• Author / Uploaded
• Kevin Rutto

Citation preview

EEE2314 TRANSMISSION LINES [TIE AND EEE] Lecturer: Kipyegon Edwin        

Transmission line parameters: analysis of distributed line constants. Phase and group velocities. Dispersive and non-dispersive circuits. Concept of position angle. Analysis of current and voltage distribution along a transmission line, reflection coefficient and voltage standing wave ratio (VSWR) Lossless line equations. Lossy short-circuit line. Measurement of reflection coefficient. Smith Chart: impedance matching using a stub. Resonant and anti-resonant lines. The quality factor of resonant and anti-resonant lines.

Prescribed Text Books 1. David K. Cheng, “Field and Wave Electromagnetics”, Addison-Wesley Publishing Company(), 2nd Edition. 2. William H. Hayt, Jr. & John A. Buck “Engineering Electromagnetics”, McGraw-Hill International Edition (2006), 7th Edition. References 1. Grant I.S. & Phillips W.R., “Electromagnetism”, John Wiley & Sons Ltd (1976). 2. Joseph A. Edminister, “Schaum’s Outline Series of Theory and Problems of Electromagnetics”, Tata McGraw-Hill Publishing Co. Ltd (1993),2nd Edition. 3. Ashutosh Pramanik, “Electromagnetism Theory and Applications”, PHI Learning Private Ltd (2008), 2nd Edition.

1 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

INTRODUCTION  A transmission line consists of two or more parallel conductors used to transmit electric energy and signals from one point to another, specifically from a source to a load e.g. 1. The connection between a transmitter and an antenna 2. Connection between computers in a network 3. Connections between a hydroelectric generating plant and substation several hundred miles away. 4. Interconnection between components of a stereo system 5. Connections between devices on a circuit board designed to operate at a high frequency  In the transmission line, electromagnetic waves have both the E and H transverse to the direction of propagation, and are generally referred to as Transverse Electromagnetic (TEM) waves.  Most common types of transmission line which support TEM waves are:a) Parallel-plate transmission line- it consists of two parallel conducting plates separated by a dielectric slab of a uniform thickness as shown in Fig.1.1. At microwaves frequencies, they can be fabricated on a dielectric substrate using printed-circuit technology and are normally called striplines.

Fig. 1.1 Parallel-plate transmission line

b) Two-wire transmission line -consists of a pair of parallel conducting wires separated by a uniform distance as shown in Fig. 1.3 e.g. overhead power and telephone line.

Fig. 1.2 Two-line transmission line

c) Coaxial transmission line – consists of an inner conductor and a coaxial outer conducting sheath separated by a dielectric medium as shown in Fig. 1.3a. This structure has the advantage of confining the electric and magnetic fields entirely within the dielectric region

2 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

Fig. 1.3 Coaxial transmission lines

d) Microstrip line – mainly used in integrated circuits where metallic strips connecting elements are deposited on dielectric substrates. An example is as shown in Fig. 1.3b.

Fig. 1.3b Microstrip line



Transmission line problems are usually solved using EM field theory and electric circuit theory. Because of simplicity, circuit theory will be used in here.

1. Transverse Electromagnetic Waves along a Parallel-Plate Transmission Line  Consider a y-polarized TEM wave propagating in the +z-direction along uniform parallel-plate transmission line as shown in Fig. 1.4.

y

d

x w

z O

Fig. 1.4 Parallel-plate transmission line





For time-harmonic fields sourceless media, the equation to satisfy is the Helmholtz’s equation [1] 2E  γ 2E  0 Solution to equ. [1a] for a wave propagating in the z-direction is E  a y E  a y E 0 e z [2a] 3

EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon



As shown the wave is polarized in the y-direction. The associated H field is obtained as follows From the equation relating the E and H field i.e. 1 1 1 H  a n  E  a z  a y E 0 e z   a x E 0 e z









[2b]



where and are the propagation constants and the intrinsic impedance respectively. Assuming a perfect conductor and a lossless dielectric, then







 

[3]

And   j  j  [4] The boundary conditions to be satisfied at the dielectric/perfect conductor interfaced are as follows: At both y  0 and y  d : [5] Et  0 Hn  0 [6] which are satisfied because E x  E Z  0 and H y  0 At y  0 (lower plate) a n  a y :

a y  D   S

or

 S  E y  E 0 e  jz

a y  H  J S

or

J S  a y  H

 a y  a x H x  a z H x  a z

[7a] E0



e  jz

[8a]

At y  d (upper plate) a n  a y :

 a y  D   Su

or

 Su  E y  E 0 e  jz

 a y  H  J S

or

J S  a y  H  a y  a x H x  a z H x  a z

[7b] E0



e  jz

[8b]

Equ. (7) and (8) shows that the surface charges and surface currents on the conducting planes vary sinusoidally with z, as does E y and H x . This illustrated in Fig. 1.5. Field phasors E and H in equ. [2a] and [2b] satisfy the two Maxwell’s curl equations:   E   jH

[9]

and

[10] jE But since E and H are given by equations [2a] and [2b], then equ. [9] and [10] becomes dE y  jH x [11] dz and H 

4 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

dH x  jE y dz

[12]

y

       E        H     0

Js

                                



 





  z

s

Fig. 1.5 Field, charge, and current distribution along a parallel-plate transmission line

 Integrating equ. [11] over y from 0 to d, have d d d E dy  j  y 0 H x dy dz 0 or dV(z)  d   jJ su (z)d  j   J su (z)w   jLI(z) dz  w where

[13]

V(z)  - E y dy  E y (z)d d

0

is the potential or voltage between the upper and lower plates.

I(z)  -Jsu (z)w is the total current flowing in the +z direction in the upper plate ( w  plate width ) and d (H/m) w is the inductance per unit length of the parallel-plate transmission line. L

 Similarly, integrating equ. [12] over x from 0 to w, have w d w H x dx  j  E y dx  0 dz 0 or dI(z)  w    j E y (z)w  j    - E y (z)d  jCV(z) dz  d where





[14]

[15]

5 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

w (F/m) [16] d is the capacitance per unit length of the parallel-plate transmission line.  Equations [13] and [15] constitute a pair of time-harmonic transmission-line equations for phasors V(z) and I(z) .They can be combined to result in a second-order differential equations for V(z) and for I(z) : dV(z)   jLI(z) dz Differentiating both side with respect to z, have, d 2 V(z) dI(z)   jL 2 dz dz dI(z) Substituting for  jCV(z) from equ. [15],have dz d 2 V(z)   2 LCV(z) [17a] 2 dz Similarly, d 2 I(z) dV(z)  jC  jC jLI(z)    2 LCI(z) [17b] 2 dz dz The solutions of equ. [17a] and [17b] are: C

V(z)  V0 e -jz

[18a]

and

I(z)  I 0 e -jz where   ω LC    (rad/m)  The impedance of the parallel-plate can be obtained as follows: V(z) V0 e -jz V0 L Z0     - jz I(z) I0 C I0e Equ. [20] can be shown using equ {13] or [15]. d μ L w  d   d  (Ω) Z0   w w  C w ε d Equ. [21] is the characteristic impedance of the line  Velocity of propagation along the line is   1 1 up      )   με LC

[18b] [19]

[20]

[21]

[22]

6 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

Lossy Parallel-Plate Transmission Line  Above evaluation was for a lossless line but in actual situations, loss occur and in this case it could be due to two causes:a) The dielectric medium may have a nonvanishingly loss tangent b) The plate may not be perfectly conducting  The two effects are characterised by defining two parameters:  Conductance , G, per unit length of the two plates  Resistance, R, per unit length of the two plate conductors

G

 C 

[23]

where G - conductance between the conductors  - permittivity of dielectric medium between the two conductors  - conductivity of the dielectric medium C - capacitance between the two conductors  For conductors with finite conductivity (  c ), power is dissipated on the plates. This is represented by a nonvashining axial electric field a z E z at plate surfaces. Thus average power dissipated in each of the conducting plates is 1 Pav  a y p   RE a z E z  a x H *x [24] 2 Due to the imperfect conductors being used, a surface impedance Zs is usually defined which given by the ratio of the tangential electric field to the surface current density at the surface of the conductor.





Et () Js For the upper plate with surface current J su  H x , Zs is E E Zs  z  z  c J su H x where  c  is intrinsic impedance of the plate conductor. But f () Z s  R s  jX s  1  j c Substituting equ. [26a] in equ. [24], have 1 1 2 Pav  a y p  RE a z J su Z s  a x J su   RE a y J su Z s or 2 2 1 1 2 2 p σ  RE J su Z s  J su R s ( W/m 2 ) 2 2 Thus, the ohmic power dissipated in a unit length of a plate having a width w Hence Zs 







[25]

[26a]

[26b]



[27] is wp σ .

