09 a lab3-nguyen thi bao tram

dddDDDASARTETE

TRƯỜNG ĐẠI HỌC BÁCH KHOA ĐÀ NẴNG

KHOA ĐIỆN TỬ – VIỄN THÔNG

BÁO CÁO THÍ NGHIỆM

KỸ THUẬT SIÊU CAO TẦN

LAB 3: Transients on Transmission Lines

Student : Nguyễn Thị Bảo Trâm

Group : 09A

Class : 06DT1

Đà Nẵng – 2010

Báo cáo TN SIÊU CAO TẦN Lab3

Nguyễn Thị Bảo Trâm – 06DT1

1. Introduction:

We have thus far focused on techniques for understanding transmission lines under sinusoidal

excitation. Powerful analytic insight is available here, permitting straightforward design of

interesting circuits.

In contrast, the analysis of transients is generally more difficult and less amenable to simple

closed-form analysis, especially when loads are reactive. This laboratory will explore the propagation

of transients on transmission lines with the aid of numerical experiments in SPICE.

2. Purely Resistive Termination:

First, let's use SPICE to investigate the propagation of pulses on a transmission line terminated

by purely resistive loads.

2.1 A step function, matched load

First, create a 50 transmission line with total length (time delay) of 25 ns, and excite it with a

Thevenin source 10u(t), with a source resistance Rg = 50 . With the load resistance RL = 50 ,

run the simulation.

To create the voltage source, use the VPWL source. Using VPWL allows various times and the

voltages at those times to be specified using the T1, T2, T3, ... and V1, V2, V3, ... parameters. The

voltage source will be piece-wise linear, connecting each specified point. Using VPULSE allows

specification of the initial voltage, V1, the voltage of the pulse, V2, the delay time, TD, the rise time,

TR, the fall time, TF, the pulse width, PW, and the period, PER.

The VPWL source is piece-wise linear and allows the user to specify voltages at specific times

using T1, T2, T3… and the corresponding V1, V2, V3 parameters. Please note that the source will

connect each specified point in the most direct way. Additionally, two voltages cannot be specified

for the same time, so that instantaneous changes must be approximated.

For Example: to specify a VPWL source that produces a 10V square pulse starting at t=0 and

lasting for 10 ns, would have the following specified parameters: T1=0, V1=0, T2=0.001n, V2=10,

T3=10n, V3=10, T4=10.001n, V4=0. (Making V3=0 would form a sawtooth wave because of the

reasons stated above.)

T1

TD = 25nsZ0 = 50

Vg

0

0

0

Vload

0

VsourceRg

50

RL

50

Vg



Using VPULSE allows specification of an initial voltage, V1, the voltage of the pulse, V2, the delay

time, TD, the rise time, TR, the fall time, TF, the pulse width, PW, and the period, PER. Single

pulses can be formed with this source. More complex signals can be formed by combining multiple

sources.

Problem 1 Plot the voltage at the source and load ends of the transmission line for t = 0…50 ns.

Using your understanding of “bounce diagrams”, explain whether this plot makes sense and shows

what you would expect to see in the “exact” answer. Do the two agree? If not, why not? A full

credit answer will describe the bounce diagram until a reasonable (whatever you consider

reasonable) number of bounces, which can explain what exactly is happening in this situation.

Please do the same in any other bounce diagram questions that may follow.

Answer:

- The voltage at the source and load ends of the transmission line for t = 0…50 ns

The reflection coefficient at the source is:

Γ𝑔 =𝑅𝑔 − 𝑍0

𝑅𝑔 + 𝑍0=

50 − 50

50 + 50= 0

The reflection coefficient at the load is:

Γ𝐿 =𝑅𝐿 − 𝑍0

𝑅𝐿 + 𝑍0=

50 − 50

50 + 50= 0

⇒ There is no reflection wave of voltage at the source and the load.

𝑉𝐿 =𝑉𝑔𝑍𝐿

𝑅𝑔 + 𝑍𝐿=

10 .50

50 + 50= 5(𝑉)

Time

0s 5ns 10ns 15ns 20ns 25ns 30ns 35ns 40ns 45ns 50ns

V(VG) V(VSOURCE) V(VLOAD)

0V

5V

10V



VL = 5 (V) after 25ns (time delay of the transmission line.) ⇒ The theory and the plot are the same.

