C:Documents and SettingsJohn A. BuckMy ... = (R +jωL)(G+jωC) = [20 +j(6 10 8)(0.4 10−6)][80...

download C:Documents and SettingsJohn A. BuckMy ...  = (R +jωL)(G+jωC) = [20 +j(6 10 8)(0.4 10−6)][80 10−3 +j(6 10)(40 10−12)] = 2.8 +j3.5m−1 = ...

of 36

  • date post

    12-May-2018
  • Category

    Documents

  • view

    229
  • download

    3

Embed Size (px)

Transcript of C:Documents and SettingsJohn A. BuckMy ... = (R +jωL)(G+jωC) = [20 +j(6 10 8)(0.4 10−6)][80...

  • CHAPTER 13

    13.1. 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 mS/m, and R = 20 /m.

    a) Find , , , , and Z0: We use

    =

    ZY =

    (R + jL)(G + jC)=

    [20 + j (6 108)(0.4 106)][80 103 + j (6 108)(40 1012)]

    = 2.8 + j3.5 m1 = + j

    Therefore, = 2.8 Np/m, = 3.5 rad/m, and = 2/ = 1.8 m. Finally,

    Z0 =

    Z

    Y=

    R + jLG + jC =

    20 + j2.4 102

    80 103 + j2.4 102 = 44 + j30

    b) If a voltage wave travels 20 m down the line, what percentage of the original amplitude remains,and by how many degrees is it phase shifted? First,

    V20

    V0= eL = e(2.8)(20) = 4.8 1025 or 4.8 1023 percent!

    Then the phase shift is given by L, which in degrees becomes

    = L(

    360

    2

    )= (3.5)(20)

    (360

    2

    )= 4.0 103 degrees

    13.2. A lossless transmission line with Z0 = 60 is being operated at 60 MHz. The velocity on the line is3 108 m/s. If the line is short-circuited at z = 0, find Zin at:

    a) z = 1m: We use the expression for input impedance (Eq. 12), under the conditions Z2 = 60and Z3 = 0:

    Zin = Z2[Z3 cos(l) + jZ2 sin(l)Z2 cos(l) + jZ3 sin(l)

    ]= j60 tan(l)

    where l = z, and where the phase constant is = 2c/f = 2(3 108)/(6 107) =(2/5) rad/m. Now, with z = 1 (l = 1), we find Zin = j60 tan(2/5) = j184.6 .

    b) z = 2 m: Zin = j60 tan(4/5) = j43.6 c) z = 2.5 m: Zin = j60 tan(5/5) = 0d) z = 1.25 m: Zin = j60 tan(/2) = j (open circuit)

    13.3. The characteristic impedance of a certain lossless transmission line is 72 . If L = 0.5 H/m, find:a) C: Use Z0 =

    L/C, or

    C = LZ20

    = 5 107

    (72)2= 9.6 1011 F/m = 96 pF/m

    213

  • 13.3b) vp:

    vp = 1LC

    = 1(5 107)(9.6 1011)

    = 1.44 108 m/s

    c) if f = 80 MHz: =

    LC = 2 80 10

    6

    1.44 108 = 3.5 rad/m

    d) The line is terminated with a load of 60 . Find and s:

    = 60 7260 + 72 = 0.09 s =

    1 + ||1 || =

    1 + .091 .09 = 1.2

    13.4. A lossless transmission line having Z0 = 120 is operating at = 5 108 rad/s. If the velocity onthe line is 2.4 108 m/s, find:

    a) L: With Z0 =

    L/C and v = 1/LC, we find L = Z0/v = 120/2.4 108 = 0.50 H/m.b) C: Use Z0v =

    L/C/

    LC C = 1/(Z0v) = [120(2.4 108)]1 = 35 pF/m.

    c) Let ZL be represented by an inductance of 0.6 H in series with a 100- resistance. Find ands: The inductive impedance is jL = j (5 108)(0.6 106) = j300. So the load impedanceis ZL = 100 + j300 . Now

