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Phasors: Sinusoidal Steady State and the Series RLC Circuit

Module by: Louis Scharf. E-mail the author

Note:

This module is part of the collection, A First Course in Electrical and Computer Engineering. The LaTeX source files for this collection were created using an optical character recognition technology, and because of this process there may be more errors than usual. Please contact us if you discover any errors.

Phasors may be used to analyze the behavior of electrical and mechanical systems that have reached a kind of equilibrium called sinusoidal steady state. In the sinusoidal steady state, every voltage and current (or force and velocity) in a system is sinusoidal with angular frequency ωω. However, the amplitudes and phases of these sinusoidal voltages and currents are all different. For example, the voltage across a resistor might lead the voltage across a capacitor by 9090 ( π2π2 radians) and lag the voltage across an inductor by 90(π2radians)90(π2radians).

In order to make our application of phasors to electrical systems concrete, we consider the series RLC circuit illustrated in Figure 1. The arrow labeled i(t)i(t) denotes a current that flows in response to the voltage applied,and the + and - on the voltage source indicate that the polarity of the applied voltage is positive on the top and negative on the bottom. Our convention is that current flows from positive to negative, in this case clockwise in the circuit.

Figure 1: Series RLC Circuit
Figure one is a circuit diagram. It is rectangular in shape. On the top of the rectangle is a resistor portion, indicated by sharp zig-zagged lines and the R label. On the right side of the rectangle is a capacitor, signified by a break in the rectangle with the ends of the broken rectangle drawn with small horizontal line segments, and the label C. On the bottom of the rectangle is an inductor portion, signified by a squiggly line and the label L. On the left side of the rectangle is a voltage source portion, signified by a circle breaking the path of the rectangle with a small zig-zag line inside the circle, the label V(t), and a plus sign above the circle and a minus sign below the circle.

We will assume that the voltage source is an audio oscillator that pro- duces the voltage

V(t)=Acos(ωt+φ).V(t)=Acos(ωt+φ).
(1)

We represent this voltage as the complex signal

V ( t ) A e j φ e j ω t V ( t ) A e j φ e j ω t
(2)

and give it the phasor representation

V(t)V;V=Aejφ.V(t)V;V=Aejφ.
(3)

We then describe the voltage source by the phasor V and remember that we can always compute the actual voltage by multiplying by ejωtejωt and taking the real part:

V ( t ) = Re { V e j ω t } . V ( t ) = Re { V e j ω t } .
(4)

Exercise 1

Show that Re [Vejωt]=Acos(ωt+φ) Re [Vejωt]=Acos(ωt+φ) when V=Aejφ.V=Aejφ.

Circuit Laws. In your circuits classes you will study the Kirchhoff laws that govern the low frequency behavior of circuits built from resistors (R)(R), inductors (L)(L), and capacitors (C)(C). In your study you will learn that the voltage dropped across a resistor is related to the current that flows through it by the equation

V R ( t ) = R i ( t ) . V R ( t ) = R i ( t ) .
(5)

You will learn that the voltage dropped across an inductor is proportional to the derivative of the current that flows through it, and the voltage dropped across a capacitor is proportional to the integral of the current that flows through it:

V L ( t ) = L d i d t ( t ) V C ( t ) = 1 C i ( t ) d t . V L ( t ) = L d i d t ( t ) V C ( t ) = 1 C i ( t ) d t .
(6)

Phasors and Complex Impedance. Now suppose that the current in the preceding equations is sinusoidal, of the form

i ( t ) = B cos ( ω t + θ ) . i ( t ) = B cos ( ω t + θ ) .
(7)

We may rewrite i(t)i(t) as

i(t)= Re {Iejωt}i(t)= Re {Iejωt}
(8)

where I I is the phasor representation of i(t)i(t).

Exercise 2

Find the phasor I I in terms of B B and θ θ in Equation 8.