2.2 A step function, mismatched load

Now, change the load impedance in the previous case to 20 .

Problem 2 Plot the voltage at the source and load ends of the transmission line for t = 0…100

ns. Using your understanding of “bounce diagrams”, compare this with what you would expect to

see in the “exact” answer. Do the two agree? If not, why not? How long does it take the answer to

settle to the final answer?

Answer:


The reflection coefficient at the source is:

Γ𝑔 =𝑅𝑔 − 𝑍0

𝑅𝑔 + 𝑍0=

50 − 50

50 − 50= 0

Rg

50

RL

20

0

Vg

Vg

0

Vload

0

T1

TD = 25nsZ0 = 50

Vsource

0

Time



0V

5V

10V



The reflection coefficient at the load is:

7

3

5020

5020

0

0

ZR

ZR

L

LL

)(55050

50.10

0

0

1 VZR

ZVV

g

g

At t = T = 25ns, the wave reaches the load (receiving end) z = l, and because 07

3L , the

mismatch generates a reflected wave with the amplitude:

7

15. 11 VV L (V)

So the voltage on line (in this case, the load voltage) is the sum of two waves:

)(86.27

20

7

15511 VVVVL

At t = 2T = 50ns, the wave reaches the sending end z = 0, and because 0g , there is no

reflection wave in form of a wave with voltage amplitude V2+ = 0. Thus, the voltage on line (in this

case, the source voltage) is:

)(86.27

20

7

15511 VVVVS

At t = 3T = 75ns, the wave reaches the load (receiving end) z = l, and because 07

3L , the

mismatch generates a reflected wave with the amplitude: 0. 22 VV L (V)

VL = 2.86 (V) Similarly, we also have: VS = 2.86 (V)

Thus, after 50ns, the voltage at two ends of the transmission line have the same final answers.

t = 0

g

z = 0

t = 0

L

z = l

V1+

T

2T

L V1+

3T

L g V1+



2.3 A step function, mismatched load and source

Now, change the load impedance in the previous case to RL=20 , and change the source

impedance to Rg=200 .

Problem 3 Plot the voltage at the source and load ends of the transmission line for t = 0…300

ns. Using your understanding of “bounce diagrams”, compare this to the “exact” answer. Do the two

agree? If not, why not? How long does it take the answer to settle down to the final answer?

Answer:


6.050200

50200

0

0

ZR

ZR

g

g

g

7

3

5020

5020

0

0

ZR

ZR

L

LL

Rg

200

RL

20

0

Vg

Vg

0

Vload

0

T1

TD = 25nsZ0 = 50

Vsource

0

Time

0s 50ns 100ns 150ns 200ns 250ns 300ns


0V

4V

8V

12V



)(250200

50.10

0

0

1 VZR

ZVV

g

g

At t = T = 25ns, the wave reaches the load (receiving end) z = l, and because 07

3L , the

mismatch generates a reflected wave with the amplitude:

86.07

62.)

7

3(. 11 VV L (V)

So the voltage on line (in this case, the load voltage) is the sum of two waves:

)(14.17

8

7

6211 VVVVL

At t = 2T = 50ns, the wave reaches the sending end z = 0, and because 6.0g , there is a

reflection wave in form of a wave with voltage amplitude V2+ =

1Vg . Thus, the voltage on line (in

this case, the source voltage) is:

)(63.0).1( 1211 VVVVVV gLLS

At t = 3T = 75ns, the wave reaches the load (receiving end) z = l, and because 07

3L , the

mismatch generates a reflected wave with the amplitude: 22 .VV L (V)

VL = 0.85 (V)

t = 0

g

z = 0

t = 0

L

z = l V1+

T

2T

L V1+

3T

4T

5T

gL V1+

gL 2

V1+

22

gL V1+



Calculate like the process above, we have the table:

Time V at sending end V at receiving end

t = 0ns VS = )(2

50200

50.10

0

0

1 VZR

ZVV

g

g

)(0 VVL

t = T =25ns VS = )(2

50200

50.10

0

0

1 VZR

ZVV

g

g

)(14.1

7

8

7

6211 VVVVL

t = 2T = 50ns

)(63.0).1( 1

211

VV

VVVV

gLL

S

)(14.17

8

7

6211 VVVVL

t = 3T=75ns

)(63.0).1( 1

211

VV

VVVV

gLL

S

)(85.0).1( 1

2 VVV gLgLLL

t = 4T=100ns )(98.0).1( 1

222 VVV gLgLgLLS

)(85.0).1( 1

2 VVV gLgLLL

t = 5T =125ns

)(98.0).

