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3. Frequency Domain Analysis

3.1 Response of Reactive Circuit Elements

In what follows, we will be discussing circuits ‘driven’ only by a sine (cosine) wave of a single frequency (if a circuit is driven by a signal composed of many frequencies we could use Fourier decomposition to analyze the effect of the circuit on a single frequency).

Generalize Ohm’s law by replacing the word ‘resistance’ with ‘impedance’ in order to describe the effect of

capacitors and inductors on current flow and voltage (usually changes in these quantities).

Inductors and capacitors exhibit ‘reactance’, i.e., they ‘react’ against changes in current or voltage (they also exhibit resistance).

Reactive elements in a circuit create a problem in the frequency domain analysis that we do not encounter with simple

resistive circuits: The voltage across a reactive circuit element

Figure 1

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and the current through it, in general, will not be in phase. We can easily see this by examining the current flowing into a single capacitor driven by a sine wave voltage source (Figure 1):

V(t) = V0 sin ωt and thusI(t) = C dV/dt = CωV0 cos ωt

Note we can write this as I(t) = CωV0 sin (ωt + φ) where φ = π/2

There is a phase shift between the voltage across the capacitor and the current through it. The current ‘leads the voltage by a phase of 900.

3.2 Phasors

In the case of more complex RLC circuits, the current through the circuit will invariably differ in phase by an angle other 900 from the voltage across it. Thus, generalizing Ohm’s law becomes more complicated than it is for a simple resistor, because now, given an input voltage, we need to know two numbers¸ the amplitude and the phase shift, in order to determine the resulting current in a circuit. The task is simplified upon realizing that one number specifies a scalar (for example, the current through…and the voltage across a resistor are specified only by a single number since there is no phase difference between the two), but a vector requires two numbers for its specification. Thus, we choose to represent voltages and currents in RLC circuits by vector quantities. Given that the voltages and currents are sinusoidal and remembering Euler’s formula…

ejφ = cos φ + j sin φwe can represent voltages and currents as vectors in the

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complex plane. Thus, we write down a general sinusoidal voltage as a complex quantity, which at time zero is applied with some arbitrary starting phase angle φ.

v(t) = V cos (ωt + φ) + jV sin (ωt + φ) The driving voltage v(t) may be pictured as a vector that rotates with frequency ω and amplitude V in the complex plane as shown in Figure 2. The real and imaginary parts of the vector represent its “(x,y) components” in the imaginary plane. The vector can be more simply represented by using Euler’s formula… v(t) = Ve j(ωt + φ) which is exactly the same formula as

the one above expressed in cosines and sines. We also note that currents can be expressed in the same way, but with a different amplitude and phase, i.e.,

i(t) = I cos (ωt + θ) + jI sin (ωt + θ)and using Euler’s formula…

i(t) = Ie j(ωt + θ)

Now, let’s consider the simplest case of the current through a single resistor R when driven by this complex sinusoidal voltage v(t). The current through the resistor is also a complex sinusoid of frequency ω, but with a different phase θ and amplitude I… From Ohm’s law… or…

Thus…

Figure 2

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And from Ohm’s law, we deduce that V = RI and φ = θ. Thus, leaving out the specific time dependence of the complex sinusoidal vectors, we can write Ohm’s law as a phasor relationship ...V = RI where

For resistors, φ will always equal θ because the voltage and current are always in phase.

Now let us find the current through the capacitor shown in the above picture and driven by this complex sinusoidal voltage. It is…

Figure 3

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Again, dropping the time dependence, which is identical for current and voltage, we have…

and we have obtained a generalization of Ohm’s law for a capacitor…

The quantity Z is called the impedance of the capacitor.

Furthermore note that since…ang(I) = ang(jωCV) = ang(jωC) + ang(V)1 → φ = θ -

900

For a capacitor, the current leads the voltage by 900.

Let’s now take a look at an inductor. As before, the voltage and the current is given by…

and

For an inductor

continuing…

Where the common time dependence has been removed leaving 1 ang(V) means the angle that a vector V makes with the x-axis (the “real” axis in this case).

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us with the phasor relationship between voltage and current… which we write as

The quantity Z is the impedance of the inductor.

Furthermore note that since…ang(V) = ang(jωLI) = ang(jωL) + ang(I) → φ = θ + 900

For an inductor, the voltage leads the current by 900.

3.3 Ohm’s Law Generalized

To summarize, we have the generalization of Ohm’s Law for an R-ohm resistor, C-farad capacitor and L-henry inductor…

The complex impedance of any of these elements in series or parallel obey the same rules as resistance.

