EEE161 Applied Electromagnetics Practical Laboratory...

Dr. Milica Markovic Applied Electromagnetics Laboratory page 1

EEE161 Applied Electromagnetics Practical Laboratory 1

Instructor: Dr. Milica MarkovicOffice: Riverside Hall 3028

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

California State University Sacramento EEE161 revised: 16. December, 2010

Chapter 1

Transmission Line Theory

1.1 Transmission Line Theory

1.1.1 What is a transmission line?

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 is a transmission line. Wheather we seethe propagation effects on a line depends on the line length. So, at lower frequencies we do notsee any difference between the signal’s phase at the generator and at the load, whereas at higherfrequencies we do.

1.1.2 Types of transmission lines

1. Coaxial Cable, Figure 1.1

2. Microstrip, Figure 1.1

3. Stripline, Figure 1.1

4. Coplanar Waveguide, Figure 1.1

5. Two-wire line, Figure 1.1

6. Parallel Plate Waveguide, Figure 1.1

7. Rectangular Waveguide, Figure 1.1

8. Optical fiber, Figure 1.1

1.1.3 Wave types

Types of waves include acoustic waves, mecahnical pressure waves, electromagnetic (EM) waves.Here we will focus on EM waves and transmission lines for EM waves.

2


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

(d) CoplanarWaveguide

(e) Two-wire Line

(f) Parallel PlateWaveguide

(g) Rect-angularWaveguide

(h) Optical Fiber

Figure 1.1: Types of transmission lines.

1.1.4 What are Transmission Line Effects?

Figure 1.3 shows a step voltage at the generator and the load of a circuit in Figure 1.4. The voltageneeds T sec to appear at the load, once the switch closes. Figure ?? 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 much time it takes for this signal to go from AA end to BB end? t = lc, where c = 3 108. If

the signal at end AA is

vAA‘(t) = Acos(ωt) (1.1)

Then at the other end the transmission line the signal is



Figure 1.2: 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 1.4, 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 1.3: Voltage along the transmission line in Figure 1.4, 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 1.4: Electronic Circuit with an emphasis on cables that connect the generator and the load.

vBB‘(t) = vAA‘(t− l

c) (1.2)

vBB‘(t) = Acos(ω(t− l

c)) (1.3)

vBB‘(t) = Acos(ωt− ω lc) (1.4)

vBB‘(t) = Acos(ωt− ω

cl) (1.5)

Since we know that angular frequency is ω = 2πf

vBB‘(t) = Acos(ωt− 2πf

cl) (1.6)

The quantity cf

is the wavelength λ. We will talk more about wavelength in Section 1.1.5

vBB‘(t) = Acos(ωt− 2π

λl) (1.7)

The quantity 2πλ

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

vBB‘(t) = Acos(ωt− βl) (1.8)

vBB‘(t) = Acos(ωt−Ψ) (1.9)

We see that at BB the signal will experience a phase shift.We will derive this equation later again from the Telegrapher’s equations 1.1.7.Now let’s see how the length of the line l affects the voltage at the end BB or a wire. Look at

Equation 1.10.



vBB‘(t) = Acos(ωt− 2πl

λ) (1.10)

The signal will experience a phase shift of 2π lλ. If this phase shift is small, there will not be much

difference between the phase of the signal at the generator and at the load. This means that wedon’t have to use transmission line theory to account for the effects of the line. If the phase shift issignificant, then we do have to use the transmission line theory. Let’s look at some numbers in thefollowing 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 for reflected signals, power loss and dispersion on the transmissionline.

Problem 1. Using Matlab plot the sinusoidal signal

y=sint as a function of time. Assume that the frequency of the signal is f=1GHz. Then repeat the exercise by plotting the sinusoidal signal as a function of angle t.

Problem 2. Using Matlab plot the sinusoidal signal

Assume that the frequency of the signal is f=1GHz. Then repeat the exercise by plotting the sinusoidal signal y=sin(t+40) as a function of angle t.

Problem 3. Using Matlab plot the sinusoidal signal y=sin(t-40)

Assume that the frequency of the signal is f=1GHz. Then repeat the exercise by plotting the sinusoidal signal as a function of angle t.

Problem 4. Plot the previous three plots on the same graph. Which signal is lagging the sinusoidal signal from problem 1? Which signal is leading? Label the graph stating which signal is leading and which is lagging.