7 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

1 1 R  2 ( W/m ) [28] J su R s w  I 2  s  2 2 w Since I  wJ su - total surface current. Equation [28] gives the power dissipated when a sinusoidal current of amplitude I flows R through a resistance of s . w Hence, the effective series resistance per unit length for both plates of a parallel-plate transmission line of width w is  R  2 f c [29] R  2 s    w  w c In summary, the formula for the distributed parameters per unit length of a parallel-plate transmission line of width w and separation d is as shown in table 1. Pσ  wp σ 



Table 1 Parameter R

Formula 2 f c w c

L

Unit /m

H /m

d w w σ d w  d



G C

S /m

F /m

MICROSTRIP LINE  Microstrip line is a practical example of a parallel-plate transmission as shown in Fig.1.6. Metal strip

Dielectric substrate

Grounded conducting plane Fig. 1.6: Mircrostrip line



The analysis done earlier assumed that the two conduction plates on each side of the substrate are of the same size as the substrate, therefore, the results obtained cannot exactly apply to the Microstrip. The approximation is closer if the width of the metal strip is much greater than the substrate thickness. 8

EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon



When the substrate has high dielectric constant, a TEM approximation is found to be reasonably satisfactory. But the exact analytical solutions complex and beyond the scope of this study.

Example 1 Neglecting losses and fringe effects and assuming the substrate of a stripline to have a thickness of 0.4mm and a dielectric constant 2.25. a) Determine the required width w of the metal strip in order for the stripline to have a characteristic resistance of 50Ω. b) Determine L and C of the line and c) Determine  p along the line d) Repeat parts (a), (b), and (c) for a characteristic resistance of 75Ω. Solution a) From equ. [21], have that d  d or w  Z0  w  Z0

 0.4  103  0 0.4  103  377    2  103 m or 2 mm  50 r 50  2.25

d 0.4 mm  4π  10 7   2.51 10 7 H/m w 2 mm w w 10 9 2 mm C     0 r   2.25   99.5  10 12 F/m d d 36π 0.4 mm

b) L  

1

c) u p  d) w 

με

d Z0



1 LC



 0.4  10   75

1 2.51 10  99.5  10 -7

3

12

 2  108 m/s

0 0.4  103  377   1.33  103 m or 1.33 mm r 75  2.25

d 0.4 mm  4π  10 7   3.77  10 7 H/m w 1.33 mm w w 109 1.33 mm C     0 r   2.25   66.2  1012 F/m d d 36π 0.4 mm 1 1 1 up     2  108 m/s -7 12 με LC 3.77  10  66.2  10 L

9 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

2. General Transmission-Line Equation    



These are equations that govern a general two-conductor uniform transmission lines. Transmission lines differ from ordinary electric networks in that its physical dimensions are a fraction of the operating wavelength or even much longer as compared with the electric networks which are very much smaller than the wavelength. Therefore, it’s convenient to describe a distributed-parameter network and must be describe by a circuit parameters that are distributed throughout its length. A differential length transmission z of transmission line is described by the following four parameters:a) Its resistance per unit length, R b) Its inductance per unit length, L c) Its conductance per unit length, G and d) Its capacitance per unit length, C NB: R and L are series elements and G and C are shunt elements. An equivalent electric circuit for as an element is as shown in Fig. 2.1 i ( z, t )

i ( z  z , t )

N R z

L z

G z

v ( z, t )

C z

v( z  z , t )

z

Fig. 2.1 Equivalent circuit of a differential length z of a two - conductortransmission lines



Applying Kirchhoff’s voltage law, have

v( z, t )  Rz i ( z, t )  Lz

[30]



i( z, t )  v( z  z, t )  0 t Rearranging equ. [30], have 

[31]



v( z  z, t )  v( z, t ) i( z, t )  R i( z, t )  L z t As the limit z  0 , equ. [31] becomes

[32]



v( z, t ) i( z, t )  R i( z, t )  L z t Similar. applying Kirchhoff’s current law to node N of Fig. 2.1, have 

10 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

v( z  z, t )  i ( z  z, t )  0 t On and rearranging, equ. [33] becomes i( z  z, t )  i( z, t ) v( z  z, t )   G v( z  z, t )  C z t i ( z, t )  Gz v( z  z, t )  Cz





[33]

[34]

As the limit z  0 , equ. [34] becomes

i( z, t ) v( z, t )  G i( z, t )  C [35] z t Equations [32] and [35] are the general transmission-line equations or telegraphist’s equations The most important case for practical problem is when v( z, t ) and i( z, t ) varies sinusoidally with time. Thus we can write them as: [36a] v( z, t )  Re[V( z)e jt ] jt [36b] i( z, t )  Re[I( z)e ] where V (z ) and I (z ) are complex amplitudes of the voltage and current along the line respectively. Substituting equ. [36a] and [36b] in equ.[32] and [35], they become For equ. [32], have    V(z)e jt  R I(z)e jt  L I(z)e jt z t   e jt V(z)   R I(z)e jt  jLI(z)e jt or z dV(z)   R  jL I(z) [37a] dz For equ. [35]    I(z)e jt  G V(z)e jt  C V(z)e jt z t   e jt V(z)   R I(z)e jt  jLI(z)e jt z dI(z)   G  jC V(z) [37b] dz 

 



















Equ. [37a] and [37b] are time-harmonic transmission-line equations. a) 

Wave Characteristics on an Infinite Transmission Line The above time-harmonic transmission-line equations can be combined to solve for V(z) and I(z) i.e. Differentiating equ. [37a] with respect to z, have d 2 V(z) dI(z)   R  jL  (i) 2 dz dz 11

EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

Substituting for

dI(z) from equ. [37b] have dz

d 2 V(z)  R  jL G  jC V(z) dz 2 d 2 V(z)  R  jL G  jC V(z) dz 2 d 2 V(z)   2 V(z) dz 2 Similarly for equ. [37b], have, 

or or [38a]

d 2 I(z) dV(z)  G  jC  2 dz dz dV(z) Substituting for from equ. [37a] have, dz d 2 I(z)   R  jL G  jC I(z) or dz 2 d 2 I(z)  R  jL G  jC I(z) or dz 2 

d 2 I(z)   2 I(z) dz 2 With  is the propagation constant define by

(ii)

[38b]

    j  R  jL G  jC  ( m 1 ) [39] with  and  being the attenuation constant ( Np m ) and phase constant ( rad m ) of the line respectively.  The solutions to equ. [38a] and [38b] are as follows: First equ. [38a], d 2 V(z) d  D , then equations becomes   2 V(z) , denoting the differentiation operator, 2 dz dz (a) D   2 V(z)  0 mz Let the solution of (a) be V(z)  e , then equ. (a) becomes m 2 e mz   2 e mz  0 or m 2   2  0 or m  





Thus, V(z)  ez or V(z)  e z Hence, the general solution of equation [38a] is of the form: V(z)  V  (z)  V  (z)  V0 e z  V0- e z Similarly for equ. [38b]

[40a]

12 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon





d 2 I(z) (b)   2 I(z) or D   2 I(z)  0 2 dz with the general solution of equ. (b) being I(z)  I  (z)  I  (z) [40b]  I 0 e z  I -0 e z where the plus and minus superscripts denote waves travelling in the +z and –z directions, respectively.  The amplitudes V0 , I 0 and V0 , I 0  are related by equs. [37a] and [37b] as follows: For the +z direction, dV(z)  V0 e z dz Substituting equ.[37a] in (c), have that R  jL I(z)  V0 e z or R  jL I 0 e z  V0 e z or

R  jL I 0

(c)

 V0 or

V0 R  jL   I 0 Similarly, V  R  jL  0   I0

(41a)

(41b)

 For an infinite line, the terms with e z must vanish i.e. there are no reflections on the transmission line. Then, V(z)  V0 e z and (42a)

I(z)  I 0 e z (42b)  The ratio of voltage to current at any z on the infinite line is called the characteristic impedance of the line given by V(z) V0 R  jL  R  jL (Ω) [43] Z0       I(z)  G  jC G  jC I0  Equs.[39] and [43] are the general expressions for the propagation constant and characteristic impedance. For three special cases of transmission lines, the expressions changes as follows: 1. Lossless line ( R  0, G  0 ), a) Propagation constant:     j  j LC [44] [45]   0,    LC [46] b) Phase velocity:  1 up   [47]  LC c) Characteristic impedance: 13 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

Z 0  R 0  jX 0 

jL L  jC C

[48]

L C

R0 

[49]

X0  0 2. Low-Loss Line ( R  L, G  C ) This conditions are usually satisfied at very high frequencies.

