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# 09 a lab3-nguyen thi bao tram

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### 09 a lab3-nguyen thi bao tram

1. 1. TRƯỜNG ĐẠI HỌC BÁCH KHOA ĐÀ NẴNG KHOA ĐIỆN TỬ – VIỄN THÔNG  BÁO CÁO THÍ NGHIỆMKỸ THUẬT SIÊU CAO TẦNLAB 3: Transients on Transmission Lines Student : Nguyễn Thị Bảo Trâm Group : 09A Class : 06DT1 Đà Nẵng – 2010 dddDDDASARTETE
2. 2. Báo cáo TN SIÊU CAO TẦN Lab31. Introduction: We have thus far focused on techniques for understanding transmission lines under sinusoidalexcitation. Powerful analytic insight is available here, permitting straightforward design ofinteresting circuits. In contrast, the analysis of transients is generally more difficult and less amenable to simpleclosed-form analysis, especially when loads are reactive. This laboratory will explore the propagationof transients on transmission lines with the aid of numerical experiments in SPICE.2. Purely Resistive Termination: First, lets use SPICE to investigate the propagation of pulses on a transmission line terminatedby 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 aThevenin source 10u(t), with a source resistance Rg = 50  . With the load resistance RL = 50  ,run the simulation. Rg T1 Vg Vsource Vload 50 TD = 25ns Vg Z0 = 50 RL 50 0 0 0 0 To create the voltage source, use the VPWL source. Using VPWL allows various times and thevoltages at those times to be specified using the T1, T2, T3, ... and V1, V2, V3, ... parameters. Thevoltage source will be piece-wise linear, connecting each specified point. Using VPULSE allowsspecification 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 timesusing T1, T2, T3… and the corresponding V1, V2, V3 parameters. Please note that the source willconnect each specified point in the most direct way. Additionally, two voltages cannot be specifiedfor 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 andlasting 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 thereasons stated above.)Nguyễn Thị Bảo Trâm – 06DT1 Page 2
3. 3. Báo cáo TN SIÊU CAO TẦN Lab3 Using VPULSE allows specification of an initial voltage, V1, the voltage of the pulse, V2, the delaytime, TD, the rise time, TR, the fall time, TF, the pulse width, PW, and the period, PER. Singlepulses can be formed with this source. More complex signals can be formed by combining multiplesources.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 showswhat you would expect to see in the “exact” answer. Do the two agree? If not, why not? A fullcredit answer will describe the bounce diagram until a reasonable (whatever you considerreasonable) 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 10V 5V 0V 0s 5ns 10ns 15ns 20ns 25ns 30ns 35ns 40ns 45ns 50ns V(VG) V(VSOURCE) V(VLOAD) Time The reflection coefficient at the source is: 𝑅 𝑔 − 𝑍0 50 − 50 Γ𝑔 = = =0 𝑅 𝑔 + 𝑍0 50 + 50 The reflection coefficient at the load is: 𝑅 𝐿 − 𝑍0 50 − 50 Γ𝐿 = = =0 𝑅 𝐿 + 𝑍0 50 + 50 ⇒ There is no reflection wave of voltage at the source and the load. 𝑉𝑔 𝑍 𝐿 10 .50 𝑉𝐿 = = = 5(𝑉) 𝑅 𝑔 + 𝑍 𝐿 50 + 50Nguyễn Thị Bảo Trâm – 06DT1 Page 3