1(

1

22

2

VV

V

gL

gLgLLS

)(92.0).

1(

1

23

222

VV

V

gL

gLgLgLLL

T = 6T =150ns VS = 0.98 (V) VL = 0.924(V)

T = 7T =175ns VS = 0.89(V) VL = 0.924(V)

T = 8T =200ns VS = 0.89(V) VL = 0.909(V)

T = 9T =225ns VS = 0.913(V) VL = 0.909(V)

- Thus, the plot and the “exact” answer agree.



- The final voltage is: )(909.020200

20.10

0

VZR

ZVV

g

Lg

⇒ It takes 225ns to settle down to the final answer.

2.4 A short pulse

Now break the transmission line into two equal pieces, with total length 25 ns (you should now have

two different transmission lines, both with the same 50 characteristic impedance, but each with a

time delay of only 12.5 ns). This permits us to sample “inside” the transmission line. With the same

transmission line, and Rg=200 and RL=20 apply a pulse of duration 10 ns to the transmission

line, namely vg(t) = 10(u(t)- u(t-10ns)).

Problem 4 Plot the voltage at the source, middle, and load ends of the transmission lines for t =

0…100 ns. Sketch the bounce diagram; do you understand the voltage plots? How long before the

“ghost” pulse (the pulse you are seeing at the middle of the transmission line) arrives at the load

end? How large is the “ghost” pulse?

Answer:

- The voltage at the source, middle, and load ends of the transmission lines for t = 0…100ns

T1

Z0 = 50TD = 12.5ns RL

20

Vsource

000

Vg

0

Rg

200

00

VloadVmiddleVgT2

Z0 = 50TD = 12.5ns

Time


V(VSOURCE) V(VMIDDLE) V(VLOAD)

-2.0V

-1.0V

0V

1.0V

2.0V



6.050200

50200

0

0

ZR

ZR

g

g

g

7

3

5020

5020

0

0

ZR

ZR

L

LL

)(250200

50.10

0

0

1 VZR

ZVV

g

g

- The bounce diagram:

- We have the table of values (by calculating):

Time V at sending end V middle V at receiving end

t = 0+ ns )(2

0

0

1 VZR

ZVVV

g

g

S

Vm = 0 (V) VL = 0 (V)

t = T/2

=12.5ns 0

50200

50.0

0

0

ZR

ZVV

g

g

S Vm = V1

+ (t = 0ns) =

2(V)

VL = 0 (V)

g

z = 0

t = 0

L

z = l V1+

T

2T

L V1+

3T

4T

5T

2lz

gL V1+

gL 2

V1+

22

gL V1+



t = T = 25ns VS = 0(V) Vm = VS (t = 12.5ns) =

0(V) )(14.1

7

8

7

62

11

V

VVVL

t = 3T/2 =

37.5ns

VS = 0(V)

)(86.07

6

11

V

VVV Lm

)(0 V

VVV mmL

(Vm at t = 2T)

t = 2T = 50ns

)(37.1

).( 1

21

V

V

VVV

gLL

S

)(0 VVV Lm VL = 0 (V)

t = 5T/2 =

62.5ns

)(050200

50.0

0

0

V

ZR

ZVV

g

g

S

)(51.026.07

3

12

V

VVV gLm

VL = 0 (V)

t = 3T=75ns VS = 0(V) Vm = VS (t = 12.5ns) =

0(V) )(29.0).( 1

2

22

VV

VVV

gLgL

L

t =

7T/2=82.5ns

VS = 0(V)

)(22.0

1

2

2

V

VVV gLm

VL = 0 (V)

t = 4T =100ns

)(35.0

).( 1

222

32

V

V

VVV

gLgL

S

)(0 VVV Lm VL = 0 (V)

It takes 12.5ns for the “ghost” pulse to get the load end. The ghost pulse has width of

12.5ns.



2.5 A longer pulse

In the previous problem, the pulse was brief (10 ns) compared to the length of the transmission line

(25 ns). Now investigate a more complicated system.

