Chapter 10: Sinusoids and Phasors

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Chapter 10: Sinusoids and Phasors 1. Motivation 2. Sinusoid Features 3. Phasors 4. Phasor Relationships for Circuit Elements 5. Impedance and Admittance 6. Kirchhoff s Laws in the Frequency Domain 7. Impedance Combinations 8. Application: Phase-Shifters 9. Summary 1

10.1 Motivation (1) How to determine v(t) and i(t)? v s (t) = 10 V DC AC How can we apply what we have learned before to determine i(t) and v(t)? 2

10.1 Motivation (2) The output of the circuit will consist of two parts: a transient part that dies out as time increases; a steady-state part that persists. Typically, the transient part dies out quickly, perhaps in a couple of milliseconds. AC circuits are the subject of this chapter. In particular, It's useful to associate a complex number with a sinusoid. Doing so allows us to define phasors and impedances. Using phasors and impedances, we obtain a new representation of the linear circuit, called the frequencydomain representation. We analyze ac circuits in the frequency domain to determine their steady-state response.

10.2 Sinusoids (1) A sinusoid is a signal that has the form of the sine or cosine function. A general expression for the sinusoid, where V m = the amplitude of the sinusoid ω = the angular frequency in radians/s ωt = the argument of the sinusoid = the phase v(t) V m sin( t ) period: T 2 4

10.2 Sinusoids (2) A periodic function is one that satisfies v(t) = v(t + nt), for all t and for all integers n. f 1 Hz T 2 f v 2 leads v 1 by or v 1 lags v 2 by Only two sinusoidal values with the same frequency can be compared by their amplitude and phase difference. If phase difference is zero, they are in phase; if phase difference is not zero, they are out of phase. 5

10.2 Sinusoids (3) Example: Given a sinusoid, 5sin(4 t 60 ), calculate its amplitude, phase, angular frequency, period, and frequency. Solution: Amplitude = 5, phase = 60 o, angular frequency = 4 π rad/s, period = 0.5 s, frequency = 2 Hz. Example: Find the phase angle between i 2 o 5cos(377t 40 ) and o i 1 4 sin(377t 25 ), does i 1 lead or lag i 2? Solution: Since sin(ωt+90 o ) = cos ωt i 1 leads i 2 by 155 o. 6

10.3 Phasor (1) A phasor is a complex number that represents the amplitude and phase of a sinusoid. Phasors may be used when the circuit is linear, the steady-state response is sought, and all independent sources are sinusoidal and have the same frequency. It can be represented in one of the following : a. Rectangular b. Polar c. Exponential j i(t) Re{I m e e where I is called a phasor. j t } I I m e z x jy r(cos j sin ) z r z re j where j r x 2 y 2 tan 1 y x I m 7

Mathematic operation of complex number: 1. Addition 2. Subtraction 10.3 Phasor (2) 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 ) 3. Multiplication 4. Division 5. Reciprocal 6. Square root 7. Complex conjugate z 1 z 2 r 1 r 2 1 2 z 1 r 1 z r 2 2 1 1 z r 1 2 z r 2 z x jy r re j 8. Euler s identity e j cos j sin 8

10.3 Phasor (3) 9

Transform a sinusoid from the time domain to the phasor domain: v(t) V m cos( t ) (time domain) 10.3 Phasor (4) V V m (phasor domain) Amplitude and phase difference are two principal concerns in the study of voltage and current sinusoids. Phasor will be defined from the cosine function in all our proceeding study. If a voltage or current expression is in the form of a sine, it will be changed to a cosine by subtracting from the phase. 10

v(t) dv dt 10.3 Phasor (5) V V j V Laplace transform f(t)e st L f (t) F(s) dt 0 vdt V j No initial cond. & set s = jω The differences between v(t) and V: v(t) is instantaneous or time-domain representation V is the frequency or phasor-domain representation. v(t) is time dependent, V is not. v(t) is always real with no complex term, V is generally complex. Note: Phasor analysis applies only when frequency is constant; when it is applied to two or more sinusoid signals only if they have 11 the same frequency.

