BÁO CÁO THÍ NGHIỆM KỸ THUẬT SIÊU CAO TẦN

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In this laboratory experiment, you will use SPICE to study sinusoidal waves on lossless transmission lines. Our goal is for you to become familiar with the basic behavior of waves reflecting from loads in transmission lines, and compare the simulations with numeric calculations and the Smith Chart.

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Nội dung Text: BÁO CÁO THÍ NGHIỆM KỸ THUẬT SIÊU CAO TẦN

TRƯỜNG ĐẠI HỌC BÁCH KHOA ĐÀ NẴNG KHOA ĐIỆN TỬ - VIỄN THÔNG  BÁO CÁO THÍ NGHIỆM KỸ THUẬT SIÊU CAO TẦN LAB 2: Basic Transmission Lines in the Frequency Domain Student : Nguyễn Thị Bảo Trâm Nguyễn Văn Hiếu Group : 09A Class : 06DT1 Đà Nẵng - 2010
In this laboratory experiment, you will use SPICE to study sinusoidal waves on lossless transmission lines. Our goal is for you to become familiar with the basic behavior of waves reflecting from loads in transmission lines, and compare the simulations with numeric calculations and the Smith Chart. 2.1 Basic Transmission Line Model There is a standard lossless transmission line model T, which is specified by several parameters. We will need to specify two of the parameters:  Z0, the characteristic impedance  TD, the time delay, which is the length of the line in time units. The length of the line L is related to the time delay through Lu Tp D (2.1) where up is the phase velocity of waves on the transmission line. As we saw in lecture and in our text, the phase velocity and characteristic impedance may be derived from the “lumped element” model of the transmission line. With L’ the inductance per unit length, and C’ the capacitance per unit length, we have up  1 (2.2) L'C' L' Z0  (2.3) C' 2.1.1 A standard coaxial cable For common RG-58 coaxial cable, the characteristic impedance is Z 0 = 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? Answer: - Transmission line is often schematically represented as a twowire line, so transmission lines (TEM wave propagation) always have at least two conductors. i(z,t) + v(z,t) - ∆z z Inductance per meter ( H/m ) : + It is the series inductance per unit length, it appears from the shape of transmission line. Inductance per meter represents the selfinductance of the two conductors per a meter, it also represents the stored magnetic energy per a meter of transmission line. - Capacitance per meter ( F/m ) : + It is the shunt capacitance per unit length, it is due to the close proximity of the two conductors, it also represents the stored electric energy per a meter of transmission line. Nguyễn Thị Bảo Trâm - Nguyễn Văn Hiếu - 06DT1 Page 2
The inductance and capacitance per meter are: Z L' We have: 0  L'. C'  L' C' u Z0 50 () 9 p L' up  2  250.10 (H/m)  250 (nH/m) 8 3 .3.10 (m/s) And: Z0.u p  L' 1 1  1 C' L'. C' C' 1 9 C'  2  0,1.10 (F/m)  0,1 (nF/m) Z0u p 50(). .3.108(m/s) 3 For lossless coaxial cables, the following formulas relate the differential inductance L’ and capacitance C’ to the radius of the inner conductor a and the outer conductor b: L'  b  2 ln  (2.4)  a C' 2 (2.5)  b  ln   a Question 2: For a different coaxial cable, μ = μ0 and ε = 3ε0. What is b/a if Z0 = 50 Ω? Answer: We can see : L'  b  b 1  2  b  ln .ln   2 ln   C' 2  a   a  2 4  a  2  b 4 2  L'  ln   a   C' 10 9 2.. 3. (F/m)  b 2  L' 2 3 0 36 2 3.10 9  ln    .Z0  .50()  .50 a   C'  0 4.10 7 (H/m) 4 2.36.107 2 3.109 35 3  7 .50  .50  2.6 10 60 6 5 3 b 6 So, a e  4,235
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Question 3: If b = 3 mm in question 2.2, what is a? Answer: If b = 3mm in question 2.2 then: b 3 (mm) a   0,708 (mm) 4,235 4,235 2.2 A SPICE model of a transmission line problem. Using SPICE, create a (matched) Thevenin source VAC with 1 Volt amplitude and 50 Ω source impedance, leading to a transmission line model T, terminated in a 100 Ω load. Edit the transmission line so that it has a characteristic impedance of 50 Ω. Also, create labels Input and Load at the ends of the transmission lines, so that you can measure the voltages conveniently. ZG T1 ZL Input Load 50 100 1Vac Z0 = 50 0Vdc VG TD = {delay } PARAMETERS: delay = 5ns 0 0 0 0 Figure 1. Circuit Schematic for Part 2.2 What we would like to do is to adjust the length of the transmission line and examine the standing wave pattern at Input over one full wavelength at a frequency of 200MHz. Question 4: At 200 MHz, and with u p = 2/3 c, what is the wavelength in the transmission line? Answer: The wavelength in the transmission line is: 2 2   3 3×3.108 / ߣ= = 1(  ) = = 200.106 Question 5: What is the time delay associated with λ/16? (Hint: Remember that TD  L  L )  f up Answer: The time delay associated with ߣ/16 is: 16 1 1 L L 10 TD      16f 16.200.10 Hz  3.12510 (s)  0.3125(ns) 6  f f up 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.
