14.06071902.Dwari.CS

download 14.06071902.Dwari.CS

of 20

Transcript of 14.06071902.Dwari.CS

  • 7/28/2019 14.06071902.Dwari.CS

    1/20

    Progress In Electromagnetics Research, PIER 64, 219238, 2006

    ANALYSIS OF LINEAR TAPERED WAVEGUIDE BYTWO APPROACHES

    S. Dwari, A. Chakraborty, and S. Sanyal

    Department of Electronics and Electrical Communication EngineeringIndian Institute of TechnologyKharagpur-721302, India

    AbstractThis paper presents the analysis of linear taperedwaveguide. Voltage-standing-wave-ratio (VSWR) is obtained fromtransmission matrix of the taper waveguide. Taper section is divided

    into number of section having uniform length. Transmission matrix oftaper waveguide is found by multiplication of transmission matrix ofeach section. Transmission matrix of each section is obtained as theproduct of three matrices. One is of the initial length of transmissionline, second one is due to discontinuity and third one is of the finallength of transmission line. Transmission matrix of discontinuityis obtained by two methods. One is by equivalent circuit of stepdiscontinuity and another is by moment method. The results are seento be in good agreement with [1, 2] and [3].

    1. INTRODUCTION

    In a tapered section the dimension of waveguide varies smoothly andfor this reason there are possibilities of providing a transition fromone impedance level to another. Ref. [4] and then [5] analyzed taperline by dividing it into a number of sections of equal length into thedirection of propagation. Ref. [1] found the transmission matrix of eachsection as the product of five matrices and multiplied the matrices ofall sections.

    This paper proposed that the transmission matrix of each sectionis obtained as the product of three matrices. They are contributedby initial length of transmission line, discontinuity and the finallength of transmission line. Transmission matrix of discontinuity isobtained by two procedures. They are by equivalent circuit of stepdiscontinuity and by moment method. The final transmission matrixof tapered waveguide is obtained by multiplication of matrices of all

  • 7/28/2019 14.06071902.Dwari.CS

    2/20

    220 Dwari, Chakraborty, and Sanyal

    sections. Voltage-standing-wave-ratio (VSWR) is obtained from finaltransmission matrix.

    2. TRANSMISSION MATRIX OF TAPER SECTION

    The linear tapered waveguide is shown in Figure 1 and it can beconsidered as if it is formed with huge number of step discontinuities.Let the number of sections be N and the length of taper be L.Let 2a0, 2b0 are initial broad and narrow dimensions respectively and2a1, 2b1 are respective final broad and narrow dimensions.

    Figure 1. (a) Tapered waveguide of length L connecting uniformwaveguides of dimensions 2a0, 2b0 and 2a1, 2b1. (b) Taper waveguideof Figure 1(a) divided into sections. (c) Expanded view of nth section.

    The nth section is blown up and shown in Figure 1(c). Eachsection consists of an initial length of waveguide, a step discontinuityand a final length of waveguide. Each of the constituents can beexpressed by a transmission matrix of order 2.

    For nth section initial and final broad dimensions are

    2a0n = 2a0 +2(a1 a0)

    N(n 1) (1)

    2a1n = 2a0 +2(a1 a0)

    Nn (2)

  • 7/28/2019 14.06071902.Dwari.CS

    3/20

    Progress In Electromagnetics Research, PIER 64, 2006 221

    Narrow wall dimensions are

    2b0n = 2b0 +2(b1 b0)

    N(n 1) (3)

    2b1n = 2b0 + 2(b1

    b0)N n (4)

    The matrix of each section is the product of three matrices. Letmatrix [T1] is for initial length, [T3] is of final length and [T2] is fordiscontinuity. So transmission matrix of each section is given by

    [T] = [T1][T2][T3] (5)

    The transmission matrices of all the sections are multiplied to obtainthe final transmission matrix of the taper.

