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Sinusoids & Phasors. A sinusoidal current is usually referred to as alternating current (ac). Circuits driven by sinusoidal current or voltage sources.

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Presentation on theme: "Sinusoids & Phasors. A sinusoidal current is usually referred to as alternating current (ac). Circuits driven by sinusoidal current or voltage sources."— Presentation transcript:

1 Sinusoids & Phasors

2 A sinusoidal current is usually referred to as alternating current (ac). Circuits driven by sinusoidal current or voltage sources are called ac circuits.

3 Sinusoids

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5 Cyclic Frequency of the sinusoid: in radians per second (rad/s), f in Hertz (Hz) ω = 2πf in radians per second (rad/s), f in Hertz (Hz) v(t) = V m sin(ωt + φ) – where (ωt + φ) is the argument and φ is the phase. – Both argument and phase can be in radians or degrees.

6 v 1 (t) = V m sin ωt and v 2 (t) = V m sin(ωt + φ) Therefore, we say that v 2 leads v 1 by φ or that v 1 lags v 2 by φ. If φ ≠ 0, we also say that v 1 and v 2 are out of phase. If φ = 0, then v 1 and v 2 are said to be in phase

7 sin(A ± B) = sinAcosB ± cosAsinB cos(A ± B) = cosAcosB ∓ sinAsinB sin(ωt ± 180 ◦ ) = −sin ωt cos(ωt ± 180 ◦ ) = −cos ωt sin(ωt ± 90 ◦ ) = ±cos ωt cos(ωt ± 90 ◦ ) = ∓ sin ωt

8 subtracting 90 ◦ from the argument of cos ωt gives sin ωt, or cos(ωt−90 ◦ ) = sin ωt. adding 180 ◦ to the argument of sin ωt gives −sin ωt, or sin(ωt − 180 ◦ ) = −sin ωt

9 Acos ωt + B sin ωt = C cos(ωt − θ) 3 cos ωt − 4 sin ωt = 5 cos(ωt + 53.1 ◦ )

10 Problems Find the amplitude, phase, period, and frequency of the sinusoid v(t) = 12 cos(50t + 10 ◦ ) Calculate the phase angle between v 1 = −10 cos(ωt + 50 ◦ ) and v 2 =12 sin(ωt − 10 ◦ ). State which sinusoid is leading.

11 Phasors Phasors provide a simple means of analyzing linear circuits excited by sinusoidal sources A complex number z can be written in rectangular form as z = x + jy

12 z = x + jy where x is the real part of z; y is the imaginary part of z. In this context, the variables x and y do not represent a location as in two-dimensional vector analysis but rather the real and imaginary parts of z in the complex plane.

13 The complex number z can also be written in polar or exponential z = r  φ = re jφ where r is the magnitude of z, and φ is the phase of z. z can be represented in three ways: – z = x + jy Rectangular form – z = r  φ Polar form – z = re jφ Exponential form

14 z = r  φ = re jφ x = r cos φ y = r sin φ Thus, z may be written as z = x + jy = r  φ = r(cos φ + j sin φ)

15 Given the complex numbers z = x + jy, z 1 = x 1 + jy 1 = r 1  φ 1 and z 2 = x 2 + jy 2 = r 2  φ 2 Addition: z 1 + z 2 = (x 1 + x 2 ) + j (y 1 + y 2 ) Subtraction: z 1 − z 2 = (x 1 − x 2 ) + j (y 1 − y 2 ) Multiplication: z 1 z 2 = r 1 r 2  (φ 1 + φ 2 )

16 Given the complex numbers z = x + jy, z 1 = x 1 + jy 1 = r 1  φ 1 and z 2 = x 2 + jy 2 = r 2  φ 2 Division: Reciprocal: Square Root:

17 Problems

18 The idea of phasor representation is based on Euler’s identity:

19 V thus the phasor representation of the sinusoid v(t), as we said earlier. In other words, a phasor is a complex representation of the magnitude and phase of a sinusoid.

20 consider the plot of the sinor Ve jωt = Vme j (ωt+φ) on the complex plane. As time increases, the sinor rotates on a circle of radius Vm at an angular velocity ω in the counterclockwise direction, as shown in Fig. 9.7(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 sinor Ve jωt on the real axis, as shown in Fig. 9.7(b). The value of the sinor at time t = 0 is the phasor V of the sinusoid v(t). The sinor may be regarded as a rotating phasor. Thus, whenever a sinusoid is expressed as a phasor, the term e jωt is implicitly present.

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27 Problems

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31 Phasor Relationships for Circuit Elements We begin with the resistor. If the current through a resistor R is i = I m cos(ωt + φ), the voltage across it is given by Ohm’s law as

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33 For the inductor L, assume the current through it is i = I m cos(ωt + φ), The voltage across the inductor is

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35 For the capacitor C, assume the voltage across it is v = V m cos(ωt + φ). The current through the capacitor is showing that the current and voltage are 90 ◦ out of phase. To be specific, the current leads the voltage by 90 ◦. Figure 9.13 shows the voltage-current relations for the capacitor; Fig. 9.14 gives the phasor diagram

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38 Problem

39 Impedance & Admittance

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47 Problem Find v(t) and i(t) in the circuit

48 KIRCHHOFF’S LAWS IN THE FREQUENCY DOMAIN

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51 Impedance Combination

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57 Problems Find the input impedance of the circuit. Assume that the circuit operates at ω = 50 rad/s.

58 Determine v o (t) in the circuit

59 Find I in the circuit:

60 Application: Phase-Shifters

61 Design an RC circuit to provide a phase of 90 ◦ leading.

62 For the RL circuit, calculate the amount of phase shift produced at 2 kHz.


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