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Figure 2. Illustration of Transmission Line Length Change for Part 2.2 One way to make this easier is to use a parameter for TD. Place the special part PARAM. Double click on it and then on New Column… Call it delay and set it to 5ns. Assign {delay} (with the curly braces) to TD on the transmission line. When you create your simulation profile, select the parametric sweep as an option. Choose Global Parameter with a parameter of delay. Set the sweep range and increment based on your TD calculations from above. Under “General Settings” set the sweep Range from Start Frequency: 200Meg to End Frequency: 200Meg and increment Total Points: 1. Nguyễn Thị Bảo Trâm - Nguyễn Văn Hiếu - 06DT1 Page 5
Using Excel, make a table of the voltage magnitudes and current magnitudes at nodes Input and Load for each length. Line Time Input Input Load Load Input No. Length Delay Voltage Current Voltage Voltage Impedance (m) (ns) (mV) (mA) (mV) (mA) (Ω) 0 0.0000 0.0000 666.667 6.667 666.667 6.667 100.00 + j 00.000 1 0.0625 0.3125 628.990 7.998 666.667 6.667 69.476 - j 36.845 2 0.1250 0.6250 527.046 10.541 666.667 6.667 40.000 - j 30.000 3 0.1875 0.9375 399.908 12.580 666.667 6.667 28.085 - j 14.894 4 0.2500 1.2500 333.333 13.333 666.667 6.667 25.000 - j 00.000 5 0.3125 1.5625 399.908 12.580 666.667 6.667 28.085 + j 14.894 6 0.3750 1.8750 527.046 10.541 666.667 6.667 4.000 + j 03.000 7 0.4375 2.1875 628.990 7.998 666.667 6.667 69.476 + j 36.845 8 0.5000 2.5000 666.667 6.667 666.667 6.667 100.00 + j 00.000 9 0.5625 2.8125 628.990 7.998 666.667 6.667 69.476 - j 36.845 10 0.6250 3.1250 527.046 10.541 666.667 6.667 40.000 - j 30.000 11 0.6875 3.4375 399.908 12.580 666.667 6.667 28.085 - j 14.894 12 0.7500 3.7500 333.333 13.333 666.667 6.667 25.000 - j 00.000 13 0.8125 4.0625 399.908 12.580 666.667 6.667 28.085 + j 14.894 14 0.8750 4.3750 527.046 10.541 666.667 6.667 4.000 + j 03.000 15 0.9375 4.6875 628.990 7.998 666.667 6.667 69.476 + j 36.845 16 1.0000 5.0000 666.667 6.667 666.667 6.667 100.00 + j 00.000 max 666.667 13.333 666.667 6.667 min 333.333 6.667 666.667 6.667 Nguyễn Thị Bảo Trâm - Nguyễn Văn Hiếu - 06DT1 Page 6
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: VSWR  V , max determine the VSWR, and from the VSWR calculate ||. V min Answer: - To examine the change of input voltage as a function of length, we can examine the change of input voltage as a function of Time Delay. We can simulate the input voltage as a function of Time Delay because Time Delay and Length of transmission line relate together from the formular : TD L  . With up is a constant, L up increases nfold as well as TD increases nfold. So, examining the change of the input voltage as a function of TD is like doing this with L (length of transmission line). - With L = λ, we have : TD  L  L  1 1 9    6  5.10 (s) 5 (ns) u p .f .f f 200.10 So, we can establish a parameter in pspice with 0 at start