Notes for course EE1.1 Circuit Analysis 2004-05 TOPIC 6 – PHASOR ANALYSIS OF AC CIRCUITS Objectives . Representations of sinusoidal voltages and currents using phasors . Using phasors to define impedance and admittance for the inductor and capacitor . AC Circuit Analysis using phasors (Ohm’s law, KCL and KVL Thevenin and Norton equivalent circuits, superposition and nodal analysis) . Maximum power transfer theorem for AC circuits

1 INTRODUCTION 1.1 General In this section we consider the general idea of describing circuits in terms of how they respond to input signals which are AC sinusoids We first look at some properties of circuits driven by sinusoidal sources and then consider some advantages of an approach to circuit analysis based on sinusoidal signals 1.2 The forced response of a circuit Phasor analysis is based on use of sinusoidal functions for voltage and current sources st Consider our 1 order RC circuit and its transient response for Vs = 0 and vco = –5 V:

0

-1

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-3 Vin(t) Vout(t) Vin(t), Vout(t)

-4

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-6 0 2 4 6 8 10 12 14 16 18 20 Time s where Vin(t) = Vs(t) and Vout(t) = vc(t) Note that the transient response decays to zero since the circuit is stable Topic 6 – Phasor Analysis

Consider now the situation when Vs = Vsinωt with V = 1 V:

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-2 Vin(t) Vout(t) Vin(t), Vout(t) -3

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-6 0 2 4 6 8 10 12 14 16 18 20 Time s This response consists of two parts, the transient response and the response due to the forcing function Vsinωt This response is called the complete response Once the transient response has decayed to zero, the output voltage becomes sinusoidal, with the same frequency as the input voltage but differing amplitude and phase This response is called the sinusoidal forced response As t → ∞, the sinusoidal forced response converges to the response the circuit would have if the excitation was sinusoidal over all time for –∞ < t < +∞ This response is called the AC steady-state response In fact for any circuit excited by sinusoidal voltage and current sources all having the same frequency ω, the AC steady-state response is that all voltages and currents are sinusoidal with frequency ω The AC steady-state analysis problem consists in finding all of these amplitudes and phases If the system is unstable, then the forced response does not converge to a sinusoidal response because the transient response does not decay to zero:

2 Topic 6 – Phasor Analysis

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-8 Vin(t) Vout(t) Vin(t), Vout(t)

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-16 Time s In this case of an unstable circuit, the AC steady state response is not defined: AC steady state response is defined only for a stable circuit 1.3 Network analysis We now consider another reason why we are interested in the response of linear circuits for sinusoidal input signals Consider a circuit which behaves in a linear fashion, such as an audio amplifier Since it is linear, the principle of superposition is obeyed This means that if a signal representing two sounds one of low frequency and one of high frequency are applied, the effect of applying them together is the same as the sum of the separate responses; ie the sound of a flute is the same whether there is also a bass guitar playing as well or not This means that we can test an amplifier to see how it behaves for all signals by applying a single sinewave and varying its frequency from the minimum frequency (10 Hz) to the maximum frequency (20 kHz); the resulting frequency response gives a good indication of how the amplifier will respond for any signal of any frequency or combination of frequencies This process is called single frequency sinusoidal testing of a circuit It involves measuring the amplitude and phase of the output signal versus frequency The instrument which performs this measurement is called a network analyser Approximate measurements can be made using a sinewave signal generator and an oscilloscope or DVM + phasemeter 1.4 Fourier analysis Another reason for working with AC steady state (sinusoidal) response is that signals which are not sinusoidal can be represented as sums of sinusoids with different frequencies This is the Fourier series As an example, consider the following harmonic sinusoidal terms: 4 4 1 4 1 4 1 y1 = sin x y3 = sin3x y5 = sin5x y7 = sin7x π π 32 π 52 π 72 Consider the effect of adding the terms successively: 3 Topic 6 – Phasor Analysis

1.5 2

1.5 1

1

0.5 0.5 y 0 y 0 -4 -3 -2 -1 0 1 2 3 -4 4 -3 -2 -1 0 1 2 3 4 -0.5 -0.5 y1 y5 -1 y3 y1+y3 -1 -1.5

-1.5 -2 x x

2 2

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1 1

0.5 0.5 y y 0 0 -4 -3 -2 -1 0 1 2 3 4-4 -3 -2 -1 0 1 2 3 4 -0.5 -0.5

y7 -1 -1 y1+y3+y5 y1+y3+y5+y7

-1.5 -1.5

-2 -2 x x As we add successive harmonic terms, the sum approximates a triangle wave more closely Since any periodic waveform can be represented as a sum of sinusoidal terms, we can do the following: 1) Express complex waveform as a sum of sinewaves 2) Determine the response of the circuit to each sinewave 3) By superposition, the response of the circuit to the complex waveform is the sum of the response to the individual sinewaves We have shown that description of circuits in terms of how they respond to sinusoidal input signals is potentially attractive The method of phasors allows us to simplify circuit analysis as much as possible for the sinusoidal signal case

2 DESCRIPTION OF SINUSOIDAL VOLATGES AND CURRNTS USING PHASORS 2.1 Relationship between sine and cosine We show the graph of two periods of x(t) = 3 cos(4πt) and two periods of y(t) = 3 sin(4πt):

4 Topic 6 – Phasor Analysis

We can describe either one - and an entire host of other sinusoids as well - by using the general sinusoid: f (t) = Acos(ωt + θ) We can use the trigonometric identity: cos(x + y) = cos(x)cos(y) − sin(x)sin(y) with x = ωt and y = θ = –90° to obtain: cos(ωt − 90) = cos(ωt)cos(−90) − sin(ωt)sin(−90) = cos(ωt).(0) − sin(ωt).(−1) = sin(ωt) The standard result is worth remembering: sin(ωt) = cos(ωt − 90) We can also generalise it: sin(α) = cos(α − 90) We can use it to express the general sinusoid in cosine form: Bsin(ωt + θ) = Bcos(ωt + θ − 90) We introduce the idea of phasors through a simple example 1.2 Example Circuit Consider the simple circuit shown:

Our aim is to determine the current i(t) and the voltages across the resistor and inductor, vR(t) and vL(t).

