EEE161 Applied Electromagneticsathena.ecs.csus.edu/~milica/EEE161/lecturenotes/master211.pdf · Dr....

Dr. Milica Markovic Applied Electromagnetics page 1

EEE161 Applied Electromagnetics

Instructor: Dr. Milica MarkovicOffice: Riverside Hall 5026

Email: [email protected]:http://gaia.ecs.csus.edu/˜milica

California State University Sacramento EEE161 revised: 5. September, 2012

Contents

1 Background Concepts Review 3

Background Concepts Review 31.1 Sinusoidal Signals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Decibels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Phasors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.5 Using Vectors to Represent Phasors in an example . . . . . . . . . . . . . . . . . . . . 13

2 Transmission Lines 152.1 Types of transmission lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 What are Transmission Line Effects? . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.3 Propagation modes on a transmission line . . . . . . . . . . . . . . . . . . . . . . . . 192.4 Wave equation on a transmission line . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.5 Visualization of Lossless Forward and Reflected Voltage Waves . . . . . . . . . . . . . 232.6 Relating forward and backward current and voltage waves on the transmission line . . 252.7 Lossless transmission line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.8 What does it mean when we say a medium is lossy or lossless? . . . . . . . . . . . . . 272.9 Low-Loss Transmission Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.10 Voltage Reflection Coefficient, Lossless Case . . . . . . . . . . . . . . . . . . . . . . . 292.11 Standing Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3 Smith Chart 35

4 Impedance Matching 414.1 Lumped Element Impedance Matching . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2

Chapter 1

Background Concepts Review

1.1 Sinusoidal Signals

Sinusoidal signals are important because all periodic signals can be represented with sinusoidal signalsof different amplitudes and phases using Fourier series.

Typical sinusoidal signal is shown in Figure 1.1. On the y-axis is the instantaneous value of thesinusoidal voltage and on the x-axis is time. Instantaneous values of voltage change from -1V to1V with time. Sinusoidal signals can be characterized by the following parameters: peak amplitude,peak-to-peak, average, rms, period, time-delay and phase. Peak amplitude, peak-to-peak, averageand rms values, are read on the y-axis in Figure 1.1, whereas period, time delay and phase are readon the x-axis.

1. Peak amplitude is measured on the y-axis as the length from the average value of the signal (inthis case zero) to the maximum value of the signal (in this case 1). For signal shown in Figure1.1, peak amplitude has a constant value of Vp = 1. Peak amplitude is NOT an instantaneousvalue of the signal, and it does not vary with time. Sometimes amplitude and peak-amplitudeare used interchangeably. Other times, amplitude is used to denote an instantaneous value ofthe signal, and peak-amplitude is meant to denote the maximum value of amplitude. We willuse amplitude to mean peak-amplitude. When we want to emphasize an instantaneous valueof the signal, we will call it an instantaneous value.

2. Peak-to-peak is measured from the minimum value of the function (in this case -1) to themaximum value of the function (in this case 1). For signal shown in Figure 1.1, peak-to-peakvoltage has a constant value of Vpp = 2.

3. RMS or root-mean-square is defined as vrms = 1T

√∫ T0v(t)2dt. For signal shown in Figure 1.1,

and other sinusoidal signals of this form, vrms = Vp√2

= 1√2

= 0.707. Root mean square value isimportant because it represents the equivalent amount of DC power.

4. Average value vave1 = 1T

∫ T0v(t)dt. For signal shown in Figure 1.1, average value is Vave1 = 0

because the function has the same area under the function in the positive and negative cycle.

3


5. Period is measured on the x-axis as the length of one full cycle of the sinusoidal signal. Forsignal shown in Figure 1.1, this value is period = T

6. Time delay represents the lag (or lead) of one function with respect to another. For example,in Figure 1.1, function cos(ωt − 90o) is time-delayed for τ = T

4with respect to cos(ωt). To

find the time delay for a sinusoidal signal from its phase, we look at the way to represent thephase 90o in terms of product of frequency and time. Since in the sinusoidal signal expressioncos(ωt + Θ) phase Θ is added to ωt term, the phase has the same units as ωt, and can berepresented as the product of ωτ = θ, τ = θ

ω, where τ represents the time delay.

7. Phase of the signal is shown in Figure 1.2(d)-(e).

Review signals shown in Figure 1.2. Signals in Figure 1.2(a)-(b) are shown as a function of time,whereas signals in Figure 1.2(c)-(d) are shown as a function of angle. See how are the graphs thesame and how are they different.

Figure 1.1: Vocabulary used in describing sinusoidal signals.



(a) sin(ωt) (b) Sinusoidal signal shifted for time delay −π/4ω

(c) Sinusoidal signal as a function of angle ωt (d) Sinusoidal signal as a function of angle ωt with aphase shift of −π/4

(e) Sinusoidal signal as a function of angle ωt with aphase shift of +π/4

Figure 1.2: Sinusoidal signal as a function of time (a)-(b) and angle (c)-(e).



(a) Sinusoidal signals of different frequencies sin(ωt) (b) Sinusoidal signals of different amplitudes sin(ωt)

Figure 1.3: Comparison of sinusoidal signals.



1.2 Decibels

We will often represent the ratio off the to voltages in decibels. Decibel is a unit that is derived fromthe unit Bell. Bells are describing an order of magnitude difference between two values. For example,one Bell represents the ratio of 10. 1B = log P1

P2. This ratio is too large to represent quantities in

microwave engineering. Decibel is defined as the ratio of two powers 10 log P1

P2. For example if we

want to say that the output power is twice the input power we say that the power gain is 3dB.

G = 10logPoutPin

G = 10 log 2

G = 3dB (1.1)

The ratio of voltages is derived from the definition above, because voltages are proportional topower.