The voltage dropped across the resistor is

V R ( t ) = R i ( t ) = R Re { I e j ω t } = Re { R I e j ω t } . V R ( t ) = R i ( t ) = R Re { I e j ω t } = Re { R I e j ω t } .
(9)

Thus the phasor representation for VR(t)VR(t) is

VR(t)VR;VR=RI.VR(t)VR;VR=RI.
(10)

We call R the impedance of the resistor because R is the scale constant that relates the “phasor voltage VR' to the “phasor current I.”

The voltage dropped across the inductor is

V L ( t ) = L d i d t ( t ) = L d d t Re { I e j ω t } . V L ( t ) = L d i d t ( t ) = L d d t Re { I e j ω t } .
(11)

The derivative may be moved through the Re [ ] Re [ ] operator (see Exercise 3) to produce the result

V L ( t ) = L Re { j ω I e j ω t } = Re { j ω L I e j ω t } . V L ( t ) = L Re { j ω I e j ω t } = Re { j ω L I e j ω t } .
(12)

Thus the phasor representation of V L ( t ) V L (t)

VL(t)VL;VL=jωLI.VL(t)VL;VL=jωLI.
(13)

We call jωLjωL the impedance of the inductor because jωLjωL is the complex scale constant that relates “phasor voltage V L V L ' to “phasor current I I.”

Exercise 3

Prove that the operators ddtddt and Re [] Re [] commute:

d d t Re { e j ω t } = Re { d d t e j ω t } . d d t Re { e j ω t } = Re { d d t e j ω t } .
(14)

The voltage dropped across the capacitor is

V C ( t ) = 1 C i ( t ) d t = 1 C Re { I e j ω t } d t . V C ( t ) = 1 C i ( t ) d t = 1 C Re { I e j ω t } d t .
(15)

The integral may be moved through the Re [ ] Re [ ] operator to produce the result

V C ( t ) = 1 C Re { I j ω e j ω t } = Re { I j ω C e j ω t } . V C ( t ) = 1 C Re { I j ω e j ω t } = Re { I j ω C e j ω t } .
(16)

Thus the phasor representation of V C ( t ) V C (t) is

VC(t)VC;VC=IjωCVC(t)VC;VC=IjωC
(17)

We call 1jωC1jωC the impedance of the capacitor because 1jωC1jωC is the complex scale constant that relates “phasor voltage V C V C " to “phasor current I I.”

Kirchhoff's Voltage Law. Kirchhoff's voltage law says that the voltage dropped in the series combination of RR, L L, and C C illustrated in Figure 1 equals the voltage generated by the source (this is one of two fundamental conservation laws in circuit theory, the other being a conservation law for current):

V ( t ) = V R ( t ) + V L ( t ) + V C ( t ) . V ( t ) = V R ( t ) + V L ( t ) + V C ( t ) .
(18)

If we replace all of these voltages by their complex representations, we have

Re { V e j ω t } = Re { ( V R + V L + V C ) e j ω t } . Re{V e j ω t }=Re{( V R + V L + V C ) e j ω t }.
(19)

An obvious solution is

V = V R + V L + V C = ( R + j ω L + 1 j ω C ) I V = V R + V L + V C = ( R + j ω L + 1 j ω C ) I
(20)

where I is the phasor representation for the current that flows in the circuit. This solution is illustrated in Figure 2, where the phasor voltages RI,jωLIRI,jωLI, and 1jωCI1jωCI are forced to add up to the phasor voltage V V.

Figure 2: Phasor Addition to Satisfy Kirchhoff's Law
Figure two is a cartesian graph with a circle and four line segments beginning at the origin. There are two line segments that begin at the origin and extend past the edge of the circle out into the first quadrant of the graph. The lower segment is labeled RI, and the second is labeled V. A third segment extends past the circe into the second quadrant, and is labeled jωLI. A fourth segment extends into the fourth quadrant, but is much shorter than the other segments and does not reach the edge of the circle. It is labeled 1/(jωC) I.

Exercise 4

Redraw Figure 2 for R=ωL=1ωC=1R=ωL=1ωC=1.