    Suppose d = LN is the length of each section, k0n is the propagationconstant of initial length and k1n is the propagation constant of finallength of nth section. Then [T1] can be written as

    [T1] =

    ejk0n

    d

    2 0

    0 ejk0nd

    2

    (6)

    [T3] can be written as

    [T3] =

    ejk1n

    d

    2 0

    0 ejk1nd

    2

    (7)

    Let 1n be the reflection co-efficient in the forward direction atdiscontinuity of nth section and 2n be the reflection co-efficient inthe backward direction at discontinuity of the nth section. Then [T2]can be written as

    [T2] =1

    1 + 1n

    1 2n

    1n 1 + 1n + 2n

    (8)

    To obtain [T2], reflection co-efficients should be determined. Reflectioncoefficient can be determined by two ways. One is using equivalentcircuit representing step discontinuities shown in Figure 2 as describedin [6]. Second is by using method of moment.

    3. DETERMINATION OF CIRCUIT PARAMETERS OFEQUIVALENT CIRCUIT FOR STEP DISCONTINUITIES

    Discontinuity of each section of linear taper waveguide is stepdiscontinuity. This step discontinuity can be represented by equivalent

  • 7/28/2019 14.06071902.Dwari.CS

    4/20

    222 Dwari, Chakraborty, and Sanyal

    Figure 2. Equivalent circuit of discontinuity.

    Figure 3. Cross section of waveguide junction for step discontinuity

    at narrow dimension.

    circuit as shown in Figure 2. Following [6] the circuit parameters of theequivalent circuit can be obtained. From circuit parameters reflectioncoefficients can be calculated. Evaluation of circuit parameters ofequivalent circuit for step discontinuities (1) in narrow dimension onlyand (2) in both narrow & broad dimension are as follows:

    3.1. Evaluation of Circuit Parameters of Equivalent Circuitfor Step Discontinuities in Narrow Dimension

    Consider the waveguide junction as shown in Figure 3. Let junction is

    at z = 0. Waveguide of dimension 2a 2b extends up to z = andwaveguide of dimension 2a 2(b + W) extends upto z = + . Thedimensions are such that only dominant mode can propagate in eachsection. Let there be an incident wave from z direction.

    In the region z < 0 transverse fields are

    Et =

    ej z + ej z V0

    1 + e0 +

    i

    Vieizei

    Ht = Y

    0

    ej z ej z

    V01 +

    h0

    i

    YiVieizhi (9)

    where ei, hi are the mode vectors, i are the cutoff mode-attenuationconstants, Yi are the characteristics admittances, and is the reflectioncoefficient for the dominant mode. The subscript 0 signifies the

  • 7/28/2019 14.06071902.Dwari.CS

    5/20

    Progress In Electromagnetics Research, PIER 64, 2006 223

    dominant mode. It is assumed that at z = matched load is there. Soin the region z > 0

    E+t = V0ejz e0 +i Vie

    iz ei

    H+t = Y+0 V0e

    jz h0 +

    i

    YiVieizhi (10)

    where the carets are used to differentiate the above parameters fromthere z < 0 counterparts. At junction region

    z=0

    E+ H+ ds =

    z=0

    E H ds (11)

    So

    Y+0 V20 +

    i

    YiV2

    i =1 1 +

    Y0 V20

    i

    YiV2

    i (12)

    Relative admittance observed from z < 0 is1 1 +

    =Y

    Y0=

    G

    Y0+j

    B

    Y0(13)

    Y0 is real and Yi are imaginary. For real Vi and Vi

    jB

    Y0=

    i

    YiV2

    i +

    i

    YiV2

    i

    Y0 V20

    (14)

    G

    Y0=

    Y+0 V20

    Y0 V20

    (15)

    For equivalent circuit of Figure 2, with matched condition at z =

    , it

    is evident thatG

    Y0= n2

    Y+0Y0

    (16)

    From Eq. (15) and Eq. (16)

    n2 =V20V20

    (17)

    For the junction shown in Figure 3, the dominant mode vectors [6] are

    e0 = uy

    1

    2absin

    2a(x + a) (18)

    e0 = uy

    12a(b + W) sin

    2a (x + a) (19)

  • 7/28/2019 14.06071902.Dwari.CS

    6/20

    224 Dwari, Chakraborty, and Sanyal

    Assumed tangential electric field in the aperture

    Eat = uy sin

    2a(x + a) (20)

    Therefore following [6]:

    V0 =

    ax=a

    by=b

    12ab

    sin2(x + a)

    2adxdy =

    2ab (21)

    V0 =

    ax=a

    by=b

    12a(b + W)

    sin2(x + a)

    2adxdy =

    2a

    b + Wb (22)

