3-Transmission Lines A transmission line is a guided structure used to efficiently transmit power or information from the source to the load. Transmission lines are used in power distribution (at low frequencies) and in communications (at high frequencies). Transmission lines basically consist of two or more parallel conductors used to connect a source to a load, example of transmission lines are coaxial cable, two –wire line, parallel plate and microstrip line. Figure 3.1 Distributed parameters of a two conductor transmission line Figure 3.1 Parameters of a short section πΏπ of a transmission line Taking a small portion, an equivalent circuit can be derived as shown above the voltage decreases as the length of the line increases πΏπ£ πΏπ = −(π + πππΏ)πΌ 3.1 The current decreases as the length increases along the line πΏπΌ πΏπ = −(πΊ + πππΆ)π 3.2 25 πΏ2 π£ πΏπ2 πΏπΌ = −(π + πππΏ) πΏπ 3.3 Put 3.2 in 3.3 πΏ2 π£ πΏπ2 = (π + πππΏ)(πΊ + πππΆ)π = Ζ2 π 3.4 Where Ζ is the propagation constant Ζ = √(π + πππΏ)(πΊ + πππΆ) 3.5 And Ζ = πΌ + ππ½, where α is the attenuation constant in Nepers/m and β is the phase constant in radians/m. The voltage V consist of the forward travelling wave and the backward travelling wave represented as π = π΄π −Ζπ + π΅π Ζπ πΏπ£ πΏπ πΏπ£ πΏπ 3.6 = −Ζπ΄π −Ζπ + Ζπ΅π Ζπ 3.7 = −Ζ(π΄π −Ζπ − π΅π Ζπ ) 3.8 Equate 3.8 to 3.3 πΏπ£ πΏπ = −Ζ(π΄π −Ζπ − π΅π Ζπ ) = −(π + πππΏ)πΌ 3.9 Ζ πΌ = π +πππΏ (π΄π −Ζπ − π΅π Ζπ ) Ζ (π +πππΏ)(πΊ+πππΆ) =√ π +πππΏ Ζ π +πππΏ 3.10 (π +πππΏ)2 πΊ+πππΆ = √π +πππΏ πΊ+πππΆ 3.11 1 πΌ = √π +πππΏ ((π΄π −Ζπ − π΅π Ζπ )) = π (π΄π −Ζπ − π΅π Ζπ ) 0 π +πππΏ π0 = √πΊ+πππΆ With the load as reference π = 0 Equation 3.6 becomes 26 3.12 π = π΄ + π΅ = ππ 3.13 And equation 3.12 π= π΄−π΅ π0 = πΌπ 3.14 Is IL ZL Vs Looking from the generator side ππΏ is seen ππΏ = ππΏ πΌπΏ = π΄π −Ζπ +π΅π Ζπ 3.15 1 (π΄π −Ζπ −π΅π Ζπ ) π0 πππ βΖπ‘ = π Ζπ‘ + π −Ζπ‘ 2 π ππβΖπ‘ = π Ζπ‘ − π −Ζπ‘ 2 π Ζπ‘ = πππ βΖπ‘ + π ππβΖπ‘ π −Ζπ‘ = πππ βΖπ‘ − π ππβΖπ‘ Substituting these in 3.15 ππΏ = ππΏ = π΄(πππ βΖπ‘ − π ππβΖπ‘) + π΅(πππ βΖπ‘ + π ππβΖπ‘) π΄ π΅ ( ) π0 πππ βΖπ‘ − π ππβΖπ‘ − π0 (πππ βΖπ‘ + π ππβΖπ‘) (π΄+π΅)πππ βΖπ‘−(π΄−π΅)π ππβΖπ‘ 3.16 π΄−π΅ π΄+π΅ πππ βΖπ‘− π ππβΖπ‘ π0 π0 Substituting 3.13 and 3.14 in 3.16 ππΏ = ππ πππ βΖπ‘−πΌπ π0 π ππβΖπ‘ 3.17 π πΌπ πππ βΖπ‘− π π ππβΖπ‘ π0 ππ πΌπ = ππ Divide numerator and denominator by ππ ππΏ = π πππ βΖπ‘− 0 π ππβΖπ‘ ππ 1 1 πππ βΖπ‘− π ππβΖπ‘ ππ π0 3.18 27 Multiply