Problem 5 Using the same transmission line and source impedance, define a new source for

which vg = +10 V for t = 0 … 20 ns, and gv V for t = 20 … 40ns. Plot the voltage at the

source, center, and load end of the transmission line for t = 0…100 ns. Is the transition from “high”

to “low” perfectly clear at the load end?

Answer:

- The voltage at the source, center, and load end of the transmission line for t = 0…100ns

- From the graph, we can see that the transition from high to low is perfectly clear at the load

end.

2.6 An Impedance Bump

Transmission lines must be protected against damage, or their impedance properties could be

compromised. In this section we’ll “damage” the transmission line by putting a weak load in the

T1

Z0 = 50TD = 12.5ns RL

20

Vsource

000

Vg

0

Rg

200

00

VloadVmiddleVgT2

Z0 = 50TD = 12.5ns

Time



-4.0V

-2.0V

0V

2.0V



middle. In particular, at the center of the transmission line, add a shunt resistance of 50 (a shunt

resistance connects the node at the middle to ground).

Problem 6 With the shunt resistance in place, repeat the previous problem. How did the

presence of the “bump” (break in the transition line) affect the voltage plots? Did any new “ghosts”

show up? Looking only at the source and end voltages, could you determine where the “bump” is?

Can you explain what you see in terms of a bounce diagram?

Answer:

- The presence of the impedance “bump” (break in the transition line) reduced the magnitude

of the middle voltage.

3. Reactive Termination

As mentioned in class, it is frequently the case that the loads at the end of a data bus are reactive

(and often capacitive).

3.1 A step into a capacitive load

Vg

T1

Z0 = 50TD = 12.5ns

VmiddleT2

Z0 = 50TD = 12.5ns

0 0

0

Rg

200

00 0

VloadVsource

0

Vg

Rshunt

50

RL

20

Time



-4.0V

-2.0V

0V

2.0V



For this exercise, again form a circuit with a source vg(t) = 10u(t), a source impedance Rg=25 ,

and a transmission line with characteristic impedance 50 and length 25ns.

Problem 7 Terminate the transmission line with a 1nF capacitor. Plot the voltage at the source

and load ends of the transmission line for t = 0…400ns. If you see any “exponential” charging or

discharging, estimate the time constant, and solve for the R. You may use the following formulas.

Vinitial = 0V

VFinal = 12.133V

V(t) = 8.4292V

t = (75 – 25) = 50ns

CL

1n0

Vsource

0

Vg

Rg

25

0

0

VloadVgT1

TD = 25nsZ0 = 50

Time

0s 50ns 100ns 150ns 200ns 250ns 300ns 350ns 400ns

V(VSOURCE) V(VLOAD)

0V

5V

10V

15V



𝜏 =−𝑡

𝑙𝑛 𝑉(𝑡) − 𝑉𝑓𝑖𝑛𝑎𝑙

𝑉𝐼𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑉𝑓𝑖𝑛𝑎𝑙

=−50. 10−9

𝑙𝑛 8.4292 − 12.133

0 − 12.133 = 4.214. 10−8

𝜏 = 𝑅𝐶 ⇒ 𝑅 =𝜏

𝐶=

4.214. 10−8

10−9= 42.14(Ω)

Problem 8 Repeat the previous problem, but with a load capacitance of 100pF.

Vinitial = 0V

VFinal = 13.33V

V(t) = 13.244V

t = (50 – 25) = 25ns

𝜏 =−𝑡



=−25. 10−9

𝑙𝑛 13.244 − 13.33

0 − 13.33 = 4.956. 10−9

𝜏 = 𝑅𝐶 ⇒ 𝑅 =𝜏

𝐶=

4.956. 10−9

100.10−12= 49.56(Ω)

Vload

CL

100pF0

0

Vg

Rg

25

0

0

VsourceVgT1

TD = 25nsZ0 = 50

Time


V(VSOURCE) V(VLOAD)

0V

5V

10V

15V



Problem 9 Repeat the previous problem, but with a load capacitance of 10nF.

Vinitial = 0V

VFinal = 10V

V(t) = 8.056V

t = (400 – 25)= 375ns

𝜏 =−𝑡



=−375. 10−9

𝑙𝑛 8.056 − 10

0 − 10 = 2.29. 10−7

𝜏 = 𝑅𝐶 ⇒ 𝑅 =𝜏

𝐶=

2.29. 10−7

10.10−9= 22.9(Ω)

Problem 10 Repeat the previous problem, but with a load inductance of 2.5μH.

