Transmission Lines. Ranga Rodrigo. January 27, Antennas and Propagation: Transmission Lines 1/72

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1 Transmission Lines Ranga Rodrigo January 27, 2009 Antennas and Propagation: Transmission Lines 1/72

2 1 Standing Waves 2 Smith Chart 3 Impedance Matching Series Reactive Matching Shunt Reactive Matching Antennas and Propagation: Transmission Lines Outline 2/72

3 Transmission Line Model Z 0 R L R L R L G C G C G C Z 0 R: series loss resistance (copper losses). G: shunt loss conductance (losses in dielectric). L: series inductance representing energy storage within the line. C: shunt capacitance representing energy storage within the line. Antennas and Propagation: Transmission Lines Summary of last Week s Lecture 3/72

4 I 1 I 2 I 3 V 1 V 2 V 3 l l If an infinitely long pair of wires were considered and the voltage and current were somehow measured at uniform spaced points along the line, then V 1 I 1 = V 2 I 2 = = V k I k = constant = Z 0 Ω. (1) This is termed the characteristic impedance of the line and is denoted by Z 0. Antennas and Propagation: Transmission Lines Summary of last Week s Lecture 4/72

5 Characteristic Impedance ( ) R + j ωl 1/2 Z 0 = Ω. (2) G + j ωc At very low frequencies: ( ) R 1/2 Z 0 = Ω. (3) G For high frequencies (ωl R and ωc G): ( ) L 1/2 Z 0 = Ω (4) C Antennas and Propagation: Transmission Lines Summary of last Week s Lecture 5/72

6 Propagation Constant d 2 V = (R + j ωl)(g + j ωc)v. (5) dx2 Equation 5 can be written as d 2 V dx 2 = γ2 V, (6) where γ = (R + j ωl)(g + j ωc) is termed the propagation constant. This is usually expressed as γ = α + j β, (7) where α represents the attenuation per unit length and β the phase shift per unit length. Antennas and Propagation: Transmission Lines Summary of last Week s Lecture 6/72

7 Equation 6 has a solution of the form V (x) = Ae γx + Be γx. (8) This suggests that the line will contain two waves, one traveling in the positive x-direction (e γx ) and the other traveling in the negative x-direction (e γx ). Antennas and Propagation: Transmission Lines Summary of last Week s Lecture 7/72

8 Sending-End Impedance I inc I ref Z 0 Z T Z S x = 0 l Antennas and Propagation: Transmission Lines Summary of last Week s Lecture 8/72

9 Z S = 1 + e 2γl ZT Z0 Z T +Z 0 Z 0 1 e 2γl Z T Z 0 Z T +Z 0 = Z T (1 + e 2γl ) + Z 0 (1 e 2γl ) Z T (1 e 2γl ) + Z 0 (1 + e 2γl ). (9) Z S = Z T cosh(γl) + Z 0 sinh(γl) Z 0 Z T sinh(γl) + Z 0 cosh(γl). (10) Z S Z 0 = Z T Z 0 + tanh(γl) 1 + Z T Z 0 tanh(γl). (11) Antennas and Propagation: Transmission Lines Summary of last Week s Lecture 9/72

10 Short Circuit Termination If a short circuit is used as the load termination, then the sending-end impedance becomes, Z S SC = Z 0 tanh(γl), (12) and, neglecting line losses for short lengths we get Z S SC = Z 0 tanh(j βl) = j Z 0 tan(βl). (13) Antennas and Propagation: Transmission Lines Summary of last Week s Lecture 10/72

11 Open Circuit Termination When the termination is open circuit, then the sending-end impedance becomes, Z S OC = Z 0 tanh(γl), (14) and, neglecting line losses for short lengths we get Z S OC = j Z 0 tanh(j βl) = j Z 0 cot(βl). (15) Antennas and Propagation: Transmission Lines Summary of last Week s Lecture 11/72

12 tan(βl) π 2 π 3π 2 2π 5π 2 βl = 2πl λ Normalized input resistance versus βl for a Antennas and Propagation: short-circuited, Transmission Lines lossless, transmission Summary of last Week sline. Lecture 12/72

13 cot(βl) π 2 π 3π 2 2π 5π 2 βl = 2πl λ Normalized input resistance versus βl for a Antennas and Propagation: open-circuited, Transmission Lines lossless, transmission Summary of last Week sline. Lecture 13/72