    Therefore

    V0

    V0= b

    b + W(23)

    n2 =b

    b + W(24)

    This is the transformation ratio of the transformer of equivalent circuit.Now the first summation in the numerator of Equation (14) is zero andsecond summation is related to aperture susceptance Ba by

    i

    YiV2

    i = j|V|2Ba = j4b2Ba (25)

    Now

    V0 =

    ax=a

    by=b

    sin (x + a)2a

    12ab

    sin (x + a)2a

    dxdy = 2ab (26)

    So

    B =2Bab

    a(27)

    3.1.1. Determination of Aperture Susceptance Ba

    Let Ex = 0 and Ey = f(x, y) be known over the cross section at z = 0.For TE to x mode at z > 0 scalar potential function is:

    =

    m=1

    n=0 Amn sin

    m

    2a (x + a)cos

    n

    2(b + W) (y + b + W)emnz

    (28)

  • 7/28/2019 14.06071902.Dwari.CS

    7/20

    Progress In Electromagnetics Research, PIER 64, 2006 225

    where Amn are mode amplitudes and mn are the mode propagationconstants. In particular Ey at z = 0 is given by

    Ey|z=0 =

    m=1

    n=0

    mnAmn sinm

    2a(x+a)cos

    n

    2(b + W)(y+b+W) (29)

    This is the form of double Fourier series: a sine series in x and a cosineseries in y. It is thus evident that mnAmn are fourier coefficients ofEy, or

    mnAmn = Emn =n

    2a(b + W)

    aa

    (b+W)(b+W)

    Ey|z=0 sin m2a

    (x + a)

    cos n2(b + W)

    (y + b + W)dxdy (30)

    where n = 1 for n = 0 and n = 2 for n > 0 (Numanns number). TheAmn and hence the field are now evaluated. The z directed complexpower at z = 0 is

    P =

    z=0

    E H

    uzds

    = a

    a

    (b+W)(b+W)

    [EyH

    x]z=0 dxdy

    =

    a

    a

    (b+W)

    (b+W)

    m,n

    Emn sinm(x + a)

    2a

    cosn(y + b + W)

    2(b

    +W

    )

    p,q

    k2

    p

    2a

    2j pq

    Epq sinp(x + a)

    2acos

    q(y + b + W)

    2(b + W)

    dxdy

    (31)

    This reduces to

    P =

    m=1

    n=0

    (Y0)

    mn|Emn|22a(b + W)

    mn(32)

  • 7/28/2019 14.06071902.Dwari.CS

    8/20

    226 Dwari, Chakraborty, and Sanyal

    Assume

    Ey|z=0 =

    sin(x + a)

    2ab < y < b

    0 |y| > b(33)

    Only non zero amplitudes are:

    E10 = 10A10 =b

    b + W(34)

    E1n = 1nA1n =2

    n

    sin

    n

    b + W

    b +

    W

    2

    sin

    nW

    2(b + W)

    (35)

    Therefore

    P =2ab2

    b + W

    (Y0)10 + 8

    n=1

    (Y0)

    1n

    sin

    nb

    b + W

    cos

    n

    2

    nbb + W

    2 (36)

    Now

    (Y0)10 =k2

    2a

    2

    =

    1

    fc

    f

    2

    (37)

    (Y0)1n =k2

    2a

    2j =

    j4(b + W)(Y0)10

    gn2 4(b + W)

    g2

    (38)

    Therefore

    P

    |V|2 = (Y0)10

    a

    2(b + W)+j

    16a

    g

    n=1

    sin2

    nb

    2(b + W)

    cos2

    n

    2

    nb

    b + W

    2n2

    4(b + W)

    g

    2

    (39)

  • 7/28/2019 14.06071902.Dwari.CS

    9/20

    Progress In Electromagnetics Research, PIER 64, 2006 227

    Figure 4. Cross section of waveguide junction for step discontinuityat both broad and narrow dimension.

    where V = 2b. The imaginary part of this is the aperture susceptanceBa.