[50]

a) Propagation constant:     j  R  jL G  jC 



 R jL1  jL 

  G  jC1  jC  



 R j2 2 LC1  jL 

 G 1  jC 

12

     

12

 R   G   1    j LC 1  jL   jC    R  G  1    j LC 1   2jL  2jC   R G RG  j LC 1   2  2jL 2jC 4 LC

[51]

    RG R G   since 0  j LC 1   4 2 LC  2jL 2jC   1 R G    j LC 1       2j  L C  

Hence,

1 C L G R  2 L C    LC b) Phase velocity:  1 up    LC c) Characteristic impedance:



Z 0  R 0  jX 0 

R  jL  G  jC

[52] [53] [54] L 1  R jL  C 1  G jC 

14 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

12

L R   G  1   1    C jL   jC  L 1 R G  1     C  2j  L C   L C L 1 R G  X0    0 C 2  L C  3. Distortionless Line ( R L  G C ) a) Propagation constant: R0 

1 2

[55] [56] [57]

    j 

R  jL  RC  jC 



R  jL  RC  jC 



R  jL  C  R  jL 



C  R  jL  L

 L



 L



L

Hence,

C L    LC

 R

[58] [59]

b) Phase velocity:  1 up    LC c) Characteristic impedance: R  jL Z 0  R 0  jX 0   RC L  jC

R0  

L C

[60]

L C

[61] [62]

[63] X0  0 From the above analysis, the characteristics of a distortionless line are the same those of a lossless line except for the attenuation constant. The phase velocity is constant because of linear dependence of the phase constant  on  . Thus, since the signal normally consist s of a band of frequencies, then it’s important that the different frequencies travel along the transmission line at the velocity to avoid distortion. This is achieved in lossless line and is approximated in the low-loss line. In lossy line, 15

EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon



different frequency components attenuate differently even if they travel at the same velocity. But with condition that R L  G C , both  and u p are made constant, hence the distortionless phase. In general,  is not linearly dependent on  leading to u p which depends of frequency. Thus, as different frequencies propagate along the line with different velocities, the signal suffers dispersion. Therefore, generally, lossy transmission line is dispersive.

Example 2 It is found that the attenuation on a 50 Ω distortionless transmission line is 0.01dB/m. The line has a capacitance of 0.1 nF/m. a) Find the resistance, inductance, and conductance per meter of the line b) Find the velocity of wave propagation c) Determine the percentage to which the amplitude of a voltage travelling wave decreases in 1km and in 5km. Solution a) For a distortionless line R G  L C But given that L R0   50 and C

 R

C 0.01  0.01dB  Np/m  1.15  10-3 Np/m L 8.69

Hence,

R 

CL

 1 LC



L  αR 0  1.15  10-3  50  57.5  10-3  m C

L  CR 02  0.1 109  502  2.5  107 H m

CR 0.1 109  57.5  10-3 G   2.3  105 S m 7 L 2.5  10 b) The velocity is given by 1 1 up    2  108 m/s 7 9 LC 2.5  10  0.1 10 c) The ratio of two voltages a distance z apart along the line is V2 V0 e z   e z V1 V0 3 V Thus, after 1km, 2  e z  e 1000  e 10001.1510  0.317,or 31.7% V1

16 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

after 1km,

3 V2  e z  e 5000  e 50001.1510  0.00318,or 0.318% V1

Example 3 A lossless transmission line is 80 cm long and operates at a frequency of 600 MHz. The line parameters are L  0.25 H/m and C  100 pF/m. Find the characteristic impedance, the phase constant, and the phase velocity. Solution For a lossless line, the characteristic impedance is given by Z0 

L 0.25  10-6   50  C 100  10-12

The phase constant is given by

     LC  2ππ(60 106 ) 0.25  10-6  100  10-12  18.85 rad/m The phase velocity is given by  2 (600  106 ) up    2  108 m/s  18.85 Exercise 1) The parameters of a certain transmission line operating at 6  108 rad/s are L  0.4 H/m, C  40 pF/m, G  80 S/m, and R  20 /m. a) Find  ,  ,  and Z0 b) If a voltage wave travels 20 m down the line, what percentage of the original wave amplitude remain, and by how many degrees is its phase shifted? 2) Compare the advantages and disadvantages of coaxial cables and two-wire transmission lines. 3) Compare the velocity of TEM-wave propagation along a parallel-plate transmission line with that in an unbounded medium. 4) Is a distortionless line lossless? Is a lossy transmission line dispersive? Explain. 5) Consider a transmission line made of two parallel brass strips,  c  1.6  107 S/m of width 20 mm and separated by a lossy dielectric slab,    0 ,  r  3,   10-3 S/m , of thickness 2.5 mm. The operating frequency is 500 MHz. a) Calculate the R,L,G and C per unit length b) How compare the magnitudes of the axial and transverse components of the electric field c) Find  and Z0

17 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

TRANSMISSION-LINE PARAMETERS  The electrical properties of a transmission line at a given frequency are completely characterized by its four distributed parameters R,L,G and C.  For parallel-plate transmission line, these parameters have been obtained above. The other types of transmission lines are considered next. a) Two-wire transmission line  Consider a line charge  l located at a distance d from the axis of a parallel, conducting, circular cylinder of radius a . We assume that both the conducting cylinder and the line charge are infinitely long. The cross-section is as shown in Fig. 2.2  To obtain a solution which satisfy the condition that at the surface of the cylinder the potential is same using method of images, the following condition must be noted (i) The image must be a parallel line charge inside the cylinder in order to make the cylinder surface at r  a an equipotential surface. Let this image line charge be  i . (ii) Because of symmetry with respect to the line OP, the image line charge must lie somewhere along OP (at point Pi ) which is a distance d i from the axis as shown in Fig. 2.3

a



P

O

l

d

Fig. 2.2 Line charge and parallel conducting cylinder

a

M ri Pi



O

r



P

i

di

l

d

Fig. 2.3 Line charge and its image





From the method of images, the electrical potential at a point M on the cylindrical surface, a distance r from a line charge of density  l is given by  r [64] VM  l ln i 2 0 r If the equipotential surface is to coincide with the cylindrical surface, then from image Fig. 2.3, the following ratio must be satisfied

18 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon







ri d i a or [65]    constant r a d a2 [66] di  d And thus the line charge   l   i , together with  l will make the dashed cylindrical surface of Fig. 2.3 equipotential. As the point M changes its location on the dashes circle, both ri and r will change; but their ratio remains a constant that equals a d . Point Pi is called the inverse point of P with respect to a circle of radius a . Therefore, the image charge   l can then replace the cylindrical conducting surface, and V and E at any point outside the surface can be determined from the line charges   l   i , and  l . Applying above principle of method of images to the two-wire transmission line with the cross-section as shown in Fig. 2.4, the equipotential surfaces of the two wires can be considered to have been generated by a pair of line charges  l and   l separated by a distance D  2di  d  d i . The potential difference between the two wires is that between any two points on the respective wires. 2

1

a

di

 l 

 l 

a

di

d

D



Fig. 2.4 Cross-section of a parallel wire transmission line and equivalent line charges Thus, applying equ. [64] and [65] to Fig. 2.4, we have that  a and V2  l ln 2 0 d

V1   

l a ln 2 0 d

NB: V1 is a positive quantity while V 2 is negative because a  d . The capacitance per unit length is given by

C

l V1  V2



 0

[67]

lna d 

19 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

a2 where d  D - d i  D d from which we get a2 or d  D - di  D d a2 d D or d 2 - a 2  Dd or d 2 d - Dd - a 2  0 With the solution being 1 d  D  D 2  4a 2 or [68] 2 1 d  D  D 2  4a 2 2 1 d  D  D 2  4a 2 is discarded because both D and d are usually much larger than a. 2 Substituting equ. [68] in [67], we have

  

  

C 

 0 ln D 2a  

Since

F m 

D 2a  1  2



[69]



ln x  x 2  1  cosh1 x

Then for x  1 , equ. [69] becomes



F m  cosh D 2a Equ. [70] applies to any medium in which the wires are.  From equ. [22], LC   And hence    D H m  L   cosh1    C   2a  cosh1 D 2a  And for an homogeneous medium, C  RC   G  Hence C 

1

[70]

[71] [72]

[73]



G

C





cosh

1

D 2a 



 cosh

1

S m

D 2a

[74]

20 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon



To determine R, we use equ. [28] and express the ohmic power dissipated per unit length of both wires in terms of p . Assuming the current J s  A m is flowing in very thin surface layer, the current in each wire is I  2aJ s and 1 2  Rs  I   W m  2  2a  Hence the series resistance per unit length for both wires is  R  1 f c  m  R  2 s    2a  a  c Pσ  2apσ 

[75]

[76]

b) Coaxial transmission line  A cross-section of a coaxial cable is as shown in Fig. 2.5.

a b

Fig. 2.5 Coaxial transmission line geometry







Assuming a current I flows in the inner conductor and returns through the outer conductor in the other direction, the due symmetry, magnetic flux density, B , has only a   component with different expressions in the two regions: (i) Inside the inner conductor (ii) Between the inner and outer conductors Also assume that the current I is distributed over the cross section of the inner conductor. Then (i) Inside the inner conductor From Ampere’s circuital law which states that the circulation of the magnetic flux density in free space around any closed path is equal to  0 times the total current flowing through the surface bounded by the path i.e. [77]  B. dl  0 I C