G = 10 logV 2out

V 2in

G = 20 logVoutVin

(1.2)

if the output voltage is 1.4 times the input voltage the voltage gain is equal to 3dB, whichrepresents the power ratio of two.



1.3 Complex Numbers

1. A complex number z can be represented in Cartesian coordinate system, Equation 1.3 or Polarcoordinate system, Equation 1.4 form.

Figure 1.4: Complex number z in rectangular and polar coordinates.

z = x+ jy (1.3)

z = |z|ejΘ (1.4)

x is the real part, y is the imaginary part, |z| is the magnitude and Θ is the angle of the complexnumber. Note that we can represent this complex number as a vector.

2. Euler’s Identity is used to convert time domain signal to phasors.

ejΘ = cosΘ + jsinΘ (1.5)

3. Cartesian and polar form representation are used in phasors.

|z| =√x2 + y2 (1.6)

Θ = arctgy

x(1.7)

4. Complex Conjugate is often seen when finding the conditions for maximum power transfer.

z∗ = (x+ jy)∗ = x− jy = |z|e−jΘ (1.8)



5. Complex number addition and subtraction is often seen went to complex impedances are placedin series and the equivalent complex impedance has to be found. the easiest way to add twocomplex numbers is to find Cartesian representation of both and then add the real partsseparately and the imaginary part separately.

z1 = x1 + jy1 (1.9)

z2 = x2 + jy2 (1.10)

z1 + z2 = x1 + x2 + j(y1 + y2) (1.11)

6. Multiplication and Division are often seen the in calculation of the transfer function of a circuit.The easiest way to divide two complex numbers is to find the polar representation of both andthen divide the amplitudes and subtract the phases.

z1 = |z1|ejΘ1 (1.12)

z2 = |z2|ejΘ2 (1.13)

z1

z2

=|z1||z2|

ejΘ1−Θ2 (1.14)

7. Some examples. Find the magnitude and phase of a complex number

−1 = ej180o (1.15)

j = ej90o (1.16)

(1.17)



1.4 Phasors

The learning objective of this section is to show how to solve a circuit in frequency domain usingphasors. we will review of fat phasers with an RC circuit example shown in Figure 1.6. To solve thiscircuit in the time domain we apply Kirchoff’s voltage law as shown in Equation 1.18 -1.19.

The circuit in Figure 1.6 is a simple RC circuit. KVL equation in the time domain is given inEquation 1.18.

vs(t) = vR(t) + vC(t) (1.18)

vs(t) = Ri+1

C

∫i(t)dt (1.19)


Where

vs(t) = Acos(ωt+ Θ) (1.20)

In order to solve this circuit we have to solve a differential equation. Use of phasors simplify theequations significantly. Differential equations become a set of linear equations. Phasor is anothername for a complex number in polar coordinate system.

In order to use phasors, the circuit has to be linear. Circuits that have only capacitors, inductorsand resistors are linear circuits. In a linear circuit, all currents and voltages are at the frequencyof the generator. That means that we don’t have to keep track of the frequency of voltages andcurrents when we are solving the circuit. We know the frequency once we know the frequency of thegenerator. The quantities that will differ for different currents and voltages is the amplitude andphase of the signal. Phasors allow us to drop the information about the frequency of the signal, andonly keep track of the magnitude and phase of the signal. In order to remove cos(ωt) term fromthe equations we have to use complex numbers. To write cos(ωt) 1 in a concise form as a complexnumber, we add to the forcing function a sinusoidal imaginary term.

A cos(ωt+ Θ) + jAsin(ωt+ Θ) (1.21)

1It is customary to use cos(ωt) for our time-domain signal. If signals in a circuit are given in terms of sin(ωt),the sin function has to be converted to a cosine. To do that, subtract 90o from the phase of the sinusoid, becausesin(ωt) = cos(ωt− 90o)



We have to make sure later when we are done with our calculations with complex number, thatwe only take the real part of the final expression. It seems that we made the above expression morecomplicated, however, if we remember Euler’s identity, the expression becomes

vs(t) = V cos(ωt+ ΘV ) = ReAcos(ωt+ Θ) + jAsin(ωt+ Θ) = ReAej(ωt+Θ) = ReAejΘejωt(1.22)

In Equation 1.22 we extracted the phase and amplitude information and separated it from thefrequency. The amplitude and phase information is called phasor VS(jω) = AejΘ. Why is thisexpression better then the one with a cos(ωt) and how can we remove t from Equation 1.19? Toanswer this question, we first have to write general expressions for voltage v(t) and current i(t) inthe Equation 1.19:

v(t) = ReV cos(ωt+ ΘV ) + jV sin(ωt+ ΘV ) = ReV ej(ωt+ΘV ) = ReV ejΘV ejωt (1.23)

If we look at the first and last expression in Equation 1.23 we see that the time domain signal isthe real part of the product of phasor and the ejωt term.

v(t) = ReV ejΘV ejωt (1.24)

Similarly for current

i(t) = I cos(ωt+ ΘI) = ReIcos(ωt+ ΘI) + jI sin(ωt+ ΘI) = ReIej(ωt+ΘI) = ReIejΘIejωt(1.25)

If we look at the first and last expression in Equation 1.25 we get a similar expression for current.

i(t) = ReIejΘIejωt (1.26)

In Equations 1.23-1.25 V and I are voltage and current amplitudes, and ΘV and ΘI are voltageand current phases. The voltage on the resistor is then given in Equation 1.27

vR(t) = R× i(t) = R×ReIejΘIejωt = ReR× IejΘIejωt (1.27)

The voltage on the capacitor is a bit more complicated. We know that i(t) = ReIejΘIejωt, butwhat is the integral of i(t)?

vC(t) =1

C

∫ReIejΘIejωtdt (1.28)