14 Standing Waves For any transmission line, a sinusoidal signal of appropriate frequency introduced at the sending end by a generator will propagate along the length of the transmission line. If the line has infinite length, then the signal never reaches the end of the line. If the signal is viewed at some distance down the line away from the generator, then it will appear to have the same frequency, but will exhibit smaller peak to peak voltage swing than at the generator. The signal, therefore, has been attenuated by the line losses due to the conductor resistance Antennas and and Propagation: dielectric Transmission Linesimperfections. Standing Waves 15/72

15 Reflection If the line is lossless, then the signal viewed at some remote point will be identical to that at the generator but time delayed by an amount dependant on the position. When a sinusoidal signal reaches the open end of a section of a lossless transmission line, it can dissipate no energy. This means that all the energy propagating along the line in the forward direction (incident) will be reflected completely, on reaching the open circuit termination. The reflected wave (backward wave) must be such that the total current at the open circuit is Antennas and zero. Propagation: Transmission Lines Standing Waves 16/72

16 Reflection As the reflected signal travels back along the line toward the generator, it reinforces the incident waveform at certain points forming maxima (nodes). Similarly, it can cancel the incident waveform at certain other points producing minima (antinodes). In an open circuit line, node voltage points will occur at the same position as the antinode current points. These waves do not represent traveling waves. They are standing waves, implying that there is no net power flow from generator to the Antennas and (open-circuit) Propagation: Transmission Lines load. Standing Waves 17/72

17 A point one-quarter wavelength away from a short circuit will have voltage and current magnitudes equivalent to those obtained for an open circuit. For a lossless line the peak value of the standing wave envelope is twice that of the incident wave. There are nulls as well due to the complete cancelation. For lossy (practical) line, the peak will be less than twice that of the incident wave, and complete cancelation rarely results. Antennas and Propagation: Transmission Lines Standing Waves 18/72

18 Voltage Standing Wave Ratio Node to antinode voltage ratio is infinity for the ideal case. Voltage standing wave ratio (VSWR) or simply standing wave ratio is defined as VSWR = V max V min = V inc + V ref V inc V ref V inc is the peak value of the incident voltage wave, similarly the others. Here the voltage values are peak values. RMS values can be used as well. (16) Antennas and Propagation: Transmission Lines Standing Waves 19/72

19 Reflection Coefficient I inc I ref A I trans Z 0 Z T A Considering the current continuity at the boundary AA, we get V inc V ref = V trans. (17) Z 0 Z 0 Z T For voltage continuity across the boundary V inc +V ref = V trans. (18) Antennas and Propagation: Transmission Lines Standing Waves 20/72

20 From Equations 17 and 18 we can obtain the reflection coefficient Γ. Γ = V ref V inc = Z T Z 0 Z T + Z 0. (19) An alternative symbol for the reflection coefficient is ρ. Sometimes ρ is used to represent Γ. From Equation 16 Therefore, 1 + V inc V ref VSWR = = 1 V inc V ref Γ = V inc V ref 1 + Γ 1 Γ. (20) = VSWR 1 VSWR + 1. (21) Antennas and Propagation: Transmission Lines Standing Waves 21/72

21 Example A certain transmission line with a characteristic impedance of 50 Ω is required to deliver 1 kw over a short distance to a load. The cable can withstand a maximum of 250 V. Determine 1 the maximum VSWR, and 2 amount of power that must be provided by the generator. Antennas and Propagation: Transmission Lines Standing Waves 22/72

22 Solution We can assume that the line is lossless. Power dissipated by the load P L is P L = P inc P ref. P L = V 2 inc V 2 ref, Z 0 Z 0 = 1 Z 0 (V inc V ref )(V inc +V ref ), = 1000W. Antennas and Propagation: Transmission Lines Standing Waves 23/72

23 (V inc V ref )(V inc +V ref ) = Because RMS voltage in the line must be less than 250 V, (V inc +V ref ) max < 250. (V inc V ref ) > 200. VSWR = V inc + V ref V inc V ref, VSWR < = Antennas and Propagation: Transmission Lines Standing Waves 24/72