5. 5. Báo cáo TN SIÊU CAO TẦN Lab3 The reflection coefficient at the load is: RL  Z 0 20  50 3 L    RL  Z 0 20  50 7 Vg Z 0 10.50 V1    5(V ) Rg  Z 0 50  50 3At t = T = 25ns, the wave reaches the load (receiving end) z = l, and because L    0 , the 7mismatch generates a reflected wave with the amplitude: 15 V1  L .V1   (V) 7So the voltage on line (in this case, the load voltage) is the sum of two waves: 15 20VL  V1  V1  5    2.86(V ) 7 7At t = 2T = 50ns, the wave reaches the sending end z = 0, and because g  0 , there is noreflection wave in form of a wave with voltage amplitude V2+ = 0. Thus, the voltage on line (in thiscase, the source voltage) is: 15 20VS  V1  V1  5    2.86(V ) 7 7 3At t = 3T = 75ns, the wave reaches the load (receiving end) z = l, and because L    0 , the 7mismatch generates a reflected wave with the amplitude: V2  L .V2  0 (V)  VL = 2.86 (V)Similarly, we also have: VS = 2.86 (V)   g   L z=l z=0 t=0 V1+ T t=0 L V1+ 2T L g V1+ 3TThus, after 50ns, the voltage at two ends of the transmission line have the same final answers.Nguyễn Thị Bảo Trâm – 06DT1 Page 5
7. 7. Báo cáo TN SIÊU CAO TẦN Lab3 Vg Z 0 10.50V1    2(V ) Rg  Z 0 200  50 3At t = T = 25ns, the wave reaches the load (receiving end) z = l, and because L    0 , the 7mismatch generates a reflected wave with the amplitude: 3 6V1  L .V1  ( ) .2    0.86 (V) 7 7So the voltage on line (in this case, the load voltage) is the sum of two waves: 6 8VL  V1  V1  2    1.14(V ) 7 7At t = 2T = 50ns, the wave reaches the sending end z = 0, and because g  0.6 , there is a reflection wave in form of a wave with voltage amplitude V2+ = gV1 . Thus, the voltage on line (inthis case, the source voltage) is:VS  V1  V1  V2  (1  L  L  g ).V1  0.63(V ) 3At t = 3T = 75ns, the wave reaches the load (receiving end) z = l, and because L    0 , the 7mismatch generates a reflected wave with the amplitude: V2  L .V2 (V)  VL = 0.85 (V)   g   L z=0 V1+ z=l t=0 t=0 T  L V1 + 2T L g V1+ 3T L  2 + g V1 4T L g V1+ 2 2 5TNguyễn Thị Bảo Trâm – 06DT1 Page 7
8. 8. Báo cáo TN SIÊU CAO TẦN Lab3 Calculate like the process above, we have the table: Time V at sending end V at receiving endt = 0ns Vg Z 0 10.50 VL  0(V ) VS = V1    2(V ) Rg  Z 0 200  50t = T =25ns Vg Z 0 10.50 6 8 VS = V1    2(V ) VL  V1  V1  2    1.14(V ) Rg  Z 0 200  50 7 7t = 2T = 50ns VS  V1  V1  V2 6 8 VL  V1  V1  2    1.14(V )  (1  L  L  g ).V1  0.63(V ) 7 7t = 3T=75ns VS  V1  V1  V2 VL  (1  L  L  g L g ).V1  0.85(V ) 2  (1  L  L  g ).V1  0.63(V )t = 4T=100ns VS  (1  L  L  g L g  L g ).V1  0.98(V ) L  (1  L  L  g L g ).V1  0.85(V ) 2 2 2 V 2t = 5T =125ns VS  (1  L  L  g L g 2 VL  (1  L  L  g L g  L g2 2 2  L g2 ).V1  0.98(V ) 2  L g2 ).V1  0.92(V ) 3T = 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. Nguyễn Thị Bảo Trâm – 06DT1 Page 8