Problem 5. Plot two periods of two sinusoidal signals y1(t) and y2(t) with different phases and at the same frequency. Estimate what is the minimal phase difference (larger than zero) at which you do not see significant difference between the signals.

Problem 6. Now assume that the sinusoidal signal is generated by a source positioned at x=0 and the source is connected to the load by a long cable. The electromagnetic signals propagate with the speed of light. The signal that has been generated at t=0 at the generator side will need a tsec time to reach the load. Calculate what is the time needed for the signal to reach the load as a function of the cable length. This will be a 2D linear graph. On y axis plot time, on x-axis cable length. The cable length should be between 1cm and 100m. Plot grid on the graph so that you can easily estimate the time from the length of cable.

Problem 7. If the voltage on the generator side is y=sin(t) then the voltage on the load side will lag for tsec. Since tsec=l/c, the voltage on the generator side will be. y=sin((t-tsec)). Assume that the length of the cable is 0.1m and the frequency of operation is 1kHz, 10kHz, 100kHz, 1MHz, 10MHz, 100MHz, 1GHz, 10GHz, 100GHz. For each frequency plot the voltage at the load y(t-tsec) and the voltage at the generator for two cycles of the period. (You will have the total of nine graphs, each with two voltage signals at the load and at the source). Comment on the graphs. For which graphs the load and source voltage look the same?



Solutions

Problem 1

Using Matlab plot the sinusoidal signal

y=sint as a function of time. Assume that the frequency of the signal is f=1GHz. Then repeat the exercise by plotting the sinusoidal signal as a function of angle t.

Code:

f=1e6;

t=0:1/100/f:4/f;

y=sin(2*pi*f*t);

plot(t,y,’Color’,’red’);

hold on

Graph of y(t)=sin(t)

In this part of the lab, we were asked to display a simple sine wave with a frequency of 1GHz. One thing to note for this part of the lab, and as for subsequent labs, is to see where the sine wave begins on our x-axis. When we phase shift our function, we should either expect to see lagging or leading of the sine wave, depending on the polarity of the phase change.

Problem 2

Using Matlab plot the sinusoidal signal

Assume that the frequency of the signal is f=1GHz. Then repeat the exercise by plotting the sinusoidal signal y=sin(t+40) as a function of angle t.

Code:

fc=1e6;

t=0:1/100/fc:4/fc;

y=sin((2*pi*fc*t)+((2*pi)/9));

plot(t,y,’Color’,’black’);

Graph of y(t)=sin(t+40)

With the phase change of +40 degrees, we see our graph shift to the left on the x-axis. This would be an example of lagging the original function y(t) = sin(t)



Problem 3

Using Matlab plot the sinusoidal signal y=sin(t-40)

Assume that the frequency of the signal is f=1GHz. Then repeat the exercise by plotting the sinusoidal signal as a function of angle t.

Code:

fc=1e6;

t=0:1/100/fc:4/fc;

y=sin((2*pi*fc*t)-((2*pi)/9));

plot(t,y);

Diagram of y(t) = sin(t-40)

With the same frequency for each plot, we notice this sin function shifts to the right. This is a lead in the sine function, as we will later see in problem 4 more vividly with all 3 graphs combined.

Problem 4

Plot the previous three plots on the same graph. Which signal is lagging the sinusoidal signal from problem 1? Which signal is leading? Label the graph stating which signal is leading and which is lagging.

Code:

f=1e6;

t=0:1/100/f:4/f;

y=sin(2*pi*f*t);

plot(t,y,’Color’,’red’);

hold on

fc=1e6;

t=0:1/100/fc:4/fc;

y=sin((2*pi*fc*t)+((2*pi)/9));

plot(t,y,’Color’,’black’);

fc=1e6;

t=0:1/100/fc:4/fc;

y=sin((2*pi*fc*t)-((2*pi)/9));

plot(t,y);



This Code allows us to plot all 3 graphs simultaneously on the same page. (different colors as well).

Plot of :

y(t) =sin(t) // RED

y(t) =sin(t-40) // BLUE

y(t) =sin(t+40) // BLACK

This graph intricately shows all three sine waves together and their lagging/leading elements can be easily tracked. The black sine wave lags while the blue one leads.

Problem 5

Plot two periods of two sinusoidal signals y1(t) and y2(t) with different phases and at the same frequency. Estimate what is the minimal phase difference (larger than zero) at which you do not see significant difference between the signals.

