Smith Chart Examples

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  • Smith chart

    The Smith Chart: allows to compute the input impedance to a transmission line

    ( ) ( )( )0 0 1 0 1

    0 1 0 1

    inin in

    in

    Z z Z Z Z + +

    = = = =

    The load reflection coefficient and the input coefficient are related as42

    jjin L Le e

    pi = =

    We write the normalized input impedance0

    1

    1inin

    inin

    Zz r jx

    Z

    + = = = +

    And the reflection coefficient as4

    jin L e p jq

    pi = = +

    Combining both equations1

    1inp jq

    z r jxp jq

    + += + =

    Solving the real and imaginary parts

    ( )2

    22

    11 1

    rp qr r

    + = + +

    ( )2

    22

    1 11p qx x

    + =

    Circles of radius ( )1

    1r +centered at 0

    1rp q

    r= =

    +

    Circles of radius1x

    centered at11p and qx

    = =

    EE 342Spring 2010 #115

  • ( )2

    22

    11 1

    rp qr r

    + = + +

    Circles of radius( )1

    1r + centered at 01rp and q

    r= =

    +

    ( )2

    22

    1 11p qx x

    + =

    Circles of radius

    1x

    centered at11p and qx

    = =

    EE 342Spring 2010 #116

    Smith chart

  • The graphs relate the real and imaginary part of the reflection coefficient at a point (p,q)

    with the real and imaginary part of the normalized input impedance (r,x)

    EE 342Spring 2010 #117

    Smith chart

  • Relation between the normalized input impedance to the line and the reflection coefficient

    We plot the normalized input impedance

    inz r jx= +

    The point defines the magnitude and

    the angle of the reflection coefficient

    2 2

    in p q = +

    ( ) 2 4in pi = =

    EE 342Spring 2010 #118

    Smith chart

    1

    1inp jq

    z r jxp jq

    + += + =

    4

    jin L e p jq

    pi = = +

  • How determine the input

    impedance

    We plot the normalized load impedance

    0

    L

    L L LZ

    z r j xZ

    = = +

    Rotate (with a compass) an angle

    ( ) 2 4in pi = =

    Clockwise TG (towards generator)

    If we know the input impedance we calculate

    0

    in

    in in inZ

    z r jxZ

    = = +

    Rotate (with a compass) an angle

    ( ) 2 4in pi = = Counterclockwise TL (towards load)

    EE 342Spring 2010 #119

    Smith chart

  • Example: a coaxial cable (r = 2.25), length 10m. Frequency source 34 MHz. The

    characteristic impedance is Z0 = 50 . The line is terminated with a load ZL = (50+j100) .

    Determine the input impedance

    Propagation velocity80 2 10

    r

    v mvs

    = =

    Wavelength5.882 1.7v mf = = =

    Normalized load impedance

    50 100 1 2

    50Lj

    z j+= = +

    Rotate 1.7 TG (3 turns plus 0.2 )

    0.29 0.82inz j=

    And unnormalizing

    0

    14.5 41in inZ z Z j= =

    EE 342Spring 2010 #120

    Smith chart

  • A B

    C

    D

    E

    Exercise:

    Match the following

    normalized impedances

    with points A,B,C,D and E

    on the Smith chart

    i) 0+j0

    ii) 1+j0

    iii) 0-j1

    iv) 0+j1

    v) +j

    vi)

    vii)

    viii) Matched load

    min

    in

    C

    ZZ

    max

    in

    C

    ZZ

    ( )0 =

    i) D

    ii) A

    iii) E

    iv) C

    v) B

    vi) D

    vii) B

    viii) A

    EE 342Spring 2010 #121

    Smith chart

  • Example:

    ( ) 20 40inZ j= ( ) 20 40LZ j= +

    0 100Z =

    ( ) 0.2 0.4inz j= We calculate( ) 0.2 0.4Lz j= +

    Determine the length

    of the line in

    wavelengths

    0.062 TG

    0.436 TG

    0.436 0.062 0.374 = =

    0.438 TL

    0.064 TL

    0.438 0.064 0.374 = =

    EE 342Spring 2010 #122

    Smith chart

  • Exercise: Determine ZL attached to

    a line with Z0 = 100 . Removing

    the load yields an input impedance

    Zin= -j80.With the unknown

    impedance attached the input

    impedance is (30 + j 40) .

    Determine ZL

    With open circuit80

    0.8100Lin Z

    jz j

    =

    = =

    0.107 TL

    0.393 TG

    Lz =

    0.25 TL

    0.25 TG

    ( )0.393 0.25 0.143 = =( )0.25 0.107 0.143 = =

    With the load attached

    ( )30 40 0.30 0.40

    100Lin Zj

    z j+ = = +

    0.065 TG

    0.435 TL

    Rotate TL 0.143

    0.435 + 0.143 = 0.578 = 0.078

    0.32 0.49 32 49L Lz j Z j= =

    EE 342Spring 2010 #123

    Smith chart

  • Exercise: Determine the load

    impedance, VSWR and load

    reflection coefficient for :

    ( ) 50 100inZ j= 0

    50Z = 0.4 =

    1 2inz j=

    0.187 TL

    450

    Rotate TL (CCW)

    0.187 0.4 0.587 0.087 + = =

    0.22 0.58Lz j= ( ) 11 29LZ j=

    -1180

    00.73 118L =

    7VSWR =

    EE 342Spring 2010 #124

    Smith chart