Now if integral and Re exchange places, and if we take all time-independent quantities in frontof the integral,



vC(t) =1

C

∫i(t)dt = Re 1

C

∫IejΘIejωtdt = Re 1

CIejΘI

∫ejωtdt (1.29)

vC(t) = Re 1

CIejΘI

1

jωejωt = Re 1

jωCIejΘIejωt (1.30)

We now replace the time-domain quantities in equation 1.19 with these newly developed expres-sions.

vs(t) = vR(t) + vC(t) (1.31)

ReAejΘejωt = ReR× IejΘIejωt+Re 1

jωCIejΘIejωt (1.32)

A common term in the previous equation is ejωt, and we can now drop Re, as long as we laterremember to take only the real part of the expresion for the phasor of voltage and current to get thetime domain expression. We can now write the equation as

VSejΘVS = R× IejΘI +

1

jωCIejΘI (1.33)

VS(jω) = RI(jω) +I(jω)

jωC(1.34)

Since this is a linear equation, we can easily solve it:

I(jω) =VS(jω)

R + 1jωC

(1.35)

Let’s say that the values are given for R, C, ω and VS such that the phasor of the current isI = 3ej45o . To obtain the signal in the time domain, we multiply the phasor I with the ejωt term,and then we find the real part of the expression to obtain its current in the time domain, as shownin Figure 1.36

i(t) = Re3ej45oejωt = Re3ejωt+45o = Re3 cos(ωt+ 45o) + j3 sin(ωt+ 45o = 3 cos(ωt+ 45o)(1.36)

In case we have an inductor in the circuit, the voltage on an inductor can be derived as shown inFigure 1.37.

vL(t) = L∂i(t)

∂t= L

ReIejΘIejωt∂t

dt = ReLIejΘI∂ejωt

∂t = ReLIejΘI jωejωt = RejωLIejΘIejωt(1.37)

Step-by-step instructions on how to solve circuits using phasors is given as follows:




cicruit element impedance low frequencies f → 0 high frequencies f → infcapacitor 1

jωC∞ 0

inductor jωL 0 ∞

Table 1.1: Impedance of the capacitor and inductors and their equivalent impedances at high andlow frequencies.

1. Adopt cosine reference for generator voltage or current.

2. replace all impedances with their phasor expressions,

3. write KVL and KCL, or use other Network Analysis techniques.

4. find the phasor expression for the required current or voltage.

5. multiply the phasor with ejωt

6. find the real part of the above expression to get the current in the time domain.

1.5 Using Vectors to Represent Phasors in an example

1. Calculate on paper the magnitude and phase of the current and voltages in a series RC circuitshown in Figure 1.7 (a) if the circuit is driven with a frequency of 1 GHz (phase is zero) and

R=1kΩ, C= 12π

10−12

F.

(a) What are the magnitude, phase and time delay of the source voltage?

(b) What are the magnitude, phase and time delay of the voltage across the resistor?

(c) What are the magnitude, phase and time delay of the voltage across the capacitor?

(d) The simulated voltage across the resistor is about 0.5V and the simulated voltage acrossthe capacitor is about 0.85V. If we use KVL 0.85V + 0.5V 6= 1V . Why? Look at Figure1.7 to help you with answer this question.

(e) If you know the current through circuit, capacitance, resistance and the frequency ofoperation, could you plot the phasor diagram to find the input voltage? How would youdifferentiate between the voltage on an inductor and a capacitor?



(a) RC circuit in ADS (b) Sinusoidal signal as a function of angle ωt

(c) Points represent polar plot of com-plex voltages in RC circuit.

(d) Vectors are drawn in the polar plotof complex voltages in RC circuit. Notehow they add up to 1V.

Figure 1.7: Adding phasors using vectors.


Chapter 2

Transmission Lines

Any wire, cable or line that guides energy from one point to another is a transmission line. Wheneveryou make a circuit on a breadboard, every wire you attach makes a transmission line with the groundwire. Whether we see the propagation (transmission line) effects on the line depends on the linelength. At lower frequencies or very short line lengths we do not see any difference between thesignal’s phase at the generator and at the load, whereas at higher frequencies we do.

2.1 Types of transmission lines

1. Coaxial Cable, Figure 2.1(a)

2. Microstrip, Figure 2.1(b)

3. Stripline, Figure 2.1(c)

4. Coplanar Waveguide, Figure 2.1(d)

5. Two-wire line, Figure 2.1(e)

6. Parallel Plate Waveguide, Figure 2.1(f)

7. Rectangular Waveguide, Figure 2.1(g)

8. Optical fiber, Figure 2.1(h)

2.2 What are Transmission Line Effects?

Figure 2.2 shows a step voltage at the generator and the load of a circuit in Figure 2.4. The voltageneeds T sec to appear at the load, once the switch closes. Figure 2.3 shows the step signal as ittraveles on the transmission line at different time steps t=0, t=T/4, t=T/2 and t=T. How muchtime it takes for this signal to go from AA’ end to BB’ end? Since electromagnetic waves propagatewith the constant speed, the speed of light, time that the signal needs to go from the generator to

15


(a) Coaxial Cable (b) Microstrip (c) Stripline

(d) CoplanarWaveguide

(e) Two-wire Line

(f) Parallel PlateWaveguide

(g) Rect-angularWaveguide

(h) Optical Fiber

Figure 2.1: Types of transmission lines.

the load depends on the length of the transmission line l. T = lc, where c = 3 108. If the signal at

the generator AA’ is

vAA‘(t) = Acos(ωt) (2.1)

Then at the other end the transmission line the signal is

vBB‘(t) = vAA‘(t− T ) (2.2)

vBB‘(t) = vAA‘(t− l

c) (2.3)

vBB‘(t) = Acos(ω(t− l

c)) (2.4)

vBB‘(t) = Acos(ωt− ω lc) (2.5)

vBB‘(t) = Acos(ωt− ω

cl) (2.6)

Since we know that angular frequency is ω = 2πf



Figure 2.2: Voltage as a function of time at the generator side z=0 (top) and the load side z=l(bottom) of the transmission line in Figure 2.4, if the switch closes at t=0 the voltage arrives att=l/c=T at the load. These graphs can be obtained by observing the voltage on an oscilloscope atthe load and at the generator side.