value and 5ns at end value. In addition, we examine L at the points which are λ/16equidistant together. Therefore, increment in parametric sweep is 0,3125ns like above value (question 5). Use PSPICE to plot the magnitude of the voltage at Input as a function of length: 720mV (2.5000n,666.667m) 600mV (1.2500n,333.333m) 400mV Length 0m 0.5m 1m 300mV 0 0.5n 1.0n 1.5n 2.0n 2.5n 3.0n 3.5n 4.0n 4.5n 5.0n M(V(INPUT)) delay V max From the Voltage Values on the plot and the relationship: VSWR  , determine the V min VSWR, and from the VSWR calculate ||: Vmax = 666.667mV Vmin = 333.333mV Nguyễn Thị Bảo Trâm - Nguyễn Văn Hiếu - 06DT1 Page 7
  666.667  =  333.333   = 2  = 2−1 1 Γ = − 1 + 1 = 2+1= 3≈0.3333 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 ||. Do the voltage and current yield the same VSWR and ||? Answer: Use PSPICE to plot the magnitude of the current at Input as a function of length: 15mA (1.2500n,13.333m) 10mA (2.5000n,6.6 66 7m ) Length 0m 0.5m 1m 5mA 0 0.5n 1.0n 1.5n 2.0n 2.5n 3.0n 3.5n 4.0n 4.5n 5.0n I(ZG) delay From the Current Values on the plot, determine the VSWR, and from the VSWR calculate ||: Imax = 13.3333mA Imin = 6.6667mA Imax 13.3333m VSWR A = = Imin 6.6667mA = 2 VSWR − 1 2−1 1 Γ= VSWR + 1 = 2+1= 3≈0.3333 The voltage and current yield the same VSWR and || 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. Answer:
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The magnitude of the impedance at Input as a function of length: 100 80 60 40 Length 0m 0.5m 1m 20 0 0.5n 1.0n 1.5n 2.0n 2.5n 3.0n 3.5n 4.0n 4.5n 5.0n M(V(INPUT)/I(ZG)) delay Plot the Real Part of the Impedance using PSPICE: 100 80 60 40 Length 0m 0.5m 1m 20 0 0.5n 1.0n 1.5n 2.0n 2.5n 3.0n 3.5n 4.0n 4.5n 5.0n R(V(INPUT)/I(ZG)) delay Nguyễn Thị Bảo Trâm - Nguyễn Văn Hiếu - 06DT1 Page 9
Plot the Imaginary Part of the Impedance using PSPICE: 40 20 0 -20 Length 0m 0.5m 1m -40 0 0.5n 1.0n 1.5n 2.0n 2.5n 3.0n 3.5n 4.0n 4.5n 5.0n IMG(V(INPUT)/I(ZG)) delay Question 9: Using the scales at the bottom of the Smith Chart, find the VSWR and ||. Do they agree with your previous answers? Answer: From the Smith Chart: Yes, these answers agree with my previous answers: VSWR = 2 and || = 1/3. Question 10: Compute  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  (2.6) 1  ZLZ 0  (2.7) ZLZ 0 Nguyễn Thị Bảo Trâm - Nguyễn Văn Hiếu - 06DT1 Page 10
Answer:  Z L  Z0 10050 1    0.3333 Z L  Z0 10050 3 1 VSWR  1 Γ 1 3   2 1 Γ 1 1 3 These answers agree with my measurements from Questions 6, 7, and 8. 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? Answer: Plot the voltage magnitude at Load as a function of length: 666.6666900mV 666.6666809mV 666.6666709mV Length 0m 0.5m 1m 666.6666609mV 0 0.5n 1.0n 1.5n 2.0n 2.5n 3.0n 3.5n 4.0n 4.5n 5.0n V(LOAD) delay The magnitude of Load Voltage does not change with length. From this, I think the power delivered to the load will not change with length, too. Plot the power at Load as a function of length: Nguyễn Thị Bảo Trâm - Nguyễn Văn Hiếu - 06DT1 Page 11