Thus inside the conductor, B1  a B 1 , dl  a rd



C1



2

B. dl   B 1 rd  2rB 1   0 I 1 0

The current through the cross-sectional area of the inner conductor is

21 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

I1 

r 2 r2 I  I a 2 a2

Thus from Ampere’s circuital law,  rI B1  a B 1  a 0 2 2a Thus in any region within the inner conductor, the magnetic flux density is  rI B1  a B 1  a 0 2 , 0  r  a 2a

[78]

[79]

(ii) Between the inner and outer conductors B2  a B 2 , dl  a rd



C2



2

B2 . dl   B 2 rd  2rB 2   0 I 0

The circumference of the outer conductor encloses the total current I. Hence,

B2  a B 2  a 

0 I ar b 2r

[80]

Consider an annular ring in the inner conductor between the radii r and r  dr . The current in a unit length of this annular ring is link by the flux that can be obtained by integrating equs. [79] and [80] i.e.

d   B 1 dr   B 2 dr a

b

r

a

0 I a  I b dr rdr  0  2 r 2 a r 2a  I  I b  0 2 a 2  r 2   0 ln 2 a 4a





[81]

But the current in the annular ring is only a fraction of the total current I i.e.

r

r  dr

Fig. 2.6 Annular ring in the inner conductor





Area of ring =  r  dr   r 2   r 2  2rdr  dr   r 2 2

2

 r  2rdr   dr   r  2rdr (Since  dr   0 ) 2

2

2

2

22 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

Hence ratio of current flowing in the annular ring is a fraction of the total current flowing in the inner conductor i.e.

2rdr 2rdr  2 a 2 a  Therefore, the flux linkage for this annular ring is 2rdr d   2 d a The total flux linkage per unit length is

  

r a

r 0

[82]

d 

r a

2rdr d r 0 a 2 r  a 2rdr   I  I b 0  a 2  r 2  0 ln   2 2 r 0 a 2 a  4a  I a  I b a  0 2  a 2  r 2 rdr  0 2 ln  rdr a 0 2a 0 a 4 4  I a a   I b  a2   0 2     0 2 ln   4  a a 2  2a  2 









0 I 0 I b  ln 8  a 0 I  1 b    ln  2  4 a











The inductance of a unit length of the coaxial transmission line is thus,   0  0 b L    ln I 8 2 a The first term

[83]

0 arises from the flux linkage internal to the solid inner conductor; known 8

as the internal inductance per unit length of the inner conductor. The second term comes the linkage of the flux that exists between the inner and outer conductors; it’s known as the external inductance per unit length of the coaxial line. In high frequency, the current in a good conductor shift to the surface of the conductor (due to skin effect), resulting in an uneven current distribution in the inner conductor and thereby changing the value of the internal inductances. In some cases the current may essentially concentrate in the skin of the inner conductor as a surface current, and the internal selfinductance is reduced to zero. Thus, since most transmission involves high frequencies, the internal inductance is assumed to be zero and hence the inductance of the coaxial transmission line is assumed to due to external inductance only i.e.

L

0 b ln 2 a

H m 

[84] 23

EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon



Using equ. [71],

C 





 L



  b ln 2 a



2 lnb a 

F m 

[85]

And from equ. [73], have 2  lnb a  C 2 S m [86] G     lnb a  where  is the equivalent conductivity of the lossy dielectric i.e.     To obtain R, we use equ.[27], where the J si on the surface of the centre conductor is different from J so on the inner surface of the outer conductor. But [87] I  2aJ si  2bJ so Thus, the power dissipated in a unit length of the centre conductor and outer conductors are respectively, 1 1  R  2 Pσi  2apσi  J si Rs  I 2  s  [88] 2 2  2a  1 1  R  2 J so Rs  I 2  s  2 2  2b  From equs. [88] and [89], the resistance per unit length is: R R R  1 1  1 f c  1 1   m  R s  s  s       2a 2b 2  a b  2  c  a b  Pσo  2bpσo 



[89]

[90]

Example 1 The dimension\s of a coaxial transmission line are a  4 mm , b  17.5 mm , and c  20 mm .The conductivity of the inner and outer conductors is 2  107 S/m , and the dielectric properties are  R  1 ,  R  3 , and  /    0.025 . Assume that the loss tangent is constant with frequency. Determine: a) L, C, R, G and Z 0 at 150 MHz b) L and R at 60 Hz

WAVE CHARACTERISTICS ON FINITE TRANSMISSION LINE  The general solutions for the time-harmonic one-dimension Helmholtz equaitions are

V(z)  V0 e z  V0- e z

[91a]

I(z)  I 0 e z  I 0- e z where

[91b]

24 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon



 

V(z) V0 [92]   Z0 I(z) I 0 As mention above, the reflected component is zero for infinite lines. For a finite the same can be achieved if the lines are matched i.e. if the line is terminated in characteristic impedance. Thus in transmission lines, a line is matched when the load impedance is equal to the characteristic impedance (not the complex conjugate of the characteristic impedance as in circuit theory) of the line. Thus, consider a finite transmission line having characteristic impedance Z 0 terminated in arbitrary load impedance of length of the line to be l as shown in Fig.2.7. A sinusoidal voltage source V g 0 0 with internal impedance Z g is connected to the line at

z  0 . Thus, 

Zg

Vg

Ii  Vi



IL

 , Z O 

Zi

ZL

 VL



 z

z  l - z

z 0 z  l

z  l z 0

Fig. 2.7 Finite Transmission line terminated with load impedance Z L



V V  [93]    L  ZL  I  z l I L which can only be satisfied by equ. [91a] and [91b] with their reflection components since the transmission line is not matched. At z  l , equs. [91a] and [91b] becomes VL  V0 e l  V0- e l [94a] IL 

V0 l V0- l e  e Z0 Z0

[94b]

Solving equ. [94a] and [94b] for V0 and V0 , have that

V0 

1 VL  I L Z 0 e l 2

[95a]

1 VL  I L Z 0 e l [95b] 2 Substituting equs. [93] in equs. [95a] and [95b] and the results in equs. [91a] and [91b], we obtain V0 

25 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

1 VL  I L Z 0 e l  1 I L Z L  I L Z 0 e l 2 2 1 1 V0  VL  I L Z 0 e l  I L ZL  I L Z 0 e l 2 2 Thus, V0 









IL [96a] Z L  Z0 e l-z   Z L  Z0 e  l-z  2 I [96b] I(z)  L Z L  Z 0 e  l-z   Z L  Z 0 e  l-z  2Z 0 Let z    z , which is the distance measured backward from the load. Hence, equs. [96a] and [96b], becomes I [97a] V( z)  L Z L  Z 0 ez  Z L  Z0 e z 2 I [97b] I(z )  L Z L  Z 0 e z  Z L  Z 0 e z 2Z 0 Since V(z) 









ez  e z  2 coshz aand ez  e z  2 sinh z , then [98a] V( z)  I L ZL coshz  Z0 sinh z I [98b] I(z )  L Z L sinh z  Z 0 cosh z  Z0 which can be used to find the voltage and current at any point along a transmission line terms of I L , Z L ,  and Z 0 . Now, Vz   is the impedance looking toward the load end of the line at a distance z from the Iz  load can be written as:Z cosh z  Z 0 sinh z Vz  [100] Zz   Z0 L Iz Z L sinh z   Z 0 cosh z Or Z  Z 0 tanh z Ω [101] Zz  Z 0 L Z 0  Z L tanh z At source end of the line, z   , the generator looking into the line sees an input impedance Zi . Z  Z 0 tanh  Ω [102] Z i  Z  Z z 0  Z 0 L Z 0  Z L tanh  z   Hence, as far as the conditions aat the generator are concerned, the terminated finite transmission line can be replaced by Z i as shown in Fig. 2.8.