9. 9. Báo cáo TN SIÊU CAO TẦN Lab3 Vg Z L 10.20 - The final voltage is: V    0.909(V ) Rg  Z 0 200  20 ⇒ It takes 225ns to settle down to the final answer.2.4 A short pulseNow break the transmission line into two equal pieces, with total length 25 ns (you should now havetwo different transmission lines, both with the same 50  characteristic impedance, but each with atime delay of only 12.5 ns). This permits us to sample “inside” the transmission line. With the sametransmission line, and Rg=200  and RL=20  apply a pulse of duration 10 ns to the transmissionline, namely vg(t) = 10(u(t)- u(t-10ns)). Rg T1 T2 Vg Vsource Vmiddle Vload 200 Vg Z0 = 50 Z0 = 50 TD = 12.5ns TD = 12.5ns RL 20 0 0 0 0 0 0Problem 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 loadend? How large is the “ghost” pulse?Answer: - The voltage at the source, middle, and load ends of the transmission lines for t = 0…100ns 2.0V 1.0V 0V -1.0V -2.0V 0s 10ns 20ns 30ns 40ns 50ns 60ns 70ns 80ns 90ns 100ns V(VSOURCE) V(VMIDDLE) V(VLOAD) TimeNguyễn Thị Bảo Trâm – 06DT1 Page 9
10. 10. Báo cáo TN SIÊU CAO TẦN Lab3 Rg  Z 0 200  50 g    0.6 Rg  Z 0 200  50 RL  Z 0 20  50 3 L    RL  Z 0 20  50 7 Vg Z 0 10.50 V1    2(V ) Rg  Z 0 200  50 - The bounce diagram:   g   L z l 2 z=0 V1+ z=l t=0 T L V1 + 2T L g V1+ 3T L  2 + g V1 4T L g V1+ 2 2 5T - We have the table of values (by calculating): Time V at sending end V middle V at receiving endt = 0+ ns Vg Z 0 Vm = 0 (V) VL = 0 (V) VS  V1   2(V ) Rg  Z 0t = T/2 Vg Z 0 0.50 Vm = V1+ (t = 0ns) = VL = 0 (V)=12.5ns VS   0 2(V) Rg  Z 0 200  50 Nguyễn Thị Bảo Trâm – 06DT1 Page 10
11. 11. Báo cáo TN SIÊU CAO TẦN Lab3t = T = 25ns VS = 0(V) Vm = VS (t = 12.5ns) = VL  V1  V1 0(V) 6 8  2   1.14(V ) 7 7t = 3T/2 = VS = 0(V) Vm  V1  LV1   VL  Vm  Vm37.5ns (Vm at t = 2T) 6  0(V )    0.86(V ) 7t = 2T = 50ns VS  V1  V2 Vm  VL  0(V ) VL = 0 (V)  (L  L  g ).V1  1.37(V )t = 5T/2 = Vg Z 0 Vm  V2  L gV1 VL = 0 (V)62.5ns VS  Rg  Z 0 3    0.6  2  0.51(V ) 0.50 7   0(V ) 200  50t = 3T=75ns VS = 0(V) Vm = VS (t = 12.5ns) = VL  V2  V2 0(V)  (L  g L g ).V1  0.29(V ) 2t= VS = 0(V) Vm  V2  L gV1 2 VL = 0 (V)7T/2=82.5ns  0.22(V )t = 4T =100ns VS  V2  V3 Vm  VL  0(V ) VL = 0 (V)  (L g  L g2 ).V1 2 2  0.35(V ) It takes 12.5ns for the “ghost” pulse to get the load end. The ghost pulse has width of 12.5ns. Nguyễn Thị Bảo Trâm – 06DT1 Page 11
12. 12. Báo cáo TN SIÊU CAO TẦN Lab32.5 A longer pulseIn 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 forwhich vg = +10 V for t = 0 … 20 ns, and vg   V for t = 20 … 40ns. Plot the voltage at thesource, 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: Rg T1 T2 Vg Vsource Vmiddle Vload 200 Vg Z0 = 50 Z0 = 50 TD = 12.5ns TD = 12.5ns RL 20 0 0 0 0 0 0 - The voltage at the source, center, and load end of the transmission line for t = 0…100ns 2.0V 0V -2.0V -4.0V 0s 10ns 20ns 30ns 40ns 50ns 60ns 70ns 80ns 90ns 100ns V(VSOURCE) V(VMIDDLE) V(VLOAD) Time - From the graph, we can see that the transition from high to low is perfectly clear at the load end.2.6 An Impedance BumpTransmission lines must be protected against damage, or their impedance properties could becompromised. In this section we’ll “damage” the transmission line by putting a weak load in theNguyễn Thị Bảo Trâm – 06DT1 Page 12