24 We can show that ( ) P L VSWR 2 1 = 1. P inc VSWR + 1 For this value of VSWR, 98.8% of the incident power is delivered to the load. Therefore, the generator must be capable of supplying at least 1012 W and must be matched to the line. In normal design, VSWR is kept below 1.1. Antennas and Propagation: Transmission Lines Standing Waves 25/72

25 Sending-End Reflection Coefficient l Γ in = Z S Z 0 Z 0, γ Z T Z S +Z 0 Γ L = Z T Z 0 Z T +Z 0 Γ in Γ L Rearranging Equation 9, we can obtain Z S = 1 + e 2γl ZT Z0 Z T +Z 0 Z 0 1 e 2γl Z T Z 0 Z T +Z 0, Γ in = Γ L e 2γl. (22) Antennas and Propagation: Transmission Lines Standing Waves 26/72

26 Impedance Matching One of the major tasks in transmission line circuit design is impedance matching, and Smith chart is a graphical procedure for solving impedance transformation problems. A mismatched transmission line (a line terminated in an impedance other than its characteristic impedance) would reflect some of the incident wave back along the line toward the generator. This interaction of the forward waves and reflected waves give rise to a resultant waveform with nodes and antinodes at fixed points along length of the transmission line. Antennas and This Propagation: way, Transmission standing Lines waves are formed. Smith Chart 28/72

27 Effects of Mismatch Adverse Effects Energy loss resulting in reduced system performance. Frequency pulling of the signal generator. Possible transmitter (and other circuitry) damage. Desirable Effect Carefully controlled mismatch can improve system performance by improving the noise characteristics. Antennas and Propagation: Transmission Lines Smith Chart 29/72

28 Controlling Impedance Mismatch There are two requirements: 1 A methodical scheme for determining the degree of mismatch between the source and the load impedance. 2 A method whereby the degree of mismatch can be varied in a known manner. Antennas and Propagation: Transmission Lines Smith Chart 30/72

29 Smith Chart Matching requires a way of handling the expressions governing the impedance transformation that occurs along a length of transmission line with known termination. In 1939, P. H. Smith, an engineer with the Bell Telephone Laboratories, developed a graphical presentation of transmission line data. He found that for passive loads (0 < Γ < 1) the graphical representation is bounded. Smith chart is a plot of the voltage reflection. Antennas and Propagation: Transmission Lines Smith Chart 31/72

30 All possible values of impedance and admittance can be can be plotted on the chart. An easy method of converting impedances to admittances is available. Smith chart provides a simple graphical method for determining the impedance transformations due to a length of transmission line. Antennas and Propagation: Transmission Lines Smith Chart 32/72

31 Normalized Resistance Loci Let Γ = Z L Z 0 Z L + Z 0 (23) Γ = u + j v and Z L Z 0 = Z n = R n + j X n. u + j v = (R n 1) + j X n (R n + 1) + j X n. (24) Equating real and imaginary parts gives R n (u 1) X n v = (u + 1), R n v + X n (u 1) = v. (25) Antennas and Propagation: Transmission Lines Smith Chart 33/72

32 X n = R n(u 1) + (u + 1). (26) v Substituting 26 in 25, R n v + [R n(u 1) + (u + 1)](u 1) v = v. (27) v 2 (R n + 1) 2uR n + u 2 (R n + 1) = 1 R n. v 2 2uR n + u 2 = 1 R n (28). 1 + R n 1 + R n ( v 2 + u R ) 2 n Rn R n (1 + R n ) = 1 R n. (29) R n Finally, the desired relationship is ( v 2 + u R ) 2 n 1 = 1 + R n (1 + R n ). (30) 2 Antennas and Propagation: Transmission Lines Smith Chart 34/72

33 When plotted on the u v plane, in Cartesian coordinates, Equation 30, ( v 2 + u R ) 2 n 1 = 1 + R n (1 + R n ), 2 represents a circle with center and radius u = R n R n R n + 1. and v = 0, For different values of R n, we can generate a family of circles. They represent normalized resistance loci. Antennas and Propagation: Transmission Lines Smith Chart 35/72

34 v Normalized Resistance u Antennas and Propagation: Transmission Lines Smith Chart 36/72

35 Normalized Reactance Loci We can repeat the above process for the normalized reactive element X n. ) ( ) (u 1) 2 + (v 1Xn 1 2 =. (31) This equation too results in a family of circles, this time, centered at u = 1 and v = 1 X n, X n and with radius 1 X n. Antennas and Propagation: Transmission Lines Smith Chart 37/72