Vg

Vg

0

0

CL

10nF

T1

TD = 25nsZ0 = 50

0

0

Vsource VloadRg

25

Time


V(VSOURCE) V(VLOAD)

0V

5V

10V



Vinitial = 13.202V

VFinal = 0V

V(t) = 4.8052V

t = (75 – 25) = 50ns

𝜏 =−𝑡



=−50. 10−9

𝑙𝑛 4.8052 − 013.202 − 0

= 4.947. 10−8

𝜏 =𝑅

𝐿 ⇒ 𝑅 =

𝐿

𝜏=

2.5. 10−6

4.947. 10−8= 50.53(Ω)

Problem 11 Repeat the previous problem, but with a load inductance of 0.25μH.

Vg

Vsource

0

L

2.5uH

1

20

VgT1

TD = 25nsZ0 = 50

0

0

VloadRg

25

Time


V(VSOURCE) V(VLOAD)

0V

5V

10V

15V



Vinitial = 13.3V

VFinal = 0V

V(t) = 1.51V

t = (36 – 25)ns

𝜏 =−𝑡



=−11. 10−9

𝑙𝑛 1.51 − 013.3 − 0

= 5.056. 10−9

𝜏 =𝑅

𝐿 ⇒ 𝑅 =

𝐿

𝜏=

0.25. 10−6

5.056. 10−9= 49.44(Ω)

Problem 12 Repeat the previous problem, but with a load composed of a parallel combination of

RL=1000 , L = 1μH and C = 100pF.

L

0.25uH

1

2

Vsource

0

Vg

0

VloadRg

25

0

0

VgT1

TD = 25nsZ0 = 50

Time


V(VSOURCE) V(VLOAD)

-5V

0V

5V

10V

15V



Problem 13 How important is the value of Rg in these exercises?

Answer:

- Rg is very important since it allows the system to reach steady state. The value of R

g helps us

to determine whether the circuit is matched or not at the input of transmission line.

4. Coupling

Many data buses are in parallel, in close proximity, such as the 32 bit and 64 bit buses found in

computers. These transmission lines will have “mutual impedance” which causes signals on one

transmission line to show up on another one.

In EE571 students analyze this coupling in great detail, but we can simulate a simplified model of a

two-wire data bus using SPICE1:

1 Note that your SPICE library contains a model for coupled lines; you could use this, but it does the modeling with an

explicit form of the Telegrapher’s Equation, and is painfully slow. I tried to use it, but realized that it would take about

one day to run every simulation!

Vg

Vsource

RL

1k

0

L

1uH

1

20

VgT1

TD = 25nsZ0 = 50

0

0

Vload

C

100pF

Rg

25

Time


V(VSOURCE) V(VLOAD)

-5V

0V

5V

10V



Here T1 and T4 represent the “actual” transmission lines, while T2 and T3 represent the cross

coupling. The coupling is slightly faster (95 ns instead of 100 ns) and has a higher characteristic

impedance.

Connect a Thevenin signal pulse (10 V, 50 ) to A and 50 loads to B, C and D2.

2 To make this model even closer to the real thing, the signal injected into T2 should have opposite sign of that injected

into T1. However, this lab will not require this.

0

0

R3

50

0

A

R4

50

0

B

0

C

0

0

0

D

Vg

0

R2

50

0

T3

Z0 = 200TD = 95ns

T1

Z0 = 50TD = 100ns

0

T2TD = 95nsZ0 = 200

T4TD = 100ns

Z0 = 50

Vg

0

R1

50



4.1 A simple pulse

Inject a 10ns pulse with the Thevenin source at A. Use a rise and fall time of 1ns.

Problem 14 Simulate the problem for 250 ns, and plot the voltage at A, B, C, and D. Describe

what you observe. How big is the VC compared to VD when the pulse arrives?

Which arrives first?

Answer:

- Plot the voltage at A, B, C, and D

- VC is about 4 times compared to the magnitude of V

D when the pulse arrives.

- VD arrives first.

Problem 15 When does a signal arrive at B? If you change the values of the load resistances at C

and D, can you eliminate the reflection? If so what value should the load have?

Answer:

- A signal arrives at B at 195ns.