13. 13. Báo cáo TN SIÊU CAO TẦN Lab3middle. In particular, at the center of the transmission line, add a shunt resistance of 50  (a shuntresistance connects the node at the middle to ground). Rg T1 T2 Vg Vsource Vmiddle Vload 200 Vg Z0 = 50 Z0 = 50 TD = 12.5ns Rshunt TD = 12.5ns RL 50 20 0 0 0 0 0 0 0Problem 6 With the shunt resistance in place, repeat the previous problem. How did thepresence 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: 2.0V 0V -2.0V -4.0V 0s 10ns 20ns 30ns 40ns 50ns 60ns 70ns 80ns 90ns 100ns V(VSOURCE) V(VMIDDLE) V(VLOAD) Time - The presence of the impedance “bump” (break in the transition line) reduced the magnitude of the middle voltage.3. Reactive TerminationAs 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 loadNguyễn Thị Bảo Trâm – 06DT1 Page 13
14. 14. Báo cáo TN SIÊU CAO TẦN Lab3For 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. Rg T1 Vg Vsource Vload 25 Vg TD = 25ns Z0 = 50 CL 0 0 1n 0 0Problem 7 Terminate the transmission line with a 1nF capacitor. Plot the voltage at the sourceand load ends of the transmission line for t = 0…400ns. If you see any “exponential” charging ordischarging, estimate the time constant, and solve for the R. You may use the following formulas. 15V 10V 5V 0V 0s 50ns 100ns 150ns 200ns 250ns 300ns 350ns 400ns V(VSOURCE) V(VLOAD) Time Vinitial = 0V VFinal = 12.133V V(t) = 8.4292V t = (75 – 25) = 50nsNguyễn Thị Bảo Trâm – 06DT1 Page 14
15. 15. Báo cáo TN SIÊU CAO TẦN Lab3 −𝑡 −50. 10−9 𝜏= = = 4.214. 10−8 𝑉(𝑡) − 𝑉𝑓𝑖𝑛𝑎𝑙 8.4292 − 12.133 𝑙𝑛 𝑉 𝑙𝑛 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑉𝑓𝑖𝑛𝑎𝑙 0 − 12.133 𝜏 4.214. 10−8 𝜏 = 𝑅𝐶 ⇒ = 𝑅= = 42.14(Ω) 𝐶 10−9Problem 8 Repeat the previous problem, but with a load capacitance of 100pF. Rg T1 Vg Vsource Vload 25 Vg TD = 25ns Z0 = 50 CL 0 0 100pF 0 0 15V 10V 5V 0V 0s 50ns 100ns 150ns 200ns 250ns 300ns 350ns 400ns V(VSOURCE) V(VLOAD) Time Vinitial = 0V VFinal = 13.33V V(t) = 13.244V t = (50 – 25) = 25ns −𝑡 −25. 10−9 𝜏= = = 4.956. 10−9 𝑉(𝑡) − 𝑉𝑓𝑖𝑛𝑎𝑙 13.244 − 13.33 𝑙𝑛 𝑉 𝑙𝑛 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑉𝑓𝑖𝑛𝑎𝑙 0 − 13.33 −9 𝜏 4.956. 10 𝜏 = 𝑅𝐶 ⇒ 𝑅= = = 49.56(Ω) 𝐶 100.10−12Nguyễn Thị Bảo Trâm – 06DT1 Page 15
16. 16. Báo cáo TN SIÊU CAO TẦN Lab3Problem 9 Repeat the previous problem, but with a load capacitance of 10nF. Rg T1 Vg Vsource Vload 25 Vg TD = 25ns Z0 = 50 CL 0 0 10nF 0 0 10V 5V 0V 0s 50ns 100ns 150ns 200ns 250ns 300ns 350ns 400ns V(VSOURCE) V(VLOAD) Time Vinitial = 0V VFinal = 10V V(t) = 8.056V t = (400 – 25)= 375ns −𝑡 −375. 10−9 𝜏= = = 2.29. 10−7 𝑉(𝑡) − 𝑉𝑓𝑖𝑛𝑎𝑙 8.056 − 10 𝑙𝑛 𝑉 𝑙𝑛 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑉𝑓𝑖𝑛𝑎𝑙 0 − 10 −7 𝜏 2.29. 10 𝜏 = 𝑅𝐶 ⇒ 𝑅= = = 22.9(Ω) 𝐶 10.10−9Problem 10 Repeat the previous problem, but with a load inductance of 2.5μH.Nguyễn Thị Bảo Trâm – 06DT1 Page 16