    Ba =16a

    gZ0

    n=1

    sin2 nb

    2(b + W) cos2

    n

    2 nb

    b + W

    2n2

    4(b + W)

    g

    2 (40)

    3.2. Evaluation of Circuit Parameters of Equivalent Circuitfor Step Discontinuities at Both Broad and NarrowDimension

    Consider the waveguide junction as shown in Figure 4. Let junction isat z = 0. Waveguide of dimension 2a 2b extends upto z = andwaveguide of dimension 2(a + W1) 2(b + W) extends upto z = + .The dimensions are such that only dominant mode can propagatein each section. Let there be an incident wave from z direction.Following similar procedure as described earlier it can be shown:

    n2 =16(a + W1)

    3ab

    2(b + W)W21 (2a + W1)2

    cos2a

    2(a + W1)(41)

    B =2Bab

    a(42)

    Ba =

    m=1

    n=0

    4a22

    k2

    m

    2(a + W1)

    2(b + W)

    n26b2(a + W1)

    1

    1 maa+W12

    2

  • 7/28/2019 14.06071902.Dwari.CS

    10/20

    228 Dwari, Chakraborty, and Sanyal

    Figure 5. Cross section of waveguide junction for step discontinuityat narrow dimension.

    sin2

    m

    2

    cos2

    n

    2

    cos2

    ma

    2(a + W1)

    sin2

    nb

    2(b + W)

    m

    2(a + W1)

    2+

    n

    2(b + W)

    2 1

    (43)

    4. REFLECTION COEFFICIENT FOR STEPDISCONTINUITIES BY METHOD OF MOMENT

    Method of moment can be used for the analysis of waveguidediscontinuities. Evaluation of reflection coefficient using momentmethod for step discontinuities (1) in narrow dimension only and (2)in both narrow & broad dimension are as follows:

    4.1. Reflection Coefficient for Step Discontinuities inNarrow Dimension by Method of Moment

    Consider the waveguide junction as shown in Figure 5. Let junction

    is at z = 0. Waveguide of dimension 2a 2b1 extends up to z = and waveguide of dimension 2a 2b2 extends upto z = + . It isassumed that b2 > b1. The dimensions are such that only dominantmode can propagate in each section. Let there be an incident wavefrom z direction. In the present analysis, the following assumptionsare made:

    the x component of the electric field at the plane of the apertureis ignored;

    only the x component of the magnetic field at the aperture planeis considered;

    the excitation in the feed waveguide is TE10 dominant wavewith the incident electric field y-directed and uniform in the

    same direction, the assumption of uniform y-directed electric

  • 7/28/2019 14.06071902.Dwari.CS

    11/20

    Progress In Electromagnetics Research, PIER 64, 2006 229

    field having no component in the x-direction follows as a naturalconsequence;

    the electric field Ey in the aperture is assumed to vary only in thex-direction and is constant in the y-direction.

    At the region of window the tangential component of the magneticfield on both sides should be identical. In this analysis two sourcesproducing these fields the source in the waveguide exciting theTE10 mode and the magnetic current source at the discontinuity areconsidered. Using the principle of superposition, the x components ofthe magnetic field at the plane of discontinuity are derived.

    When aperture is shortened (no magnetic current source) the x

    component of magnetic fields are denoted by HI/1x and H

    II /1x on guides

    with narrow dimensions 2b1 & 2b2 respectively.When the generator is shortened the x component of magnetic

    fields are denoted by HI/2x and H

    II /2x on guides with narrow dimensions

    2b1 & 2b2 respectively.

    At the plane of discontinuity using principle of superposition it ispossible to write

    HI/1x + HI/2x = H

    II /1x + H

    II /2x (44)

    Electric field at plane of discontinuity is considered as

    Edis = uy

    Mp=1

    Epe

    p, p = 1, 2, . . . . . . , M . (45)

    Where basis function ep, (p = 1, 2, . . . . . . , M ) are defined by

    e

    p = sin

    p(x + a)

    2a,

    a < x < a

    b1 < y < b10, otherwise

    (46)

    As in [6] internally scattered electric field is given by

    Escat =

    m

    n

    Vemneemn + V

    mmne

    mmn (47)

    Magnetic fields are related to the electric fields as follows:

    Hex = Ye

    mn1Eey

    Hmx = Ym

    mn1Emy

    for wave propagating in z direction. (48)

    Hex =

    Yemn2E

    ey

    Hmx = Ymmn2Emy for wave propagating in +z direction. (49)

  • 7/28/2019 14.06071902.Dwari.CS

    12/20

    230 Dwari, Chakraborty, and Sanyal

    Where Yemn1 & Ym

    mn1 are the characteristic admittances of TEmn &TMmn modes for the waveguide of z direction Yemn2 & Ymmn2 are thecharacteristic admittances of TEmn & TMmn modes for the waveguideof +z direction.

