Nothing Special   »   [go: up one dir, main page]
Scribd Logo
Upload

Welcome to Scribd!

  • 3 views

    Lecture

    Uploaded by

    eror4700

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd

    Lecture

    Uploaded by

    eror4700
    3 views13 pages

    Original Title

    Lecture (4)

    Copyright

    © © All Rights Reserved

    Available Formats

    PDF, TXT or read online from Scribd

    Did you find this document useful?

    Is this content inappropriate?

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    3 views13 pages

    Lecture

    Uploaded by

    eror4700

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 13

    Communications and Computers Engineering Department The Higher Institute of Engineering- El-Shorouk City

    Electromagnetic Fields (ECE 271)


    The Academic year 2022/2023- First Semester
    Lecture (4)
    Third year Electronics and Communications Engineering
    Program 1
    Terminated Lossless Transmission Line
    -The total voltage on the line is
    − j z j z
    V ( z ) = V1 e + V2 e
    -The total current on the line is
    V1 − j z V2 j z
    I ( z) = e − e
    ZO ZO
    -The total voltage and current at the load are related by the load impedance
    (Zr) at z=0 V (0) V1 + V2
    Zr = = ZO
    I (0) V1 − V2
    Zr − Zo
    V2 = V1
    Zr + Zo
    -The amplitude of the reflected voltage wave normalized by the amplitude
    of the incident voltage wave is defined as the voltage reflection coefficient 
    − j l
    V2 Z r − Z o V2 e −2 j  l
     (0) = = &  (l ) = j l
    =  (0)e & -1   (0)  1
    V1 Z r + Z o V1 e 2
    Terminated Lossless Transmission Line
    ➢ The total voltage and current waves on the line can be written as
    − j z j z
    V ( z ) = V1 (e + e )
    V1 − j  z j z
    I ( z) = (e − e )
    Zo
    ➢ It is seen that the voltage and current on the line consists of a
    superposition of an incident and a reflected wave, such waves are
    called Standing Waves.

    ➢ At  = 0 , there is no reflected wave (the load impedance Zr must be


    equal to the characteristic impedance Zo). And the line is called flat.

    3
    Terminated Lossless Transmission Line
    - When the load is mismatched,
    − j z j z
    V ( z ) = V1 (e + e )
    2 j z
    V ( z ) = V1 1 +  e
    −2 j  l
    at a distance z = -l , V ( −l ) = V1 1 +  e
    j ( − 2  l )
    V (−l ) = V1 1 +  e
    where,  is the phase of the reflection coefficient.
    the voltage magnitude oscillates with position along the line

    Vmax = V1 (1 +  ) The distance between successive maxima (or minima) is l = .
    2

    Vmin = V1 (1 −  ) The distance between a maximum and minimum is l = .
    4
    − the voltage standing wave ratio (VSWR) is a measure of mismatch of the line and can be expressed as:
    Vmax 1 + 
    VSWR = =
    Vmin 1 − 
    - VSWR is real number, lies in the range of 1  VSWR   4
    Special Cases of Lossless terminated Lines
    ❑Case 1: Short Circuit case: Zr=0

    Zr − Zo 0 − Zo
    = = = −1
    Zr + Zo 0 + Zo
    1+  Voltage
    VSWR = =
    1− 

    The voltage and current on the line can be obtained as:


    − j z j z − j z j z
    V ( z ) = V1 (e + e ) = V1 (e −e ) Current
    V1 − j  z j z V − j z j z
    I ( z) = (e −  e ) = 1 (e +e )
    Zo Zo
    2j j l − j l
    V (−l ) = V1 (e − e ) = 2 jV1 sin(  l )
    2j
    Impedance
    2 V1 j l − j l V
    I (−l ) = (e + e ) = 2 1 cos(  l )
    2 Zo Zo
    input impedance: Z input = jZ o tan(  l ) 5
    Special Cases of Lossless terminated Lines
    ❑Case 2: Open Circuit case: Zr=∞
    Zr − Zo 1 − Zo Zr
    = = =1
    Zr + Zo 1 + Zo Zr
    1+ 
    VSWR = =
    1−  Voltage
    The voltage and current on the line can be obtained as:
    − j z j z − j z j z
    V ( z ) = V1 (e + e ) = V1 (e +e )
    V1 − j  z j z V1 − j  z j z
    I ( z) = (e − e ) = (e −e ) Current
    Zo Zo
    2 j l − j l
    V (−l ) = V1 (e + e ) = 2V1 cos(  l )
    2
    2 j V1 j l − j l V
    I (−l ) = (e − e ) = 2 j 1 sin(  l )
    2 j Zo Zo Impedance
    input impedance: Z input = − jZ o cot(  l )
    6
    Special Cases of Lossless terminated Lines
    Z r + jZ o tan(  l )
    Z input = Z o
    Z o + jZ r tan(  l )

    ➢ For a transmission line, l=
    2 , we get Zinput= Zr .

    ➢A half-wavelength line (or any multiple of 2 ) does not transform the


    load impedance, regardless of its characteristic impedance.

    n Zo2
    ➢For a transmission line with l=,we get Zinput .
    , for n=1,3,5,... =
    4 Zr
    Such transmission lines are known as quarter-wave transmission lines and
    can transform the load impedance in an inverse manner depending on its
    characteristic impedance.
    7
    Example:
    Calculate the reflection coefficient of lossless T.L with characteristic
    impedance Zo=50 Ω for the following load impedance
    (a) Zr=25 Ω (b) Zr=75 Ω (c) Zr=30-j40 Ω

    Solution:
    Z r − Z o 25 − 50
    (a)  = = = −0.33
    Z r + Z o 25 + 50
    Z r − Z o 75 − 50
    (b)  = = = 0.2
    Z r + Z o 75 + 50
    Z r − Z o 30 − j 40 − 50 −20 − j 40
    (c)  = = = = − j 0.5
    Z r + Z o 30 − j 40 + 50 80 − j 40

    8
    Example:
    A 50-Ω lossless transmission line is to be matched to a resistive load
    impedance with Zr=100Ω via a quarter-wave section as shown,
    thereby eliminating reflections along the feedline. Find the
    characteristic impedance of the quarter-wave transformer.

    9
    Solution:
    • To eliminate reflections at terminal AA’, the input impedance Zin looking
    into the quarter-wave line should be equal to Z01 (the characteristic
    impedance of the feedline). Thus, Zin = 50Ω .

    Z 022
    Z in =
    Zr
    Z 02 = 50  100 = 70.7
    • Since the lines are lossless, all the incident power will end up getting
    transferred into the load Zr .

    10
    Example:
    A source with 50  source impedance drives a 50  transmission line
    that is 1/8 of wavelength long, terminated in a load Zr = 50 – j25 .
    Calculate:
    (i) The reflection coefficient
    (ii) VSWR
    (iii)The input impedance seen by the source.

    11
    SOLUTION:
    (i) The reflection coefficient,
    Zr − Z0
    =
    Zr + Z0

    =
    ( 50 − j 25) − 50 = 0.242 − 76o
    ( 50 − j 25) + 50

    (ii) VSWR 1+ 
    VSWR = = 1.64
    1− 

    (iii) The input impedance seen by the source, Zin


    2   
     = =  tan =1
    4
     8 4 Zr +
    Z in = Z 0
    jZ 0 tan 
    Z 0 + jZ r tan 
    50 − j 25 + j 50
    = 50
    50 + j 50 + 25
    = 30.8 − j 3.8
    12
    Power Reflection Coefficient and Transmission
    coefficient
    ➢Power reflection coefficient = 
    2

    ➢Power Transmission coefficient = 1- 


    2

    Example:
    If the power reflection coefficient is -20dB, find the voltage reflection
    coefficient and Power transmission coefficient.
    Solution: 10 log  = −20dB
    2

    20
    −
     = 10 = 0.01
    2
    10

     = 0.1
    Power Transmission coefficient = 1 −  = 0.99
    2
    13

    You might also like