    = ZL Z0ZL + Z0 =

    100 + j300 120100 + j300 + 120 = 0.62 + j0.52 = 0.808

    40

    Then

    s = 1 + ||1 || =

    1 + 0.8081 0.808 = 9.4

    13.5. Two characteristics of a certain lossless transmission line are Z0 = 50 and = 0 + j0.2 m1 atf = 60 MHz.

    a) Find L and C for the line: We have = 0.2 = LC and Z0 = 50 =

    L/C. Thus

    Z0= C C =

    Z0= 0.2

    (2 60 106)(50) =1

    3 1010 = 33.3 pF/m

    Then L = CZ20 = (33.3 1012)(50)2 = 8.33 108 H/m = 83.3 nH/m.b) A load, ZL = 60 + j80 is located at z = 0. What is the shortest distance from the load to a

    point at which Zin = Rin + j0? I will do this using two different methods:The Hard Way: We use the general expression

    Zin = Z0[ZL + jZ0 tan(l)Z0 + jZL tan(l)

    ]

    We can then normalize the impedances with respect to Z0 and write

    zin = ZinZ0

    =[

    (ZL/Z0) + j tan(l)1 + j (ZL/Z0) tan(l)

    ]=

    [zL + j tan(l)

    1 + jzL tan(l)]

    where zL = (60 + j80)/50 = 1.2 + j1.6.

    214

  • 13.5b. (continued) Using this, and defining x = tan(l), we find

    zin =[

    1.2 + j (1.6 + x)(1 1.6x) + j1.2x

    ] [(1 1.6x) j1.2x(1 1.6x) j1.2x

    ]

    The second bracketed term is a factor of one, composed of the complex conjugate of the denomi-nator of the first term, divided by itself. Carrying out this product, we find

    zin =[

    1.2(1 1.6x) + 1.2x(1.6 + x) j [(1.2)2x (1.6 + x)(1 1.6x)](1 1.6x)2 + (1.2)2x2

    ]

    We require the imaginary part to be zero. Thus

    (1.2)2x (1.6 + x)(1 1.6x) = 0 1.6x2 + 3x 1.6 = 0

    So

    x = tan(l) = 3

    9 + 4(1.6)22(1.6)

    = (.433, 2.31)

    We take the positive root, and find

    l = tan1(.433) = 0.409 l = 0.4090.2

    = 0.65 m = 65 cm

    The Easy Way: We find

    = 60 + j80 5060 + j80 + 50 = 0.405 + j0.432 = 0.59

    0.818

    Thus = 0.818 rad, and we use the fact that the input impedance will be purely real at the locationof a voltage minimum or maximum. The first voltage maximum will occur at a distance in frontof the load given by

    zmax = 2

    = 0.8182(0.2)

    = 0.65 m

    13.6. The propagation constant of a lossy transmission line is 1 + j2 m1, and its characteristic impedanceis 20 + j0 at = 1 Mrad/s. Find L, C, R, and G for the line: Begin with

    Z0 =

    R + jLG + jL = 20 R + jL = 400(G + jC) (1)

    Then 2 = (R + jL)(G + jC) = (1 + j2)2 400(G + jC)2 = (1 + j2)2 (2)

    where (1) has been used. Eq. 2 now becomes G + jC = (1 + j2)/20. Equating real and imaginaryparts leads to G = .05 S/m and C = 1/(10) = 107 = 0.1 F/m.

    215

  • 13.6. (continued) Now, (1) becomes

    20 =

    R + jL1 + j2

    20 20 = R + jL

    1 + j2 20 + j40 = R + jL

    Again, equating real and imaginary parts leads to R = 20 /m and L = 40/ = 40 H/m.

    13.7. The dimensions of the outer conductor of a coaxial cable are b and c, c > b. Assume = c and let = 0. Find the magnetic energy stored per unit length in the region b < r < c for a uniformlydistributed total current I flowing in opposite directions in the inner and outer conductors: First, fromthe inner conductor, the magnetic field will be

    H1 = I2

    a

    The contribution from the outer conductor to the magnetic field within that conductor is found fromAmperes circuital law to be:

    H2 = I2

    2 b2c2 b2 a

    The total magnetic field within the outer conductor will be the sum of the two fields, or