Code:



fc=1e6;

t=0:1/100/fc:4/fc;

y=sin(2*pi*fc*t);

plot(t,y);

hold on

fc=1e6;

t=0:1/100/fc:4/fc;

y=sin((2*pi*fc*t)+.025);

plot(t,y);

Plots both sine functions simultaneously with the minimal amount of phase change that displays 1 wave. In reality 1 wave would be a 2 waves combined to generate a thick wave. When the thick wave separates into two individual waves, the phase is too large. The phase chosen for this sine wave is 0.25 degrees.

Diagram of stacked waves:

y(t)=sin(t)

y(t)=sin(t+.25)

Problem 6

Now assume that the sinusoidal signal is generated by a source positioned at x=0 and the source is connected to the load by a long cable. The electromagnetic signals propagate with the speed of light. The signal that has been generated at t=0 at the generator side will need a tsec time to reach the load. Calculate what is the time needed for the signal to reach the load as a function of the cable length. This will be a 2D linear graph. On y axis plot time, on x-axis cable length. The cable length should be between 1cm and 100m. Plot grid on the graph so that you can easily estimate the time from the length of cable.

Code:

c=3e8

t=y./c

y=1:.001:100

plot(t,y);

Diagram of signal generator(t) and cable length

Graph shows the signal generated at specific lengths of cable with respet to time. As the cable length gets larger, the signal grows dramatically.

Problem 7

If the voltage on the generator side is y=sin(t) then the voltage on the load side will lag for tsec. Since tsec=l/c, the voltage on the generator side will be. y=sin((t-tsec)). Assume that the length of the cable is 0.1m and the frequency of operation is 1kHz, 10kHz, 100kHz, 1MHz, 10MHz, 100MHz, 1GHz, 10GHz, 100GHz. For each frequency plot the voltage at the load y(t-tsec) and the voltage at the generator for two cycles of the period. (You will have the total of nine graphs, each with two voltage signals at the load and at the source). Comment on the graphs. For which graphs the load and source voltage look the same?

Code:

c=3e8;

fc=1e11; %change this number for different frequencies

t=0:1/100/fc:1/fc;



y=sin(2*pi*fc*(t-(1/c)));

plot(t,y);

hold on

F=1000Hz

Dispersion is an effect where different frequencies travel with different speeds on the transmissionline.

Example Find what is the length of the cable at which we need to take into account transmissionline effects if the frequency of operation is 10 GHz.

1.1.5 What is wavelength?

HERE EXPLANATION OF WAVE ON TRANSLINE 1 FIGURE

1.1.6 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

1.1.7 Derivation of the voltage on a transmission line

In this section we will derive what is the expression for the signal along a wire as a function of spacez. In previous classes, you learned about voltage or current as a function of time. In this section,we will see that the voltages and currents propagating on a cable may also be a function of distancefrom the generator.

We want to derive the equations for the case when the transmission line is longer then the fractionof a wavelength. To make sure that we don’t encounter the transmission line effects to start with,we can look at the piece of a transmission line that is much smaller then the fraction of a wavelenth.In other words we cut the transmission line into small pieces to make sure there are no transmissionline effects, as the pieces are shorter then the fraction of a wavelength.

Plan:

• Look at an infinitensimal length of a transmission line ∆z.

• Model infinitesimal transmission line lenght with an equivalent circuit.

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

• Apply phasors (equations become linear)



Figure 1.5: Section of a coaxial cable.



• 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.

Figure 1.6: Sinusoidal voltage propagation on a coaxial cable.

Figure 1.7: Circuit Model for a section of a coaxial cable.

KVL

−v(z, t) +R′∆z i(z, t) + L

′∆z

∂i(z, t)

∂t+ v(z + ∆z, t) = 0

KCL

i(z, t) = i(z + ∆z) + iCG(z + ∆z, t)

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

′∆z

∂v(z + ∆z, t)

∂t

Rearrange the KCL and KVL Equations 1.11, 1.14 divide them with ∆z Equations 1.12, 1.15 let∆z → 0 and recognize the expression for the derivative Equations, 1.13, 1.16.