The source voltage vs(t) is known:  vs (t) = 4 2 cos 2t + 45 ( ) In order to introduce the idea of phasors, we work initially from a given solution for the AC forced response (we shall derive this solution later using phasors):

5 Topic 6 – Phasor Analysis

i(t) = cos(2t) A Let us check whether this solution is correct by checking whether Kirchhoff’s voltage law (KVL) is satisfied: First, we find the sum of the element voltages: di(t) d v (t) + v (t) = Ri(t) + L = 4cos(2t) + 2 cos(2t) = 4cos(2t) − 4sin(2t) R L dt dt = 4cos(2t) − 4cos(2t − 90) = 4cos(2t) + 4cos(2t + 90) = 4 2 cos 2t + 45 ( ) Hence we have:

vs(t) = vR (t) + vL (t) KVL is satisfied, so we have confirmed that the solution we have been given is correct We have used: d cos(at) = -asin(at) dt The sum of a sine function and a cosine function with the same frequency is equivalent to a single sine or cosine function with specified phase angle: The initial form can be expressed as: f = Acos x − Bsin x We first scale numerator and denominator by a common factor: ⎡ A B ⎤ f = A2 + B2 ⎢cos x − sin x ⎥ ⎣ A2 + B2 A2 + B2 ⎦ We can now equate the factors of cos x and sin x to cos y and sin y: cos(x + y) = cos(x)cos(y) − sin(x)sin(y) Hence

f = A2 + B2 cos x cos y − sin x sin y [ ] = A2 + B2 cos(x + y) where

A B −1⎡ B⎤ cos y = sin y = y = tan ⎢ ⎥ A2 + B2 A2 + B2 ⎣ A⎦ (For the tan-1 expression, the correct quadrant must be used depending on the signs of both A and B) 2 2 ⎛ −1⎛ B⎞⎞ f = Acos x − Bsin x = A + B cos⎜ x + tan ⎜ ⎟⎟ ⎝ ⎝ A⎠⎠ Our task now is to generate the given solution for ourselves; in order to do this, we will first introduce the idea of phasors to describe voltages and currents 6 Topic 6 – Phasor Analysis

1.3 Introducing phasors Let us write out the KVL equation for our example:

vs(t) = vR(t) + vL (t) 4 2 cos 2t + 45 = 4cos(2t) + 4cos 2t + 90 ( ) ( ) We make use of Euler's identity:

re jθ = r cosθ + j sinθ = r cosθ + jr sinθ ( ) Using this we may state that:

r cos Re ⎡re jθ ⎤ θ = ⎣ ⎦ Using this result, we can write the KVL equation as follows: ⎡ ⎛ π ⎞ ⎤ ⎡ ⎛ π ⎞ ⎤ j⎜ 2t+ ⎟ j⎜ 2t+ ⎟ ⎢ ⎥ ⎡ j 2t ⎤ ⎢ ⎥ Re 4 2e ⎝ 4 ⎠ = Re 4e ( ) +Re 4e ⎝ 2 ⎠ ⎢ ⎥ ⎣⎢ ⎦⎥ ⎢ ⎥ ⎣⎢ ⎦⎥ ⎣⎢ ⎦⎥ or

Re[v˜ s(t)] = Re[v˜ R(t)] +Re[v˜ L (t)] where x˜ denotes the vector whose real part is equal to x This equation can be represented graphically as follows: ~ vs

~ vL

~ vR

2t

vL vs vR Note that all three vectors are rotating at a rate of 2 rad/sec. Note that the vectors are complex quantities. The procedure we have used to derive the vectors from the real parts also identifies imaginary parts according to Euler's identity An equation which is true for complex quantities must be true also separately for the real parts and for the imaginary parts of those complex quantities It follows that our KVL equation in terms of the real parts of the vectors must be true also for the vectors themselves Hence, we may write:

7 Topic 6 – Phasor Analysis

v˜ s(t) = v˜ R (t) + v˜ L (t) ⎛ π ⎞ ⎛ π ⎞ j⎜ 2t+ ⎟ j⎜ 2t+ ⎟ j 2t 4 2e ⎝ 4 ⎠ = 4e ( ) + 4e ⎝ 2 ⎠ The final step in developing phasors is to take out from both sides of the vector equation the common factor ej2t; this is tantamount to removing the common rotation of all the vectors: π π j j 4 2e j2te 4 = 4e j2te j0 + 4e j2te 2 π π j j 4 2e 4 = 4e j0 + 4e 2

Vs = VR + VL

The final quantities Vs, VR and VL are referred to as phasors The removal of ejωt is tantamount to the statement that the relationship between the vectors is independent of their common rotation: It is customary to indicate phasors by use of upper-case letters; in these notes, at least at the beginning, we will use the bar as well This emphasises the fact that phasors are transformed voltages and currents no longer directly observable on an oscilloscope – we need instruments such as a network analyser or gain and phase meter. Note that the phasors are complex numbers which may be represented in a phasor diagram For the above example the phasor diagram is simply obtained by putting t = 0 in the vector diagram: __ __ VL Vs

__ VR Let us summarise the steps we have taken to turn a voltage or current into a phasor:

Time domain voltage or current x(t) = Xm cos(ωt + θx ) ⎡ j ωt+θ ⎤ Express as real part of rotating vector x t = Re X e ( x ) ( ) ⎣⎢ m ⎦⎥

j(ωt+θ x ) Use rotating vector in place of real voltage or current x˜ (t) = Xme

jθ x Remove rotation of vector by setting t = 0 X = Xme

Once the principle of deriving phasors is accepted, the procedure may be carried out directly in a single step:

8 Topic 6 – Phasor Analysis

jθx x(t) = Xm cos(ωt + θx ) ⇒ X = Xme Note that, in general, phasors are complex quantities and therefore may be expressed in either polar or rectangular form: jθ x X = Xme = Xm∠θx = Xm cosθx + jXm sinθx In special cases, a phasor may be wholly real or wholly imaginary Note that the phasor does not contain the frequency of a signal and therefore the frequency (2 rad/sec on our example circuit and the same for all voltages and currents) must be supplied separately We have derived the phasor concept through working with cosine functions and realising that they may be expressed as the real parts of rotating vectors