Figure 2.3: Voltage along the transmission line in Figure 2.4, for four different time intervals t=0,switch closes, t=T/4, t=T/2 and t=T. It is assumed that the length of the transmission line is equalto l=T/c.



Figure 2.4: Electronic Circuit with an emphasis on cables that connect the generator and the load.

vBB‘(t) = Acos(ωt− 2πf

cl) (2.7)

The quantity cf

is called the wavelength λ.

vBB‘(t) = Acos(ωt− 2π

λl) (2.8)

The quantity 2πλ

is the propagation constant βFinally the expression for the voltage at BB end is

vBB‘(t) = Acos(ωt− βl) (2.9)

vBB‘(t) = Acos(ωt−Ψ) (2.10)

We see that at BB’ the signal will experience a phase shift. We will derive this equation againlater from the Telegrapher’s equations. Now let’s see how the length of the line l affects the voltageat the end BB’. Look at Equation 2.8. The signal will experience a phase shift of 2π l

λ. If this phase

shift is small, there will not be much difference between the phase of the signal at the generator andat the load. This means that we don’t have to use transmission line theory to account for the effectsof the line. If the phase shift is significant, then we do have to use the transmission line theory. Let’slook at some numbers in the following example.

1. If lλ< 0.01 then the angle 2π l

λis of the order of 0.0314 rad or about 20. In this case, the phase

is obviosly something that we don’t have to worry about. When the length of the transmissionline is much smaller than λ, l << λ

100the wave propagation on the line can be ingnored.

2. If lλ> 0.01, say l

λ= 0.1, then the phase is 200, which is a significant phase shift. In this case

it may be necessary to account fro transmission line effects.



2.3 Propagation modes on a transmission line

Coax, two wire line, microstrip etc can be approximated as TEM up to the 30-40 GHz (unshielded),up to 140 GHz shielded.

1. TEM E, M field is entirely transverse to the direction of propagation

2. TE, TM E or M field is in the direction of propagation

Transmission lines that we are discussing here are TEM lines.

2.4 Wave equation on a transmission line

In this section we will derive the expression for voltage and current along a transmission line. Thisexpression will have two variables, time t and space z. So far we have only seen voltages and currentsas a function of time. This is because all circuit elements seen so far were lumped elements. Indistributed systems we want to derive the equations for voltage and current for the case when thetransmission line is longer then the fraction of a wavelength. To make sure that we don’t encounterany transmission line effects to start with, we can look at the piece of a transmission line that ismuch smaller then the fraction of a wavelength. In other words we cut the transmission line intosmall pieces to make sure there are no transmission line effects, as the pieces are shorter then thefraction of a wavelength. We then represent each piece with an equivalent circuit as shown in Figure2.5 (a). To derive expressions for current and voltage on transmission line we will use the followingfive-step plan

1. Look at an infinitesimal length of a transmission line ∆z.

2. Represent that piece with an equivalent circuit.

3. Write KCL, KVL for the piece in the time domain (we get differential equations)

4. Apply phasors (equations become linear)

5. Solve the linear system of equations to get the expression for the voltage and current on thetransmission line as a function of z.

Let’s follow the plan now. Look at a small piece of a transmission line and represented it withan equivalent circuit. What is modeled by the circuit elements?

Write KVL and KCL equations for the circuit above.KVL

−v(z, t) +R∆zi(z, t) + L∆z∂i(z, t)

∂t+ v(z + ∆z, t) = 0

KCL



(a) Coaxial cable is cut in shortpieces.

(b) Equivalent circuit of a section oftransmission line.

(c) Equivalent circuit of transmission line.

Figure 2.5: Section of a coaxial cable.



i(z, t) = i(z + ∆z) + iCG(z + ∆z, t)

i(z, t) = i(z + ∆z) +G∆zv(z + ∆z, t) + C∆z∂v(z + ∆z, t)

∂t

Rearrange the KCL and KVL Equations 2.11, 2.15 and divide them with ∆z. Equations 2.12,2.16. let ∆z → 0 and recognize the definition of the derivative Equations, 2.13, 2.17.

KVL

−(v(z + ∆z, t)− v(z, t)) = R∆zi(z, t) + L∆z∂i(z, t)

∂t(2.11)

−v(z + ∆z, t)− v(z, t)

∆z= Ri(z, t) + L

∂i(z, t)

∂t(2.12)

lim∆z→0

−v(z + ∆z, t)− v(z, t)

∆z = lim

∆z→0Ri(z, t) + L

∂i(z, t)

∂t (2.13)

−v(z, t)

∂z= Ri(z, t) + L

∂i(z, t)

∂t(2.14)

KCL

−(i(z + ∆z, t)− i(z, t)) = G∆zv(z + ∆z, t) + C∆z∂v(z + ∆z, t)

∂t(2.15)

−i(z + ∆z, t)− i(z, t)∆z

= Gv(z + ∆z, t) + C∂v(z + ∆z, t)

∂t(2.16)

lim∆z→0

−i(z + ∆z, t)− i(z, t)∆z

= lim∆z→0

Gv(z + ∆z, t) + C∂v(z + ∆z, t)

∂t (2.17)

−i(z, t)∂z

= Gv(z + ∆z, t) + C∂v(z + ∆z, t)

∂t(2.18)

We just derived Telegrapher’s equations in time-domain:

−v(z, t)

∂z= Ri(z, t) + L

∂i(z, t)

∂t

−i(z, t)∂z

= Gv(z + ∆z, t) + C∂v(z + ∆z, t)

∂t

Telegrapher’s equations are two differential equations with two unknowns, i(z, t), v(z, t). It is notimpossible to solve them, however we would prefer to have linear equations. So what do we do now?Express time-domain variables as phasors!

v(z, t) = ReV (z)ejωti(z, t) = ReI(z)ejωt



And we get the Telegrapher’s equations in phasor form

−∂V (z)

∂z= (R + jωL)I(z) (2.19)

−∂I(z)

∂z= (G+ jωC)V (z) (2.20)

Two equations, two unknowns. To solve these equations, we first take a derivative of bothequations with respect to z.