36 v Normalized Reactance u Antennas and Propagation: Transmission Lines Smith Chart 38/72

37 v u Antennas and Propagation: Transmission Lines Smith Chart 39/72

38 For any passive impedance, Re(Z ) 0 and z 0 real, Γ 1. Thus all possible values of Γ can be plotted on this polar chart having a maximum radius value of unity. It can accommodate reflection coefficient values for all impedances from short (Γ = ) to an open (Γ = 1 0 ) Antennas and Propagation: Transmission Lines Smith Chart 40/72

39 Constant VSWR Circles Smith chart is really a plot of the reflection coefficient Γ = u + j v. Constant VSWR circles are actually Γ = constant circles. From Equation 20 we have the relation VSWR = 1 + Γ 1 Γ. We can, therefore, relate Γ and VSWR to plot the circles. For a matched system, the reflection coefficient will be zero and VSWR will be unity, the center of the Smith chart. For al reflection coefficient of one, the VSWR will be infinity and will overlay the contour for zero resistance. Antennas and Propagation: Transmission Lines Smith Chart 41/72

40 v Constant VSWR u Antennas and Propagation: Transmission Lines Smith Chart 42/72

41 v Normalized resistance Normalized reactance Constant VSWR u Antennas and Propagation: Transmission Lines Smith Chart 43/72

42 Points and Properties of Smith Chart At point u = 1, v = 0 normalized resistance and reactance terms are large: position for an open circuit. At point u = 0, v = 1 the value is 0 + j 1: position for perfect inductive reactance. At point u = 0, v = 1 the value is 0 j 1: position for perfect capacitive reactance. At point u = 1, v = 0 zero resistance and zero reactance contours intersect: position for the short circuit. Point u = 0, v = 0, the center of the Smith chart, represents the normalized impedance 1 + j 0, which under normal circumstances, represents Antennas and the Propagation: transmission Transmission Lines line characteristic Smithimpedance. Chart 44/72

43 Point Comment Representation u = 1, normalized resistance open circuit v = 0 and reactance terms are large u = 0, v = j 1 perfect inductive reactance u = 0, 0 j 1 perfect capacitive v = 1 reactance u = 1, v = 0 zero resistance and zero reactance con- short circuit u = 0, v = 0 tours intersect center of the Smith chart, normalized impedance 1 + j 0 characteristic impedance Antennas and Propagation: Transmission Lines Smith Chart 45/72

44 Example Consider a load impedance Z L = j 11.8 Ω connected to a 50 Ω line of length l = λ/6. Plot Γ L, and graphically obtain Γ in. Compare the result with the computed value. Antennas and Propagation: Transmission Lines Smith Chart 46/72

45 Solution Form Equation 18 Γ L = Γ Γ = V in ref = Γ L e = Z 2γl T. Z 0. V inc Z T + Z j j = We notice that if the line is lossless α = 0, and γ = j β. So, 2βl = 2 2π/λ (λ/6) = 2π/3 = 120. We can use Equation 22 to obtain the input reflection coefficient. Γ in = = The same result can be graphically obtained. Because Γ L = Γ in for a lossless line, we merely have to rotate clockwise from Γ L an angle 2βl (120 ). Antennas and Propagation: Transmission Lines Smith Chart 47/72

46 v Normalized resistance Normalized reactance Constant Γ u Antennas and Propagation: Transmission Lines Smith Chart 48/72

47 Impedance Transformation Using Smith Chart The above example showed us that the graphical solution of the reflection coefficient transformation is simple. However, we had to compute Γ L using the impedances. It is possible to solve impedance transformations completely graphically. We just need to replace every point on the polar reflection coefficient chart by it normalized equivalent impedance. We can do this by using the following equations: Z = Z n = 1 + Γ Z 0 1 Γ, and Y = Y n = 1 Γ Y Γ (32) Antennas and Propagation: Transmission Lines Smith Chart 49/72

48 Example Plot the following on the Smith chart. 1 Γ 1 = Γ 2 = Γ 3 = Γ 4 = Γ 5 = Γ 6 = Antennas and Propagation: Transmission Lines Smith Chart 50/72