17. 17. Báo cáo TN SIÊU CAO TẦN Lab3 Rg T1 Vg Vsource Vload 25 1 Vg TD = 25ns Z0 = 50 L 2.5uH 0 0 2 0 0 15V 10V 5V 0V 0s 50ns 100ns 150ns 200ns 250ns 300ns 350ns 400ns V(VSOURCE) V(VLOAD) Time Vinitial = 13.202V VFinal = 0V V(t) = 4.8052V t = (75 – 25) = 50ns −𝑡 −50. 10−9 𝜏= = = 4.947. 10−8 𝑉(𝑡) − 𝑉𝑓𝑖𝑛𝑎𝑙 4.8052 − 0 𝑙𝑛 𝑉 𝑙𝑛 13.202 − 0 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑉𝑓𝑖𝑛𝑎𝑙 −6 𝑅 𝐿 2.5. 10 𝜏= ⇒ 𝑅= = = 50.53(Ω) 𝐿 𝜏 4.947. 10−8Problem 11 Repeat the previous problem, but with a load inductance of 0.25μH.Nguyễn Thị Bảo Trâm – 06DT1 Page 17
18. 18. Báo cáo TN SIÊU CAO TẦN Lab3 Rg T1 Vg Vsource Vload 25 1 Vg TD = 25ns Z0 = 50 L 0.25uH 0 0 2 0 0 15V 10V 5V 0V -5V 0s 50ns 100ns 150ns 200ns 250ns 300ns 350ns 400ns V(VSOURCE) V(VLOAD) Time Vinitial = 13.3V VFinal = 0V V(t) = 1.51V t = (36 – 25)ns −𝑡 −11. 10−9 𝜏= = = 5.056. 10−9 𝑉(𝑡) − 𝑉𝑓𝑖𝑛𝑎𝑙 1.51 − 0 𝑙𝑛 𝑉 𝑙𝑛 13.3 − 0 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 − 𝑉𝑓𝑖𝑛𝑎𝑙 −6 𝑅 𝐿 0.25. 10 𝜏= ⇒ 𝑅= = = 49.44(Ω) 𝐿 𝜏 5.056. 10−9Problem 12 Repeat the previous problem, but with a load composed of a parallel combination ofRL=1000  , L = 1μH and C = 100pF.Nguyễn Thị Bảo Trâm – 06DT1 Page 18
19. 19. Báo cáo TN SIÊU CAO TẦN Lab3 Rg T1 Vg Vsource Vload 25 1 Vg TD = 25ns RL L C Z0 = 50 1k 1uH 100pF 0 0 2 0 0 10V 5V 0V -5V 0s 50ns 100ns 150ns 200ns 250ns 300ns 350ns 400ns V(VSOURCE) V(VLOAD) TimeProblem 13 How important is the value of Rg in these exercises?Answer: - R is very important since it allows the system to reach steady state. The value of R helps us g g to determine whether the circuit is matched or not at the input of transmission line.4. CouplingMany data buses are in parallel, in close proximity, such as the 32 bit and 64 bit buses found incomputers. These transmission lines will have “mutual impedance” which causes signals on onetransmission line to show up on another one.In EE571 students analyze this coupling in great detail, but we can simulate a simplified model of atwo-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 anexplicit form of the Telegrapher’s Equation, and is painfully slow. I tried to use it, but realized that it would take aboutone day to run every simulation!Nguyễn Thị Bảo Trâm – 06DT1 Page 19
20. 20. Báo cáo TN SIÊU CAO TẦN Lab3Here T1 and T4 represent the “actual” transmission lines, while T2 and T3 represent the crosscoupling. The coupling is slightly faster (95 ns instead of 100 ns) and has a higher characteristicimpedance.Connect a Thevenin signal pulse (10 V, 50  ) to A and 50  loads to B, C and D2. Z0 = 50 TD = 100ns T1 A C R1 R3 50 0 0 Vg Z0 = 200 50 Vg TD = 95ns T2 0 0 0 0 Z0 = 200 TD = 95ns T3 0 0 Z0 = 50 TD = 100ns T4 B D R4 R2 50 50 0 0 0 02 To make this model even closer to the real thing, the signal injected into T2 should have opposite sign of that injectedinto T1. However, this lab will not require this.Nguyễn Thị Bảo Trâm – 06DT1 Page 20