10.3 Phasor (6) Example: Use phasor approach, determine the current i(t) in a circuit described by the integro-differential equation. di 4i 8 idt 3 dt 50cos(2t 75 ) a1 50; a2 4.642 Solution: 1 75 ; 2 143.2 12

10.3 Phasor (7) We can derive the differential equations for the following circuit in order to solve for v o (t) in phase domain V o. 2 d vo 5 dv 0 dt 2 3 dt 20v0 400 sin(4t 15 o ) 3 However, the derivation may sometimes be very tedious. Instead of first deriving the differential equation and then transforming it into phasor to solve for V o, we can transform all the RLC components into phasor first, then apply the KCL laws and other theorems to set up a phasor equation involving V o directly. 13

10.4 Phasor Relationships for Circuit Elements (1) Resistor: Inductor: Capacitor: V leads I by 90 I leads V by 90 14

10.4 Phasor Relationships for Circuit Elements (2) Example: If voltage v(t) = 6 cos(100t 30 o ) is applied to a 50 μf capacitor, calculate the current, i(t), through the capacitor. Answer: i(t) = 30 cos(100t + 60 o ) ma 15

16

10.5 Impedance and Admittance (1) The impedance Z of a circuit is the ratio of the phasor voltage V to the phasor current I, Z V R jx I measured in ohms Ω. where R = Re(Z) is the resistance and X = Im(Z) is the reactance. Positive X is for L (or lagging) and negative X is for C (or leading). The admittance Y is the reciprocal of impedance, Y 1 I G jb Unit: siemens (S) Z V 17

10.5 Impedance and Admittance (2) After we know how to convert RLC components from time to phasor domain, we can transform a time domain circuit into a phasor/frequency domain circuit. Hence, we can apply the KCL laws and other theorems to directly set up phasor equations involving our target variables for solving. Example: Determine v(t) and i(t). v s 5cos(10t) V Answers: i(t) = 1.118 cos(10t 26.56 o ) A; v(t) = 2.236 cos(10t + 63.43 o ) V 18

10.6 Kirchhoff s Laws in the Freq. Domain (1) Both KVL and KCL are hold in the phasor domain or more commonly called frequency domain. Moreover, the variables to be handled are phasors, which are complex numbers. All the mathematical operations involved are now in complex domain. same frequency complex numbers 19

10.6 Kirchhoff s Laws in the Freq. Domain (2) 20

10.7 Impedance Combinations (1) The following principles used for DC circuit analysis all apply to AC circuit. For example: a. voltage division b. current division c. circuit reduction d. impedance equivalence 21

10.7 Impedance Combinations (2) Example: Find the input impedance at ω = 50 rad/s. Example: Determine the input impedance at ω = 10 rad/s. Answer: Z in = 32.38 j73.76 22

10.8 Application: Phase-Shifters Series RC shift circuits: - Leading output - Lagging output

10.9 Summary (1) With the pervasive use of ac electric power in the home and industry, it is important for engineers to analyze circuits with sinusoidal independent sources. The steady-state response of a linear circuit to a sinusoidal input is itself a sinusoid having the same frequency as the input signal. Circuits that contain inductors and capacitors are represented by differential equations. When the input to the circuit is sinusoidal, the phasors and impedances can be used to represent the circuit in the frequency domain. In the frequency domain, the circuit is represented by algebraic equations. The steady-state response of a linear circuit with a sinusoidal input is obtained as follows: 1. Transform the circuit into the frequency domain, using phasors and impedances.

10.9 Summary (2) 2. Represent the frequency-domain circuit by algebraic equation, for example, mesh or node equations. 3. Solve the algebraic equations to obtain the response of the circuit. 4. Transform the response into the time domain, using phasors. A circuit contains several sinusoidal sources, two cases: When all of the sinusoidal sources have the same frequency, the response will be a sinusoid with that frequency, and the problem can be solved in the same way that it would be if there was only one source. When the sinusoidal sources have different frequencies, superposition is used to break the time-domain circuit up into several circuits, each with sinusoidal inputs all at the same frequency. Each of the separate circuits is analyzed separately and the responses are summed in the time domain.