    For waveguide ofz direction modal vectors are given by

    e emn =1

    ab1mn

    (mb1)2 + (na)2

    ux

    n

    2b1

    cos

    m

    2a(x + a)

    sin

    n

    2b1(y + b1)

    uy

    m

    2a

    sin

    m

    2a(x + a)

    cos

    n

    2b1(y + b1)

    (50)

    e mmn = 2

    ab1

    (mb1)2 + (na)2

    ux

    m

    2a

    cos

    m

    2a(x + a)

    sinn

    2b1 (y + b1)

    + uy n

    2b1

    sinm

    2a (x + a)

    cos

    n

    2b1(y + b1)

    (51)

    Similarly for waveguide of +z direction modal vectors are given by

    e emn =1

    ab2mn

    (mb2)2 + (na)2

    ux

    n

    2b2

    cos

    m

    2a(x + a)

    sin

    n

    2b2(y + b2)

    uy

    m

    2a

    sin

    m

    2a(x + a)

    cos n

    2b2(y + b2) (52)

    e mmn = 2

    ab2

    (mb2)2 + (na)2

    ux

    m

    2a

    cos

    m

    2a(x + a)

    sin

    n

    2b2(y + b2)

    + uy

    n

    2b2

    sin

    m

    2a(x + a)

    cos

    n

    2b2(y + b2)

    (53)

    where m, n are the Neumanns number satisfying

    i = 1 for i = 0

    2 otherwise(54)

  • 7/28/2019 14.06071902.Dwari.CS

    13/20

    Progress In Electromagnetics Research, PIER 64, 2006 231

    Following [6] modal voltages of waveguide ofz direction are given by

    Vemn =

    aperture

    Edis e emnds

    =mb1 sin(n)

    n

    ab1mn

    (mb1)2+(na)2

    Mp=1

    Ep

    sin(p+m)

    (p+m) sin(pm)

    (pm)

    (55)

    Vmmn =

    aperture

    Edis e mmnds

    =2a sin(n)

    ab1

    (mb1)2+(na)2

    Mp=1

    Ep

    sin(p+m)

    (p+m) sin(pm)

    (pm)

    (56)

    Similarly modal voltages of waveguide of +z direction are given by

    Vemn =mb2

    n

    ab2mn

    (mb2)2 + (na)2

    Mp=1

    Ep

    sin(p + m)

    (p + m) sin(p m)

    (p m)

    sinn(b2 + b1)

    2b2 sin n(b2 b1)

    2b2

    (57)

    Vmmn =2a

    ab2

    (mb2)2 + (na)2

    Mp=1

    Ep

    sin(p + m)

    (p + m) sin(p m)

    (p m)

    sinn(b2 + b1)

    2b2 sin n(b2 b1)

    2b2

    (58)

    NowHI/1x = 2H

    incx = 2Y0 cos x2a (59)

    Where Hincx is the x component of magnetic field due to incident TE10mode and Y0 is the characteristic admittance of the line due to TE10mode.

    Now

    HII /1x = 0 (60)

    HI/2x =

    m

    n

    Mp=1

    Ep[sin c(p + m)sin c(pm)]

    ab1

    (mb1)2+(na)2

    (sin c(n)) m2b1mnY

    emn1

    2a +

    2n2a

    b1 Y

    m

    mn1

  • 7/28/2019 14.06071902.Dwari.CS

    14/20

    232 Dwari, Chakraborty, and Sanyal

    sin

    m

    2a(x + a)

    cos

    n

    2b1(y + b1)

    (61)

    HII /2x = m

    n

    M

    p=1

    Ep [sin c(p+m)

    sin c(p

    m)]

    ab2

    (mb

    2)

    2

    +(na

    )

    2

    1

    n

    m2b2mnY

    emn2

    2a+

    2n2a

    b2Ymmn2

    sinn(b2 + b1)

    2b2 sin n(b2 b1)

    2b2

    sin

    m

    2a(x + a)

    cos

    n

    2b2(y + b2)

    (62)

    Therefore continuity equation will be

    HI/1x + HI/2x = H

    II /2x (63)

    The weighting function Wq is selected to be the same form as the basisfunction.