Impedance. We call the complex number R+jωL+1jωCR+jωL+1jωC the complex impedance for the series RLC network because it is the complex number that relates the phasor voltage V V to the phasor current I I:

V = Z I Z = R + j ω L + 1 j ω C . V = Z I Z = R + j ω L + 1 j ω C .
(21)

The complex number Z Z depends on the numerical values of resistance (R)(R), inductance (L)(L), and capacitance (C)(C), but it also depends on the angular frequency (ω)(ω) used for the sinusoidal source. This impedance may be manipulated as follows to put it into an illuminating form:

Z = R + j ( ω L - 1 ω C ) = R + j L C ( ω L C - 1 ω L C ) . Z = R + j ( ω L - 1 ω C ) = R + j L C ( ω L C - 1 ω L C ) .
(22)

The parameter ω 0 = 1 L C ω 0 = 1 L C is a parameter that you will learn to call an "undamped natural frequency" in your more advanced circuits courses. With it, we may write the impedance as

Z = R + j ω 0 L ( ω ω 0 - ω 0 ω ) . Z = R + j ω 0 L ( ω ω 0 - ω 0 ω ) .
(23)

The frequency ω ω 0 ω ω 0 is a normalized frequency that we denote by ν ν. Then the impedence, as a function of normalized frequency, is

Z ( ν ) = R + j ω 0 L ( ν - 1 ν ) . Z ( ν ) = R + j ω 0 L ( ν - 1 ν ) .
(24)

When the normalized frequency equals one ( ν = 1 ) (ν=1), then the impedance is entirely real and Z = R Z=R. The circuit looks like it is a single resistor.

| Z ( ν ) | = R [ 1 + ( ω 0 L R ) 2 ( ν - 1 ν ) 2 ] 1 / 2 . arg Z(ν)=tan-1ω0LR(ν-1ν). | Z ( ν ) | = R [ 1 + ( ω 0 L R ) 2 ( ν - 1 ν ) 2 ] 1 / 2 . arg Z(ν)=tan-1ω0LR(ν-1ν).
(25)

The impedance obeys the following symmetries around ν=1ν=1:

Z ( ν ) = Z * ( 1 ν ) | Z ( ν ) | = | Z ( 1 ν ) | a r g Z ( ν ) = - a r g Z ( 1 ν ) . Z ( ν ) = Z * ( 1 ν ) | Z ( ν ) | = | Z ( 1 ν ) | a r g Z ( ν ) = - a r g Z ( 1 ν ) .
(26)

In the next paragraph we show how this impedance function influences the current that flows in the circuit.

Resonance. The phasor representation for the current that flows the current that flows in the series RLC circuit is

I = V Z ( ν ) = 1 | Z ( ν ) | e - j arg Z ( ν ) V I = V Z ( ν ) = 1 | Z ( ν ) | e - j arg Z ( ν ) V
(27)

The function H ( ν ) = 1 Z ( ν ) H(ν)= 1 Z ( ν ) displays a "resonance phenomenon." that is, | H ( ν ) | |H(ν)| peaks at ν = 1 ν=1 and decreases to zero and ν = 0 ν=0 and ν = ν=:

| H ( ν ) | = 0 , ν = 0 1 R ν = 1 0 , ν = . | H ( ν ) | = 0 , ν = 0 1 R ν = 1 0 , ν = .
(28)

When | H ( ν ) | = 0 |H(ν)|=0, no current flows.

The function |H(ν)||H(ν)| is plotted against the normalized frequency ν=ν=ωω0ωω0 in Figure 3.14. The resonance peak occurs at ν=1ν=1, where |H(ν)|=|H(ν)|=1R1R meaning that the circuit looks purely resistive. Resonance phenomena underlie the frequency selectivity of all electrical and mechanical networks.

Figure 3: Resonance in a Series RLC Circuit
Figure three is a bell curve plotted on a cartesian graph. The curve is centered at a horizontal value of 1, and its tails extend past 10 to the right and 0.1 to the left. The curve's maximum value is labeled as 1/R. To the right of the graph is an equation that reads v = ω/ω_0. Above the graph is a title that reads, | H (v) | versus v.

Exercise 5

(MATLAB) Write a MATLAB program to compute and plot |H(ν)||H(ν)| and argH(ν)argH(ν) versus ν ν for ν ν ranging from 0.1 to 10 in steps of 0.1. Carry out your computations for ω0LR=10,1,0.1ω0LR=10,1,0.1, and 0.01, and overplot your results.