−∂2V (z)

∂z2= (R + jωL)

∂I(z)

∂z(2.21)

−∂2I(z)

∂z2= (G+ jωC)

∂V (z)

∂z(2.22)

Rearange the previous equations:

− 1

(R + jωL)

∂I(z)

∂z=∂2V (z)

∂z2(2.23)

− 1

(G+ jωC)

∂V (z)

∂z=∂2I(z)

∂z2(2.24)

Substitute Eq.2.23 into Eq.2.20 and Eq.2.24 into Eq.2.19 and we get

−∂2V (z)

∂z2= (G+ jωC)(R + jωL)V (z) (2.25)

−∂2I(z)

∂z2= (G+ jωC)(R + jωL)I(z) (2.26)

Or if we rearrange

∂2V (z)

∂z2− (G+ jωC)(R + jωL)V (z) = 0 (2.27)

∂2I(z)

∂z2− (G+ jωC)(R + jωL)I(z) = 0 (2.28)

The above Eq.2.27 and Eq.2.28 are called the wave equation, and they represent current andvoltage wave on a transmission line. γ = (G+ jωC)(R+ jωL) is the complex propagation constant.This constant has a real and an imaginary part.

γ = α + jβ



where α is the attenuation constant and β is the phase constant.

α = Re√

(G+ jωC)(R + jωL)β = Im

√(G+ jωC)(R + jωL)

The general solution of the differential equation of the type 2.27 or 2.28? From differentialequation class, we know that the solution to this kind of equation is:

V (z) = V +0 e−γz + V −0 e

γz

I(z) = I+0 e−γz + I−0 e

γz

In this equation V +0 and V −0 are the phasors of forward and reflected voltage waves, and I+

0

and I−0 are the phasors of forward and reflected current wave. The time domain expression for thecurrent and voltage on the transmission line we can get by multiplying the phasor of the voltage andcurrent with ejωt and taking the real part of it.

v(t) = Re(V +0 e

(−α−jβ)z + V −0 e(α+jβ)z)ejωt

v(t) = |V +0 |e−αz cos(ωt− βz + 6 V +

0 ) + |V −0 |eαz cos(ωt+ βz + 6 V −0 ) (2.29)

We’ll look at the Matlab program the next class to see that if the signs of the ωt and βz are thesame the wave moves in the forward +z direction. If the signs of ωt and βz are opposite, the wavemoves in the −z direction.

2.5 Visualization of Lossless Forward and Reflected Voltage

Waves

We will show next that if the signs of the ωt and βz are the same, as in Equation 2.31, the wavemoves in the forward +z direction. If the signs of ωt and βz are opposite, as in Equation 2.30, thewave moves in the −z direction. In order to see this, we will visualize Equations 2.30 and 2.31 usingMatlab code below.

vf (t) = |V +0 |e−αz cos(ωt− βz + 6 V +

0 ) (2.30)

vr(t) = |V −0 |eαz cos(ωt+ βz + 6 V −0 ) (2.31)

Figure 2.6 shows forward and reflected waves on a transmission line. On x-axis is the spatialcoordinate z from the generator to the load, where the transmission line is connected, and on y-axisis the magnitude of the voltage on the line. The red line on both graphs is the voltage signal at atime .1 ns. We would obtain Figure 2.6 if we had a camera that can take picture of voltages, andwe took the first picture at t1 = .1 ns on the entire transmission line. The blue dotted line on both



graphs is the same signal .1 ns later, at time t2 =.2 ns. We see that the signal has moved to the rightin the time of 1 ns, or from the generator to the load. On the bottom graph we see that at a time.1 ns, the red line represents the reflected signal. Dashed blue line shows the signal at a time .2 ns.We see that the signal has moved to the left, or from the load to the generator.

Figure 2.6: Forward (top) and reflected (bottom) waves on a transmission line.

clear all

clc

f = 10^9;

w = 2*pi*f

c=3*10^8;

beta=2*pi*f/c;

lambda=c/f;

t1=0.1*10^(-9)

t2=0.2*10^(-9)

x=0:lambda/20:2*lambda;

y1=sin(w*t1 - beta.*x);

y2=sin(w*t2 - beta.*x);

y3=sin(w*t1 + beta.*x);

y4=sin(w*t2 + beta.*x);

subplot (2,1,1),

plot(x,y1,’r’),...



hold on

plot(x,y2,’--b’),...

hold off

subplot (2,1,2),

plot(x,y3,’r’)

hold on

plot(x,y4,’--b’)

hold off

Using matlab code above, repeat the visualization of signals in the previous section for a lossytransmission line. Assume that α = 0.1 Np, and all other variables are the same as in the previoussection. How do the voltages compare in the lossy and lossless cases?

2.6 Relating forward and backward current and voltage waves

on the transmission line

In equations 2.32-2.33 V +0 and V −0 are the phasors of forward and reflected going voltage waves,

and I+0 and I−0 are the phasors of forward and reflected current waves. In this section, we will see

how the phasors of forward and reflected voltage and current waves are related to transmission lineimpedance.