Also, Kirchoff’s laws apply for multiply connected networks containing any type of circuit element…we just have to use the complex representation for V and I:(i) the sum of complex voltage drops around a loop is zero (ii) the sum of complex currents into a point is zero.

3.4 Representation of Phasors

Z = Z1 + Z1 + Z1 + …(series)

(parallel)

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Here we introduce a new notation and fairly elegant notation to represent phase vectors. The representation conveys the same information of a phasor as does its complex exponential representation.

First, let’s consider a phasor that represents the impedance ZR of a resistor of value R ohms. It is a vector in the complex plane that points along the real axis and its length is R. The complex representation of this resistor would be ZR = R ejθ where θ = 00. (Remember… R ejθ = R (cos θ + j sin θ) = R at θ = 00).

The impedance of a capacitor is…ZC = 1/ωC ejθ = 1/ωC (cos θ + j sin θ) = -j/ωC at θ = -900.

It’s a phasor that points down the imaginary axis at angle -900.

The impedance of an inductor is…ZL = ωL ejθ = ωL (cos θ + j sin θ) = jωL at θ = 900.

It’s a phasor that points up the imaginary axis at angle 900.

We can get rid of the complex exponential and denote these phasors instead using the following notation…

ZR = R∟00 ZC = 1/ωC∟-900 ZL = ωL∟900

The ∟ symbol stands for the angle relative to the real axis at which the phasor points. Indeed, phasors representing any complex voltage, current or impedance may be denoted this way. Now let’s give an example…

Example 3.1

I’m going to violate

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my dictum of using only realistic circuit components in this example because I swiped it from a stupid electrical engineering book…but it does illustrate phasor frequency domain analysis using the elegant notation introduced above.

Consider the low pass R-C circuit shown in Figure 4. The value of the components used are shown in the figure. The network is driven by a sinusoidal voltage source of amplitude A = 6 volts and angular frequency ω = 2 rad/s. Thus, vS = 6 sin 2t.

At time t = 0, the voltage is vS = 0, Now the sine function “lags behind” the cosine function by 900, so we can re-write the voltage function as…

vS = 6 cos (2t - 900)which we’ll write in the new notation as… vS = 6∟-900. It shows that the applied voltage can be represented by a phasor that points downward at –900 along the imaginary axis in the complex plane.

We wish to find the output voltage v0 and the current i through the capacitor. The impedance of the resistor is simply ZR = 1 (leaving off the Ω unit) and the impedance of the capacitor is ZC = 1/jωC = -j (again leaving off the Ω unit). Thus, Ohm’s law is:

VS = Z I = 1I – jIUsing our new notation…

6∟-900 = (1 – j)I = √2 ∟-450 →I = (6∟-900)/( √2 ∟-450) = 3 √2 ∟-450 A

And the output voltage is…V0 = -j I = (1∟-900)( 3 √2 ∟-450) = 3 √2 ∟-1350 V

(Note that the factor 1∟-900 in the above expression is the phasor representation for –j).

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The phasor diagram for this circuit is shown in Figure 5.

Figure 5

The capacitor voltage V0 lags the applied voltage VS by 450. Thus, if the output voltage is taken across the capacitor (as in the Figure) the circuit is a “lag” network. Note, though, that the current leads the applied voltage by 450. Since VR = 1I = I, the voltage across the resistor leads the applied voltage VS by 450. If the output voltage was taken across the resistor, then the network would be a “lead” network.

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Figure 6

We show in Figure 6 a plot of the input and output voltage as a function of time. Compare the plot with the results of the phasor analysis which was carried out in the frequency domain. I could give you the problem of reproducing the time domain plot by analyzing the circuit using differential equations in the time domain.

3.5 Frequency Response of High & Low Pass RC Networks

Continuing with the frequency domain analysis we’ve just discussed, we analyze the high and low pass networks in terms of their ability to pass certain frequencies while blocking others.

We first consider the high-pass network shown in Figure 7a. This network can be viewed as a voltage

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divider whose output voltage V2 is given by…

and…where τ = RC.

The magnitude of M = (called the ‘gain’) can be calculated by multiplying M by its complex conjugate M* and then taking the square root…

Note, that for ωτ << 1, M rises linearly with ω at 6 db/octave2 as shown in Fig 7b. The output at frequencies in this region of the spectrum are attenuated. For ωτ >> 1, M is flat and these frequencies are passed without attenuation. The transition region

occurs at and close to the ‘corner frequency’ ωC = 1/τ. This frequency represents the “breakpoint” of the filter and the gain factor there is -3 db.