KVL

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

∂t(1.11)

−v(z + ∆z, t)− v(z, t)

∆z= Ri(z, t) + L

∂i(z, t)

∂t(1.12)

−∂v(z, t)

∂z= Ri(z, t) + L

∂i(z, t)

∂t(1.13)

KCL

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

∂t(1.14)

−i(z + ∆z, t)− i(z, t)∆z

= Gv(z + ∆z, t) + C∂v(z + ∆z, t)

∂t(1.15)

−∂i(z, t)∂z

= Gv(z + ∆z, t) + C∂v(z + ∆z, t)

∂t(1.16)

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

These are two differential equations with two unknowns. It is not impossible to solve them,however we would prefer to have linear differential 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) (1.17)

−∂I(z)

∂z= (G+ jωC)V (z) (1.18)

Two equations, two unknowns. To solve these equations, we first integrate both equations overz,

−∂2V (z)

∂z2= (R + jωL)

∂I(z)

∂z

−∂2I(z)

∂z2= (G+ jωC)

∂V (z)

∂z



Now rearange the previous equations

− 1

(R + jωL)

∂I(z)

∂z=∂2V (z)

∂z2(1.19)

− 1

(G+ jωC)

∂V (z)

∂z=∂2I(z)

∂z2(1.20)

now substitute Eq.1.19 into Eq.1.17 and Eq.1.20 into Eq.1.18 and we get

−∂2V (z)

∂z2= (G+ jωC)(R + jωL)V (z)

−∂2I(z)

∂z2= (G+ jωC)(R + jωL)I(z)

Or if we rearrange

−∂2V (z)

∂z2− (G+ jωC)(R + jωL)V (z) = 0 (1.21)

−∂2I(z)

∂z2− (G+ jωC)(R + jωL)I(z) = 0 (1.22)

The above Eq.1.21 and Eq.1.22 are the equations of the current and voltage wave on a transmissionline. γ = (G + jωC)(R + jωL) is the complex propagation constant. This constant has a real andan 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)

What is the general solution of the differential equation of the type 1.21 or 1.22?

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 phasors1 of forward and backward going voltage waves, and

I+0 and I−0 are the phasors of forward and backward going current waves

1complex numbers having an amplitude and the phase



The time domain expression for the current and voltage on the transmission line we get

v(t) = Re(V +0 e

(−α+jβ)z + V −0 e(α+jβ)z)ejωt (1.23)

v(t) = |V +0 |e−αzcos(ωt+ βz + 6 V +

0 ) + |V −0 |e−αzcos(ωt− βz + 6 V −0 ) (1.24)

(1.25)

Equation 1.24 can be separated into two Equations 1.27-reffws.

vf (t) = |V +0 |e−αzcos(ωt− βz + 6 V −0 ) (1.26)

vr(t) = |V −0 |eαzcos(ωt+ βz + 6 V +0 ) (1.27)

Visualisation of Lossless Forward and Reflected Voltage Waves

We will show next that if the signs of the ωt and βz are the same, the wave moves in the forward+z direction. If the signs of ωt and βz are opposite, the wave moves in the −z direction. In orderto see this, we will visualise Equations 1.26-1.27 using Matlab code below.

Figure 1.8 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. Figure shows red and a dashed blue lines that representthe signal at two time points on the transmission line.Top figure shows a forward, and the bottomfigure shows the reflected signal. On the top graph, we will plot the Equation 1.26, and assume thatthe phase 6 V +

0 = 6 V −0 =0, and the magnitude of both voltages is |V +0 | = |V −0 |=1 V, the frequency of

this signal is 1 GHz, and there is no loss on the line, α = 0 Np. In Matlab code below, the top figurein Figure 1.8 shows equation cos(ωt− βz), and the bottow figure shows equation cos(ωt+ βz). Thered line on both graphs is the voltage signal at a time .1 ns. We would obtain Figure 1.8 if we hada camera that can take picture of voltages, and took a picture of the signal at the time t1 = .1 nson 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 right in the time of 1 ns, or from thegenerator to the load. On the bottom graph we see that at a time .1 ns, the red line represents thereflected signal. Dashed blue line shows the signal at a time .2 ns. We see that the signal has movedto the left, or from the load to the generator.

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;



Figure 1.8: Forward (top) and reflected (bottom) waves on a transmission line.

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

Visualisation of Lossless Forward and Reflected Voltage Waves

Repeat the visualization pocedure in the previous section for a lossy transmission line. Assume thatα = 0.1 Np, and all other variables are the same as in the previous section. How do the voltagescompare in the lossy and lossless cases?