26 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

Zg Ii Zi

Vg

Vi

Fig. 2.8 Equivalent circuit for finite transmission line of Fig. 2.7 at generator end. With the input voltage Vi and input current I i being obtained as Zi Vi  Vg Zg  Zi

[103a]

Vg

Ii 

[103b] Zg  Zi The power delivered by the generator to the input terminal of the line is Pav i  1 Re Vi I*i z0,z [104] 2 The average power delivered to the load is Pav L  1 Re VL I*L z,z0 [105] 2 If Z L  Z0 (i.e. the line is terminated with its characteristic impedance), then from equ [102], Zi  Z 0 or the impedance of the line looking toward the load at any distance z from the load is [106] Zz  Z0 And equs. [97a] and [97b], reduces to Vz   I L Z 0 e  e z  Vi e z [107a]

  







[107b] Iz  I L e e z  I i e z Now, equs. [107a] and [107b] are similar to equs. [42a] and [42b] - representing waves travelling in +z direction, and thus no reflected waves. Hence, when a finite transmission line is terminated with its own characteristic impedance (when a finite transmission line is matched), the voltage and current distribution on the line are exactly the same as though the line has been extended to infinity. Example: A signal generator having an internal resistance 1 Ω and an open-circuit voltage Vg t   0.3 cos 2ππ18 t V is connected to a 50 Ω lossless transmission line. The lne is 4 m long, and the velocity of the wave propagation on the line is 2.5 108 m / s . For a matched load, find a) Instantaneous expression for the voltage and current at an arbitrary location on the line b) The instantaneous expressions for the voltage and current at the load, and 27 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

c) The average power transmitted to the load Solution: a) To get the voltage and current at an arbitrary location on the line, its necessary to first obtained those at the input end z  0, z   . Thus, given Vg  0.30 0 V , a phasors with a cosine reference Zg  R g  1 Ω

Z0  R 0  50 Ω

  2π 108 rad/s u p  2.5  108 m/s

  4m Since the line is terminated with a matched load, Zi  Z0  50Ω , then the input voltage and current can be evaluated from the equivalent circuit of Fig. 2.8. Thus, from [103a] and [103a], have Vi 

Zi 50 Vg   0.30 0  0.2940 0 V Zg  Zi 1  50

0.30 0 Ii    0.00590 0 A Zg  Zi 1  50 Since only forward waves exist in a matched line, we use the equations, Vg

Vz   V0 e   j z

V0   j z e Z0 Now, for the given line,   0, and Iz  







2π  108  0.8π rad/m 2.5  108

up Hence, Vz  0.94e j0.8z V Iz  0.0059e j0.8z A These are phasors. The corresponding instantaneous expressions are, 8 vz, t   Re 0.294ej2 10 0.8z 









 0.294cos 2  10  0.8z V 8 iz, t   Re 0.0059ej2 10 0.8z 



8







 0.0059cos 2π 108  0.8π. A b) At the load , z  0, z    4 m , v4, t   0.294cos 2π 108  3.2π V





28 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon





i4, t   0.0059cos 2π 108  3.2π A c) The average power transmitted to the load on a lossless line is eqeal to that at the input terminal . Pav L  Pav i  1 Re VL z I*L z  2 1  0.294  0.0059  8.7  10 4 W 2





TRANSMISSION LINES AS CIRCUIT ELEMENTS  Transmission lines at ultrahigh frequencies (frequencies from 300 MHz to 3 GHz), wavelength from 1 m t0 0.1 m, they serve as circuit elements. At these frequencies, ordinary lumped-circuit elements are difficult to make, and stray fields become important. Thus, sections of transmission lines can be designed to give an inductive or capacitive impedance and are use to match an arbitrary load to the internal impedance of a generator for maximum power transfer.  At higher than 3GHz frequency, they become physically small and normally waveguides components are used.  In most cases, transmission-line segments can be considered lossless:   j , Z0  R 0 , and tanh   j tanh  j   j tan   . Thus, equ.[102], for the input impedance Z i of the lossless line of length  terminated in Z L becomes

Z L  jR 0 tanh  [108] Ω R 0  jZ L tanh  The special cases of the transmission line include: 1) Open-circuit termination Z L    . Equ. [108] becomes: jR 0 Z i  jX io     jR 0 cot  [109] tan   Equ. [109] shows that the input impedance of an open-circuited lossless line is purely reactive. It can either be capacitive or inductive depending on the value of  2   . Fig. 2.9, shows a plot X io  R 0 cot  verse  .  When   1 , the capacitive reactance can be obtained by noting that tan    . Hence from equ. [109], LC R 1 Z i  jX io   j 0   j  j [110]  C  LC which is the impedance of a capacitance of C farads. Zi  R 0

2) Short-circuit termination Z L  0 . Here, equ. [108] reduces to jR 0 tan  Z i  jX is  R 0  jR 0 tan  R0

[111]

29 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

X io  R 0 cot 

Since tan  can range from   to   , the input impedance of a short-circuited lossless line can also be either purely inductive or capacitive, depending on the value of   . A graph of X is verse  is as shown in Fig.3.0.

0

(Inductive)





2



3 4

4



5 4 (Capacitive)

X is  R 0 tan 

Fig. 2.9 Input reactance of open-circuited transmission line

(Inductive)

 0

3 4

4





5 4



2 (Capacitive)

 

Fig. 3.0 Input reactance of short-circuited transmission line Comparing Fig. 2.9 and 3.0, have that in the range where X io is capacitive X is is inductive and vice versa. For short-circuited line, if  l  1 , then equ. [111] becomes

jR0 βl L  jR0 βl  j ω LC l  jωωL R0 C which is the impedance of an inductance of Ll henries. Z i  jX is  R0

[112]

30 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

3) Quarter-wave section l   4,l   2  . When the length of the line is an odd multiple of  4 i.e. l  2n  1 λ 4 , n  1,2,3,..., then 2π λ π   2n  1  2n  1 and λ 4 2 π   tan   tan 2n  1    2  Thus, equ. [108] becomes ZL  jR 0 Z L  jR 0 tanh  R2 tanh  [113] Zi  R 0  R0  0 R0 R 0  jZ L tanh  ZL  jZ L tanh  Hence, a quarter-wave lossless line transforms the load impedance to the input terminals as its inverse multiplied by the square of the characteristic resistance. It acts as an impedance inverter and is often called a quarter-wave transformer. Thus, an open-circuited, quarter-wave line appears as a short circuit at the input terminals and vice versa. 4) Half -wave section    2,     . When the length of the line is an integral multiple of λ 2 ,   n 2 , n  1,2,3,... , then 2π  nλ        n , λ  2  tan   0 and equ. [108] reduces to

Z L  jR 0 tanh  Z [114]  R 0 L  ZL R 0  jZ L tanh  R0 Thus, a half-wave lossless line transfers the load impedance to the input terminals without change. By measuring the input impedance of a line section under open- and short-circuit conditions, the characteristic impedance and the propagation constant of the line can be determined. Now from equ. [102], have that Open-circuited line, Z L   : [115] Z io  Z 0 coth  Short-circuited line, Z L  0 : [116] Zis  Z0 tanh   Therefore from the two equations, Zi  R 0

Z 0  Zio Zis And 1 

  tanh 1



Z is Z io

[117]

m -1

[118]

31 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

Example The open-circuti and short-circuit impedances measured at the input terminals of a lossless transmission line of length 1.5m, which is less than a quarter wavelength, are  j54.6 and j103 , respectively. a) Find Z 0 and  of the line b) Without changing the operating frequency, find the input impedance of a shortcircuited line that is twice the given length c) How should the short-circuited line be in order for it to appear as an open circuit at the input terminal? Solution: Given Z i 0   j54.6 , Z is   j103 , l  1.5m, then a) Z 0  Z io Z is 

 j54.6 j103  75 

Z is 1 1 j103 j  tanh 1  tan 1 1.373  j0.628 rad/m l Z io 1.5 - j54.6 1.5 b) For a short-circuited line twice as long , l  3 m , l j0.628  3  j1.883 rad Thus, the input impedance is given by Z is  Z 0 tanh    75tanh j1.883  j75tan 1080  j75- 3.08   j231  c) In order for a short circuited line to appear as an open circuit at the input terminals, it should be an odd multiple of a quarter-wavelength long: 2 2π    10 m  0.628 Hence the required line length is

  tan 1



l  2n  1





4  2.5n  1, n  1,2,3,... Above consideration having that of a lossless transmission line, but if the line is a lossy, then, when the line length is a multiple of λ 2 with a short-circuit termination, equ. [116], becomes sinh   j l Z is  Z 0 tanh    Z 0 cosh  j l sinh l cosh jl  coshl sinh jl  Z0 coshl cosh jl  sinh l sinh jl

sinh l cos l  j coshl sin l coshl cos l  j sinh l sin l For, l   2,l  nπ and sin l  0 , hence equ. [119] becomes sinh l cos l sinh l Z is  Z 0  Z0  Z 0 tanh l cosh l cos l cosh l Assuming a low-loss line i.e. l  1 and thus, tanh l  l and Z is  Z 0 l   Z0





[119]

[120]

32 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon



 



which gives a small value of Z is but not zero as in the previous analysis. At l  n  2 in this case is a condition of a series resonance. If l  n  4 ,( n  odd ), l  n 2 and cos l  0 and equ. [119] becomes Z0 Z j coshl sin l coshl Z is  Z 0  Z0   0 [121] j sinh l sin l sinh l tanh l l which is large but not infinite. This is a condition for a parallel-resonance. From above analysis, behaves as a frequency-selective circuit with a quality factor, Q, which can be determined by first obtaining the half-power bandwidth. A bandwidth of a parallel-resonant circuit is the frequency at range f  f 2  f 1 around the resonant frequency f 0 , where f 2  f 0  f and f 1  f 0  f are half-power frequencies at which the voltage across the parallel circuit is 1 2 or 70.7% of the its maximum value at f 0 . Let f  f 0  f , f - small frequency shift from the resonant frequency. Then,

l 

2 π f 0  f  2 πf l l up up 

2 π f 0  f  n nπ  f 0  f    up 4 2 up



nπ  f 0  f  nπf 0  2 up 2u p

 f  1  f0 

  

but

f0   u p 2

2 nπ   f  n  f   1     1   , but    2 2  f0  4  f0  nπ   f  n 2  f  n  f  1     1     1   2 2  f0  4   f0  2  f0 nπ nπ  f    , n  odd   2 2  f 0  

 nπ nπ  f    nπ  f   nπ  f      sin       cos l  cos     2  f0  2  f0  2  2  f0   nπ nπ  f    nπ  f       cos     1 sin l  sin   2 f  2  f 0    2   0 