It would be equally valid to have worked with sine functions and realised that they may be expressed as the imaginary parts of rotating vectors; in fact the vectors obtained would be the same in both cases, although the phase values in the sine function would differ from those in the cosine functions

In order to avoid confusion when converting between voltages as functions of time and phasors it is necessary to be clear about whether the cosine/real part convention is being used or the sine/imaginary part one

Here, we shall always assume the cosine/real part convention

1.4 Phasor Examples 1.4.1 Example 1 A voltage in a circuit has the form: v(t) =10cos 2t − 45 V ( ) Find the corresponding phasor in polar and rectangular form: Solution: We can immediately write the Euler form by inspection:

V =10∠ − 45 V Notice the standard notation: for the phasor, we use the same symbol as that for the time-varying voltage, only in uppercase and with an overbar We next simply use Euler's formula to write: V = 10cos(−45 ) + j10sin(−45 ) = 10cos(45 ) − j10sin(45 ) 10 10 = − j 2 2

= 5 2 − j5 2 V The unit for a phasor is the same as the unit for the time quantity it represents

9 Topic 6 – Phasor Analysis

1.4.2 Example 2 If I = 6 + j8 A and ω = 5 rad/s find i(t) Solution: Phasor I is given in rectangular form, so we must convert it to Euler form: I = 6 + j8 = 2(3+ j4) 2(5∠53.1)  =10∠53.1 A We can now simply write down the time-varying form: i(t) =10cos 5t + 53.1 A ( ) We just identify the magnitude of the phasor with the amplitude of the sinusoid, and the angle of the phasor with the phase of the sinusoid We note that we have used cosine functions in each case rather than sine functions This is a convention, but one we will stick to: Each and every phasor represents a cosine function of time (not a sine function) We now consider an example to illustrate this: 1.4.3 Example 3 Let x(t) = 4 2 sin 3t + 45 ( ) Find the phasor X in rectangular form Solution: Here we must perform the preliminary step of expressing x(t) as a cosine function Using the trigonometric identity sin(ωt) = cos(ωt − 90), we have: x(t) = 4 2 cos 3t − 45 ( ) The Euler form for the phasor is:  X = 4 2∠ − 45 We next convert to rectangular form:

X = 4 2 ⎡cos −45 + j sin −45 ⎤ ⎣ ( ) ( )⎦ = 4 − j4

10 Topic 6 – Phasor Analysis

Remember that the cosine is an even function and the sine is an odd one We can now express sinusoidal voltages and currents in the form of phasors The next step is to define the concepts of impedance and admittance which are necessary to carry out phasor analysis of a circuit We approach the concepts of impedance and admittance through a very useful concept: the system function

2 THE SYSTEM FUNCTION We show a circuit, represented by a rectangle, having only one independent source whose waveform x(t) is sinusoidal with amplitude Xm and phase θ:

A voltage or current variable in the circuit which is of interest, perhaps the output signal, is denoted y(t)

We know that this response will be sinusoidal, having some amplitude Ym and phase β as shown

If we can find the amplitude Ym and phase β of y(t) we will have succeeded in our analysis The circuit with its time-varying sources is referred to as the time domain representation To solve for the circuit response, we represent the input and response waveforms by their phasors as shown in the following figure:

Now consider the following ratio: Y Y ∠β Y = m = m ∠β − θ X Xm∠θ Xm where φ = β −θ We refer to Y X as a system function, and write it (in polar form) as: Y = H( jω) = H( jω)∠φ(ω) X We have:

Ym H ( jω ) = Xm φ(ω ) = β − θ We express H(jω) as a function of ω in order to allow for the fact that the system function might (and generally will) change if we alter the frequency We can rearrange the equation defining H(jω) in order to use it in an analysis procedure as follows: Y = H( jω)X

11 Topic 6 – Phasor Analysis

This is equivalent to the statement that: Y = H ( jω ) X ∠Y = ∠H ( jω ) + ∠X We may re-write this as:

Ym = Xm H ( jω ) β = θ + ∠H ( jω )

We merely multiply the magnitudes of H(jω) and X and add their angles to find the polar form of Y Note that the system function is commonly described as the transfer function or the frequency response function of a circuit Having used phasors to define the transfer function of a system, we can now use phasors to describe basic circuit elements

3 IMPEDANCE AND ADMITTANCE 3.1 General Consider the time-domain representation of a 2-terminal element shown with voltage v(t) and current i(t):

The phasor representation is shown below:

We define the impedance of the sub-circuit to be a system function with i(t) as input and v(t) as output: V Z( jω) = = Z( jω)∠φ I Hence we have that:

Vm Z( jω) = Im ∠Z( jω) = φ = ∠V −∠I = β −θ We can reverse the situation and consider v(t) to be the input and i(t) as the output:

12 Topic 6 – Phasor Analysis

I 1 I Y( jω) = = = m ∠ − φ V Z( jω) Vm In this case, the system function describing the element is called the admittance The unit of impedance is the Ohm The unit of admittance the Siemens Impedance is the generalization of resistance Admittance is the generalization of conductance. We will now see what forms impedance and admittance have for resistors, inductors, and capacitors 3.2 The resistor In the time-domain the resistor can be represented as follows:

In the time domain equation we use Ohm's law to write the voltage in terms of the current:

v(t) = Ri(t) = RIm cos(ωt + θ) W also have:

v(t) = Vm cos(ωt + β) We can simply equate magnitudes and phases:

Vm = RIm and β = θ The impedance is: V V ∠β RI ∠θ Z ( jω ) = = m = m = R I Im∠θ Im∠θ Thus the phasor representation is as follows:

Thus, Ohm's law continues to hold for phasor description of a resistor: V = RI Clearly, the admittance of a resistor is easily obtained: I I ∠θ I ∠θ 1 Y ( jω ) = = m = m = = G V Vm∠β RIm∠θ R Using conductance, Ohm's law becomes:

13 Topic 6 – Phasor Analysis

I = GV 3.3 The inductor The time domain representation for the inductor is shown:

The time domain relationship is: di(t) v(t) = L dt d(Im cos(ωt + θ)) = L dt = −ωLIm sin(ωt + θ)  = −ωLIm cos(ωt + θ − 90 )  = ωLIm cos ωt + θ + 90 ( ) Where we followed the rule of expressing all sinusoidal functions as cosines: Since also: v(t) = Vm cos(ωt + β) we obtain:

Vm = ωLIm and β = θ + 90 Hence, the impedance of the inductor is:

V V ∠β ωLI ∠θ + 90 Z ( jω ) = = m = m = ωL∠90 = jωL I I I m∠θ m∠θ In the last step, we have simply converted from Euler to rectangular form We now see the important result that when phasors are used to represent voltage and current, the inductor obeys Ohm's law: V = jωLI The only difference is that the impedance is purely imaginary and frequency-dependent The resulting phasor representation for the inductor is as follows:

14 Topic 6 – Phasor Analysis

3.4 The capacitor The time domain representation for the capacitor is as follows:

The time-domain relationship is: dv(t) i(t) = C dt d(Vm cos(ωt + β)) = C dt = −ωCVm sin(ωt + β)  = −ωCVm cos(ωt + β − 90 )  = ωCVm cos ωt + β + 90 ( ) Since also:

i(t) = Im cos(ωt + θ) we can equate magnitudes and angles, to obtain:

Im = ωCVm and

θ = β + 90 Hence, the impedance of the capacitor is: V V ∠β 1 1 1 Z ( jω ) = = m = ∠ − 90 = − j = I  ωC ωC jωC ωCVm∠β + 90 In the last step, we have simply converted from Euler to rectangular form and noted that –j = 1/j The capacitor, too, obeys Ohm's law when we use phasors: 1 V = I jωC The resulting phasor representation for the capacitor is as follows:

The results for impedance and admittance of the 2-terminal elements can be summarised as follows:

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V I Impedance Z( jω) = Admittance Y( jω) = I V Resistor R 1 = G R Inductor jωL 1 jωL Capacitor 1 jωC jωC Units Ohms Ω Siemens S Here is the net result of what we have accomplished We know that each of the elements R, L, and C obeys Ohm's law, provided we use the impedance of the appropriate element in the place of resistance and phasors for the defining voltages and currents We also know that KVL holds for phasor voltages and KCL holds for phasor currents In fact, all of our DC analysis techniques – superposition, Thevenin and Norton equivalents, nodal analysis, etc. – are all based upon only these facts and linearity Thus, all of our DC analysis techniques continue to hold for AC forced response with impedances replacing resistances and phasors replacing time-varying voltages and currents We are now ready to solve the problem of the simple RL circuit which we considered at the outset:

4 SOLUTION OF EXAMPLE USING PHASOR ANALYSIS Solve for the forced response of the currents and voltages for the circuit shown using phasor techniques:

Solution We convert all of the voltages and currents to phasors and represent the inductor and the resistor by their impedances:

 V s = 4 2∠45

Since the frequency of the signals is not contained in the phasor representations of the elements, it has to be stated separately in the circuit diagram The frequency is deduced as the factor of t in the argument for the cosine in the expression for the voltage of the source; hence ω = 2 radians/second (rad/s)

16 Topic 6 – Phasor Analysis

Note that the impedance of the inductor is in general jωL The resistor and the inductor are connected in series, so we do as we would with resistors – we simply add their impedances: Z = 4 + j4 Then we use Ohm's law with phasors to find the phasor current:

V 4 2∠45 4 2∠45 I = s = = =1∠0  Z 4 + j4 4 2∠45 Thus: i(t) = cos(2t) A We have now derived using phasors the solution for the current which we assumed earlier We have used:  1+ j = 2∠45 The general form is of course: b a + jb = a2 + b2 ∠tan−1 a When phase is obtained from real and imaginary parts in this way there is an ambiguity in the phase; we explore this through the following example: Consider the problem of finding the Euler form of the phasor X = – 4 – j4

We could calculate the angle as:

1⎛ −4⎞ 1 θ = tan− ⎜ ⎟ = tan− (1) = 45 ⎝ −4⎠ But this is wrong! It is the phase angle for the phasor X = 4 + j4 The problem is that the tangent function has period 180° (π radians), unlike the sine and cosine whose periods are both 360° (2 π radians) The best way to guard against such difficulties is to retain the signs of the real and imaginary parts, and possibly make a sketch, as above The correct angle is 180° + 45° = 225° or –180° + 45° = – 135° In computer packages, such as Excel, this problem can be solved by use of a special function ATAN2(Ximag Xreal) which takes separate real and imaginary parts as arguments rather than ATAN(Xi/Xr) which takes the single argument Ximag/Xreal A less serious issue that arises in working with phasors is that angles are ambiguous with respect to multiples of 360° because of the periodic nature of the cosine function

17 Topic 6 – Phasor Analysis

This problem can be solved by restricting angles of phasors to be within the range 0 to 360° or between -180° and +180° We now easily find the inductor and resistor voltages using Ohm's law for impedances   VR = IR = 4 × 1∠0 = 4∠0 = 4 V and     VL = IZ = I jωL = j4 × 1∠0 = 4∠90 × 1∠0 = 4∠90 Therefore, we have:

 vL (t) = 4cos 2t + 90 A ( ) and

vR (t) = 4cos(2t) V

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vs 0 vR vL Vs, VR and VL

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4.1 Summary of the Phasor Method Notes: 1) When we converted to the phasor equivalent circuit, we made a note of the frequency in a small box It is assumed that all voltages and currents in the circuit are cosine functions at this frequency with amplitude and phase given by the phasors Having the frequency clearly stated on the diagram makes it easily accessible 2) It can happen that there is confusion between impedances and phasors Both are complex numbers, but only phasors are representative of sinusoidal time-varying voltages and currents 18 Topic 6 – Phasor Analysis