1.1.8 Relating forward and backward current and voltage waves on thetransmission line -Transmissio Line Impedance Z0

V (z) = V +0 e−γz + V −0 e

γz (1.28)

I(z) = I+0 e−γz + I−0 e

γz (1.29)

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

I+0 and I−0 are the phasors of forward and backward going current waves

We will relate the phasors of forward and backward going voltage and current waves.When substitute the voltage wave equation into Telegrapher’s Eq. 1.17. The equation is repeated

here Eq.1.30.

−∂V (z)

∂z= (R + jωL)I(z) (1.30)

γV +0 e−γz − γV −0 eγz = (R + jωL)I(z) (1.31)

We now rearrange Eq.1.31

I(z) =γ

R + jωL(V +

0 e−γz + V −0 e

γz)

I(z) =γV +

0

R + jωLe−γz − γV −0

R + jωLeγz (1.32)

Now we compare Eq.1.32 with the Eq.1.29.

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



1.1.9 Voltage Reflection Coefficient Γ, 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 (1.33)

I(z) =V +

0

Z0

e−jβz − V −0Z0

ejβz (1.34)

Figure 1.9: Transmission Line connects generator and the load.

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

ZL =V (0)

I0

Substitute the boundary condition in Eq.1.46

ZL = Z0V +

0 + V −0V +

0 − V −0(1.35)

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

(1.36)

The quantityV −0V +0

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

voltage phasor. Voltage reflection coefficient isi n 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 reacive 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.

1.1.10 Lossless transmission line

In many applications it is sufficient to assume that the transmission line is lossless, R → 0 2andG→ 03. This is a lossless transmission line.

In this case the transmission line parameters are

• Propagation constant

γ =√

(R + jωL)(G+ jωC)

γ =√jωL jωC

γ = jω√LC = jβ

• Transmission line impedance

Z0 =

√R + jωL

G+ jωC

Z0 =

√jωL

jωC

Z0 =

√L

C

• Phase velocity

v =ω

β

v =ω

ω√LC

v =1√LC

2metal resistance is low3dielectric conductance is low



• Group velocity

• Wavelength

λ =2π

β

λ =2π

ω√LC

λ =2π

√ε0µ0εr

λ =c

f√εr

λ =λ0√εr

1.1.11 Low-Loss Transmission Line

In some practical applications, losses are small, but not negligible. R << ωL 4and G << ωC5.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 1.38, RG << ω2LC.

γ =√

(R + jωL)(G+ jωC) (1.37)

γ = jω√LC

√1 − j

(R

ωL+

G

ωC

)− RG

ω2LC(1.38)

γ ≈ jω√LC

√1 − j

(R

ωL+

G

ωC

)(1.39)

Taylor’s series for function√

1 + x =√

1 − j(RωL

+ GωC

)in Equation 1.39 is shown in Equa-

tions 1.40-1.41.

√1 + x = 1 +

x

2− x2

8+x3

16− ... for |x| < 1 (1.40)

γ ≈ jω√LC

√1 − j

(R

ωL+

G

ωC

)= jω

√LC

(1− j

2

(R

ωL+

G

ωC

))(1.41)

4metal resistance is lower than the inductive impedance5dielectric conductance is lower than the capacitive impedance



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

α =ω√LC

2

(R

ωL+

G

ωC

)(1.42)

β = ω√LC (1.43)

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 1.1.11(a).At a 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 thegroup velocity at this freqency. These diagrams are usefull, 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 1.1.11(b).

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

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

Figure 1.10: Omega-Beta diagrams.



• Transmission line impedance Transmission line impedance is the impedance that the forwardvoltage and current see as they propagate down the line from the generator to the load.

Z0 =

√R + jωL

G+ jωC

Z0 =

√jωL

jωC

Z0 =

√L

C

• Phase velocity Phase velocity is the velocity of the carrier frequency.

v =ω

β

v =ω

ω√LC

v =1√LC

• Group velocity Group velocity is the velocity of the modulated signal impressed on the carrierfrequency.

vg =1∂β∂ω

(1.44)

vg =1√LC

(1.45)

• Wavelength Wavelenght is how far the signal travels on the line while the carrier signal goesthrough one whole period in time, see Section??.