   [122]

  

[123] [124]

nπ  f     1 2  f 0  Substituting equs. [122], [123], and [124] in [119], noting that l  1 , and retaining only small terms of the first order, we get where it has assumed that

33 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

Z is 

Z0 nπ  f l  j  2  f0

[125]

  

and

Z is



2

2

Z0 





l 2   nπ  f  

[126]

2

 2  f0  2

At f  f 0 , f  0 , Z is is a maximum and Z is 2

Z is

2 max



Z0

2

l 2

. Thus,

1



[127] 2     n π  f max    1    2  l  f0   When f   f 2 ,that is the half-power frequencies at which the ratio of equ. [127] is 1 or 2 nπ  f    f       1 , n  odd [128] 2l  2f0  2  f 0  Therefore, the Q of a parallel-resonant circuit (a shorted lossy line having a length equal to an odd multiple of  4 ) is f  Q 0  [129] Δf 2 But from equs. [52] and [53] for a low-loss line, have 1 C L   R G  2 L C

Z is

2

   LC Hence,

Q

  2

 LC 1 C L 2  R G  2 L C



L R  GL C



1 R L  G C 

[130]

For a well insulated line GL C  R , L Q [131] R Example The measured attenuation of an air-dielectric coaxial transmission line at 400 MHz is 0.01 dB/m . Determine the Q and the half-power bandwidth of a quarter-wavelength section of the line with a short-circuit termination. 34 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

Solution: Given f  4  10 8 Hz ,

c 3  10 8    0.75 m f 4  10 8 2 2 8     rad/m or 8.38 rad/m  0.75 3 0.01   0.01 dB/m  Np/m 8.69 Therefore,  8.38  8.69 Q   3640 2 0.01 The half-power bandwidth f 0 4  10 8 Δf    109.89 KHz Q 3640

LINES WITH RESISTIVE TERMINATION  When a line is terminated in a load impedance Z L  Z 0 , both an incident wave and reflected wave exist. Equ. [96a] gives the phasors expression for the voltage at any distance z   l - z from the load end.  NB: the term e z represents the incident voltage wave and the term with e z represents the reflected voltage wave.  Thus, I V(z )  L Z L  Z 0 e γz  Z L  Z 0 e γz 2







  IL Z L  Z 0 e z 1  Z L  Z 0 e 2γz  2  Z L  Z0 









[132a]

IL Z L  Z 0 ez 1  e 2γz 2 where Z  Z0 [133]  L   e j Z L  Z0  is called the voltage reflection coefficient of the load impedance Z L . Its generally a complex quantity with a magnitude   1 . The current equation corresponding to V(z) from equ. [97b] is I [132b] I z   L Z L  Z 0 e z 1  e 2 γz 2Z0 For a lossless line,   j , thus equ. [132a] and [132b] becomes 







35 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon





IL Z L  R0 e jz 1  e  j2z 2 I  L Z L  R0 e jz 1  e j 2 z  2

V(z ) 





[134a]

and









 





IL Z L  R0 ez 1  e j 2 z 2Z0 But from equ. [98a] and [98b], the voltage and current phasors can be written as I  z  

[134b]

[135b] V(z )  VL cos βz   jI L R0 sin βz  V [135b] I(z )  I L cos βz   j L sin βz  R0 If the termination is purely resistive, Z L  RL and VL  I L RL , the voltage and current magnitudes are given by 2 V( z )  VL cos2 βz   R 0 R L  sin 2 βz 

[136a]

2 I(z )  I L cos2 βz   R L R 0  sin 2 βz 

[136b]

Plots of V( z ) and I(z ) as functions of z are standing waves with their maxima and minima occurring at fixed locations along the line. Thus, a ratio of the maximum to minimum voltages along the line can be defined called the standing-wave ratio (SWR), S: V 1 Γ [137] S  max  Vmin 1  Γ or S 1  [138] S 1 Hence, from equs. [137] and [138], for a lossless transmission line have that 0, S  1 when Z L  Z 0 (matched load) S   when Z L  0 (short circuit)   -1 , S   when Z L   (open circuit)   1 , NB: A high standing wave ratio is on a line is undesirable because it results in a large power loss. From equs. [134a] and [134b], the Vmax and I min occur together when

   2βz M   2nπ,,

n  0,1,2

[139]

and for the V min and I max , they occur together when

   2βz M   2n  1π,,



n  0,1,2 For a resistive termination on a lossless line, Z L  RL , Z 0  R0 and equ. [133],

[140]

36 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

RL  R0 RL  R0 The voltage reflection coefficient is therefore purely real. Two cases are possible. 

[141]

(i) RL  R0 : In this case  is  ve and    0 . At termination, z   0 and condition in equ. [139] is satisfied (for n  0 ) . This means Vmax (or I min ) will occur at the termination resistance. Other Vmax (or I min ) occur at z   n 2 (n  1,2,...) from the load. (ii) RL  R0 : equ. [141] shows that  is negative real and    - . At termination, z   0 and condition [140] is satisfied (for n  0 ) This means V min (or I max ) will occur at the 

termination resistance. Other Vmax (or I min ) occur at z   n 2 (n  1,2,...) from the load. This is illustrated in Fig. 3.1. V  z  for RL  R0

I z  for RL  R0

I z  for RL  R0 V z  for RL  R0 z



0 3 4  2  4 Fig. 3.1 Voltage and current standing waves on resistance-terminated lossless lines



Open circuit line Standing wave is similar to a resistance terminated line with RL  R0 , except that V( z ) and I( z ) curves are magnitudes of sinusoidal functions of the distance z from the load. This can

be verified by letting RL   in equ. [136a] and [136b]. i.e. I L  0 and VL   . Thus, V(z )  V L cos βz  [142a]

I(z ) 







S

VL sin βz  R0

[142b]

RL  R0 1  R0 RL   1 and RL  R0 1  R0 RL 1 Γ 1 Γ



11  11

37 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

Short-circuited line The standing-waves are similar to those on a resistance-terminated line with RL  R0 . Here, RL  0 , V L  0 and I L is finite. Hence, equ. [136a] and [136b] reduces to V(z )  I L R0 sin βz  [143a] I( z )  I L cos βz 



[143b]

Typical standing waves for open- and short-circuited lines are shown in Fig. 3.2.

V z  for open - circuited line I z  for short - circuited line

I z  for open - circuited line V z  for short - circuited line z



3 4

 2

 4

0

Fig. 3.2 Voltage and current standing waves on open- and short-circuited lossless lines

LINES WITH ARBITRARY TERMINATION Let the terminating impedance be Z L  RL  jX L . Assume the voltage standing wave on the line to be as shown Fig. 3.2.

38 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon



Thus, nether a Vmax nor V min appear at z   0 . V min will occur at an extra distance lm i.e. V min is where its suppose to be if the original terminating impedance Z L is replaced by a



line section of length lm terminated by a pure resistance. Since any complex impedance can be obtained as the input impedance of a section of a lossless line terminated in a resistive load as shown in equ. [108], then using Rm for Z L and lm for l , we get R  jR 0 tan βlm [144] Ri  jX i  R0 m Ω R 0  jR m tan βlm And thus, we can solve for Rm and lm