numerator and denominator of 3.18 by ππ ππΏ = ππ πππ βΖπ‘−π0 π ππβΖπ‘ ππ πππ βΖπ‘− π ππβΖπ‘ π0 3.19 Multiply numerator and denominator by π0 ππΏ = π0 (ππ πππ βΖπ‘−π0 π ππβΖπ‘) 3.20 π0 πππ βΖπ‘−ππ π ππβΖπ‘ To find the input impedance of the network, make ππ the subject From 3.20 ππΏ π0 πππ βΖπ‘ − ππΏ ππ π ππβΖπ‘ = π0 ππ πππ βΖπ‘ − ππ 2 π ππβΖπ‘ π0 (ππΏ πππ βΖπ‘ + π0 π ππβΖπ‘) = ππ (π0 πππ βΖπ‘ − ππΏ π ππβΖπ‘) ππ = πππ = π0 (ππΏ πππ βΖπ‘+π0 π ππβΖπ‘) 3.21 π0 πππ βΖπ‘−ππΏ π ππβΖπ‘ Equation 3.21 is the general formula for input impedance of transmission lines. The wavelength of the wave is π= 2π π½ The wave velocity is π’= π = ππ π½ Lossless Transmission Line π =0=πΊ Ζ = πΌ + π½π½ Ζ = √(π + πππΏ)(πΊ + πππΆ) Ζ = √ππ 2 πΏπΆ = ππ√πΏπΆ = πΌ + π½π½ This means for a lossless line πΌ = 0, and β= π√πΏπΆ π= 2π π½ , π’ = πλ 28 π’= 2ππ π½ = π π½ π = π√πΏπΆ = 1 √πΏπΆ π0 = √ π + πππΏ πΊ + πππΆ But π =0=πΊ πΏ ∴ π0 = √ = π 0 + ππ0 πΆ πΏ π 0 = √πΆ , π0 = 0 Distortionless Line π πΏ πΊ π πΏ = πΆ or πΊ = πΆ Ζ = √(π + πππΏ)(πΊ + πππΆ) πΏ πΆ Ζ = √(1 + ππ ) (1 + ππ ) π πΊ π πΊ Since πΏ πΆ = π πΊ Then πΏ Ζ = √π πΊ (1 + ππ ) = πΌ + π½π½ π Ζ = √π πΊ + ππ πΏ √π πΊ π πΏ2 Ζ = √π πΊ + ππ√ 2 π πΊ π Ζ = √π πΊ + ππ√πΏ2 ⋅ 29 πΊ π But πΊ πΆ = π πΏ Ζ = √π πΊ + ππ√πΏ2 ⋅ πΆ πΏ Ζ = √π πΊ + ππ√πΏπΆ For a distortionless line πΌ = √π πΊ and π½ = π√πΏπΆ πΏ (1 + ππ π ) π π + πππΏ π πΏ π0 = √ =√ =√ =√ πΆ πΊ + πππΆ πΊ πΆ (1 + ππ πΊ ) πΊ π= π’= 2π π½ , π’ = πλ 2ππ π π 1 = = = π½ π½ π√πΏπΆ √πΏπΆ Example 3.1 An air line has characteristics impedance of 50Ω and phase constant of 5rad/m, at frequency of 200MHz. Calculate the inductance per meter and the capacitance per meter of the line. Solution An air line can be regarded as a lossless line since π = 0 πΏ π0 = √ πΆ β = π√πΏπΆ π0 πΏ πΏ 1 = √ ÷ π√πΏπΆ = √ × π½ πΆ πΆ π√πΏπΆ π0 1 πΏ 1 1 1 1 = ⋅√ × = √ 2= π½ π πΆ πΏπΆ π πΆ ππΆ 30 50 1 = = 10 5 ππΆ πΆ= 1 1 = = 79.57ππΉ/π 10π 2 × π × 200 × 106 × 10 β = π√πΏπΆ π½ 5 = √πΏπΆ = = 3.97 × 10−9 π 2 × π × 200 × 106 πΏπΆ = (3.97 × 10−9 )2 1.5831 × 10−17 πΏ= = 198.93 × 10−9 π» = 198.93ππ»/π 79.57 × 10−12 Example 3.2 A distortionless line has π0 = 60πΊ, πΌ = 20πππ/π π’ = 0.6π, where c is the speed of light in a vacuum. Find R, L, G, C and λ at 100MHz. Solution πΊ πΆ = π πΏ π0 = 60πΊ, πΌ = 20πππ/π, π’ = 0.6π, f = 100MHz β = π√πΏπΆ πΌ = √π πΊ πΏ π π0 = √ = √ πΆ πΊ Ζ = √π πΊ + ππ√πΏπΆ πΌ = 20 × 10−3 = √π πΊ π π0 = 60 = √ πΊ π πΌπ0 = 20 × 10−3 × 60 = √ × π πΊ = π πΊ π = 1.2πΊ/π 31 π πΊ = πΌ 2 πΊ= πΌ 2 (20 × 10−3 )2 = = 333ππ/π π 1.2 2 × π × 100 × 106 3.49πππ π 8 π½ = = 0.6 × 3 × 10 π’ π πΏ π0 = √ πΆ β = π√πΏπΆ πΏ π0 π½ = π√ ⋅ πΏπΆ = ππΏ πΆ πΏ= πΏ= πΏ= π0 π½ π 60 × 3.49 2 × π × 100 × 106 209.4 = 333 × 10−9 = 333ππ»/π 638318530.7 πΏ π0 = √ πΆ πΆ= πΏ π0 2 333 × 10−9 = = 92.5 × 10−12 πΉ/π = 92.5ππΉ/π 602 Reflection Coefficient and Standing Wave Ratio Since π(π) = π΄π −Ζπ + π΅π Ζπ π(π) = π΄(π −Ζπ + ππΏ π Ζπ ) A is the forward travelling voltage π0 + A is the backward travelling voltage π0 − 32 ππΏ = π΅ π0 − = π΄ π0 + ππΏ is the voltage reflection coefficient (at the load) defined as the ratio of the reflected wave to the incident wave. From equation 3.12 πΌ(π) = π΄ π΄ 0 π0 πΌ(π) = π (π −Ζπ − ππΏ π Ζπ ), π΄ −Ζπ π΅ Ζπ π − π π0 π0 π΅ = πΌ0 + , − π = πΌ0 − 0 πΌ0 − = ππΏ πΌ0 + Generally the voltage reflection coefficient at any point on the line can be defined as the ratio of the magnitude of the reflected voltage wave to that of the incident wave. π(π) = π0 − π Ζπ π0 − 2Ζπ = π = ππΏ π 2Ζπ π0 + π −Ζπ π0 + At the load π = 0 π(0) π0 + (1 + ππΏ ) 1 + ππΏ ππΏ = = + = π0 ( ) πΌ(0) π0 1 − ππΏ π0 (1 + ππΏ ) ππΏ = π0 ( 1 + ππΏ ) 1 − ππΏ ππΏ − ππΏ ππΏ = π0 + π0 ππΏ ππΏ − π0 = (ππΏ + π0 )ππΏ ππΏ = |ππΏ | = ππΏ − π0 ππΏ + π0 ππππ₯ − ππππ ππππ₯ + ππππ ππππ₯ ππππ − 1 π − 1 |ππΏ | = = ππππ₯ π +1 + 1 ππππ 33 π − 1 = |ππΏ |π + |ππΏ | π − |ππΏ |π = 1 + |ππΏ | π (1 − |ππΏ |) = 1 + |ππΏ | π = 1 + |ππΏ | 1 − |ππΏ | Shorted Line (ππ³ = π) ππ π = π0 (ππΏ πππ βΖπ + π0 π ππβΖπ) π0 (ππΏ + π0 π‘ππβΖπ) = π0 πππ βΖπ + ππΏ π ππβΖπ π0 + ππΏ π‘ππβΖπ ππ π = π0 (π0 π ππβΖπ) = π0 π‘ππβΖπ π0 πππ βΖπ ππ π = π0 π‘ππβΖπ ππΏ = ππΏ − π0 = −1 ππΏ + π0 π = 1 + |ππΏ | =∞ 1 − |ππΏ | Open Circuited Line (ππ³ = ∞) πππ = π0 (∞πππ βΖπ + π0 π ππβΖπ) π0 πππ βΖπ + ∞π ππβΖπ πππ = π0 ∞πππ βΖπ = π0 πππ‘βΖπ ∞π ππβΖπ ππΏ = 1 π =∞ Since ππΏ = 1 it means π0 + = π0 − Observe that ππ π × πππ = ππ 2 34 Matched Line (ππ³ = ππ ) πππ = π0 (π0 πππ βΖπ + π0 π ππβΖπ) π0 πππ βΖπ + π0 π ππβΖπ πππ = π0 ππΏ = 0 π =1 There is no reflection, the incident power is fully absorbed by the load, and so maximum power transfer is possible when a transmission line is matched to load. 35 Use of Smith Chart Determination of reflection coefficient ππ³ Example 3.3 If π0 = 80πΊ and ππΏ = 35 + π50πΊ. Use the Smith Chart provided to calculate the reflection coefficient of the line. Indicate OP and OQ on the chart. Solution To confirm