λ =2π

β

λ =2π

ω√LC

λ =2π

√ε0µ0εr

λ =c

f√εr

λ =λ0√εr



1.1.12 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 (1.46)

I(z) =V +

0

Z0


ejβz (1.47)

1.1.13 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

(1.48)

If we substitute this expression to Eq.1.466 we get for the voltage wave

V (z) = V +0 e−jβz + V −0 e

jβz (1.49)

V (z) = V +0 e−jβz + ΓV +

0 ejβz

V (z) = V +0 (e−jβz − Γejβz) (1.50)

since Γ = |Γ|ejΘr Eq.1.51 becomes

V (z) = V +0 (e−jβz − |Γ|ejβz+Θr) (1.51)

and for the current wave

I(z) =V +

0

Z0


ejβz (1.52)

I(z) =V +

0

Z0

e−jβz + ΓV +

0

Z0

ejβz

I(z) =V +

0

Z0

(e−jβz − Γejβz) (1.53)

The voltage and the current waveform on a transmission line are therefore given by Eqns.1.51,1.53. 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.

6repeated here as 1.49



In Eq.1.51, 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 constant7, Z0 is the impedance of thetransmission line8. 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∗9.

|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 + Γ2)

|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) (1.54)

The magnitude of the total voltage on the transmission line is given by Eq.1.54. It seems like acomplicated function.

• 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.

HERE PICTURE OF THE FLAT LINE

• Let’s look at another case, Γ = 0.5 and Θr = 0. The equation for the magnitude

|V (z)| = V +0

√5

4+ cos2βz (1.55)

The function 1.55 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

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:

PICTURE OF STANDING WAVE PATTERN WITH SHORT

7imaginary part of the complex propagation constant8defined as the ratio of forward going voltage and current9Prove this in Carthesian and Polar coordinate system



• General Case.

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

|V (z)max| = V +0

√1 + ‖Gamma|2 + 2|Γ|

|V (z)max| = V +0

√(1 + ‖Gamma|)2

|V (z)max| = V +0 (1 + |Γ|) (1.56)

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

|V (z)min| = V +0

√1 + ‖Gamma|2 − 2|Γ|

|V (z)min| = V +0

√(1− ‖Gamma|)2

|V (z)min| = V +0 (1− |Γ|) (1.57)

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|

(1.58)

The voltage maximum position on the line is where

cos(2βz) = 1

2βz + Θr = 2nπ

z =2nπ −Θr

2β

z =2nπ −Θr

4π(1.59)


Chapter 2

Artificial Transmission Line Lab

2.1 Instrumentation Needed

1. One Artificial Transmission Line

2. One Low-Frequency Signal Generator

3. One Multimeter and/or Oscilloscope

Connect to the experimental setup as shown in Figure 2.2(a). The artificial transmission line boxis shown in Figure 2.2(b).

2.2 Transmission Line Impedance and Voltage on a Low-

Loss matched line.

Understanding the equivalent model

Equivalent model of the transmission line consists off 50 sections of an RLC filter. The number ofsections will affect the bandwidth off the artificial transmission line and how close it simulates theactual cable. The bandwidth of the model can be calculated from the following equation

BW ≈ N1

10 τ(2.1)

This means that a 50 section artificial transmissio line with a delay ofWhere τ =

√Ctotal Ltotal is the time delay of the transmission line.

2.2.1 Question 1

How many wavelengths long is the transmission line at f = 8MHz frequency if L = 3.55H/mile,C = .1µF/mile, and the length of the whole transmission line is miles?

Each capacitor in the

27


(a) Experimental Setup

(b) Artificial Tranmission Line Box

Figure 2.1: Artificial Transmission Line Laboratory

2.2.2 Question 2

What is the characteristic impedance of the transmission line?To calculate the characteristic impedance of the line, use the equations for the lossy transmission

line impedance given in Equation ??. Comment on the value of the imaginary part of the transmissionline impedance. Is this a low-loss line? Check out the section about low-loss lines, and see if you can

2.3 Matched Line

2.3.1 Question 3

Terminate the line with the matched load. What is the value of the matched load ZL?Since the characteristic impedance of the transmission line is a complex number, you will need

to use both resistors and capacitors for ZL.

2.3.2 Measure the voltage on the line.

Using a multimeter measure the line voltage and plot it as a function of distance along the line.Collect data, then use Matlab to plot it.



2.3.3 Question 4

Are there any losses in the transmission line? Was your line well matched?

2.3.4 Question 5

Calculate the attenuation coefficients α from your measurements, and from the theory. How do theycompare? How will you find the current from the measured voltage along the line?

2.4 Unmatched Line

We will now look at what happens when the line is not well matched.