    HT = H1 + H2 = I2

    [c2 2c2 b2

    ]a

    The energy density is

    wm = 120H

    2T =

    0I2

    82

    [c2 2c2 b2

    ]2J/m3

    The stored energy per unit length in the outer conductor is now

    Wm = 1

    0

    20

    cb

    0I2

    82

    [c2 2c2 b2

    ]2 d d dz = 0I

    2

    4(c2 b2)2 c

    b

    [c4

    2c2 + 3

    ]d

    = 0I2

    4

    [c4

    (c2 b2)2 ln( cb

    )+ b

    2 (3/4)c2(c2 b2)

    ]J

    13.8. The conductors of a coaxial transmission line are copper (c = 5.8 107 S/m) and the dielectric ispolyethylene (R = 2.26, / = 0.0002). If the inner radius of the outer conductor is 4 mm, findthe radius of the inner conductor so that (assuming a lossless line):

    a) Z0 = 50 : Use

    Z0 = 12

    ln

    (b

    a

    )= 50 ln

    (b

    a

    )=

    2

    R(50)

    377= 1.25

    Thus b/a = e1.25 = 3.50, or a = 4/3.50 = 1.142 mm

    216

  • 13.8b. C = 100 pF/m: Begin with

    C = 2

    ln(b/a)= 1010 ln

    (b

    a

    )= 2(2.26)(8.854 102) = 1.257

    So b/a = e1.257 = 3.51, or a = 4/3.51 = 1.138 mm.c) L = 0.2 H/m: Use

    L = 02

    ln

    (b

    a

    )= 0.2 106 ln

    (b

    a

    )= 2(0.2 10

    6)4 107 = 1

    Thus b/a = e1 = 2.718, or a = b/2.718 = 1.472 mm.

    13.9. Two aluminum-clad steel conductors are used to construct a two-wire transmission line. Let Al =3.8107 S/m, St = 5106 S/m, and St = 100 H/m. The radius of the steel wire is 0.5 in., and thealuminum coating is 0.05 in. thick. The dielectric is air, and the center-to-center wire separation is 4 in.Find C, L, G, and R for the line at 10 MHz: The first question is whether we are in the high frequencyor low frequency regime. Calculation of the skin depth, , will tell us. We have, for aluminum,

    = 1f 0Al

    = 1(107)(4 107)(3.8 107)

    = 2.58 105 m

    so we are clearly in the high frequency regime, where uniform current distributions cannot be assumed.Furthermore, the skin depth is considerably less than the aluminum layer thickness, so the bulk of thecurrent resides in the aluminum, and we may neglect the steel. Assuming solid aluminum wires ofradius a = 0.5 + 0.05 = 0.55 in = 0.014 m, the resistance of the two-wire line is now

    R = 1aAl

    = 1(.014)(2.58 105)(3.8 107) = 0.023 /m

    Next, since the dielectric is air, no leakage will occur from wire to wire, and so G = 0 mho/m. Nowthe capacitance will be

    C = 0cosh1(d/2a)

    = 8.85 1012

    cosh1 (4/(2 0.55)) = 1.42 1011 F/m = 14.2 pF/m

    Finally, the inductance per unit length will be

    L = 0

    cosh(d/2a) = 4 107

    cosh (4/(2 0.55)) = 7.86 107 H/m = 0.786 H/m

    217

  • 13.10. Each conductor of a two-wire transmission line has a radius of 0.5mm; their center-to-center distanceis 0.8cm. Let f = 150MHz and assume = 0 and c (note error in problem statement). Findthe dielectric constant of the insulating medium if

    a) Z0 = 300 : Use

    300 = 1

    0

    R0cosh1

    (d

    2a

    )

    R =

    120

    300cosh1

    (8

    2(.5)

    )= 1.107 R = 1.23

    b) C = 20 pF/m: Use

    20 1012 =

    cosh1(d/2a) R =

    20 10120

    cosh1(8) = 1.99

    c) vp = 2.6 108 m/s:

    vp = 1LC

    = 100

    R

    = cR

    R =(

    3.0 1082.6 108

    )2= 1.33

    13.11. Pertinent dimensions for the transmission line shown in Fig. 13.4 are b = 3 mm, and d = 0.2 mm.The conductors and the dielectric are non-magnetic.

    a) If the characteristic impedance of the line is 15 , find R: We use

    Z0 =

    (d

    b

    )= 15 R =

    (377

    15

    )2.04

    9= 2.8

    b) Assume copper conductors and operation at 2 108 rad/s. If RC = GL, determine the