Download - Report Lab 2 EE 361

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Page 1: Report Lab 2 EE 361

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Report Lab 2 EE 361

2.1 Basic Transmission Line Model

For common RG-58 coaxial cable, the characteristic impedance is Z0 = 50 Ω and the phase velocity up = 2/3 c. (Note: c = speed of light = 3e8 m/s)Question 1: For such a transmission line, what are the inductance and capacitance per meter?

up=1

√L' C 'L'C '=2.5×10−17 (1)

Z0=√ L 'C '

L 'C '

=2500L'=2500C ' (2)

From (1) and (2), we have: 2500C '2=2.5×10−17C '=1×10−10F /mFrom (2), we have: L'=2500C'=2.5×10−7 H /m

Question 2: For a different coaxial cable, μ = μ0 and ε = 3ε0. What is b/a if Z0 = 50 Ω?L 'C '

= µ2πln (b/a ) 1

2πεln (b/a )= µ

4 π2 εln2¿¿

So :

ln 2(ba )=4 π2 εµ

. L'C '

=12π 2ε 0

µ0.Z0

2=12π2 (8.854×10−12 )

4 π ×10−7.(50)2=2.086

ln (ba )=1.444Thus b/a = 4.235

Question 3: If b = 3 mm in question 2.2, what is a?

We have b /a=4.235a= b4.235

Thus, if b = 3 mm, a 0.708mm

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2.2 A SPICE model of a transmission line problem.

Question 4: At 200 MHz, and with up= 2/3 c, what is the wavelength in the transmission line?

λ=up

f=2 /3c

f=

23

×(3×108)

200×106=1m

Thus, at 200 MHz and up = 2/3 c, the wavelength in the transmission line is 1m.

Question 5: What is the time delay associated with λ/16?

TD= Lu p

= Lλf

=

λ16λf

= 116 f

= 116 (200×106)

=3.125×10−10 s=0.3125ns

Thus the time delay associated with λ/16 is 0.3125 ns.Use SPICE to simulate the steady state AC response of this transmission line for length 0, λ/16, 2λ/16, …, 15λ/16, λ. Center your sweep on the frequency of interest and sweep linearly.

Time (ps) Input V (mV) input I (mA) Output V (mV) Output I (mA)0 666.667 6.6667 666.667 6.6667

312.5 629.114 7.9982 666.667 6.6667625 527.046 10.541 666.667 6.6667

937.5 399.908 12.580 666.667 6.66671250 333.333 13.328 666.667 6.6667

1562.5 399.690 12.584 666.667 6.66671875 526.213 10.548 666.667 6.6667

2187.5 628.990 8.0023 666.667 6.66672500 666.667 6.6667 666.667 6.6667

2812.5 627.988 7.9982 666.667 6.66673125 527.046 10.541 666.667 6.6667

3437.5 339.908 12.580 666.667 6.66673750 333.333 13.328 666.667 6.6667

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4062.5 399.690 12.584 666.667 6.66674375 526.213 10.548 666.667 6.6667

4687.5 628.990 8.0023 666.667 6.66675000 666.667 6.6667 666.667 6.6667 Thus, Vmin = 333.333 mV and Vmax = 666.667 mV

Question 6: Use PSPICE, Excel, or Matlab to plot the magnitude of the voltage at Input as a function of

length. From the Voltage Values on the plot and the relationship:SWR=V max

V min, determine the VSWR,

and from the VSWR calculate |Γ|.Plot:

From this plot, we can see that Vmax = 666.667 mV and Vmin = 333.333 mVThus, we have

VSWR=V max

V min=666.667mV333.333mV

=2

And

|Γ|=VSWR−1VSWR+1

=2−12+1

=13=0.333

Question 7: Use PSPICE, Excel, or Matlab to plot the magnitude of the current at Input as a function of length. From the Current Values on the plot, determine the VSWR, and from the VSWR calculate |Γ|. Do the voltage and current yield the same VSWR and |Γ|?Plot:

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From this plot, we see that Imax= 13.333 mA and Imin = 6.6667 mAThus, we have

VSWR=Imax

Imin=13.333mA6.6667mA

=2

And

|Γ|=VSWR−1VSWR+1

=2−12+1

=13=0.333

Thus, the voltage and current yield the same VSWR and |Γ|

Question 8: Plot the magnitude of the impedance at Input as a function of length using the data you collected with PSPICE. Plot the Real and Imaginary Parts of the Impedance using PSPICE and also plot impedance using a Smith Chart.

Because the impedance is the division of total voltage over total current. So to plot the real and imaginary parts of the impedance, we can plot the real and imaginary parts of the division of total voltage over total current.

Plot of real part of impedance:

Plot of imaginary part of impedance:

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Question 9:Using the scales at the bottom of the Smith Chart, find the VSWR and |Γ|. Do they agree with your previous answers? (Ignore Smith Chart questions)

Question 10: Compute Γ and VSWR directly using equations (2.6) and (2.7) below. Do these agree with your measurements from question 6, 7 & 8?From class recall that:

VSWR=1+¿ Γ∨ ¿1−¿ Γ∨¿¿

¿

And

Γ=ZL−Z0Z L+Z0

So, we have

Γ=ZL−Z0ZL−Z0

=100Ω−50Ω100Ω+50Ω

=13=0.333

And

VSWR=1+¿ Γ∨ ¿

1−¿ Γ∨¿=1+1 /31−1/3

=2¿¿

Thus, the results agree with the previous answers.