1. Terminate the line with an impedance equal to 2Z0, where Z0 in the characteristic impedanceof the lossless line.

2. Choose a frequency so that the transmission line is λ/2 long.

3. What frequency is this?

4. Measure the voltage along the line in the same manner you did before, and make a new graphwith Matlab.

5. Was the output voltage smaller than the input voltage?

6. Now select the frequency so that the line is a quarter of a wavelength long and plot the linevoltage on the same graph as in the previous part. As the last part, terminate the lossless linein the short and open circuit and plot the line voltage for both cases.

7. What would happen to the output voltage if λ/4 transmission line was a perfect one? How dothe currents and voltages compare the case of half and quarter long transmission lines? Whatstandard circuit element is the quarter wavelength long line remind you off? Explain.

2.5 Transmission Line Impedance Measurement

Short/Open etc.


Chapter 3

The Slotted Line Laboratory

There are 12 different measurements in this lab:6 the coaxial slotted line and six for the waveguide slotted line..

3.1 Coaxial Slotted Line

1. Explain a block diagram of the slotted line. What are the variables you’re trying to measure?SWR meter is a narrow band hi again audio amplifier with the arson Val in parenthesis rectified,

metered out for display. SWR scale calibration assumes a perfect square load detector (diode detectorhaving a fee out proportional to input power). Coaxial line may be connected to the 1000 Hz output ofthe linear SWR meter amplifier,, to monitored the date demodulated signal coming from the diode isthe out. SWR meter amplifier is tuned to their maximum gain at about 1000 Hz with approximately25 Hz bandwidth for noise reduction purposes, when the signal coming from the diode detector isquite small.

2. The UHF generator frequency has to be set at 500 MHz. With a short-circuit at the and offslotted line, produces 100% reflection. This reflection produces knowledge that are spaced lambdaby 2. SWR in that case is infinite. To locate nulls the SWR meter should be set to cry again 50or 60 DB’s. Then mouse can be located with good accuracy by locating to equal voltage points toeither side of the no.

3. measured the impedance of an unknown loads by use of slotted line to measure the SWRproduced by an unknown loads,, first move the probe carriage along the slotted line to the observableVmax location along the slot, adjusting the SWR meter again (preferably under 40 DB gain position)to produce a full-scale deflection on the SWR scale of the outputs meter. Then the probe carriageis moved to the the minimum location a quarter wave away at which location the SWR scale givesa direct reading of the design SWR value. Check this measurement a couple of times to confirm.

4. To find the proxy load position and that the minimum location along the slotted line, usethe location reference of the short-circuit at the load plane. The difference between the minimumof the location of the short and the minimum when the unknown load is connected corresponds tothe rotation needed on the Smith chart to find unknown loan-to-value. this business is in a fractionof the wavelength.. Wavelength can be found from the measurement of halfway decent is betweendenials whenever the slotted line length permits locating two adjacent dolls. Otherwise Landa may

30


be calculated using CRF assuming ideally lossless airline in assuming that the operating frequenciesknown..

Draw the SWR circle on the Smit chart.. This is the circle whose radius is equal to the appropriateSWR ratio on the chart. The point of the minimum for short-circuit on the chart is at the intersectionof SWR circle in the negative GR axis.

6. To measure the antenna input impedance versus frequency,, first obtain proxy load positionsbeta over the desired frequency range by putting their reference short at the desired antenna loadplane (which is at the end of the cable used to connect the slotted line and the antenna).. Take dataon the proxy load plane locations versus the frequency.. Then with a monopole antenna replacing theshort, take SWR in the minimum location data in each of the same Dial set frequencies.. obtainedthe input impedance for each of those frequencies and label them on the Smith charts. Plot also realand imaginary parts of the antennas impedance versus frequency on using Matlab.

3.2 Waveguide Slotted Line

3.2.1 Newer Labs

the last time we looked at how micropower is measured. Now we will look at another importantmeasurement column data voltages, impedances and reflection coefficients. One way of doing this isby using the slotted line conflagration, which enables direct sampling of the electric field amplitudeof the standing wave. It is made of a cortex are waveguide section that has a longitudinal slot inwhich a probe with a diode detectors inserted. There is a generator and one end of the line, and theunknown load terminates the line at the other end. The probe is a need like small pose that acts asa receiving antenna and samples the electric field. (You’ll understand better how this works whenwe study antennas later.)