- We cannot eliminate the reflection by changing the values of the load resistances at C and

D. We just can reduce VC and VD to zero.

- The plot when the load resistances at C and D are 0.001Ω:

Time

0s 50ns 100ns 150ns 200ns 250ns

V(VG) V(A) V(B) V(C) V(D)

-5V

0V

5V

10V



5. Impedance Matching

5.1. Pre-lab Assignment

5.1. A. Design a quarter-wave transformer to match a 150 Ω load to a source resistance of 75 Ω.

State the length of your transmission line(s) in terms of the wavelength.

Answer:

𝑍𝐿 = 150Ω

𝑍𝑖𝑛 = 75Ω

The input impedance is:

𝑍𝑖𝑛 = Z0

ZL + jZ0tgβl

Z0 + jZLtgβl

With 𝑙 =𝜆

4⇒ 𝛽𝑙 =

2𝜋

𝜆.𝜆

4=

𝜋

2

We can divide the numerator and the denominator by tgβl and take the limit as βl →π

2 to get:

𝑍𝑖𝑛 = 𝑍0

2

𝑍𝐿

The characteristic impedance of the matching section is:

⇒ 𝑍0 = 𝑍𝐿.𝑍𝑖𝑛 = 150.75 = 106.066(Ω)

Time

0s 50ns 100ns 150ns 200ns 250ns

V(VG) V(A) V(B) V(C) V(D)

-5V

0V

5V

10V



5.1. B. Using the Smith Chart, if the characteristic impedance is given as 75 Ω, design a stub-

matching network to match a 150 Ω load to a 75 Ω source. Do this for both a shorted and an open

circuited stub. State the length of your transmission line(s) in terms of wavelength.

Answer:

𝑍0 = 75Ω

𝑍𝑖𝑛 = 75Ω

𝑍𝐿 = 150Ω

- The normalized load impedance is: 𝑧𝐿 =ZL

Z0=

150

75= 2Ω (point A on the Smith chart)

- Construct the appropriate SWR circle through point A. From the Smith chart, the normal load

admittance: yL = 0.5 (at point B l 0 )

- The SWR circle intersects the 1+jb circle at two points, denoted as y1 , y2. Thus the distance

d, from the load to the stub, is given by either of these two intersections. Reading the WTG

scale, we obtain:

d2 = (0.348– 0) = 0.348

At the two intersection points, the normalized admittances are:

y1 = 1 + j0.7

y2 = 1 – j0.7

- Thus, the first tuning solution requires a stub with a susceptance of – j0.7. The length of an

open-circuited stub that gives this susceptance can be found on Smith chart by starting at y

= 0 (the open circuit) and moving along the outer edge of the chart (g=0) toward the

generator to the – j0.7 point. The length is then:

lo1= 0.402

Similarly, the required open-circuited stub length for the second solution is:

lo2= 0.097

- The length of an shorted-circuited stub that gives this susceptance can be found on Smith

chart by starting at y = ∞ (the shorted circuit) and moving along the outer edge of the chart

(g=0) toward the generator to the – j0.7 point. The length is then:

ls1=(0.402 – 0.25) = 0.152

Similarly, the required shorted-circuited stub length for the second solution is:

ls2= (0.25 + 0.097) = 0.347

5.2. Lab Assignment

5.2.A. Find the actual length of the quarter-wave matching network you designed in part 5.1.A if up

= 2E+8 (m/s) and frequency = 1 GHz. Simulate the frequency response of the circuit by sweeping

the frequency from 1 MHz to 3 GHz using a 5Vpp sine wave source with a source resistance of 75 Ω.

Plot the input and load voltage over frequency. Plot the magnitude of the reflection coefficient of



the matching network and find the bandwidth where |Γ| is less than 0.2. What is the input

impedance of the matching network at 1 GHz?

Answer:

up = 2.108 m

s ; f = 1GHz

)(2.010

10.29

8

mf

u p

)(05.04

2.0

4ml

nsssm

m

u

lT

p

D 25.0)(10.5.2/10.2

05.0 10

8

- The input and load voltage over frequency:

Vload

RL

150

0

0

Vinput

Vs2.5Vac0Vdc

T1

TD = 0.25nsZ0 = 106.066

0

0

Rs

75



- The magnitude of the reflection coefficient of the matching network:

|Γ| is less than 0.2 from 608.1 MHz to 1.388 GHz and frequencies from 2.607 GHz to 3 GHz

- The input impedance of the matching network at 1 GHz is 75Ω.