Circle Criterion and Power Factor. Our study of the impedance Z(ν)Z(ν) and the function H(ν)=1Z(ν)H(ν)=1Z(ν) brings insight into the resonance of an RLC circuit and illustrates the ffequency selectivity of the circuit. But there is more that we can do to illuminate the behavior of the circuit.

V = R I + j ( ω L - 1 ω C ) I . V = R I + j ( ω L - 1 ω C ) I .
(29)

This equation shows how voltage is divided between resistor voltage RI and inductor-capacitor voltage j ( ω L - 1 ω C ) I . j(ωL- 1 ω C )I.

V = R I + j ω 0 L ( ω ω 0 - ω 0 ω ) I V = R I + j ω 0 L ( ω ω 0 - ω 0 ω ) I
(30)

or

V = R I + j ω 0 L R ( ν - 1 ν ) R I . V = R I + j ω 0 L R ( ν - 1 ν ) R I .
(31)

In order to simplify our notation, we can write this equation as

V = V R + j k ( ν ) V R V = V R + j k ( ν ) V R
(32)

where V R V R is the phasor voltage RIRI and k(ν)k(ν) is the real variable

k ( ν ) = ω 0 L R ( ν - 1 ν ) . k(ν)= ω 0 L R (ν- 1 ν ).
(33)

Equation 32 brings very important geometrical insights. First, even though the phasor voltage V R V R in the RLC circuit is complex, the terms V R V R and jk(ν)VRjk(ν)VR are out of phase by π2π2 radians. This means that, for every allowable value of V R V R , the corresponding jk(ν)VRjk(ν)VR must add in a right triangle to produce the source voltage V V. This is illustrated in Figure 4(a). As the frequency ν ν changes, then k(ν)k(ν) changes, producing other values of V R V R and jk(ν)VRjk(ν)VR that sum to V V. Several such solutions for V R V R and jk(ν)VRjk(ν)VR are illustrated in Figure 3.15(b). From the figure we gain the clear impression that the phasor voltage V >R V >R lies on a circle of radius V2V2 centered at V2V2 Let's try this solution,

V R = V 2 + V 2 e j ψ = V 2 ( 1 + e j ψ ) , V R = V 2 + V 2 e j ψ = V 2 ( 1 + e j ψ ) ,
(34)

and explore its consequences. When this solution is substituted into Equation 32, the result is

V = V 2 ( 1 + e j ψ ) + j k ( ν ) V 2 ( 1 + e j ψ ) V = V 2 ( 1 + e j ψ ) + j k ( ν ) V 2 ( 1 + e j ψ )
(35)

or

2 = ( 1 + e j ψ ) [ 1 + j k ( ν ) ] . 2 = ( 1 + e j ψ ) [ 1 + j k ( ν ) ] .
(36)
Figure 4: The Components of V;(a)V;(a) Addition of VR and jk(ν)VRjk(ν)VR to Produce V, and (b) Several Values of VR and jk(ν)VRjk(ν)VR that Produce V
(a) (b)
Figure four part a is a cartesian graph with three line segments. Two segments point out from the origin, one moving far into the first quadrant, with a label V at its end, and another moving a smaller distance into the fourth quadrant, with a label V_R at its end. A third line segment connects beginning from the end of the line segment in the fourth quadrant and ending at the end of the line segment in the first quadrant. The third line segment is labeled jk(v) V_R.Figure four part b is a cartesian graph consisting of an assortment of line segments mostly encompassed by a circle whose center is somewhere in the first quadrant and its edge passing through the origin. Inside the circle, from the origin, are five line segments that extend to the edge of the circle at various unevenly spaced points on the edge of the circle. Out of these five segments, the segment with the strongest positive slope is labeled V_R, and the segment with the second-strongest negative slope is labeled V_R. The middle of the five segments extends furthest to the right, and its end at the edge of the circle is labeled V. Two more line segments from this point follow, both connecting this end point to the end points of the line segments labeled V_R. Both of these line segments are labeled jk(v)V_R. A final line segment is drawn tangent to the circle and passing through the origin from the second quadrant into the fourth.