If Saudi line measurements, we want to find the unknown load. We measured the SWR on theline and the distance from the low to the first voltage minimum aliment. We need to measure toquantities, says the load impedance is a complex number with both in amplitude and phase. Fromthe SWR, the magnitude of the reflection coefficient is

SWR formulawe know that the volume minimum occurs for formula from here, we can find the unknown

complex load impedanceimpedance formulathe slotted line is now a story construing, except that high millimeter wave frequencies. The

modern instrument used for measurement impedances is a vector network analyzer, which we willstudy next time

experiment for the slotted lineclick the slotted line setup is shown in figure 41. The HP 864B RF signal generator goes up to

500 MHz of frequency. It has the possibility of being amplitude modulated with a 1 kHz signal. Theoutput power you will uses 20 DVM. The output from the generator goes to a 500 MHz low passfilter. The purpose of this filter is to get rid of all the higher harmonics, in order to maintain ispure for sine wave as possible. The output of the future is the input coaxial slotted line, which isabout 60 cm long. The other end of the slotted line is terminated with the unknown load we wish



to measure. All connect tutors in this system are GR (general radio) connectors. The slotted lineprobe has the diode detector built-in one of the two lands. Connect the send to the HP 415 ESWart meter. The diode detects the amp to the of the 1 kHz low-frequency signal that modulates theart signal picked up by the probe. The SWR meter is just a narrow band high gain amplifier for the1 kHz that the 500 MHz is modulated with. It it’s high gain assures good measurement sensitivity.The other GR connectors on the probe is used for optimizing the match of the probe to the line.Connect the adjustable stop tuna to this end. This unit is just a section of quarks with a short bedthe other and so that this impedance can change by physically moving this short.

Number one in this first part of the experiment you will use a short circuit termination. Set thefrequency at 500 MHz.find and also the standing wave pattern.

Question one: why are the nulls of shorted circuit standing wave pattern sharp?the sharp nullswill be buried in noise. The best way to find their position is to move towards the zero from both

sides, locate points a and B from figure 42, and then take them zero to be between them. measureand record the position of several successive zeros along an arbitrary position scale.

question two: what is the wavelength in the line equal to?Number two: what is the wavelength in the line equal to?Increase the generator power and repeat the measurement.Question four: why is this measurement not as good as the previous one? What gets worse: the

minimum or the maximum of the voltage?Number three in this part of the experiment, you will measure two loads at both 500 and 250

MHz. The first load is the unknown load from part two, and the second one is along to sister of thesame value, mounted in the metal jig from experiment to. First determine the position of the proxyplane by connecting a short termination. Then connect the load, and measure the distance betweenthe first the next of the proxy plane in the proxy plane. You can either move towards or away fromthe load, but make sure you are consistent on the Smith chart. Repeat this measurement for thetwo loads at F1 equals 250 MHz and after equals 500 MHz. Feel a table as the one shown in table41. The distance fell in the fifth column is the electrical vistas between columns 3 and four. Use thisthis this and this may chart to find the normalized load impedance Z from the measures SWR. Thenfind the load impedance Z by multiplying by the characteristic impedance of the Coax. include yourSmith chart.

Four at 500 MHz measured the impedance of a car X 50 ohm termination, repeating the procedurefrom part three. Include the Smith chart.

Question 5Y this the harder measurement than for badly matched loads?In the past part of this experiment you will measure the impedance of the monopole antenna

above a metal plane at 500 MHz. The antennae 15 cm long, which is about a quarter wavelength ofthis frequency. This is called the quarter wave monopole. They impedance of the monopole with orwithout the ground plane is very different.

Question six: this antenna can be viewed as a current flowing through a Short St., Finn Y whichis located about an incident perfectly conducting plane. What basic fear massage among medicswould you use to find electric and magnetic field of such a system? Illustrate your answer with asimple drawing you do not have time to find the fields.

Question seven: the impedance of an antenna is defined as the voltage across its terminal by



current.ConclusionSolve all of the above examples by hand and compare the results with the ones you found

during the lab. In addition, for each of the problems, write what you have learned. Do not write anessay, just in a few sentences write what you have done in each problem, and specifically what youhave learned from the work done.

Due DateSubmit the printout of the lab work and the conclusion by NEXT FRIDAY. LAB IS DUE BY

noon (stamped) next Friday in the main office. No late labs. Do not bring previous lab writeup forthe next Lab session.


EEE161 Applied Electromagnetics Practical Laboratory...

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