49 Fortunately, we do not +j 1 have to use the equations as, on the Smith chart, the nor- +j 2 malized impedance grid 7 has been superim- +j 0.4 posed. Normalized resistance Normalized reactance Constant VSWR Γ 5 : Z = j j 4 +j Γ 1 : Z = 0 = Z : Γ 2 Γ 3 : Z = 1/3 Γ 4 : Z = j.2 Γ 6 : Z = 0.6 j 0.8 j 4 j.4 j 2 j 1 Antennas and Propagation: Transmission Lines Smith Chart 51/72

50 Example A 5.2 cm length of lossless 100 Ω line is terminated in a load impedance Z L = 30 + j 50 Ω. 1 Calculate Γ L and SWR along the line. 2 Determine the impedance and admittance at the input and at a point 2.0 cm from the load end. The signal frequency is 750 MHz. Solution Z L = 30+ j 50, and Z 0 = 100. Therefore, Z Ln = 0.3+ j 0.5. Antennas and Propagation: Transmission Lines Smith Chart 52/72

51 j 0.5 +j 1 Normalized resistance Normalized reactance Constant VSWR +j 2 +j j 0.2 Z Ln 3 +j Z 1n = R n = 4.23 j.2 j 4 j.4 Γ = j 2 j 1 Antennas and Propagation: Transmission Lines Smith Chart 53/72

52 1 Γ = The constant Γ circle is the SWR circle. We obtain 4.23 as the SWR value from the point of intersection between the SWR circle and the right half of the resistance axis. This point is marked as Z 1n = R n. At this point R n = 1+ Γ 1 Γ. 2 λ = / = 40 cm. In order to find the impedance at the input, we must travel 5.2/40 = 0.130λ toward the generator, along a constant VSWR circle. Antennas and Propagation: Transmission Lines Smith Chart 54/72

53 123.42,0.0786λ +j 0.5 +j 0.4 +j Normalized resistance Normalized reactance Constant VSWR +j 2 +j 0.2 Z Ln 3 Z inn +j Z 1n = R n = 4.23 j.2 Y Ln j 4 j.4 Γ = j 2 j 1 Antennas and Propagation: Transmission Lines Smith Chart 55/72

54 1 We read off the chart that Z in n = 2 + j 2. Therefore, Z in = j Admittance points are directly opposite (across (1 + j 0)) in the Smith chart. We read off the chart that Y Ln = 0.88 j Do the other calculations on your own. Antennas and Propagation: Transmission Lines Smith Chart 56/72

55 Some Impedance Matching Techniques There are two common types of impedance matching: Conjugate Matching: The matching of a load impedance to a generator for maximum transfer of power. Z 0 Matching: The matching of a load impedance to a transmission line to eliminate wave reflection at the load. Antennas and Propagation: Transmission Lines Impedance Matching 58/72

56 Conjugate Matching Maximum power is delivered to a load when Z L is set equal to the complex conjugate of the generator impedance Z G. That is Z L = Z G = R G j X G. In situations where the load impedance is not adjustable, a matching network may be placed between the generator and the fixed load. Antennas and Propagation: Transmission Lines Impedance Matching 59/72

57 Z 0 Matching This type matches a load impedance to the characteristic impedance of a transmission line, i.e., Z L = Z 0. In this case Γ L = 0 and hence SWR along the line is unity. If Z L Z 0, a matching network may be used to eliminate the standing waves on the line. With Z L connected to the transmission line Γ = V ref = Z T Z 0 through the matching network, VZ inc 0 matching Z T + Z 0 requires that the input impedance of the network VSWR = 1 + Γ equal to Z 0 or Z inn = Z in /Z 0 = 1 + 1j 0. Γ For a Smith chart normalized to Z 0, this means that Z inn must be at the center of the chart. Antennas and Propagation: Transmission Lines Impedance Matching 60/72

58 For a well-designed source, Z G = Z 0, the characteristic impedance of its output line. With Z 0 real, matching the load to the line (Z L = Z 0 ) results in a conjugate match between the generator and the load. Antennas and Propagation: Transmission Lines Impedance Matching 61/72

59 Series Reactive Matching Example Find the reactance X in the following figure to achieve matching. j X Matching network To generator Z 0 = 50 Ω Z 01 = Z 0 Z L = 25 + j 30 in A l L Antennas and Propagation: Transmission Lines Impedance Matching 62/72