    Wq =

    sin

    q(x + a)

    2a,

    a < x < ab1 < y < b1

    0, otherwise(64)

    where q = 1, 2, . . . . . . , M .Inner multiplication is defined as

    < H, Wq >=

    Aperture at discontinuity

    H Wq dxdy (65)

    So

    HI/1x , Wq

    +

    HI/2x , Wq

    =

    HII /2x , Wq

    (66)

    orL1 + L2 = L3 (67)

    Therefore

    L1 =4Y0ab1

    1

    q+ 1cos

    2(2q+ 1) +

    1

    q 1 cos

    2(2q 1)

    (68)

    L2 = 2M

    p=1

    Ep

    m

    n

    m2b1mnY

    emn1

    2a+

    2n2aYmmn1b1

    a2b21

    (mb1)2+(na)2

    (sin(n))

    2

    [sin c(p + m)sin c(p m)]

  • 7/28/2019 14.06071902.Dwari.CS

    15/20

    Progress In Electromagnetics Research, PIER 64, 2006 233

    [sin c(q+ m) sin c(q m)]] (69)

    L3 = 2M

    p=1

    Ep

    m

    n

    m2b2mnY

    mmn2

    2a+

    2n2aYmmn2b2

    a2b22(mb2)2 + (na)2

    1

    n

    2

    sinn(b2+b1)

    2b2sin n(b2b1)

    2b2

    2

    [sin c(p+m)sin c(pm)][sin c(q+m)sin c(qm)]

    (70)

    To solve the unknown coefficients Ep(p = 1, 2, . . . . . . , M ), Equation(67) can be written in the matrix form for all p and q as below:

    [A1] + [A2]

    Ep

    = [A3]

    Ep

    (71)

    or,Ep

    = {[A3] [A2]}1 [A1] (72)

    Ei = Incident electric field = cosx

    2a(73)

    ER = Reflected electric field

    = cos x2a

    M

    p=1

    Ep [sin c(p + 1) sin c(p 1)]sin(x+a)

    2a

    (74)

    Reflection coefficient is given by

    1 = 1 M

    p=1

    Ep [sin c(p + 1) sin c(p 1)] (75)

    Similarly if there be an incident wave from +z direction then

    Ei = Incident electric field = cosx

    2a(76)

    ER = Reflected electric field

    = cos x2a

    b1b2

    Mp=1

    Ep [sin c(p+1) sin c(p1)]sin(x+a)

    2a

    (77)

  • 7/28/2019 14.06071902.Dwari.CS

    16/20

    234 Dwari, Chakraborty, and Sanyal

    Figure 6. Cross section of waveguide junction for step discontinuityat both broad and narrow dimension.

    Reflection coefficient is given by

    2 =

    1

    b1

    b2

    M

    p=1

    E

    p

    [sin c(p + 1)

    sin c(p

    1)] (78)

    4.2. Reflection Coefficient for Step Discontinuities at BothBroad and Narrow Dimension by Method of Moment

    Consider the waveguide junction as shown in Figure 6. Let junction isat z = 0. Waveguide of dimension 2a1 2b1 extends up to z = and waveguide of dimension 2a2 2b2 extends upto z = + . It isconsidered that a1 > a2 and b2 > b1. The dimensions are such thatonly dominant mode can propagate in each section. Following similarprocedure as described in previous section it can be shown:

    If there be an incident wave from

    z direction then reflection

    coefficient:

    1 = 1 a2a1

    Mp=1

    Ep

    cos

    (p + 1)

    2sin c

    (pa1 + a2)

    2a1

    cos (p 1)2

    sin c(pa1 a2)

    2a1

    (79)

    If there be an incident wave from +z direction then reflectioncoefficient:

    2 = 1 b1b2

    M

    p=1Ep [sin c(p + 1) sin c(p 1)] (80)

  • 7/28/2019 14.06071902.Dwari.CS

    17/20

    Progress In Electromagnetics Research, PIER 64, 2006 235

    Figure 7. VSWR versus number of sections for tapering in narrowdimensions.