the result to be obtained from the Smith chart, calculate the value of ππΏ using the π −π formula ππΏ = ππΏ+π0 πΏ 0 ππΏ = 35 + π50 − 80 −45 + π50 = 35 + π50 + 80 115 + π50 ππΏ = 0.5364∠108.40 Using the Smith chart Step 1. Obtain the normalized impedance ππΏ 35 + π50 ππΏ π = = = 0.4375 + π0.625 π0 80 Step 2. Locate the real value of the normalised impedance on the chart using the scale on the horizontal line at the centre and identify the circle it belongs. Step 3. Locate the arc with value equal to the imaginary part of the normalizes impedance Step 4. Locate the point where the circle in step 2 crosses the arc in step 3, mark this as point P. Step 5. The origin ‘O’ is at the centre of the chart at ‘1.0’ and point ‘Q’ is located at the edge of the chart where a straight line OP passes through the ‘0.0’ circle. Step 6. Measure OP and OQ ππ 4.2 ππΏ = = = 0.538461 ππ 7.8 The angle is shown on the edge where line OP passes through. Here it is 1080 ∴ ππΏ = 0.5384∠1080 36 Example 3.3 If π0 = 30πΊ and ππΏ = 42 + π24πΊ. Use the Smith Chart to find ππΏ . Solution ππΏ π = ππΏ = ππΏ 42 − π24 = = 1.4 + π0.8 π0 30 2.75 = 0.3525∠−450 7.8 Determination of Standing wave ratio using the Smith chart To obtain the standing wave ratio, draw a circle with radius OP and centre at 0. Locate point S where the S circle meets the ππΏ π axis. The value of r at this point is s π = π for ( π ≥ 1) 1+|π | Do this for the previous examples and confirm answer using π = 1−|ππΏ| πΏ Determination of input impedance π distance on the line corresponds to a movement of 7200 on the chart. π 2 distance on the line corresponds to a movement of 3600 on the chart. Step 1. π’ Calculate the wavelength using the information given π = π Step 2. Determine how many wavelengths the length of the line corresponds to, then multiply by 7200 Step 3. Rotate along the line (clockwise i.e. towards generator) mark the point corresponding to the angle obtained. Step 4. Find the value of the normalized impedance at that point. Step 5. Finally multiply by π0 to obtain the actual value of the input impedance πππ . 37 Example 3.4 A 30m long lossless transmission line with π0 = 50πΊ operating at 2Mhz is terminated with a load ππΏ = 60 + π40πΊ. If π’ = 0.6π on the line find a. The reflection coefficient ππΏ . b. The standing wave ratio. c. The input impedance. Solution π = 0.3523∠560 π = 2.1 π = (2400 ) πππ = 50(0.47 + π0.035) = 23.5 + π1.75πΊ 38