60 +j j λ Normalized resistance Normalized reactance Constant VSWR +j 2 +j Z 2.87 Ln 1.67 Z E +j λ Effect of adding j X n = j j.2 Effect of adding j X n = j Z A j 4 j.4 j j 1 Antennas and Propagation: Transmission Lines Impedance Matching 63/72

61 If we assume that the design frequency is 2000 MHz and the wavelength within the line is equal to the free space wavelength, we find that l = λ = cm. l = cm. This is if we pick the point E, the first intersection point between the constant VSWR circle and R n = 1 circle. If we pick point A, we get l = = cm. Since ωl/z 0 = 1.1, we have ωl = 55 Ω. The we get L = 55/2πf = 4.38 nh. With the matching network, Z inn = 1+ j 0 and thus the SWR on the 50Ω line to the left of Z in is unity. Antennas and Propagation: Transmission Lines Impedance Matching 64/72

62 The matching network that we designed will not perfectly match at frequencies other than the design frequency. This is because both l/λ 0 and ωl/z 0 are frequency dependant. λ 0 is the free space wavelength. ωl Freq. λ 0 (cm) l/λ 0 Z 0 Z An Z in SWR (MHz) j j j j j Antennas and Propagation: Transmission Lines Impedance Matching 65/72

63 Shunt Reactive Matching Example Find the susceptance B in the following figure to achieve matching. Matching network To generator Z 0 = 50 Ω j B Z 01 = Z 0 Z L = 20 in A l L Antennas and Propagation: Transmission Lines Impedance Matching 66/72

64 Smith chart solution must be carried out in admittances. Z Ln = 50/Z 0 =.40 Ω. We can find Y Ln by rotating the Z Ln point halfway around the chart on its SWR circle. From the chart, we get Y Ln = j 0. Since matching requires Y inn = 1, the length l is chosen so that the real part of Y An is equal to unity. As before, there are two possibilities since the SWR circle intersects the G n = 1 circle at two points, Y An = 1 + j 0.95 and Y An = 1 j Antennas and Propagation: Transmission Lines Impedance Matching 67/72

65 +j 0.4 +j 1 7 Normalized resistance Normalized reactance Constant VSWR +j 2 3 +j 4 +j Z Ln Y Ln = j Effect of adding j X n = j 0.95 j.2 Y An = 1 j 0.95 j 4 j λ j 2 j Antennas and Propagation: Transmission Lines Impedance Matching 68/72

66 For l = 0.09λ, Y An = 1 j 0.95, and with Y 0 = 0.01 S, Y A = 0.02 j 0.19 S. This means that a shunt capacitance of admittance value +j S is required to cancel the inductive effect of Y A. Thus adding a capacitive susceptance B = ωc = S at plane A results in Y in = 0.02 S or Y inn = 1.0, a matched condition. Short-circuit or open-circuit stubs may be used to obtain the required susceptance. It is best to use a high impedance shorted line for inductive susceptance and a low impedance open-circuited line for capacitive susceptance. Antennas and Propagation: Transmission Lines Impedance Matching 69/72

67 Shunt Reactive Matching with a Shorted Stub l Z 0 Z 0 Z L Z 0s l s short Antennas and Propagation: Transmission Lines Impedance Matching 70/72

68 Shunt Reactive Matching with a Open-Circuited Stub l Z 0 Z 0 Z L Z 0s l s open Antennas and Propagation: Transmission Lines Impedance Matching 71/72

69 Stub Tuners We can do stub tuning using either series or shunt stubs. However, series and shunt reactance matching techniques usually result in low SWR values over narrow range (typically, less than 20%). Improved broadband performance can often be realized by utilizing series and parallel resonant circuits. Antennas and Propagation: Transmission Lines Impedance Matching 72/72

Transmission Lines. Ranga Rodrigo. January 13, Antennas and Propagation: Transmission Lines 1/46

Transmission Lines. Ranga Rodrigo. January 13, Antennas and Propagation: Transmission Lines 1/46 Transmission Lines Ranga Rodrigo January 13, 2009 Antennas and Propagation: Transmission Lines 1/46 1 Basic Transmission Line Properties 2 Standing Waves Antennas and Propagation: Transmission Lines Outline