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									                                                                            DESIGN NOTE
TRIUMF                                                                       TRI-DN-96-2
                                                                              March 1996




   From Maxwell’s equations for a cavity resonator to a parallel
                LCR circuit using simple maths


                                      S. Koscielniak
                                         TRIUMF


                                      ABSTRACT

The purpose of these calculations is to demonstrate that a resonant cavity behaves like
a parallel LCR circuit in the neighbourhood of resonance. This property has been well
known for a long time, but the rigorous development is very lengthy and mathematical. The
argumenmts used here are rough; but nonetheless results are useful and easily comprehended.




TRIUMF        4004 WESBROOK MALL, VANCOUVER, B.C. CANADA V6T 2A3
              From Maxwell’s equations for a cavity resonator
                to a parallel LCR circuit using simple maths

    The purpose of these calculations is to demonstrate that a resonant cavity behaves like
a parallel LCR circuit. This property has been well known for a long time, but the rigorous
development[1] is very lengthy and mathematical. The arguments used here are very rough; but
nontheless results are useful. We shall consider the TM010 mode in a pill-box cavity. In section 1
we find the resonance frequency. In section 2 we find the magnetic field in terms of a driving
current. In section 3 we include the effect of losses and find the quality factor. In section 4 we
find the shunt resistance.



1     Self-excited, lossless

We postulate magnetic and electric fields of the form:

                        H = H h(x)e+jωc t ,
                            ˜                               E = E e(x)e+jωc t .
                                                                ˜                                (1)
     ˜ ˜
Here H, E are phasors (i.e. complex quantities) and the vectors h, e specify the field shapes.
Here x is a general coordinate vector.

    Take cylindrical polar coordinates r, φ, z and right-handed orthogonal unit vectors nr , nφ , nz .
Pick field shapes which satisfy the boundary conditions of a pill-box resonator, of length d and
radius a.
                            h = nφ J1 (kr) ,            e = nz J0 (kr) ,                       (2)
with J0 (ka) = 0 or ka = P01 ≈ 2.405. Applying Stokes’ theorem1 to Maxwell’s equation for the
‘curl’ of magnetic field, we find

                                      H · dl = jωc          E · dS .                             (3)

 We take a circular path (see figure 1) close to the circumference of the cavity, that is just inside
the dielectric.

                            h · dl = 2πaJ1 (ka)                                                  (4)
                                               a
                           e · dS = 2π             J0 (kr)rdr = 2π(a/k)J1(ka)                    (5)
                                           0

Hence
                                        ˜           ˜
                                        H = j(ωc /k)E .                                          (6)

    Applying Stokes’ theorem to Maxwell’s equation for the ‘curl’ of electric field, we find

                                  −µjωc       H · dS =        E · dl .                           (7)

                                                     1
       path                         z

                     E

                             H




                             resonator wall

Figure 1: integration path for magnetic field




                 integration path



                         H
                                               z
                 E



Figure 2: integration path for electric field




                     2
We take a rectangular path (see figure 2) along the centre line, and along the cavity walls.
                                                            a
                                       h · dS = d               J1 (kr)dr = d/k                               (8)
                                                        0

                                           e · dl = d                                                         (9)

Hence
                                                        ˜   ˜
                                              −µj(ωc /k)H = E .                                              (10)
We have two equations (6, 10) for two unknowns: the resonance frequency, ωc , and the field
      ˜ ˜
ratio H/E . We obtain
                           ωc = ±kc      and      ˜       ˜
                                                 E = ∓jη H ,                          (11)
where
                                   √
                          c = 1/       µ       and          η=       µ/ = 376.7 ohm .                        (12)
η is known as the free-space impedance. Let us find the charges and current on the cavity walls,
which can be considered the field sources.


1.1     Surface current and charge

The current density, Jc , flows longitudinally over the cylindrical side wall at r = a. If we take
an integration contour similar to that for equation (6) but just inside the conducting material,
then we shall find:
                                 H · dl = 0 = (Jc + jωc E) · dS .                            (13)
                              ˜
Hence the total wall current, Ic , is

                                              E˜
                                          ˜
                                Jc · dS = Ic = 2πaJ1 (P01 )                 ampere .                         (14)
                                              jη
Here we replaced ωc /k by c.

