Transmission Lines

EELE 3332 – Electromagnetic II Chapter 11 Transmission Lines

Islamic University of Gaza Electrical Engineering Department Dr. Talal Skaik

2012 1 11.5 The Smith Chart

The Smith Chart is a graphical tool for analyzing transmission lines.

The Smith Chart is made up of circles and arcs of circles.

The circles are called “constant resistance circles”

The arcs are “constant reactance circles”

The combination of intersecting circles and arcs inside the chart allow us to locate the normalized impedance and then to find the impedance anywhere on the line.

We will assume lossless transmission lines. (Zo=R0)

2 The Smith Chart

The Smith Chart is constructed within a circle of unit radius (|Γ|≤1).

 ||    ri j 

Where Γr and Γi are the real and imaginary parts of Γ. 3 The Smith Chart Smith Chart is normalized so that all impedances are normalized to the characteristic impedance Z0. (Hence, it can be used for any line)

For the load impedance ZL , the normalized impedance zL is:

ZL zL   r  jx Z0

Resistance Reactance Circles Circles r-circles x-circles

4 The Smith Chart

Plot normalized impedance

zL=1+j2

5 The Smith Chart

6 The Smith Chart

• Impedance: – Z1 = 100 + j50 – Z2 = 75 -j100 – Z3 = j200 – Z4 = 150 – Z5 = infinity (an open circuit) – Z6 = 0 (a short circuit) – Z7 = 50 – Z8 = 184 -j900

• Normalized impedance (line impedance (50 Ohms) : – z1 = 2 + j – z2 = 1.5 -j2 – z3 = j4 – z4 = 3 – z5 = infinity – z6 = 0 – z7 = 1 – z8 = 3.68 -j18 7 The Smith Chart

8 The Smith Chart •A complete revolution (3600) around Smith Chart represents a distance of (λ/2) on the line. λ distance on the line corresponds to a 7200 movement on the chart. λ→ 7200 •Clockwise movement represents moving toward the Generator. •Counter-clockwise movement represents moving toward the

Load. 9 The Smith Chart

Three scales: •Outermost scale: distance on line in terms of wavelength, moving toward Generator. •Middle scale: distance on line in terms of wavelength, moving toward Load. •Innermost scale : to determine θΓ (in degrees)

Given Z0, ZL, λ and length of line, We can determine Zin, Yin, s, and Γ. 10 Example 11.4

•A lossless transmission line with Z0=50 Ω is 30 m long and operates at 2 MHz. The line is terminated with a load ZL=60+j40 Ω. If u=0.6c on the line, find •(a) The reflection coefficient Γ •(b) The standing wave ratio s

•(c) The input impedance. Solution Method 1 (Without the Smith Chart) Z Z 60 jj 40  50 10  40 (a)  =L0    0.3523  560 ZL0 +Z 60jj 40  50 110  40 1 |  | 1  0.3523 (b ) s= 2.088

1 |  | 1  0.3523 11 Example 11.4- Solution continued

2  (2 106 ) 2  (c) ll = (30)   1200 (electrical length) u 0.6(3 108 ) 3

ZL  jZ0 tan  l ZZin  0  Z0  jZL tan  l 60jj 40 50tan1200  500 50jj (60 40) tan120  23.97 j 1.35  24.01  3.220

12 Example 11.4

ZL (a) Normalized : zL  Z0 60 j 40  =1.2+j0.8 50

Locate zL at point P 0P z | |   0.3516 L L 0Q

 is between 0SP and 0 : 0  angle PS 0 56

0  L 0.3516  56 ______(b ) To find standing wave (c) to find Zin , express l interms of  or in degrees. ratio s , draw s-circle u 0.6(3 1080 ) 30 720   90 m, l =30 m=    2400 f 2 106 90 3 3 (radius 0P and center 0) 0  Move clockwise from load toward Generator 240 on the s-circle meets r -axis s-circle from point PG to point . At G , We obtain: zin  0.47 j 0.03 at s  2.1

Hence Zin Z0 z in  50(0.47  j 0.03)  23.5  j 1.5  13 Example 11.5

A load of 100+j150 Ω is connected to a 75 Ω lossless line. Find: a) The reflection coefficient Γ b) The standing wave ratio s c) The load admittance YL d) Zin at 0.4λ from the load. e) The locations of Vmax and Vmin with respect to the load if the line is 0.6 λ long. f) Zin at the generator.

14 Example 11.5

Z L 100 j 150 (a) zL  Z0 75 =1.33+j2 (point P ) 0P | |   0.659 L 0Q 0  angle PS 0 40 0  L 0.659  40

ZZL  0 Check : L ZZL  0 100j 150 75  L 100j 150 75 0 L 0.659  40 ______(b ) Draw s-circle passing through P and obtain s=4.82 1 | | check: s= 4.865 1 | | 15 Example 11.5

YYLL 0 y 1  (0.228j 0.35) 75

YL 3.04 j 4.67 mS Check : 11 YL  ZjL 100 150  3.07j 4.62 mS  (d ) The 0.4 corresponds to 0.4 72000 288  Move 2880 on s-circle from P toward generator (clockwise) to reach point R .

zjin 0.3 0.63

Zin Z0 z in 75(0.3  j 0.63)

=22.5+j47.2516  Example 11.5 – Solution continued

()d Check : 2 l (0.4 )  360 (0.4)  1440  0

ZL  jZ0 tan l ZZin  0  Z0  jZL tan tan  l 100jj 150 75tan1440 =750 75jj (100 150) tan tan144 =54.41 65.250 or

Zin  21.9j 47.6

17 Example 11.5 (e) 0.6  0.6  7200  4320 1 revolution 720 Start from P (Load), move alone s-circle 4320 , or one revolution +720 , and reach Generator at point G .  We have passed through

point T (location of Vmin ) once, and point S (location

of Vmax ) twice. From load:

1st Vmax is located at : 400  0.055 7200

2nd Vmax is located at :  0.055 0.555 2

The only Vmin is located at : 0.055  / 4 0.3055 18 Example 11.5

(f) At G (Generator end), zjin  1.8 - 2.2

Zjin  75(1.8 - 2.2)  135-j 165