    5. RESULTS

    From final transmission matrix of taper waveguide reflection co-efficient and the VSWR are calculated. For tapering in narrowdimension considering 2a0 = 5.81cm, 2b0 = 1.20cm, 2a1 = 5.81cm,2b1 = 2.91cm, f = 3.96 GHz, and L = 4 cm, VSWR is plottedagainst number of sections N in Figure 7 using equivalent circuitrepresentation of step discontinuities. For large number of sectionsVSWR is converged. In convergence region the plot of VSWR withtaper length is shown in Figure 8 along with previously publishedresults.

    Calculating reflection coefficient of step discontinuities by momentmethod, for tapering in narrow dimension with same dimensionalparameter as above, the plot of VSWR against length is shownin Figure 9 along with previously published results. Similarly for

    tapering in both dimension with 2a0 = 0.90inch, 2b0 = 0.40 inch,2a1 = 0.75inch, 2b1 = 0.60 inch, f = 8.7GHz and L = 2.85 inch,VSWR is plotted against number of sections N in Figure 10 usingequivalent circuit representation of step discontinuities. In convergenceregion the plot of VSWR with frequency is shown in Figure 11 alongwith previously published results.

    Calculating reflection coefficient of step discontinuities by momentmethod, for double taper section with same dimensional parameter asabove, the plot of VSWR against frequency is shown in Figure 12 alongwith previously published results.

  • 7/28/2019 14.06071902.Dwari.CS

    18/20

    236 Dwari, Chakraborty, and Sanyal

    Figure 8. VSWR versus taper length of a taper at a fixedfrequency of 3.96 GHz. A- Theoretical curve by Johnsons [3] method.B- Theoretical curve by chakraborty & Sanyals [1] method. C-Experimental points as reported by Matsumaru [2]. D- By equivalentcircuit.

    Figure 9. VSWR versus taper length of a taper at a fixedfrequency of 3.96 GHz. A- Theoretical curve by Johnsons [3] method.B- Theoretical curve by chakraborty & Sanyals [1] method. C-Experimental points as reported by Matsumaru [2]. D- By moment

    circuit.

  • 7/28/2019 14.06071902.Dwari.CS

    19/20

    Progress In Electromagnetics Research, PIER 64, 2006 237

    Figure 10. VSWR versus number of sections for double taper section.

    Figure 11. VSWR versus frequency for a linear double taper. A-Theoretical curve by Johnsons [3] method. B- Theoretical curveby chakraborty & Sanyals [1] method. C- Experimental points asreported by Johnson [3]. D- By equivalent circuit.

    Figure 12. VSWR versus frequency for a linear double taper. A-Theoretical curve by Johnsons [3] method. B- Theoretical curveby chakraborty & Sanyals [1] method. C- Experimental points as

    reported by Johnson [3]. D- By moment circuit.

  • 7/28/2019 14.06071902.Dwari.CS

    20/20

    238 Dwari, Chakraborty, and Sanyal

    6. CONCLUSION

    The results obtained in the presented methods are to be in goodagreement with results of [1, 2] and [3]. Here it is assumed that only

    TE10 mode is propagating along the tapered line. Even if other modesare generated they are not supported by the structure.

    REFERENCES

    1. Chakraborty, A. and G. S. Sanyal, Transmission matrix of alinear double taper in rectangular waveguides, IEEE Trans.Microw. Theory Tech., Vol. MTT-28, No. 6, 577579, Jun. 1980.

    2. Matsumaru, K., Reflection coefficient of E-plane taperedwaveguides, IRE Trans. Microw. Theory Tech., Vol. MTT-6,143149, Apr. 1958.

    3. Johnson, R. C., Design of linear double tapers in rectangular

    waveguides,IRE Trans. Microw. Theory Tech.,

    Vol. MTT-7,374378, July 1959.

    4. Ghosh, R. N., Microwave Circuit Theory and Analysis, Ch. 12,352355, McGraw-Hill, New York, 1963.

    5. Mallick, A. K. and G. S. Sanyal, Electromagnatic wavepropagation in a rectangular waveguide with sinusoidally varyingwidth, IEEE Trans. Microw. Theory Tech., Vol. MTT-26, 243249, Apr. 1978.

    6. Harrington, R. F., Time Harmonic Electromagnetic Fields,McGraw-Hill, New York, 1961.