V (z) = V +0 e−γz + V −0 e

γz (2.32)

I(z) = I+0 e−γz + I−0 e

γz (2.33)

When substitute the voltage wave equation into Telegrapher’s Equations ??. The equation isrepeated here Eq.2.34-2.35.

−∂V (z)

∂z= (R + jωL)I(z) (2.34)

γV +0 e−γz − γV −0 eγz = (R + jωL)I(z) (2.35)

We now rearrange Eq.2.35

I(z) =γ

R + jωL(V +

0 exp−γz + V −0 exp

γz)

I(z) =γV +

0

R + jωLe−γz − γV −0

R + jωLeγz (2.36)

Now we compare Eq.2.36 with the Eq.2.33. In order for two transcendental equations to be equal,the coefficients next to exponential terms have to be the same.



I+0 =

γV +0

R + jωL

I−0 = − γV −0R + jωL

We can define the characteristic impedance of a transmission line as the ratio of the voltage tocurrent amplitude of the forward going wave.

Z0 =V +

0

I+0

Z0 =R + jωL

γ

Z0 =

√R + jωL

G+ jωC

2.7 Lossless transmission line

In many practical applications R→ 0 1and G→ 02. This is a lossless transmission line.In this case the transmission line parameters are

• Propagation constant

γ =√

(R + jωL)(G+ jωC)

γ =√jωLjωC

γ = jω√LC = jβ

• Transmission line impedance

Z0 =

√R + jωL

G+ jωC

Z0 =

√jωL

jωC

Z0 =

√L

C

1metal resistance is low2dielectric conductance is low



• Wave velocity

v =ω

β

v =ω

ω√LC

v =1√LC

• Wavelength

λ =2π

β

λ =2π

ω√LC

λ =2π

√ε0µ0εr

λ =c

f√εr

λ =λ0√εr

2.8 What does it mean when we say a medium is lossy or

lossless?

In a lossless medium electromagnetic wave power is not turning into heat, there is no loss of amplitude.In lossy medium electromagnetic wave is heating up the medium, therefore its power is decreasingas e−αx.

medium attenuation constant α [dB/km]coax 60

waveguide 2fiber-optic 0.5

In guided wave systems such as transmission lines and waveguides the attenuation of power withdistance follows approximately e−2αx. The power radiated by an antenna falls off as 1/r2.

2.9 Low-Loss Transmission Line

In some practical applications, losses are small, but not negligible. R << ωL 3and G << ωC4.

3metal resistance is lower than the inductive impedance4dielectric conductance is lower than the capacitive impedance



In this case the transmission line parameters are

• Propagation constant

We can re-write the propagation constant as shown below. In somel applications, losses aresmall, but not negligible. R << ωL and G << ωC, then in Equation 2.38, RG << ω2LC.

γ =√

(R + jωL)(G+ jωC) (2.37)

γ = jω√LC

√1 − j

(R

ωL+

G

ωC

)− RG

ω2LC(2.38)

γ ≈ jω√LC

√1 − j

(R

ωL+

G

ωC

)(2.39)

Taylor’s series for function√

1 + x =√

1 − j(RωL

+ GωC

)in Equation 2.39 is shown in Equa-

tions 2.40-2.41.

√1 + x = 1 +

x

2− x2

8+x3

16− ... for |x| < 1 (2.40)

γ ≈ jω√LC

√1 − j

(R

ωL+

G

ωC

)= jω

√LC

(1− j

2

(R

ωL+

G

ωC

))(2.41)

The real and imaginary part of the propagation constant γ are:

α =ω√LC

2

(R

ωL+

G

ωC

)(2.42)

β = ω√LC (2.43)

We see that the phase constant β is the same as in the lossless case, and the attenuationconstant α is frequency independent. This means that all frequencies of a modulated signalare attenuated the same amount, and there is no dispersion on the line. When the phaseconstant is a linear function of frequency, β = const ω, then the phase velocity is a constantvp = ω

β= 1

const, and the group velocity is also a constant, and equal to the phase velocity.

In this case, all frequencies of the modulated signal are propagated with the same speed, andthere is no distortion of the signal. This is the case only when the losses in the transmissionline are small.

We usually represent phase and group velocity on ω − β diagrams, shown in Figure 2.9(a). Ata frequency ω1, the ratio of ω

βgives the phase velocity (graphically, this is the slope of the red

line on the graph), whereas the slope of the ω − β curve (blue line on the graph) gives the



group velocity at this frequency. These diagrams are useful, as they show how phase constantβ varies with frequency, and it also shows how phase and group velocities vary with frequency.We can see that in this case, both group and phase velocity for this line are positive quantities,which is a representation of what is called a forward wave. In a forward wave, both signal andthe energy propagate in the forward direction. Backward waves are waves where the signalpropagates forward, however energy propagates backwards. In backward waves, phase andgroup velocities have opposite signs.

When the phase constant is a linear function of frequency, then, phase velocity and groupvelocity are equal, and do not depend on frequency. Both velocities are equal to the slope ofthe line in Figure 2.9(b).

(a) Phase constant β is a nonlin-ear function of frequency.

(b) Phase constant β is a linearfunction of frequency.

Figure 2.7: Omega-Beta diagrams.

2.10 Voltage Reflection Coefficient, Lossless Case

The equations for the voltage and current on the transmission line we derived so far are

V (z) = V +0 e−jβz + V −0 e

jβz (2.44)

I(z) =V +

0

Z0

e−jβz − V −0Z0

ejβz (2.45)

At z = 0 the impedance of the load has to be

ZL =V (0)

I0

Substitute the boundary condition in Eq.2.44

ZL = Z0V +

0 + V −0V +

0 − V −0(2.46)



Figure 2.8: Transmission Line connects generator and the load.

