig. 8-2: Complex exponential 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 SJTU

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Construct the phasors for the following signals: EXAMPLE 8-1 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 SJTU

The waveform corresponding to this phasor sum is (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 j V2 1 V V1 SJTU

Construct the phasors representing the following signals: EXAMPLE 8-2 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 SJTU

(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 SJTU

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

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

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

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

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. SJTU

The Impedance Concept V=ZI or Z= V/I Eq.(8-16) 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 SJTU

The Impedance Concept SJTU

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: SJTU

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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 ? SJTU

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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. SJTU

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

(a) Transform the circuit into the phasor domain. EXAMPLE 8-6 Fig. 8-8 The circuit in Fig. 8 - 8 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) SJTU

SOLUTION: SJTU

PARALLEL EQUIVALENCE AND CURRENT DIVISION Rest of the circuit Y1 Y2 YN I V I1 I2 I3 phasor version of the current division principle SJTU

(a) Transform the circuit into the phasor domain. EXAMPLE 8-9 Fig. 8-13 The circuit in Fig. 8-13 is operating in the sinusoidal steady state with iS(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). SJTU

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 SJTU

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

EXAMPLE 8-10 Fig. 8-15   Find the steady-state currents i(t), and iC(t) in the circuit of Fig. 8-15 (for Vs=100cos2000t V, L=250mH, C=0.5  F, and R=3k ). SOLUTION: Vs=100∠0° SJTU

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Y←→△ TRANSFORMATIONS The equations for the △ to Y transformation are SJTU

The equations for a Y-to- △ transformation are when Z1=Z2=Z3=ZY or ZA=ZB=ZC=ZN. ZY=ZN /3 and ZN =3ZY balanced conditions SJTU

EXAMPLE 8-12 Use a △ to Y transformation to solve for the phasor current IX in Fig. 8-18. SOLUTION: ABC △ to Y Fig. 8-18 SJTU

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