The impedance (or admittance) is simply a complex constant that takes the place of resistance in DC circuit analysis The impedances (or admittances) define the relationships between the amplitudes and phases of the phasors Summary of phasor analysis method: 1 Convert all sine voltages and currents to cosines 2 Draw the phasor equivalent circuit, making a note of the common frequency of all independent sources. Represent each voltage and each current by a phasor and each passive element by its impedance or admittance 3 Solve for the desired phasor(s) 4 Convert phasors to Euler form and write the time domain form To illustrate the importance of the first step, we will work one more example: Example 5 Solve for the forced response of the voltage v(t) in the circuit shown using phasor techniques:

Assume that is(t) = 10 sin(3t) A Solution: We first convert the current source to cosine form:

 is(t) =10cos 3t − 90 A ( ) Then convert voltages and currents to phasors and the passive elements to impedances, resulting in the phasor equivalent circuit:

where  Is = 10∠ − 90 = − j10 A The two passive elements are connected in parallel, with an equivalent impedance: 3 × (− j4) 12∠ − 90 Z = = =2.4∠ − 36.9 Ω  3 − j4 5∠ − 53.1 Thus, the phasor voltage is given by:    V = Z I = 2.4∠ − 36.9 × 10∠ − 90 = 24∠ − 126.9 V The corresponding time-domain sinusoidal waveform is given by:

19 Topic 6 – Phasor Analysis

v(t) = 24cos 3t −126.9 V ( ) 5 METHODS OF AC CIRCUIT ANALYSIS

The circuit analysis methods which we have used already – Ohm's law, KCL and KVL, Thevenin and Norton equivalents, superposition and nodal analysis – apply for AC analysis using phasors We illustrate these methods in circuit analysis using phasors by means of some examples 5.1 Example 6 Find the forced response for the circuit shown using equivalent impedance and voltage division:

Solution: We first show the circuit in phasor form:

The first step is to convert the sine function to a cosine; only then do we convert the voltage source to phasor form Note that the frequency is 2 rad/s There are many ways to solve this circuit, but we choose to find the equivalent impedance of the capacitor and its parallel resistor: 1× (− j2) − j2 1+ j2 4 − j2 Z = = × = = 0.8 − j0.4 Ω 1+ (− j2) 1− j2 1+ j2 5 Note the method for rationalising a rational function: a + jb a + jb c − jd ac + bd bc − ad z = = × = + j c + jd c + jd c − jd c 2 + d2 c 2 + d2 Using this equivalent impedance, we can redraw the circuit in the equivalent form shown:

Next, we simply use the voltage divider rule (in phasor form) to compute the phasor V associated with the voltage v(t): Z 0.8 − j0.4 −1.6 − j3.2 V = Vs = × (− j4) = = −2 + j0 V Z + j2 0.8 − j0.4 + j2 0.8 + j1.6

20 Topic 6 – Phasor Analysis

Finally, we convert to Euler form: v(t) = −2cos(2t) = 2cos 2t −180 V ( ) 5.2 Example 7 Find the forced response for i(t) in the circuit shown using element combining:

Note that ω = 3 radians/s We compute the complex impedance of each element, convert the current source to its phasor form, and draw the phasor equivalent circuit:

This equivalent, we stress, is only valid at the single frequency of 3 rad/s As this frequency does not appear explicitly anywhere on the circuit, we have made a note of it in the circuit diagram The 1 Ω resistor is connected in parallel with the capacitor and the series combination of the inductor and the 3 Ω resistor, so we compute the impedance of the subcircuit made up of all the passive elements: 1 3+ 3j 3+ j3 Z j3 = = = Ω ( ) 1 1 1+ + 3+ 3j −1+ j +1 3+ j4 − j3 3+ j3 The current through the 1 Ω resistor is the same, by Ohm's law, as the voltage across it Thus: V 3+ j3 3 2∠45 I I Z j3 5 0 5 0 3 2 8.1 = = s ( ) = × ∠ =  × ∠ = ∠ − Ω 1 3+ j4 5∠53.1 Thus, the time-domain waveform for the forced response of the current i(t) is: i(t) = 3 2 cos 3t − 8.1 A ( ) 5.3 Example 8 Use KCL to determine the forced response for v(t) in the following circuit:

21 Topic 6 – Phasor Analysis

Solution By now, the process should be familiar: we represent the sinusoidal current sources with their phasors and convert all passive elements to impedances This results in the circuit shown below:

We have chosen the bottom node as the reference and have labeled the other nodes with symbols for unknown phasor voltages V and Va The KCL equation for the node labeled V is:

V −Va V −Va V  + + = 4∠ − 90 = − j4 j4 8 − j6 KCL for the node labeled Va gives: V −V V −V V a + a + a = 8∠0 = 8 j4 8 4 We rationalize by multiplying the first equation by j24 and the second by j8:

6 V −Va + j3 V −Va − 4V = 96 ( ) ( ) 2(Va −V) + j(Va −V ) + j2(Va) = j64 Grouping like terms:

(2 + 3j)V + (−6 − 3j)Va = 96 (−2 − j)V + (2 + 3j)Va = j64 We can assemble the two equations into matrix form: ⎡2 + 3j −6 − 3j⎤ ⎡ V ⎤ ⎡ 96 ⎤ ⎢ ⎥ ×⎢ ⎥ = ⎢ ⎥ ⎣− 2 − j 2 + 3j ⎦ ⎣V a⎦ ⎣ j64⎦ We can solve this by inverting the (2 × 2) coefficient matrix: The general form of the matrix equation is:

22 Topic 6 – Phasor Analysis

⎡ a11 a12⎤ ⎡ x1⎤ ⎡ d1⎤ ⎢ ⎥ ×⎢ ⎥ = ⎢ ⎥ ⎣a 21 a22⎦ ⎣ x2⎦ ⎣d 2⎦ which we may write: A × x = d where

⎡ a11 a12⎤ ⎡ x1⎤ ⎡ d1⎤ A = ⎢ ⎥ x = ⎢ ⎥ d = ⎢ ⎥ ⎣a 21 a22⎦ ⎣ x2⎦ ⎣d 2⎦ The solution is: A−1A × x = A−1d x = A−1d where