We can now solve the above equation for V −0

ZLZ0

(V +0 − V −0 ) = V +

0 + V −0

(ZLZ0

− 1)V +0 = (

ZLZ0

+ 1)V −0

V −0V +

0

=ZL

Z0− 1

ZL

Z0+ 1

V −0V +

0

=ZL − Z0

ZL + Z0

(2.47)

The quantityV −0

V +0

is called voltage reflection coefficient Γ. It relates the reflected to incident

voltage phasor. Voltage reflection coefficient is in general a complex number, it has a magnitude anda phase.

Examples

1. 100 Ω transmission line is terminated in a series connection of a 50 Ω resistor and 10 pF capac-itor. The frequency of operation is 100 MHz. Find the voltage reflection coefficient.

2. For purely reactive load ZL = jXL find the reflection coefficient.

The end of this lecture is spent in the lab making a Matlab program to make a movie of a wavemoving left and right.

2.11 Standing Waves

In the previous section we introduced the voltage reflection coefficient that relates the forward toreflected voltage phasor.

Γ =V −0V +

0

=ZL − Z0

ZL + Z0

(2.48)



If we substitute Equation 2.48 to Eq.2.49 we get for the voltage wave

V (z) = V +0 e−jβz + V −0 e

jβz (2.49)

V (z) = V +0 e−jβz + ΓV +

0 ejβz

V (z) = V +0 (e−jβz − Γejβz) (2.50)

since Γ = |Γ|ejΘr Eq.2.51 becomes

V (z) = V +0 (e−jβz − |Γ|ejβz+Θr) (2.51)

and for the current wave

I(z) =V +

0

Z0

e−jβz − V −0Z0

ejβz (2.52)

I(z) =V +

0

Z0

e−jβz + ΓV +

0

Z0

ejβz

I(z) =V +

0

Z0

(e−jβz − Γejβz) (2.53)

The voltage and the current waveform on a transmission line are therefore given by Eqns.2.51,2.53. Now we have two equations and one unknown V +

0 ! We will solve these two equations in Lecture7. Now let’s look at the physical meaning of these equations.

In Eq.2.51, Γ is the voltage reflection coefficient, V +0 is the phasor of the forward going wave, z

is the axis in the direction of wave propagation, β is the phase constant5, Z0 is the impedance of thetransmission line6. V (z) is a complex number, phasor. We will find the magnitude and phase of thevoltage on the transmission line.

The magnitude of a complex number can be found as |z| =√zz∗.

|V (z)| =√V (z)V (z)∗

|V (z)| =√V +

0 (e−jβz − |Γ|ejβz+Θr)V +0 (ejβz − |Γ|e−(jβz+Θr))

|V (z)| = V +0

√(e−jβz − |Γ|ejβz+|Θr)(ejβz − |Γ|e−(jβz+Θr))

|V (z)| = V +0

√1 + |Γ|e−(2jβz+|Θr) + |Γ|ej2βz+Θr + |Γ|2)

|V (z)| = V +0

√1 + |Γ|2 + |Γ|(e−(2jβz+Θr) + e(j2βz+Θr))

|V (z)| = V +0

√1 + |Γ|2 + 2|Γ| cos(2βz + Θr) (2.54)

The magnitude of the total voltage on the transmission line is given by Eq.2.54. It seems like acomplicated function.

5imaginary part of the complex propagation constant6defined as the ratio of forward going voltage and current



• Let’s start from a simple case when the voltage reflection coefficient on the tranmission line isΓ = 0 and draw the magnitude of the total voltage. The magnitude is the green line in Figure2.9. To see the movie of this transmission line go to the class web page under InstructionalVideos. Forward voltage is shown in red, reflected voltage in pink, and the magnitude of thevoltage is green. Magnitude of the voltage is constant everywhere on the transmission line, andso the line is called flat.

Figure 2.9: Flat line.

• Let’s look at another case, Γ = 0.5 and Θr = 0. Equation 2.55 represents the magnitude of thevoltage on the transmission line, and Figure 2.10 shows in green how this function looks on atransmission line. This case is shown in Figure 2.10.

|V (z)| = V +0

√5

4+ cos2βz (2.55)

Figure 2.10: Voltage on a transmission line with reflection coefficient magnitude 0.5, and zero phase.

The function 2.55 is at it’s maximum when cos(2βz) = 1 or z = k2λ, and the function value is

V (z) = 1.5V +0 . It is at it’s minimum when cos(2βz) = −1 or z = 2k+1

4λ and the function value

is V (z) = 0.5V +0



Figure 2.11: Shorted Transmission Line.

It is important to mention here that the function that we see looks like a cosine with an averagevalue of V +

0 , but it is not. The minimums of the function are sharper then the maximums, sowhen the reflection coefficient is at it’s maximum of Γ = 1 the function looks like this:

• General Case.

In general the voltage maximums will occur when cos(2βz) = 1

|V (z)max| = V +0

√1 + |Γ|2 + 2|Γ|

|V (z)max| = V +0

√(1 + |Γ|)2

|V (z)max| = V +0 (1 + |Γ|) (2.56)

In general the voltage minimums will occur when cos(2βz) = −1,

|V (z)min| = V +0

√1 + |Γ|2 − 2|Γ|

|V (z)min| = V +0

√(1− |Γ|)2

|V (z)min| = V +0 (1− |Γ|) (2.57)

The ratio of voltage minimum on the line over the voltage maximum is called the VoltageStanding Wave Ratio (VSWR) or just Standing Wave Ratio (SWR).