Frequency

1.0MHz 3.0MHz 10MHz 30MHz 100MHz 300MHz 1.0GHz 3.0GHz 10GHz

V(VINPUT) V(VLOAD)

1.2V

1.4V

1.6V

1.8V

Frequency


(V(VINPUT)/ I(Rs)-75)/(V(VINPUT)/I(Rs)+75)

0

100m

200m

300m

400m



5.2.B. Using the assumptions of section 5.2.A, simulate your pre-lab design in part 5.1.B and find

the Γ bandwidth (the portion of the signal less than 0.2) of the matching network as well as the

input impedance at 1 GHz. Include the same plots as in section 5.2.A.

Answer:

up = 2.108 m

s ; f = 1GHz

)(2.010

10.29

8

mf

u p

d1 =0.152𝜆 = 0.152×0.2 = 0.0304 (m) ⇒ nsssm

m

u

dT

p

D 152.0)(10.52.1/10.2

0304.0 10

8

1

lo1= 0.402𝜆 = 0.402×0.2 = 0.0804 (m) ⇒ nsssm

m

uT

p

D 402.0)(10.02.4/10.2

0804.0l 10

8

o1

ls1=0.152𝜆 = 0.152×0.2 = 0.0304 (m) ⇒ nsssm

m

uT

p

D 152.0)(10.52.1/10.2

0304.0l 10

8

s1

Open-circuited stub:

Frequency


V(VINPUT)/ I(Rs)

50

100

150

(1.0000G,75.000)





|Γ| is less than 0.2 from 924 MHz to 1.11 GHz.

0

R1

1000G

Vs2.5Vac0Vdc

Vload

T2

TD = 0.402nsZ0 = 75

T1

TD = 0.152nsZ0 = 75

VinputRs

75

RL

150

000

00

0

Frequency


V(VINPUT) V(VLOAD)

0V

0.5V

1.0V

1.5V

2.0V



- The input impedance of the matching network at 1 GHz is 75.029Ω.

Shorted-circuited stub:

Frequency


(V(VINPUT)/I(Rs)-75)/(V(VINPUT)/I(Rs)+75)

0

0.5

1.0

Frequency


V(VINPUT)/ I(Rs)

0

50

100

150

(1.0000G,75.029)





0

Rs

75

Vinput

T2

TD = 0.152nsZ0 = 75

RL

150

Vload

Vs2.5Vac0Vdc

T1

TD = 0.152nsZ0 = 75

0

0 0

0 0

0

Frequency


V(VINPUT) V(VLOAD)

0V

0.5V

1.0V

1.5V

2.0V



|Γ| is less than 0.2 from 850 MHz to 1.21 GHz and from 2.07 GHz to 2.44 GHz

- The input impedance of the matching network at 1 GHz is 75.029Ω.

Problem 16 Compare the results for the matching networks that you designed in the lab (quarter

wave, open and short stub). Which one is a better choice. Why?

Answer:

- Compare the results for the matching networks that we designed in the lab (quarter wave,

open and short stub), the quarter wave is the best choice among of these. Because: it has

the largest Γ bandwidth (the portion of the signal less than 0.2) about 1.102GHz while the

open and short stub’s are quite less than 1.0GHz. Besides, it also has an input impedance of

exactly 75 ohms.

Frequency


(V(VINPUT)/I(Rs)-75)/(V(VINPUT)/I(Rs)+75)

0

0.5

1.0

Frequency


V(VINPUT)/ I(Rs)

0

50

100

(1.0000G,75.029)



Problem 17 For stub matching networks compare the performance of an open stub versus a

shorted stub.

Answer:

- For stub matching networks compare the performance of an open stub versus a shorted

stub: Γ bandwidth (the portion of the signal less than 0.2) about 0.186GHz and 0.73GHz

respectively. So the shorted stub circuit is the better one.

Problem 18 Suppose you had a lossless line terminated by a complex load, but wanted to carry

out the match using a quarter-wave transformer. How could you accomplish this?

Answer:

- Suppose we had a lossless line terminated by a complex load, but wanted to carry out the

match using a quarter-wave transformer. To accomplish this we could use a transmission

line with a complex characteristic impedance.

09 a lab3-nguyen thi bao tram

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