Bao cao thi nghiem sieu cao tan

EE361 LAB 4 REPORT June 28, 2013

Group 05

Le Quang Hoa

Vo Quang Tuyen

Bui Van

Part I: Electrical length

Question 1 The transmission line of length l connects a load RL to a sinusoidal voltage source with an oscillatory frequency f as shown in Fig. 1. Assuming the velocity of wave propagation on the line is c, for which of the situations is it reasonable to ignore the presence of the transmission line in the solution of the circuit? Why or why not? (a) f = 60 Hz, l = 50 km (b) f = 20 kHz, l = 20 cm (c) f = 600 MHz, l = 20 cm (d) f = 10 GHz, l = 1 mm

(a) f = 60 Hz, l = 50 km 𝐶

ʎ=𝑓= 𝑙

𝑚 𝑠

3×108 ( ) 60 (𝐻𝑧)

= 5 × 106 (𝑚) = 5000(𝑘𝑚)

50

= 5000 = 0.01  Safe to ignore the presence of transmission line ʎ (b) f = 20 kHz, l = 20 cm 𝐶

ʎ=𝑓= 𝑙 ʎ

=

𝑚 𝑠

3×108 ( ) 20 (𝑘𝐻𝑧)

20×10−2 𝑚 15000𝑚

= 15000(𝑚)

= 1.33 × 10−5 < 0.01 Safe to ignore the presence of

transmission line (c) f = 600 MHz, l = 20 cm 𝐶

ʎ=𝑓= 𝑙 ʎ

=

𝑚 𝑠

3×108 ( )

600 (𝑀𝐻𝑧) 20×10−2 𝑚 0.5𝑚

= 0.5(𝑚)

= 0.4 > 0.01 Not reasonable to ignore the presence of

transmission line EE361 LAB 4 REPORT - JUNE 28, 2013

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(d) f = 10 GHz, l = 1 mm 𝐶

ʎ=𝑓= 𝑙 ʎ

=

𝑚 𝑠

3×108 ( ) 10 (𝐺𝐻𝑧)

1×10−3 𝑚 0.03𝑚

= 0.03(𝑚)

= 0.033 > 0.01 Not reasonable to ignore the presence of

transmission line

Part II: Lumped element modeling of transmission lines

Question 2 The transmission line AB in Fig. 1 is lossless and has a per-unit-length inductance L’= 0.25 uH/m and a per-unit-length capacitance C’= 100 pF/m. The length of the transmission line, l, is 10 cm. (a) Compute the characteristic impedance, Z0, of the line. (b) Compute the velocity of propagation, up. (c) Compute the time delay, TD, of the line. (d) Find the maximum frequency of the excitation source for which the effects of L’ and C’ are negligible.

(a) Compute the characteristic impedance, Z0, of the line. 0.25×10−6 𝐻

𝐿′

Zo = √𝐶′ = √100×10−12 𝐹 = 50 (Ω) Where L’: The combined inductance of both conductors per unit length, in H/m C’: The capacitance of the two conductors per unit length, in F/m (b) Compute the velocity of propagation, up. up =

1 √𝐿′𝐶′

=

1 √0.25×10−6 ×100×10−12

= 2 × 108 (𝑚/𝑠)

(c) Compute the time delay, TD, of the line. TD = EE361 LAB 4 REPORT - JUNE 28, 2013

𝑙 𝑢𝑝

=

10𝑐𝑚 2×108 𝑚/𝑠

=

10×10−2 𝑚 2×108 𝑚/𝑠

= 5×10−10 (𝑠) 2

(d) Find the maximum frequency of the excitation source for which the effects of L’ and C’ are negligible. Effects of L’ and C’ are negligible when: 𝑙 ʎ

≤ 0.01 or

𝑙 2𝜋 ⍵√𝐿′𝐶′

=

10×10−2 1 𝑓√𝐿′𝐶′

= 5 × 10−10 𝑓 ≤ 0.01 .So f

≤ 20 MHz

Question 3 We will model the transmission line in Question 2 using 20 sections of equivalent circuits. Find the capacitance and inductance values in one equivalent circuit.

C =C’× l = 10×10−2 (𝑚) × 100 × 10−12 (𝐹/𝑚) = 10pF L = L’ × l = 10×10−2 (𝑚) × 0.25 × 10−6 (𝐻/𝑚) = 2.5×10−8 𝐻 In one equivalent circuit: 𝐶 10𝑝𝐹 Ci = 20 = 20 = 5×10−13 𝐹 𝐿

Question 4

Li = 20 =

2.5×10−8 𝐻 20

= 1.25×10−9 𝐻

Set up the circuit in Fig. 1, with RS= 50 Ω, RL= 50 Ω, and the transmission line AB using 20 sections model as in Question 3. Use a sinusoidal voltage source with an amplitude of 2 V. We want to analyze the system’s transient response over a range from 0 ns to 5 ns. Run transient simulations for the source frequency of (i) 1 MHz (ii) 10 MHz (iii) 100 MHz (iv) 1 GHz. (a) Submit the circuit schematic. For each case (i) – (iv). (b) Plot the voltages at points A and B (beginning and end of the transmission line). (c) Plot the currents through RS and RL. (d) From the plots, find the maximum voltage difference and maximum current difference within 5 ns.