We can apply a similar analysis to the low pass network shown in Figure 8a. This network can be viewed as a voltage

2 An octave is a frequency span of 2:1. 6 db (decibels) represents a voltage ratio of 2:1 (#db = 20 log (V2/V1)).

Figure 7

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divider whose output voltage V2 is given by…

This reduces to…

The magnitude M is …

The correlation between various aspects of STC networks is summarized in Table 2-1.

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Example 3.2

Find the Thevenin equivalent of the circuit shown in Figure 9 from the perspective of an attached load ZL. Show the equivalent in both the frequency domain and the time domain.

Figure 9

We need to calculate both the open circuit voltage V0, and the output impedance Z0. We start by noting that the same current i flows through all components with the output open-circuited.

i = iC = iL = iR (3.2.1)Note that the impedances of the (i) capacitor, (ii) inductor and (iii) resistor respectively are: (i) 1/jωC = -j50, (ii) jωL = j200 and (iii) R = 200 so using Ohm’s law we get…

(VC – V0) / ZC = (V0 – VR) / ZL = (VR – 0) / R(3.2.2)Where what the symbols stand for should be obvious. The first of the equations (3.2.2) yields…

[(5∟-300) – V0] / -j50 = (V0 – VR) / j200 (3.2.3)and this reduces to…

3 V0 + VR = 20∟-300 (3.2.4)The second equation of (3.2.2) yields…

(V0 – VR) / j200 = VR / 200 → VR = V0 /(1 + j) (3.2.5)

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Substituting this expression for VR into (3.2.4) yields the open-circuit voltage…

V0 = [(1 + j)( 20∟-300)] / (4 + j3) (3.2.6)Note that (1 + j) and (4 + j3) are phasors in the complex plane and may be written as…

1+ j = √2 ∟450 and 4 + 3j = 5 ∟36.90

So Equation (3.2.6) reduces to…V0 = (√2 ∟450) ( 20∟-300) / (5 ∟36.90) = 4√2∟-21.90 (3.2.7)

Now we have to find Z0. We do so by looking back into the circuit from the perspective of the load (Figure 10). The output impedance that we “see” is…

Z0 = (ZL + ZR)║ ZC = (j200 + 200)(-j50) / (j200 + 200 -j50) = 200(1 – j) / (4 + j3) = 8 – j56 = 40√2∟-81.90 Ω

We show the Thevenin equivalent circuit in Figure 11.

Figure 10

Figure 11 Thevenin equivalent in the frequency domain

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The fact that the output resistance is Z0 = 8 – j56 Ω tells us that the Thevenin equivalent in the time domain can represented as an 8 Ω resistor in series with a 4.46 mF capacitor (the impedance of the capacitor at ω = 4 rad/s is 56 Ω → 1/ωC = 56 → C = 4.46 mF) as shown in Figure 12.

Now, suppose ZL = R = 48 Ω. (a) What is the voltage VL across ZL?

Figure 12 indicates that Z0 and ZL (= R) can be considered to be a voltage divider in order to calculate VL. Thus…

VL = [R / (Z0 + R)] V0 (3.2.1)VL = (48)( 4√2∟-21.90) / (40√2∟-81.90 + 48∟00)

Here we draw a phasor diagram (Figure 13) to show that phasor arithmetic in the complex plane is exactly like vector arithmetic in the x-y plane. We’ll represent these phasors in terms of their (x,y) components.

Z0 = (x,y) = (40√2 cos (-81.90), 40√2 sin (-81.90))

Figure 12 Thevenin equivalent in the time domain.

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= (7.971, -56.004)R = (48, 0)Z0 + R = (55.971, -56.004)φ = -450 (phase of Z0 + R)

= 79.18

Substituting these values into Equation 3.2.1 above, we obtain…

VL = (48∟00)( 4√2∟-21.90) / 79.18∟-450

= 3.43∟23.10

= 3.43 cos (4t + 23.10)

3.6 Power

The instantaneous power delivered to any circuit element is given by the product P = VI. In dc circuits containing only resistances, the instantaneous power and the average power3 are identical. But when reactive components are involved and subjected to sinusoidal signals, some rather funny things can happen; for instance, the sign of the VI product can reverse over a cycle. For example, let’s consider a single capacitor driven by a sinusoidal voltage as discussed in section 3.1 and illustrated again in Figure 14.

3 where T is some arbitrary time over which the average is to be computed.