Z L can be obtained can be determined if S and z m in the figure above know. [note that z m  l m   2 ] . The procedure is:S 1 (i) Find  from S. Use   S 1 (ii) Find   from z m . Use    2z m   for n  0 (iii)Find Z L i.e. 1   e j  [144] Z L  RL  jX L  R0 1   e j  Example: The standing-wave ratio on a lossless 50 Ω transmission line terminated in an unknown load impedance is found to be 3.0. The distance between successive voltage minima is 20 cm, and the first minimum is located at 5 cm from the load. Determine (i) The reflection coefficient  (ii) The load impedance Z L (iii)In addition, find the equivalent length and terminating resistance of a line such that the input impedance is equal to Z L .. Solution: (i) The distance between successive voltage minima is half a wavelength. Hence,   2  0.2  0.4 m and 2 2    5 rad/m  0.4 S 1 31   0.5 Step 1:   S 1 31 Step 2:    2z m    2  5  0.05    0.5 rad



   e j   0.5e  j0.5π   j0.5

(ii) The load impedance Z L 1  j0.5 Z L  50  500.06 - j0.8  30  j40  1  j0.5 R  j50 tan βlm (iii) 30  j40  50 m or 50  jRm tan βlm 39 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

z m  l m   2 or lm   2  z m  0.2  0.05  0.15 m R 50 Rm  0   16.7  S 3 THE SMITH CHART  Transmission-line calculations such as the determination of input impedance, reflection coefficient and load impedance often involve tedious manipulations of complex numbers. This can be eliminated using graphical method. One of the most widely used graphical chart is the Smith chart.  A Smith chart is a graphical plot of normalized resistance and reactance functions in the reflection-coefficient plane.  Consider a lossless transmission line with reflection coefficient being given by Z  R0 [145]  L   e j Z L  R0  Normalizing the load impedance with respect to characteristic impedance of the line, have Z X [146] z L  L  j L  r  jx R0 R0  Thus z 1 [147]   r  ji  L zL  1  Rearranging equ. [147], have that j  1 1 e  zL   1   1   e j or

r  jx 



1  r   ji 1  r   ji

[148]



1  r   ji  1  r   ji 1  r   ji  1  r   ji

1  r2  i2  j2i

1  r 2  i2

 Hence 1  r2  i2 and r 1  r 2  i2 2i x 1  r 2  i2  Rearranging equ. [149], have

[149] [150]

40 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

2

r    1  2  r    i    1 r   1 r 

2

[151]

r r and centred at r  and i  0 1 r 1 r Different values of r yields circles of different radii with centres at different positions on r  axis . which is an equation of a circle having a radius



i

r

0



Thus, the properties of the r-circles are:(i) The centres of all r-circles lie on the i  axis . (ii) The r  0 circle, having a unity radius and centred at the origin, is the largest. (iii)The r-circles become progressively smaller as r increases from 0 to  , ending at the r  1, r  0  point for open circuit. 41

EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

(iv) All r-circles pass through the r  1, r  0  point 

Similarly, equ. [150] can be written as 2

r  1   i  1    1  x   x

2

2

[152]

which is an equation of circle having a radius 

1 1 and centered at r  1 and i  . x x

Different valued of x yields circles of different radii with centres at different positions on the r  1 line. The properties of these x -circles are:(i) The centres of all x -circles lie on the r  1 ; those for x  0 (inductive reactance) lie above the r  axis and those for x  0 (capacitive reactance) lie below the r  axis . (ii) The x  0 circle becomes the r  axis . (iii)The x -circles becomes progressively smaller x increases from 0 to  , ending at the

r













 1, r  0  point for open circuit. (iv) All x -circles pass through the r  1, r  0  point. The Smith chart above marked with r and i uses rectangular coordinates. The same chart can be marked with polar coordinates, such that every point in the   plane is specified by a magnitude  and a phase angel   as shown below.

These   circles are not normally shown on the chart but can be drawn once the point representing a certain z L  r  jx is located by simply drawing a circle centred at origin through the point. The fractional distance from the centre to the the point (compared to with the unit radius to the edge of the chart) is equal to the magnitude  of the load reflection coefficient; and the angle that the line makes with the real axis is   . Since x  0 along the real axis, PM and Pm both represents situation with a purely resistive load Z L  RL . Thus, RL  R0 at PM , where r  1 ; and RL  R0 at Pm , where r  1 . Therefore, since S  RL R0  r for RL  R0 , then the value of the r -circles passing through the point PM is numerically equal to the standing-wave ratio while the value passing through Pm on the negative- real axis is numerically equal to 1 S . If the input impedance looking towards the load at distance z from the load is used, then for a lossless line, [from equs. [132a] and [132b]], have  1  e  j2 z  V  z  [153] Z i  z    Z0   j 2 z   I  z  1   e   The normalized input impedance is j Z i 1  e  j2 z 1   e   Z 0 1  e  j2 z 1   e j where

zi 

[154]

42 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

     2 z  

[155]

Equ. [154] is exactly the same form as equ. [148]. Therefore  and hence S are not changed by the additional line length z . Thus, on Smith chart we find  and   for a given z L at the load and keeping  constant, we subtract (rotate in the clockwise direction) from   an angle equal to 2 z   4z   . This will locate the point  e j which determines z i .

i

0

r

43 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

Example 1: Use the Smith chart to find the input impedance of a section of a 50  lossless transmission line that is 0.1 wavelength long and is terminated in a short-circuit. Solution: zL  0 R0  50  z   0.1 (i) Enter the smith chart at the intersection of r  0 and x  0 (point Psc ) (ii) Move along the perimeter of the chart (   1 ) by 0.1 “wavelengths toward generator” in a clockwise direction to P1 . (iii)At P1 , read r  0 and x  0.725 or z i  j0.725 . Thus the input impedance is Z i  R0 zi  50  j0.725  j36.25  (which is purely inductive) This can be verified using equ.[111] i.e.  2  Z i  jX is  jR0 tan l  j50 tan  0.1  j50 tan 0.2   j36.33     Example 2: A lossless transmission line of length 0.434 and characteristic impedance 100  is terminated in an impedance 260  j180  . Find a) The voltage reflection coefficient b) The standing-wave ratio c) The input impedance, and d) The location of a voltage maximum on the line. Solution: Given Z L  260  j180  R0  100  z   0.434 a) The voltage reflection coefficient is obtained using the following steps: 260  j180 Ω  2.6  j1.8 [point P2 ]. 1. Enter the Smith chart at z L  Z L R 0  100 2. With the centre at the origin, draw a circle of radius OP 2    0.60 .(The radius of the chart OP sc  1 ). I.e. on measuring, OP 2  4.7cm and OP sc  7.7cm ,hence, OP 2 4.7cm   0.61 OP sc 7.7cm 3. Draw a straight line OP2 and extend it to P2 on the periphery. Read 0.220 on “wavelengths toward generator” scale. The phase angle    0.25  0.22  4π  0.12 rad or 210 .[we multiply the change in wavelength by 4π because angles on the Smith chart are measured in 2z or 4πz  . A halfwavelength change in line length corresponds to complete revolution on the Smith chart.] The answer then is 44 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

   e j   0.61210.

b) The standing-wave ratio The   0.61circle intersects with the positive-real axis OPoc at PM i.e. r  S  4 . Hence, the voltage standing-wave ratio is 4. c) The input impedance 1. Move P2 at0.22 by a total of 0.434 “wavelengths toward the generator” first to 0.500 (same as 0.000) and then further to 0.154 [ 0.500  0.220  0.154  0.434 ] to P3 . 2. Join O to P3 by a straight line which intersects the   0.61circle at P3 . 3. Read r  0.69 and x  1.2 . Hence, Zi  R 0 z i  1000.69  j1.2  69  j120 Ω d) The location of a voltage maximum on the line. In going from P2 to P3 , the   0.61 circle intersects the positive-real axis OPoc at

PM where the voltage is a maximum. Thus, a voltage maximum appears at 0.250  0.220  0.03 from the load. Example 3: The standing-wave ratio on a lossless 50 Ω transmission line terminated in an unknown load impedance is found to be 3.0. The distance between successive voltage minima is 20 cm, and the first minimum is located at 5 cm from the load. Determine (i) The reflection coefficient  (ii) The load impedance Z L (iii)In addition, find the equivalent length and terminating resistance of a line such that the input impedance is equal to Z L .. Solution: Given R0  100  S  3.0   2  0.2  0.4m First voltage minimum at zm  0.05m a) The reflection coefficient  On the positive-real axis OPoc , locate the point PM at which r  S  3.0 . Then

OP M 3.9cm   0.5 OP sc 7.7cm We cannot find   until we have located the point that represents the normalized load impedance. b) The load impedance Z L 1. Draw a circle centred at the origin with radius OPM , which intersects with the negativereal axis OPsc at Pm where there will be a voltage minimum. 2. Since z m   0.05 0.4  0.125, move from Psc ”wavelength toward load in the counterclockwise direction to PL . OPM    0.5 i.e.

45 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

3. Join O to PL by a straight line , intersecting the   0.5 circle at PL . This is the point representing the normalized load impedance. 4. Read the angle Poc OPL  900   2 rad i.e. Poc OPL  4π0.25  0.125   2 rad . Hence,    - π 2 rad or

  0.5 - 900   j0.5. 5. Read at PL , z L  0.6  j0.8 , thus, Z L  500.6  j0.8  30  j40  c) In addition, find the equivalent length and terminating resistance of a line such that the input impedance is equal to Z L ..



 z m  0.2.  0.05  0.15 m 2 R 50 Rm  0   16.7  S 3

lm 

SMITH CHART CALCULATIONS FOR LOSSY LINES  Above analysis have been based on lossless transmission-lines but practically low-loss line are normally used.  For a lossy line of length l , equ. [154] is modified such that 2z  j e 1  e 2z e  j2 z 1   e [156] zi      1  e 2z e  j2 z 1   e 2z e j where φ  θ Γ  2βz  

Hence, to find z i and z L , we cannot simply move along the  -circle; auxiliary calculations are necessary to account for the e 2z factor.