−1 1 ⎡ a22 −a12⎤ 1 ⎡ a22 −a12⎤ A = ⎢ ⎥ = ⎢ ⎥ A ⎣− a21 a11 ⎦ a11a22 − a21a12 ⎣− a21 a11 ⎦ Hence

⎡ x1⎤ 1 ⎡ a22 −a12⎤ ⎡ d1⎤ ⎢ ⎥ = ⎢ ⎥ ×⎢ ⎥ ⎣ x2⎦ a11a22 − a21a12 ⎣− a21 a11 ⎦ ⎣d 2⎦ A = (2 + 3j)(2 + 3j) − (−2 − j)(−6 − 3j) = (4 − 9 + j6 + j6) − (12 − 3+ j6 + j6) = −14

−1 1 ⎡− 2 − 3j −6 − 3j⎤ A = ⎢ ⎥ 14 ⎣ −2 − j −2 − 3j⎦ ⎡ V ⎤ 1 ⎡− 2 − 3j −6 − 3j⎤ ⎡ 96 ⎤ ⎢ ⎥ = ⎢ ⎥ ×⎢ ⎥ ⎣V a⎦ 14 ⎣ −2 − j −2 − 3j⎦ ⎣ j64⎦ 1 − j672 V = 96(−2 − 3j) + j64(−6 − 3j) = = − j48 = 48∠ − 90 14 [ ] 14 1 − j224  Va = [96(−2 − j) + j64(−2 − 3j)] = = − j16 =16∠ − 90 14 14 Thus, the time-domain response we are seeking is:

v t = 48cos 2t − 90 V ( ) ( ) 48sin 2t V = ( ) MATLAB code and output for solving these equations is as follows: A=[2+3j -6-3j;-2-j 2+3j] b=[96 64j].' v=A\b mag=abs(v)

23 Topic 6 – Phasor Analysis phase=180*atan2(imag(v),real(v))/pi OUTPUT A = 2.0000 + 3.0000i -6.0000 - 3.0000i -2.0000 - 1.0000i 2.0000 + 3.0000i b = 96.0000 0 +64.0000i v = 0.0000 -48.0000i -0.0000 -16.0000i mag = 48.0000 16.0000 phase = -90.0000 -90.0000 5.4 Example 9 Find the forced response for the current i(t) in the circuit shown using nodal analysis:

Solution Choosing the ground reference at the bottom results in the phasor equivalent circuit shown: There is only one node with an unknown voltage V and it is a super-node encompassing the floating voltage source:

24 Topic 6 – Phasor Analysis

The voltage at node 6 is known Thus we only need to apply KCL to the super-node: V V V − 2 V − 2 V − 6 V − 2 − 6 + + + + + = 0 j4 4 8 − j2 8 j4 The solution for V is: V = 2 − j2 = 2 2∠ − 45 Hence, we have: V 2 2∠ − 45 1 I 135 = =  = ∠ − j4 4∠90 2 Hence, the time-domain form for i(t) is:

1  i(t) = cos(4t −135 ) 2 6 SERIES AND PARALLEL EQUIVALENT SUB-CIRCUITS

6.1 General Let's now investigate the idea of an equivalent sub-circuit Consider a two-terminal passive sub-circuit in phasor form:

Let us suppose that we have computed or measured the impedance to be: Z( jω) = R(ω) + jX(ω) where R(ω) = Re[Z( jω)] X(ω) = Im[Z( jω)] We call R(ω) the resistance and X(ω) the reactance of the sub-circuit Note that the resistance of a circuit with inductors and/or capacitors can be a function of frequency Because Z( jω) = R(ω) + jX(ω), and because we know that the impedances of elements connected in series add, we have the series equivalent sub-circuit:

25 Topic 6 – Phasor Analysis

Now let's compute the admittance: 1 1 1 R(ω ) − jX (ω ) Y ( jω ) = = = Z (ω ) R(ω ) + jX (ω ) R(ω ) + jX (ω ) R(ω ) − jX (ω ) R(ω ) X (ω ) = − j R2 (ω ) + X2 (ω ) R2 (ω ) + X2 (ω )

We refer to the real and imaginary parts of Y(jω) as the conductance G(ω) and susceptance B(ω): Y( jω) = G(ω) + jB(ω) where G(ω ) = Re ⎡Y ( jω )⎤ B(ω ) = Im ⎡Y ( jω )⎤ ⎣ ⎦ ⎣ ⎦ Thus we have: R(ω ) X (ω ) G(ω ) = B(ω ) = − R2 (ω ) + X2 (ω ) R2 (ω ) + X2 (ω ) Thus we see that the parallel sub-circuit is also a valid equivalent for our two-terminal sub-circuit:

Note that G(ω) and B(ω) have the unit S (Siemens) Alternatively, we can start with an admittance model of the circuit given by: Y( jω) = G(ω) + jB(ω) Then the impedance of the parallel sub-circuit can be determined as follows: 1 Z ( jω ) = Y (ω ) 1 = G(ω ) + jB(ω ) 1 G(ω ) − jB(ω ) = G(ω ) + jB(ω ) G(ω ) − jB(ω ) G(ω ) B(ω ) = − j G2 (ω ) + B2 (ω ) G2 (ω ) + B2 (ω ) Since: Z( jω) = R(ω) + jX(ω) We have: G(ω ) B(ω ) R(ω ) = X (ω ) = − G2 (ω ) + B2 (ω ) G2 (ω ) + B2 (ω )

26 Topic 6 – Phasor Analysis

Example 11 Find the equivalent parallel sub-circuit for the two-element sub-circuit shown at a frequency of ω = 2 rad/s

Solution The impedance of this sub-circuit is: Z( j2) = 4 + j4 Ω Computing the admittance: 1 Y( j2) = 4 + j4 4 − j4 = 42 + 42 1 1 = − j S 8 8 This represents a resistor having a conductance of 1/8 S (a resistance of 8 Ω) connected in parallel with an inductance having an admittance of 1/(jωL) = –j 1/8 S The value of the inductance for ω = 2 rad/s comes out to be L = 4 H The parallel equivalent sub-circuit is as follows:

The equivalence we have shown between elements is only valid for the specified frequency of ω = 2 rad/s To illustrate, we change the frequency to ω = 4 rad/s The impedance of the series circuit is now Z = 4 + j8 Ω We can calculate Y = 1/Z = 0.05 – j0.l S The latter is the admittance of the parallel connection of a 20 Ω resistor and a 2.5 H inductor Hence, both the resistor and inductor in the parallel circuit have different values for equivalence at ω = 4 rad/s compared to what they had for equivalence at 2 rad/s 6.2 Inductive and Capacitive Sub-circuits The passive elements, resistor, capacitor and inductor, themselves form two-terminal sub-circuits with complex impedances For the resistor, Z(jω) = R + j0, so its resistance is R and its reactance is zero The inductor has Z(jω) = 0 + jωL-so its resistance is zero and its reactance is ωL

27 Topic 6 – Phasor Analysis

The capacitor has Z(jω) = 1/(jωC) = 0 + j (-1/(ωC)), so its resistance is zero and its reactance is - 1/(ωC) Thus, the resistor is purely resistive, and both the inductor and the capacitor are purely reactive The inductor has a positive reactance and the capacitor a negative reactance For any passive two-terminal network with impedance Z(jω) = R + jX(ω), we say that it is inductive if X(ω) > 0 and capacitive if X(ω) < 0 Because X(ω) is a function of ω, a general two-terminal sub-circuit can be inductive at one frequency and capacitive at another 6.3 Resonant sub-circuits Consider the series LC sub-circuit shown:

Its impedance is given by the sum of two impedances in series: 1 ⎡ 1 ⎤ Z( jω) = + jωL = j ωL − Ω jωC ⎣⎢ ωC⎦⎥ Z(jω) is zero under the condition: 1 ωoL = ωoC 1 ω 2 = o LC 1 ωo = LC

This phenomenon is called series resonance and ωo is the series resonant-frequency

For frequencies ω > ωo, the reactance ωL −1 (ωC) is positive and the sub-circuit is inductive; for values of ω < ωo, the reactance is negative and so the sub-circuit is capacitive A plot of reactance of the inductor, the capacitor and the series tuned circuit has the form shown:

4

XL = w*L 2

X=XL + XC

0 XC=-1/(w*C)

-2 Reactance

-4

-6

-8 0 0.5 1 1.5 2 2.5 3 3.5 4 Frequency (rad/s)

28 Topic 6 – Phasor Analysis

At ω = ωo =1 LC , the series sub-circuit has zero impedance and therefore is equivalent to a short circuit at that frequency Consider now the parallel LC sub-circuit:

The impedance of the two parallel branches is given by: 1 × jωL jωC ωL 1 Z jω = = j = j Ω ( ) 1 2 1 + jωL 1− ω LC − ωC jωC ωL

The impedance is infinite at ωo =1 LC

Here ωo is the parallel resonant frequency The phenomenon itself is called parallel resonance The parallel LC sub-circuit is inductive for ω < ωo and capacitive for larger values of ω > ωo A plot of reactance of the inductor, the capacitor and the parallel tuned circuit has the form shown:

10

8

6

4

2 XL = w*L

0

Reactance -2 XC = -1/(w*C) -4 X = 1/(1/XL + 1/XC)

-6

-8

-10 0 0.5 1 1.5 2 2.5 3 3.5 4 Frequency (rad/s)

The parallel LC sub-circuit, , has infinite impedance at the resonant frequency and so is equivalent to an open circuit at that frequency

7 SUPERPOSITION FOR AC FORCED RESPONSE When a circuit has two or more independent sources, some care must be taken; it is only if all the independent sources in a circuit are sinusoidal and have the same frequency that we can draw a single phasor equivalent Also, the impedances of inductors and capacitors depend on frequency and therefore there must be a single operating frequency However, analysis by superposition allows us to analyse circuits with sources having different frequencies

29 Topic 6 – Phasor Analysis

Since in superposition, we carry out a separate circuit analysis each with just one source, it is possible to handle sources with different frequencies However, it is necessary to draw a separate phasor circuit diagram for each analysis with the frequency of the source clearly labeled; furthermore in each analysis we must use the applicable frequency to calculate the impedances of the inductors and capacitors Then, we convert each solution phasor back to the time domain Then we can add the time domain expressions using superposition to obtain the total response 7.1.1 Example 12 Consider the following circuit:

The voltage sources are given by:

vs1(t) =10cos(3t) V

vs2(t) =10cos(4t) V Find the resulting forced response for v(t)

Repeat the solution if vs1(t) = vs2(t) =10cos(3t) V In the first situation the sources are at different frequencies, so we are forced to use superposition To see why on an intuitive basis, consider this: when the two sources have different frequencies, which frequency do we use in computing the inductor's impedance? There must be a separate circuit and a separate set of impedances for each frequency Thus, deactivating the two sources one at a time produces the two partial response equivalent circuits shown:

Note that the different frequencies are clearly stated and that the impedance of the inductor differs in the two cases because of the change in frequency It can easily be shown that:  V1 = 5 2∠45  v1(t) = 5 2 cos(3t + 45 ) V  V2 = 4∠36.9  v2(t) = 4cos 4t + 36.9 V ( )

30 Topic 6 – Phasor Analysis

Thus, the solution is:

  v(t) = v1(t) + v2(t) = 5 2 cos 3t + 45 + 4cos 4t + 36.9 V ( ) ( )

When both sources are identical vs1 = vs2 = 10 cos(3t) V, we can draw a single phasor equivalent in- corporating both sources:

Using nodal analysis, the single nodal equation is: V −10 V −10 V + + = 0 6 6 j3 Solving, we get: j10 10∠90 V 5 2 45 = =  = ∠ 1+ j 2∠45 so that v(t) is given by: v(t) = 5 2 cos 3t + 45 ( ) 8 THEVENIN AND NORTON EQUIVALENTS IN PHASOR FORM