21. 21. Báo cáo TN SIÊU CAO TẦN Lab34.1 A simple pulseInject 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. Describewhat 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 10V 5V 0V -5V 0s 50ns 100ns 150ns 200ns 250ns V(VG) V(A) V(B) V(C) V(D) Time - V is about 4 times compared to the magnitude of V when the pulse arrives. C D - V arrives first. DProblem 15 When does a signal arrive at B? If you change the values of the load resistances at Cand 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:Nguyễn Thị Bảo Trâm – 06DT1 Page 21
22. 22. Báo cáo TN SIÊU CAO TẦN Lab3 10V 5V 0V -5V 0s 50ns 100ns 150ns 200ns 250ns V(VG) V(A) V(B) V(C) V(D) Time5. Impedance Matching5.1. Pre-lab Assignment5.1. 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: 𝑍 𝐿 = 150Ω 𝑍 𝑖𝑛 = 75ΩThe input impedance is: ZL + jZ0 tgβl 𝑍 𝑖𝑛 = Z0 Z0 + jZL tgβl 𝜆 2𝜋 𝜆 𝜋With 𝑙 = 4 ⇒ 𝛽𝑙 = . 𝜆 4 = 2 πWe can divide the numerator and the denominator by tgβl and take the limit as βl → to get: 2 𝑍0 2 𝑍 𝑖𝑛 = 𝑍𝐿The characteristic impedance of the matching section is: ⇒ 𝑍0 = 𝑍 𝐿. 𝑍 𝑖𝑛 = 150.75 = 106.066(Ω)Nguyễn Thị Bảo Trâm – 06DT1 Page 22
23. 23. Báo cáo TN SIÊU CAO TẦN Lab35.1. 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 opencircuited stub. State the length of your transmission line(s) in terms of wavelength.Answer: 𝑍0 = 75Ω 𝑍 𝑖𝑛 = 75Ω 𝑍 𝐿 = 150Ω ZL 150 - The normalized load impedance is: 𝑧 𝐿 = = = 2Ω (point A on the Smith chart) Z0 75 - 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 Assignment5.2.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 sweepingthe 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 ofNguyễn Thị Bảo Trâm – 06DT1 Page 23
24. 24. Báo cáo TN SIÊU CAO TẦN Lab3the matching network and find the bandwidth where |Γ| is less than 0.2. What is the inputimpedance of the matching network at 1 GHz?Answer: mup = 2.108 ; f = 1GHz s 2.108 up   0.2(m) f 10 9  0.2l   0.05 (m) 4 4 l 0.05 mTD   8  2.5.10 10 ( s)  0.25ns u p 2.10 m / s Rs T1 Vinput Vload 75 Vs 2.5Vac TD = 0.25ns RL 0Vdc Z0 = 106.066 150 0 0 0 0 - The input and load voltage over frequency:Nguyễn Thị Bảo Trâm – 06DT1 Page 24
25. 25. Báo cáo TN SIÊU CAO TẦN Lab3 1.8V 1.6V 1.4V 1.2V 1.0MHz 3.0MHz 10MHz 30MHz 100MHz 300MHz 1.0GHz 3.0GHz 10GHz V(VINPUT) V(VLOAD) Frequency - The magnitude of the reflection coefficient of the matching network: 400m 300m 200m 100m 0 1.0MHz 3.0MHz 10MHz 30MHz 100MHz 300MHz 1.0GHz 3.0GHz 10GHz (V(VINPUT)/ I(Rs)-75)/(V(VINPUT)/I(Rs)+75) Frequency |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.Nguyễn Thị Bảo Trâm – 06DT1 Page 25