EE361 LAB 4 REPORT - JUNE 28, 2013

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a) The circuit schematic :

b) Plot the voltages at points A and B: i) 1 MHz

ii)

10 MHz

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iii)

100MHz

iv)

1 GHz

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c) Plot the currents through RS and RL: i)

1 MHz

ii)

10 MHz

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iii)

100 MHz

iv)

1 GHz

d) From the plots, find the maximum voltage difference and maximum current difference within 5 ns. i) 1 MHz The maximum voltage difference: 3.1416 mV The maximum current difference: 1193.626 uA ii) 10 MHz The maximum voltage difference: 31.411 mV The maximum current difference: 11.7601 mA EE361 LAB 4 REPORT - JUNE 28, 2013

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iii)

100 MHz The maximum voltage difference: 309.015 mV The maximum current difference: 39.021 mA 1 GHz The maximum voltage difference: 2 V The maximum current difference: 20 mA

iv)

Part III: KVL and Faraday’s Law

Question 5 (a) Write a KVL and KCL for the circuit in Fig. 1. (b) Compute the electrical length l/𝜆 of the transmission line for the cases (i) – (iv) in Question 4. Is it reasonable to ignore the presence of the transmission line? (c) Referring to the simulations performed in Question 4 (i) – (iv), does KCL/KVL hold for each case? Explain?

(a) Write a KVL and KCL for the circuit in Fig. 1. KVL: Vs = VRs + VRL KCL: IRs = IRL (b) Compute the electrical length l/ʎ of the transmission line for the cases (i) – (iv) in Question 4. (i)

1 MHz 𝑐 2 × 108 (𝑚/𝑠) ʎ= = = 200(𝑚) 𝑓 106 (𝐻𝑧) 𝑙 ʎ

(ii)

10×10−2 200

= 5 × 10−4 ≤ 0.01  Can ignore the presence of TM

10 MHz 𝑐 2 × 108 (𝑚/𝑠) ʎ= = = 20(𝑚) 𝑓 10 × 106 (𝐻𝑧) 𝑙 ʎ

(iii)

=

=

10×10−2 20

= 5 × 10−3 ≤ 0.01  Can ignore the presence of TM

100 MHz 𝑐 2 × 108 (𝑚/𝑠) ʎ= = = 2(𝑚) 𝑓 100 × 106 (𝐻𝑧)

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𝑙

= ʎ (iv)

10×10−2 2

= 0.05 > 0.01  Cannot ignore the presence of TM

1 GHz. 𝑐 2 × 108 (𝑚/𝑠) ʎ= = = 0.2(𝑚) 𝑓 109 (𝐻𝑧) 𝑙

= ʎ

10×10−2 0.2

= 0.5 > 0.01  Cannot ignore the presence of TM

(c) Referring to the simulations performed in Question 4 (i) – (iv), does KCL/KVL hold for each case? Explain? In cases (i) & (ii): KVL & KCL hold. Based on the graph, we see that the voltage at points A and B changes in nearly a straight line from t = 0ns to t = 5ns. Thus, hold true. In cases (iii) & (iv): KVL &KCL is false. On the contrary, from t = 0ns to t = 5ns, the voltages at point A and B change in the shape of sinusoids. Thus there will be

Question 6 (a) Write the Faraday’s Law in integral form. Explain clearly the Faraday’s Law in your own words. (b) Using Faraday’s Law in integral form and a clear diagram, explain the exact relation between KVL and Faraday’s Law. (c) Describe the voltage and current conditions under which KVL break.

(a) Write the Faraday’s Law in integral form. Explain clearly the Faraday’s Law in your own words.

Where: + ∂B/∂t is the rate of change of B, it’s how much B is changing (∂B) in a given time (∂t). + dA is the area inside the closed loop dl.  So the total E around a loop is just equal to the minus of the changing B through the loop. What happens if there is no B? Well there is no ∂B so ∂B/∂t is zero, EE361 LAB 4 REPORT - JUNE 28, 2013

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which makes the integral 0, so no E. What happens if you have a constant B? Well again ∂B is 0. So ∂B/∂t is zero, which makes the integral 0, so again no E. You can only induce an E field from a changing B field. The importance of the minus sign comes from the fact that E fields create B fields and B fields create E fields (As shown in Faradays and Amperes Laws). If the minus wasn’t there then the fields would just keep building and building eventually giving infinite energy, and that is not allowed! (b) Using Faraday’s Law in integral form and a clear diagram, explain the exact relation between KVL and Faraday’s Law.

For any static system: ∮𝑐 𝐸𝑑𝑙 = 0 for any closed path C − ∮𝑐 𝐸𝑑𝑙 = 0 ∑𝑖=1(− ∮𝑐𝑖 −𝐸𝑖𝑑𝑙 ) = 0 for any close path So ∑(∇Vi) = 0  This is KVL (c) Describe the voltage and current conditions under which KVL break. KVL is false for time-varying systems Why? 𝜕𝑩

∇×E = − 𝜕𝑡 ≠0 in general ∫ ∇ × 𝐄 𝐝𝐬 = ∫ − 𝑆

𝑺

𝜕𝑩 ≠𝟎 𝜕𝑡

 ∮𝑐 𝐸𝑑𝑙 ≠ 0

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