We illustrate this by means of two examples: Example 16 Find the Thevenin and Norton phasor equivalents for the subcircuit shown:

Solution The phasor equivalent circuit is as follows:

Remember from our previous work that one option is to find two out of three things: the open circuit voltage, the short circuit current, and the equivalent impedance of the deactivated sub-circuit We choose to derive the first and the last of these The open circuit voltage is found using the voltage divider rule:

31 Topic 6 – Phasor Analysis

j2   Voc = × 2∠ − 45 =1∠0 V 2 + j2 The equivalent impedance with voltage source de-activated is: 2 × j2 2∠90 Z 2 45 1 j = =  = ∠ = + Ω 2 + j2 2∠45 We can draw the Thevenin equivalent as shown:

We can recognize that, at the given frequency only, the j Ω impedance is equivalent to a 0.5 H inductor:

To get the Norton equivalent, we need only find the short circuit current, which we know is the ratio of open circuit voltage to equivalent impedance:

V 1∠0 1 I = oc = = ∠ − 45 A sc Z j2  ( ) 2∠45 2 The Norton equivalent is the parallel connection of a current source having this value and the equivalent impedance:

We can interpret the equivalent impedance as a series RL circuit:

We can also use the equivalent parallel model for the equivalent impedance: 1 1 1− j 1 1 Y = = = = − j S Z 1+ j 2 2 2

This represents the parallel connection of a 2 Ω resistor and a j2 Ω inductor; this equivalence is only valid at the specified frequency

32 Topic 6 – Phasor Analysis

At the stated frequency, the inductor value is 1 H Example 17 Solve the circuit in the figure for the steady state sinusoidal response for i(t) using Thevenin equivalent sub-circuits:

Solution The phasor equivalent circuit is as follows:

We see that the two current sources have elements connected in parallel with them; hence, we can derive the Thevenin equivalent sub-circuit for each This results in the equivalent circuit shown:

We have combined the two series Thevenin impedances (+j Ω and –j Ω) with the j4 Ω inductive impedance and the 4 Ω resistive impedance This gives: 24(1− j) I = = 6∠ − 90 A 4 1 j ( + ) Thus: i(t) = 6cos 4t − 90 A ( )

33 Topic 6 – Phasor Analysis

9 ELEMENT COMBINING USING IMPEDANCE AND ADMITTANCE Parallel and series connections of elements:

Expression for v Series Parallel Expression for i 1 1 Impedance = 1 V = ZI Zeq = ∑Zi Z ∑ Z I = V Or eq i Z Admittance Yeq = ∑Yi I = YV Voltage division and current division:

Z1 I V1

V Z2 V2

Z3 V3

V Z1 V1 = IZ1 = Z1 = V Z1 + Z2 + Z3 Z1 + Z2 + Z3 I

I1 I2 I3

I Y1 Y2 Y3 V

I Y1 I1 = VY1 = Y1 = I Y1 + Y2 + Y3 Y1 + Y2 + Y3

10 MAXIMUM POWER TRANSFER INCLUDING REACTIVE ELEMENTS We addressed this question earlier for the case in which the source and load were purely resistive Consider the circuit shown:

34 Topic 6 – Phasor Analysis

We suppose that there is a load impedance, specified by Z, to which we are to deliver power from the source (represented by a ‘black box’) We assume operation in the AC steady state We next derive the Thevenin equivalent circuit for the source:

The load voltage (working with phasors) is given by voltage division: Z V = Voc Z + Zeq Current I is given by: V I = oc Z + Zeq We now let: Z = R + jX Zeq = Req + jXeq

where R and X are the resistance and the reactance of the load and Req and Xeq are the resistance and reactance of the Thevenin equivalent impedance When working with voltage and current in phasor form, the product V × I gives what is called complex power absorbed by the load (the concept of complex power will be covered later); we are interested in real power because it is real power which produces heat in the load ⎡ ⎤ Real power is given by Re ⎣V × I ⎦ Real power absorbed by the load is given by: ⎡ ⎤ Z 2 R 2 R 2 P = Re ⎢ V ⎥ = V = V L ⎢ 2 oc ⎥ 2 oc 2 2 oc Z Z Z Z R R X X ⎣⎢( + eq ) ⎦⎥ + eq ( + eq ) + ( + eq )

where we have written Z = R + jX and Zeq = Req + jXeq We can now determine the conditions for maximum power transfer between the source and the load Whether the source is given and we are to find the load or vice versa, the power in the load is maximised if X and Xeq cancel

Hence, the condition on X and Xeq for maximum power transfer is:

X = −Xeq Once this condition is satisfied, we have: R 2 P V L = 2 oc (R + Req ) 35 Topic 6 – Phasor Analysis

This is the same expression that we encountered when we found the condition for maximum power for resistor circuits

We showed earlier that if the source Req is given and the load R for maximum power transfer is to be determined, the power in the load is maximised for:

R = Req The power absorbed by the load under the maximum power condition is: 1 2 PL = Voc 4Req Consider the following example: Example Consider the following load circuit:

It is to be driven from the following source circuit:

Determine the reactance Xa which maximises the poser in the load? What is the maximum power? Solution We first determine the impedance of the load: 1 4 4 + j4 4 + j4 Z = R + jX = j4 + = j4 + = j4 + = j4 + = 2 + j6 Ω 1 1 1 j 2 2 − j − 4 4

Since X = 6 Ω, the condition for maximum power transfer is Xeq = –6 Ω The power in the load is given by: 2 2 1 2 P V V W L = 2 oc = oc (2 + 2) 8

11 CONCLUSIONS In this topic, we have introduced a method based on phasors for the analysis of circuits with sinusoidal voltages and currents Using phasors we were able to define impedance and admittance for the inductor and capacitor allowing Ohm's law to be formulated for these elements

36 Topic 6 – Phasor Analysis

Previous methods for AC Circuit Analysis using phasors (Ohm’s law, KCL and KVL, Thevenin and Norton equivalent circuits, superposition and nodal analysis) were shown to apply to phasors Finally, we considered maximum a power transfer theorem for circuits containing inductors and capacitors

37