Example 4: The input impedance of a short-circuited lossy transmission line of length 2 m and characteristic impedance 75  (approximately real) is 45  j225  . a) Find  and  of the line b) Determine the input impedance if the short-circuit is replaced by a load impedance Z L  67.5  j45 Ω . Solution: a) The short-circuit load is represented by Psc on the Smith chart. 45  j225  0.6  j3.0 in the chart as P1 . 1. Enter z i1  75 2. Draw a straight line from the origin O through P1 to P1 . 3. Measure

OP 1 6.9cm   0.89  e 2z . Hence, OP1 7.75cm

46 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

1  1  1 ln   ln1.124  0.029 Np/m 2l  0.89  4 4. Record that the arc Psc PL is 0.20 “wavelength toward the generator”. Thus, l   0.20 and 2βl  4π l   4π  0.20  0.8π . Hence, 0.8π 0.8π β  4π l   4π  0.20    0.2π rad/m 2l 4 b) To find the input impedance for Z L  67.5  j45 Ω 67.5  j45 Ω 1. Enter z L  Z L Z 0   0.9  j0.6 on the Smith chart as P2 . 75 2. Draw a straight line from O through P2 to P2 where the “wavelengths toward the generator” reading is 0.364. 3. Draw a  - circle centred at O with radius OP 2 .



4. Move P2 along the perimeter by 0.20 “wavelength towards generator” to P3 at 0.364  0.2  0.564 or 0.064 . 5. Join P3 and O by a straight line, intersecting the  - circle at P3 . 6. Mark on the line OP3 a point Pi such that OP i OP 3  e 2αl  0.89 i.e. OP i  0.89  OP 3  0.89  2.4cm  2.1cm 7. At Pi read zi  0.64  j0.27 . Hence,

Z i  750.64  j0.27  48  j20.25 Ω

TRANSMISSION-LINE IMPEDANCE MATCHING a) Impedance Matching by Quarter-wave Transformer  A simple method for matching a resistive load R L to a lossless transmission line of characteristic impedance R 0 is to insert a quarter-wave transformer with a characteristic impedance R 0 such that

R 0  R 0 R L 

[157]

Since the length of the quarter-wave line depends on wavelength, this matching is frequencysensitive. Example 5:A signal generator is a to feed equal power through a lossless air transmission line with a characteristic impedance 50  to two separate resistive loads, 64  and 25  . Quarter-wave transformers are used to match the loads to the 50  line, as shown below. a) Determine the required characteristic impedance of the quarter-wave lines b) Find the standing-wave ratios on the matching line sections. Solution: a) To feed equal power to the two loads, the input resistance at the junction with the main line looking towards each load must be equal to 2R 0 i.e. R i1  R i2  2R 0  100 Ω .[such that the equivalent resistance at the junction is equal to R 0 ]: 47

EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

R 01  R i1R L1  100  64  80 Ω

R 02  R i2 R L2  100  25  50 Ω

 4

R L1  64 Ω

R 01 R 0  50 Ω

R 02

 4

R L2  25 Ω

b) Under matched conditions there are no standing waves on the main transmission line ( S  1 ). The standing-wave ratios on the two matching line sections are as follows: Matching section 1: R  R 01 64  80 1  L1   0.11 R L1  R 01 64  80

S1 

1  Γ1 1  Γ1



1  0.11  1.25 1  0.11

Matching section 2: R  R 02 25  80 2  L2   0.33 R L2  R 02 25  80

S2  



1 Γ2 1 Γ2



1  0.33  1.99 1  0.33

Above have analysis has been for lossless lines but for lossy lines, quarter-wave cannot be used. Instead a method for matching an arbitrary load impedance to a line by using singleopen or short-circuited line section (a single stub)in parallel with the main line and an appropriate distance from the load is used. 1 In this case, on the Smith chart, admittance is normally used. Let YL  denote the load ZL impedance. The normalized load impedance is 48

EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

ZL 1 1 [158]   R 0 R 0 YL y L Where Y [159] y L  L  R0 YL  g  jb Y0 is the normalized load admittance have normalized conductance g and normalized susceptance b. Equation [158] implies that a quarter-wave line with a unity normalized characteristic impedance will transform z L to y L , and vice versa. On the Smith chart we only move the point representing z L along the  -circle by a quarter-wavelength to locate the point zL 



representing y L . 

1  Since a  4 -change in the line length  z     corresponds to a change of  radians 4  2π λ    2z   2    π  on the Smith chart, the points representing z L and y L are then λ 4   diametrically opposite each other on the  -circle.

Example 6: Given Z L  95  j20  , find YL Solution: Choosing R 0  50 Ω

Z L 95  j20   1.9  j0.4 R0 50 On the Smith enter z L as point P1 . The point P2 on the other side of the line joining P1 and O represents y L : y L  0.5  j0.1 Hence, 1 1 0.5  j0.1  10  j2 mS YL  yL  R0 50 zL 

Example 7: Find the input admittance of an open-circuited line of characteristic impedance 300  and length 0.04λ . Solution: 1. Starting from the Poc , move along the perimeter of the chart by 0.04 “wavelengths toward generator” to point P1 i.e. 0.25  0.04  0.29 . 2. Draw a straight line from P1 through O, intersecting at P1 on the opposite side. 3. Read at P1 y i  0  j0.26 Hence,

49 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

Yi 

1 1 0  j0.26  j0.87 mS yL  R0 300

SINGLE-STUB MATCHING  This is matching load impedance Z L to a lossless line that has characteristic impedance R 0 by placing a single short-circuited stub in parallel with the line and shown below. d B yi yL yB R0

yS

ZL

B

R0

l 









The length of the stub l and the distance from the load, d needs to be determined, such that the impedance of the parallel combination to the right of the points B - B equals R 0 . A short-circuited stub is are usually used in preference to the open-circuited stubs because an infinite terminating impedance is more difficult to realize than a zero terminating impedance for reasons of radiation from an open end and coupling effects with neighbouring objects. Since Z L and the stub are in parallel, its more appropriate to analyse the matching requirements in terms of admittances. The basic requirement is [160] Yi  YB  Ys 1  Y0  R0 In terms of normalized admittances, equ. [160] becomes [161] 1  yB  ys where y B  R 0 YB -for load section y s  R 0 Ys - for short-circuited stub. Since the input admittance of a short-circuited stub is purely susceptive, y s is purely imaginary. Thus, equ. [161] become can only be satisfied if y B  1  jb B [162] and [163] y s   jb B 50

EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon





Therefore, its our objective to find the length d such that the admittance, y B , of the load section looking to the right of terminals B - B has a unity real part and to find the length l B of the stub required to cancel the imaginary part. Using the Smith chart as an admittance chart, the following steps are followed: 1. Enter the point representing the normalized load admittance y L . 2. Draw the  -circle for y L , which will intersect the g  1 circle at two points. At these points, y B1  1  jbB1 and y B2  1  jb B2 . Both are possible solutions. 3. Determine load-section lengths d1 and d 2 from the angles between the point representing y L and the points representing y B1 and y B2 . 4. Determine stub lengths l B1 and l B2 from the angles between the short-circuit point on the extreme right of the chart to the points representing - jbB1 and - jbB2 respectively.

Example 7: A 50  transmission line is connected to a load impedance Z L  35  j47.5  . Find the position and length of a short-circuited stub required to match the line. Solution: Given: R0  50  Z L  35  j47.5  35  j47.5 Ω z L  Z L R0   0.70  j0.95 50 1. Enter z L on the Smith chart as P1 2. Draw a  -circle centred at O with radius OP1 3. Draw a straight line from P1 through O to point P2 on the perimeter, intersecting the  -circle at P2 , which represents y L . Note 0.109 at P2 on the “wavelengths towards generator”scale. 4. Note the two points of intersection of the  -circle with the g  1 -circle. At P3 : y B1  1  j1.2  1  jbB1 ; At P4 : y B2  1  j1.2  1  jbB2 ; 5. Solutions for the positions of the stub: For P3 (from P2 to P3 ): d 1  0.168  0.109  0.059λ For P4 (from P4 to P3 ): d 2  0.332  0.109  0.223λ 6. Solutions for the length of short-circuited stub to provide y s   jb B . For P3 (from Psc on the extreme right of chart to P3 , which represents  jbB1   j1.2 ) l B1  0.361  0.250λ  0.111λ . 51 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon

For P4 (from Psc to P4 , which represents  jbB2  j1.2 ) l B2  0.139  0.250λ  0.389λ . In general, the solution with the shorter lengths is preferred unless there are other practical constraints.

52 EEE 2314 TRANSMISSION LINES

JKUAT-Kipyegon