SWR =V (z)maxV (z)min

SWR =1 + |Γ|1− |Γ|

(2.58)



The voltage maximum position on the line is where

cos(2βz) = 1

2βz + Θr = 2nπ

z =2nπ −Θr

2β

z =2nπ −Θr

4π(2.59)


Chapter 3

Smith Chart

Smith Chart is a very useful tool that is used to visualize impedances and reflection coefficients.Lumped element and transmission line impedance matching would be very difficult to understand ifwe couldn’t use Smith Charts. Simulation software such as ADS and measurement equipment, suchas Network Analyzers use Smith Chart to represent simulated or measured data. Smith Chart atfirst looks like Black Magic, but it is a very simple and useful tool that will help you understandimpedance/admittance transformations and transmission lines better. In essence, Smith Chart is aunit circle centered at the origin with a radius of 1. Smith Chart is used to graphically representreflection coefficient. Real and imaginary axis of reflection coefficient (Cartesian coordinates) arenot shown on the actual Smith Chart, but the center of the Smith Chart is where the origin of thecoordinate system would be. We usually represent the reflection coefficient in polar coordinates, witha magnitude and an angle. Magnitude is the distance between the point and the origin, and theangle is measured from the x-axis. An example location of several reflection coefficients is given inFigure 3.1. If you don’t see why are the points positioned as shown, review polar representation ofcomplex numbers.

In Figure 3.2 circle and line shown represent all points on the Smith Chart that have constant mag-nitude or angle of the reflection coefficient. At high frequencies it is difficult to measure impedancesdirectly, as it is difficult to measure (or sometimes even define) voltage and current. To measureimpedances, engineers use Network Analyzer shown in Figure 3.3.

Since reflection coefficient and impedance are related through Equation 3.1, we can find impedancethat corresponds to the reflection coefficient, Equation 3.2. In other words, every point on the SmithChart represents one reflection coefficient Γ and one impedance ZL.

Γ =ZL − ZZL + Z

(3.1)

zL =ZLZ

=1 + Γ

1− Γ(3.2)

Figure 3.4 shows circles on the Smith Chart that represent constant (normalized) reactances, andresistances. Figure 3.5, shows how to find zL = 1 + j1. This impedance is at a point where thecircle of constant resistance rL = 1 crosses the circle of constant reactance xL = 1. Figure 3.6 shows

35


Figure 3.1: Examples of location of reflection coefficient on the Smith Chart.

(a) |Γ|=0.5, |Γ|=1 (b) < Γ = 45, < Γ = 120

Figure 3.2: Points of constant magnitude (a) of the reflection coefficient and phase (b) of the reflectioncoefficient.



Figure 3.3: HP8510 Network Analyzer in Microwave Laboratory measures impedances up to 26.5GHz.This piece of equipment is on permanent loan curtesy of Defense Micro Electronic Activity (DMEA),Sacramento

how to find the reflection coefficient if normalized load impedance zL = 1 + j1 is given. Measurethe distance bewteen the origin using the scale ”Reflection Coefficient E or I”on the bottom of theSmith Chart to find the magnitude of the reflection coefficient. To find the angle of the reflectioncoefficient, we read the scale ”Angle of Reflection Coefficient” on the perimeter of the Smith Chart,shown in green. The reflection coefficient is therefore 0.5ej62o , which is close to the actual value0.5ej64o . If you have done this using ruler and compass, and nicely sharpened pencil, you would getexactly the right answer. Try it out!



(a) Normalized resistance circles. (b) Normalized reactance circles.

Figure 3.4: All points on the circle have the constant imaginary part of the impedance.

Figure 3.5: Red circle that represents all points on Smith Chart with normalized resistance r = 1and blue circle that represents all points on Smith Chart with normalized reactance x = 1 cross atpoint where ZL = 1 + j1.



Figure 3.6: To find the magnitude of the reflection coefficient from impedance ZL = 1 + j1, wemeasure the distance bewteen the origin using the scale ”Reflection Coefficient E or I”on the bottomof the Smith Chart. To find the angle of the reflection coefficient, we read the scale ”Angle ofReflection Coefficient” on the perimeter of the Smith Chart.

Brief review of impedance and admittance

Impedance’s real part is called resistance R, and imaginary part is called reactance X. Admit-tance’s real part is called conductance, and imaginary part is called susceptance. It is easier toadd impedances when elements are in series, and it is easier to add admittances when elements arein parallel, see Figure 3.7, because we just add real and imaginary parts separately. This is thereason we have special Smith Chart shown in Figure 3.8 that has both impedance and admittancecircles on it. This way, we can use Smith Chart to read off the values for equivalent impedance oradmittance when we add impedances or admittances in parallel or series. This comes in handy forlumped-element impedance matching that we will talk about next.



Figure 3.7: It is easier to use admittance when the circuit elements are in parallel and impedancewhen the circuit elements are in series.

Figure 3.8: On Smith Chart Best impedances are shown in red, and admittances in green.


Chapter 4

Impedance Matching

Impedance matching is a technique that ensures maximum power transfer between the generator Vgand the load ZL. The maximum power transfer is achieved when the input impedance looking intothe circuit looking from he generator is a complex conjugate of the impedance of the load. In thecase shown in Figure 4.1, the impedance of the generator and the line is Z, so to perform maximumpower transfer, the input impedance of the impedance matching circuit has to be equal to Z. Weperform impedance matching to remove the reflected wave on the transmission line, and maximizethe power delivery to the load. When there is only the forward wave on the line, the line is called”flat”, because the magnitude of the forward wave is the same everywhere on the line (and equal toV + ). When we insert the impedance matching circuit between the load ZL and the transmission lineZ, the input impedance presented to the generator and the transmission line from the impedancematching circuit will be Z, and the impedance presented to the load impedance ZL will be Z∗L. Ourtask is therefore to add capacitors and inductors to the load ZL to make the input impedance looklike Z. We do not want to use resistors in an impedance matching circuit because they will usesome of the power that we need at ZL. If we represent ZL on the Smith Chart Best as a point, ourtask will be to take that point to the center of the Smith Chart, where the impedance is equal to Z.

Figure 4.1: The result of impedance matching.

41


4.1 Lumped Element Impedance Matching


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