If we multiply the left-hand side by its complex conjugate and the right-hand side by its complex conjugate, we obtain the identity

4=2(1+cosψ)[1+k2(ν)].4=2(1+cosψ)[1+k2(ν)].
(37)

This equation tells us how the angle ψ ψ depends on k(ν)k(ν) and, conversely, how k(ν)k(ν) depends on ψψ:

cos ψ = 1 - k 2 ( ν ) 1 + k 2 ( ν ) cos ψ = 1 - k 2 ( ν ) 1 + k 2 ( ν )
(38)
k 2 ( ν ) = 1 - cos ψ 1 + cos ψ k 2 ( ν ) = 1 - cos ψ 1 + cos ψ
(39)

The number cosψcosψ lies between -1-1 and +1+1, so a circular solution does indeed work.

Exercise 6

Check -1cosψ1-1cosψ1 for -<k<-<k< and -<k<-<k< for -πψπ-πψπ. Sketch k k versus ψψ and ψψ versus kk.

The equation VR=V2(1+ejψ)VR=V2(1+ejψ) is illustrated in Figure 5. The angle that V R V R makes with V V is determined from the equation

2 φ + π - ψ = π φ = ψ 2 2 φ + π - ψ = π φ = ψ 2
(40)
Figure 5: The Voltages V and VR, and the Power Factor cosφcosφ
Figure five is a cartesian graph containing five line segments and one circle. The circle is centered somewhere in the first quadrant a point of its edge on the origin of the graph. At the origin is also a tangent line to this circle that begins in the second quadrant and moves with a constant negative slope to the fourth quadrant. Inside the graph are the other four line segments. Two begin from the origin, one extending with a strong positive slope to the edge of the circle at a spot in the first quadrant. This segment has an arrow pointing away from the origin, and is labeled V_R. The other extends with a shallower positive slope and stops at approximately (not designated) the center of the circle. The angle between these two line segments from the origin is measured as Φ. The third line segment continues from the same point that the second ends from (approximately the center of the circle) and with the same slope extends to the opposite side of the circle, with the end of this segment marked by an arrow pointing away from the origin and labeled V. Also from the center of the circle is a segment that meets the end of the first segment at the end of the circle near the top of the figure, thus having a negative slope. The angle between this line segment and the line segment from the center to the far-right edge of the circle is measured as ψ. The angle between the last described line segment and the first line segment from the origin towards the top of the circle is also labeled Φ.

In the study of power systems, cos φ cosφ is a "power factor" that determines how much power is delivered to the resistor. We may denote the power factor as

η=cosφ=cosψ2.η=cosφ=cosψ2.
(41)

But cosψcosψ may be written as

η=cosφ=cosψ2η=cosφ=cosψ2
(42)

But cosψcosψ may be written as

cos ψ = cos ( φ + φ ) = cos 2 φ - - sin 2 φ = cos 2 φ - ( 1 - cos 2 φ ) = 2 cos 2 φ - 1 = 2 η 2 - 1 . cos ψ = cos ( φ + φ ) = cos 2 φ - - sin 2 φ = cos 2 φ - ( 1 - cos 2 φ ) = 2 cos 2 φ - 1 = 2 η 2 - 1 .
(43)

Therefore the square of the power factor η is

η2=cosψ+12=11+k2(ν)η2=cosψ+12=11+k2(ν)
(44)

The power factor is a maximum of 1 for k(ν)=0k(ν)=0, corresponding to ν=1ν=1(ω=ω0)(ω=ω0). It is a minimum of 0 for k(ν)=±k(ν)=±, corresponding to ν=0,ν=0,(ω=0,)(ω=0,).

Exercise 7

With k k defined as k(ν)=ω0LR(ν-1ν)k(ν)=ω0LR(ν-1ν), plot k2(ν)k2(ν), cosψcosψ, and η 2 η 2 versus ν ν.

Exercise 8

Find the value of ν ν that makes the power factor η=0.707η=0.707.

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