Figure 13

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Figure 14

During time intervals A and C, power is being delivered to the capacitor, causing it to charge up; it’s storing more and more energy. During intervals B and D, the IV product is negative and the capacitor is discharging and delivering energy back to the sine wave generator! The average power consumed over a complete cycle is zero. This is, in fact, always true for any purely reactive elements (capacitors, inductors or any combinations thereof). We can prove this for purely sinusoidal signals as follows…

For any arbitrary element or combination thereof, as shown in Figure 15, the instantaneous power absorbed by the element is p(t) = v(t) i(t). For a

Figure 154

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sinusoidal voltage, say v(t) = V cos (ωt + φ1), the current will also be sinusoidal, say i(t) = I cos (ωt + φ2). The instantaneous power absorbed by the element is then…

p(t) = [V cos (ωt + φ1)][ I cos (ωt + φ2)]Using the trigonometric identity…

cosα cosβ = ½ cos(α – β) + ½ cos(α + β)we obtain…

p(t) = ½ VI cos(φ1 – φ2) + ½ VI cos(2ωt + φ1 + φ2)

What, then, is the average power absorbed by the element? The 2nd term on the right in the last equation is a sinusoid of angular frequency 2ω and thus the average of this term over a complete cycle is identically zero. The average power consumed by the element is then

P = ½ VI cos(φ1 – φ2) = ½ VI cos Δφwhere Δφ = φ1 – φ2 is the phase difference between the voltage and the current. For a resistor (or circuit elements consisting only of resistors), the voltage and current are always in phase and cos Δφ = cos 00 =1 and the average power absorbed is

P = ½ VI = RI2 = V2/Rresults I’m sure that you learned in elementary physics.

For either an inductor or a capacitor (or circuit elements consisting only of these reactive elements) the voltage and current will always be 900 out of phase and since cos Δφ is zero, no power will be consumed on average.

Example 3.3

For the circuit shown in Figure 9, when ZL = R = 48 Ω, calculate

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the average power absorbed by each element in the circuit.

From Example 3.2 above, we calculated the voltage VL across the 48 Ω load resistor. That is also the voltage across the series combination of the 50 H inductor and the 200 Ω resistor, so first let’s now calculate the currents through each of these three elements…(i) Current through the load resistor iZL…

iZL = VL / ZL = 3.43∟23.10 / 48 = 71.46∟23.10 mA

(ii) Current through the 50H inductor and the 200 Ω resistor (which we’ll call R1) iLR1…

iLR1 = VL / (Z50H + R1) = 3.43∟23.10 / (200∟900 + 200∟00) = 3.43∟23.10 / (200√2∟450) = 12.13∟-21.90 mA

(iii) Current through the 5 mF capacitor iC…iC = iZL + iLR1 = 71.46∟23.10 + 12.13∟-21.90

iZL = 71.46 (cos 23.10, sin 23.10) = (65.73, 28.036)iLR1= 12.13 (cos -21.90, sin -21.90) = (11.255, -4.524)Adding components, we get…iC = (76.985, 23.512) = 80.5∟170 mA

Now, we calculate voltages across the circuit elements…

(i) Voltage across the 200 Ω resistor…VR1 = iLR1 R1 = (12.13∟-21.90)(200) mA Ω = 2.43∟-21.90 V

(ii) Voltage across the 50H inductor ΔV50H …ΔV50H = VL – VR1 = 3.43∟23.10 - 2.43∟-21.90

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VL = 3.43(cos 23.10, sin 23.10) = (3.155, 1.346)VR1 = 2.43(cos -21.90, sin -21.90) = (2.255, -0.906)ΔV50H = (0.900, 2.252) = 2.425∟68.20

(iii) The voltage across the 5 mH capacitor is…ΔVC = V – VL = 5∟-300 – 3.43∟23.10

V = 5(cos -300, sin -300) = (4.330, -2.5)VL = (3.155, 1.346)ΔVC = (1.175, 3.846) = 4.021∟-730

We can now calculate the average power absorbed by each element according to P = ½ IV cos θ…(i) The load resistor:

PL = ½ (71.46∟23.10)( 3.43∟23.10) = 122 mW(ii) The 200 Ω resistor:

PR1 = ½ (12.13 ∟-21.90) (2.43∟-21.90) = 14.7 mW(iii) The 5 mF capacitor:

PC = ½ (80.5∟170)( 4.021∟-730) = 162 cos(900) = 0(iv) The 50H inductor:

P50H = ½ (12.13∟-21.90)( 2.425∟68.20)=14.7 cos(900) = 0

I’m sure that you could have guessed that the power absorbed by both the capacitor and inductor was 0, since they both are reactive elements.

Enough of this stuff. Let’s move on to some non-linear elements add a lot of spice to electrical circuits.