Question 11: Plot the voltage magnitude at Load as a function of length. How does the voltage change with length? From this, how do you think the power delivered to the load will change with length?

Plot:

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From the plot, we can see that the voltage changes with legnth sinusoidally.

The power can be expressed by P=VI . Thus, the voltage is a sinusoidal function of length. So the the power delivered to the load changes with legnth sinusoidally.

2.3 A shortcut, and more load impedances

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Question 12: If you have 1 meter of the coaxial cable described in question 4, at what frequency does it have length λ/2? At what frequency does it have length 2.5λ? (Note that we are NOT changing the physical length of the line, only it’s “electrical length” as defined above.) We have f = 200MHz

at what frequency does it have length λ/2? f’= 12f = 100MHz

what frequency does it have length 2.5λ? f’ ’= 2.5f = 500MHz

Question 13: Plot the magnitude of the voltage at Input for the different “lengths” (remember that you are really just adjusting the frequency) properly relabeling the horizontal axis. (You can do this by hand or by using text boxes in Pspice.) Does this agree with your plot in question 6? What is the VSWR?

Does this agree with your plot in question 6? YesWhat is the VSWR?

Vmax = 666.667 mV and Vmin = 333.333 mVThus, we have

VSWR=V max

V min=666.667mV333.333mV

=2

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Replace the 100 Ω load with a 25 Ω load.Question 14: Plot the magnitude of the voltage at Input, and compare to the previous case of 100 Ω. From the plot, what is the VSWR? On a Smith Chart, what similarity is there between the 100 Ω and 25 Ω cases?

This is similar to the previous case of 100 ΩVmax = 666.667 mV and Vmin = 333.333 mVThus, we have

VSWR=V max

V min=666.667mV333.333mV

=2

Replace the load with a “short circuit,” namely 0.001 Ω.Question 15: Plot the magnitude of the voltage at Input. From the plot, find the VSWR. From equations (2.6) and (2.7) calculate the VSWR. Do these two results agree?

From the plot Vmax=1, Vmin=0

VSWR=V max

V min=1mV0mV

=∞

From equation(2.6) (2.7)

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Γ=ZL−Z0ZL−Z0

=0.001Ω−50Ω0.001Ω+50Ω

≃−1

Thus, VSWR=1+¿ Γ∨ ¿

1−¿ Γ∨¿=1+11−1

=쨨

Replace the load with an “open circuit,” namely 1 MΩ. (remember that in PSPICE, MEG = “mega”, M = “milli”)Question 16: Plot the magnitude of the voltage at Input. Find the VSWR. Also, calculate the VSWR. Do these two results agree?

From the plot Vmax=1, Vmin=0

VSWR=V max

V min=1mV0mV

=∞

From equation(2.6) (2.7)

Γ=ZL−Z0ZL−Z0

=106Ω−50Ω

106Ω+50Ω≃1

Thus, VSWR=1+¿ Γ∨ ¿

1−¿ Γ∨¿=1+11−1

=쨨

Question 17: How are the plots from Question 15 and Question 16 similar? How are these two impedances related on the Smith Chart?The frequency and the amplitude in question 15 are the same as those in the question 16

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2.4 Power Calculations

Question 18: Using the equations above and below (and possibly equations from the book), calculate V+, V-, Vin and VL. How do the calculated Vin and VLcompare to your Pspice results?l=1.65m, f=200MHz

We have

Γ=ZL−Z0ZL−Z0

=100Ω−50Ω100Ω+50Ω

=13

And

Z¿=Z01+Γ Lexp(2 jβl)1−Γ Lexp (2 jβl)

=501+ 13exp (2 j 2π 1.65)

1−13exp (2 j 2π 1.65)

=33.74+ j 24.07

Hence,

V ¿=V gZ¿

Z¿+Zg=1 33.74+ j24.07

(33.74+ j 24.07 )+50=0.45+ j 0.16

So, |Vin| = 0.477 V = 477 mV

V+¿=V ¿

1exp (− jβl )+Γ exp ( jβl)

=(0.45+ j0.16) 1exp (−2 j2π1.65 )+1/3exp (2 j2π 1.65)

¿

Hence,

V +¿=−0.50+ j 0.38¿

So, |V+| = 0.629 V = 629 mV

PSPICE plot

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We can see that Vi measure = 666.667 mV and Vin measure = 475.626 mV. Thus, Vi and Vin approximate to the PSPICE result.

Question 19: Calculate Pav (average power delivered to the load), Piav and Pr

av (average power in the incident and reflected waves, respectively). (Equations for Pav, Pi

av, and Prav are listed in Chapter 2 of the

text.) Show that conservation of power been satisfied.Pav

i =¿¿¿¿

Pavr =−|Γ L

2|Pavi =−0.44mW

So

Pav=Pavi +Pav

r =3.96mW−0.44mW=3.52mWQuestion 20: Use SPICE to find the magnitude of V+and V- at Input. Use these values to compute Pi

av and Pr

av. Do these match your answers in Questions 18 and 19?