26. 26. Báo cáo TN SIÊU CAO TẦN Lab3 150 100 (1.0000G,75.000) 50 1.0MHz 3.0MHz 10MHz 30MHz 100MHz 300MHz 1.0GHz 3.0GHz 10GHz V(VINPUT)/ I(Rs) Frequency5.2.B. Using the assumptions of section 5.2.A, simulate your pre-lab design in part 5.1.B and findthe Γ bandwidth (the portion of the signal less than 0.2) of the matching network as well as theinput impedance at 1 GHz. Include the same plots as in section 5.2.A.Answer: mup = 2.108 ; f = 1GHz s up 2.108   0.2(m) f 10 9 d1 0.0304 md1 =0.152𝜆 = 0.152×0.2 = 0.0304 (m) ⇒ TD    1.52.10 10 ( s)  0.152ns u p 2.108 m / s l o1 0.0804 mlo1= 0.402𝜆 = 0.402×0.2 = 0.0804 (m) ⇒ TD    4.02.10 10 ( s)  0.402ns u p 2.108 m / s l s1 0.0304 mls1=0.152𝜆 = 0.152×0.2 = 0.0304 (m) ⇒ TD    1.52.10 10 ( s)  0.152ns u p 2.108 m / s  Open-circuited stub:Nguyễn Thị Bảo Trâm – 06DT1 Page 26
27. 27. Báo cáo TN SIÊU CAO TẦN Lab3 Rs T1 Vinput Vload 75 Vs 2.5Vac TD = 0.152ns RL 0Vdc Z0 = 75 150 0 0 T2 0 0 TD = 0.402ns R1 Z0 = 75 1000G 0 0 0 - The input and load voltage over frequency: 2.0V 1.5V 1.0V 0.5V 0V 1.0MHz 3.0MHz 10MHz 30MHz 100MHz 300MHz 1.0GHz 3.0GHz 10GHz V(VINPUT) V(VLOAD) Frequency - The magnitude of the reflection coefficient of the matching network: |Γ| is less than 0.2 from 924 MHz to 1.11 GHz.Nguyễn Thị Bảo Trâm – 06DT1 Page 27
28. 28. Báo cáo TN SIÊU CAO TẦN Lab3 1.0 0.5 0 1.0MHz 3.0MHz 10MHz 30MHz 100MHz 300MHz 1.0GHz 3.0GHz 10GHz (V(VINPUT)/I(Rs)-75)/(V(VINPUT)/I(Rs)+75) Frequency - The input impedance of the matching network at 1 GHz is 75.029Ω. 150 100 (1.0000G,75.029) 50 0 1.0MHz 3.0MHz 10MHz 30MHz 100MHz 300MHz 1.0GHz 3.0GHz 10GHz V(VINPUT)/ I(Rs) Frequency  Shorted-circuited stub:Nguyễn Thị Bảo Trâm – 06DT1 Page 28
29. 29. Báo cáo TN SIÊU CAO TẦN Lab3 Rs T1 Vinput Vload 75 Vs 2.5Vac TD = 0.152ns RL 0Vdc Z0 = 75 150 0 0 T2 0 0 TD = 0.152ns Z0 = 75 0 0 0 - The input and load voltage over frequency: 2.0V 1.5V 1.0V 0.5V 0V 1.0MHz 3.0MHz 10MHz 30MHz 100MHz 300MHz 1.0GHz 3.0GHz 10GHz V(VINPUT) V(VLOAD) Frequency - The magnitude of the reflection coefficient of the matching network:Nguyễn Thị Bảo Trâm – 06DT1 Page 29
30. 30. Báo cáo TN SIÊU CAO TẦN Lab3 |Γ| is less than 0.2 from 850 MHz to 1.21 GHz and from 2.07 GHz to 2.44 GHz 1.0 0.5 0 1.0MHz 3.0MHz 10MHz 30MHz 100MHz 300MHz 1.0GHz 3.0GHz 10GHz (V(VINPUT)/I(Rs)-75)/(V(VINPUT)/I(Rs)+75) Frequency - The input impedance of the matching network at 1 GHz is 75.029Ω. 100 (1.0000G,75.029) 50 0 1.0MHz 3.0MHz 10MHz 30MHz 100MHz 300MHz 1.0GHz 3.0GHz 10GHz V(VINPUT)/ I(Rs) FrequencyProblem 16 Compare the results for the matching networks that you designed in the lab (quarterwave, 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.Nguyễn Thị Bảo Trâm – 06DT1 Page 30
31. 31. Báo cáo TN SIÊU CAO TẦN Lab3Problem 17 For stub matching networks compare the performance of an open stub versus ashorted 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 carryout 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.Nguyễn Thị Bảo Trâm – 06DT1 Page 31