2.3 A shortcut, and more load impedances SPICE has a nice mechanism for scanning in frequency, but does not directly scan the length of the transmission line. The “electrical length” of a transmission line is βl, l  2 2f l l (2.8)  up Thus, changing the length of a transmission line from l to 10l achieves the same effect as scanning the frequency from 10f to f. Or to put it differently, if a transmission line is 1 λ at f 0, then it is 0.5 λ long at 0.5f0 and 2 λ long at 2f0. 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.) Answer: The coaxial cable in question 4 has frequency f = 200MHz Thus it has length 1ߣ at 1 2  1. 200100 =  = 2 2 And it has length 2.5ߣ at 2.5 = 2.5 × 200500 =  Using a 1meter length of transmission line, adjust your SPICE simulation, sweeping linearly in frequency from 0.5 to 2.5 wavelengths. In this simulation we are not adjusting the Length of the Line. We are adjusting the frequency of the system so as to produce similar effects to adjusting the length of the line. Nguyễn Thị Bảo Trâm - Nguyễn Văn Hiếu - 06DT1 Page 12
ZG T1 ZL Input Load 50 100 1Vac Z0 = 50 0Vdc VG TD = 5ns 0 0 0 0 Figure 3. Circuit Schematic for Question 13 (Fixed Length) 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? Answer: Plot the magnitude of the voltage at Input for the different “lengths” properly relabeling the horizontal axis. 700mV 600mV 500mV 400mV Length 5m 0. 0. 75m 1m 1. 1.5m 1. 75m 2m 2.25m 2.5m 300mV 25m 100MH 150MHz 200MHz 300MHz 350MHz 400MHz 450MHz z 250MHz 500MHz M(V(INPUT)) Frequency This plot agrees with the plot in Question 6. Calculate the VSWR: Vmax = 666.667mV Vmin = 333.333mV Vmax 666.667 VSWR = Vmin mV = 333.333 mV = 2
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Replace the 100 Ω load with a 25 Ω load. ZG T1 ZL Input Load 50 25 VG TD = 5ns 1Vac Z0 = 50 0Vdc 0 0 0 0 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? Answer: Plot the magnitude of the voltage at Input: 700mV 600mV 500mV 400mV Length 0.5m 1m 1.5m 2m 2.5m 300mV 250MHz 300MHz 350MHz 400MHz 450MHz 500MHz 100MHz 150MHz 200MHz M(V(INPUT)) Frequency Calculate the VSWR: Vmax = 666.667mV Vmin = 333.333mV Vmax 666.667mV VSWR Vmin = 333.333mV = 2 = ⇒ VSWR are the same for the 100Ω and 25Ω cases.
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Replace the load with a “short circuit,” namely 0.001 Ω: ZG T1 ZL Input Load 50 0.001 VG TD = 5ns 1Vac Z0 = 50 0Vdc 0 0 0 0 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? Answer: Plot the magnitude of the voltage at Input: 1.0V 0.5V 0V 100MHz 150MHz 200MHz 250MHz 300MHz 350MHz 400MHz 450MHz 500MHz M(V(INPUT)) Frequency Calculate the VSWR: o From the plot: Vmax = 1V Vmin = 0V ⇒ VSWR = V max 1V = = ∞ Vmin 0V o From equations (2.6) and (2.7):  ZLZ 0 0,001() 50()  0,001 ()  50() 1 ZLZ 0 1 Γ 11 ⇒ = = ∞ VSWR= 1 Γ 11 These two results agree. Nguyễn Thị Bảo Trâm - Nguyễn Văn Hiếu - 06DT1 Page 15