    The longitudinal electric field starts and terminates at the surface charges on the end walls
at z = 0 and z = d. If we apply Gauss’ theorem to an infinitesimal cylindrical volume element
(in contact with the conductor) we shall find the surface charge density is
                                        ˜
                               σc (r) = EJ0 (kr)                coulomb/(metre)2 .                           (15)



2       Driven, off resonance, lossless

Suppose the cavity is driven by a charged particle beam current. For simplicity, we shall suppose
the wavelength between bunches is long compared to the cavity gap-width, and so can ignore
    1
    The spatial relation of dl to dS is analogous to that of nr and nφ to nz ; if l is taken anti-clockwise in the
plane of the page, then dS points out of the page.

                                                        3
that the beam component is a travelling wave.
                            Jb = Jb j(x)e+jωt ,
                                 ˜                   where       j = nz jb (r) .                (16)
We take the magnetic and electric fields to have the same time dependence as the beam current,
that is e+jωt with ω = ωc .

    Applying Stokes theorem to Maxwell’s equation for the ‘curl’ of magnetic field, we find

                                         H · dl =   (Jb + jωE) · dS .                           (17)
                        ˜
The total beam current, Ib , is given by
                                                          ˜
                                                Jb · dS = Ib .                                  (18)

We take a circular path close to the circumference of the cavity, that is just inside the dielectric.
                                              ˜
                                             Ib           ω˜
                                   ˜
                                  H =                + j E.                                     (19)
                                          2πaJ1 (ka)       k

    Applying Stokes’ theorem to Maxwell’s equation for the ‘curl’ of electric field, we take a
rectangular path along the centre line, and along the cavity walls.
                                                    ˜   ˜
                                            −µj(ω/k)H = E .                                     (20)
                                                  ˜      ˜                              ˜
We have two equations (19, 20) for two unknowns: E and H in terms of the total current Ib . We
eliminate E˜ by taking (19) − j(ω/k) × (20). The physical analogue of our ability to eliminate
 ˜
E by differencing the simultaneous equations is that the peak electrical and magnetic stored
                                                             2
energies are (almost) equal. We use the fact that µ/k 2 = 1/ωc to find
                                               2        ˜
                                        ˜     ωc        Ib
                                        H =                                                (21)
                                         (ωc − ω 2) 2πaJ1 (P01 )
                                            2

                                 ˜             ω ˜
                                E = −jη H .                                                (22)
                                              ωc
Note, the magnetic and electric stored energies are equal at resonance. Evidently, if ω > ωc
then more energy is stored in the electric field than in the magnetic field; this is the analogue
of the fact that more energy is stored in the capacitance of a parallel LCR circuit if driven
above its resonance frequency. The excess energy is alternately passed back and forth between
the beam and the cavity each quarter cycle of the field oscillation.


2.1    Relation of voltage to current

The peak, instantaneous gap voltage is
                                    d                         ωωc          ˜
                                                                           Ib
                       ˜
                       V =−             E · nz dz = +jηd                            .           (23)
                                0                          (ωc − ω 2 ) 2πaJ1 (P01 )
                                                             2

Evidently, if ωc > ω then the gap voltage (induced by the beam current) leads the beam current
by 90◦ ; whereas if ωc < ω then the gap voltage lags by 90◦ .

                                                     4
2.2         Wall current

Again, we may try to find the surface current that flows in the cavity walls. We take an
integration contour similar to that for equation (17) but just inside the conducting material,
then we shall find:
                                 ˜ ˜              ˜
                            0 = Ic + Ib + j(ω/k)E × 2πaJ1 (P01 ) .                       (24)
                ˜
If we eliminate E (by using equations 21, 22) , we shall find
                                                               2
                                               ˜    ˜         ωc
                                              −Ic = Ib                .                     (25)
                                                          (ωc − ω 2 )
                                                            2


                                                               ˜
Incidentally, in equation (24) we may associate the first term, Ic , with the current through
                                                    ˜
the lumped inductance, and the last term (∝ j(ω/k)E) with the current through the lumped
capacitance of the equivalent LC circuit.



2.3         generator current

From Ampere’s theorem (relating the line integral of H to current sources) it is evident that a
current Ig flowing in a coupling loop2 and moving parallel to the cavity axis will have exactly
         ˜
the same effect as the beam current, provided that both currents flow in the same sense and
have the same magnitude. Consequently, one may use superposition to find the effect of cavity
excitation by generator and beam together.



3         Self-excited, lossy

We postulate fields of the form:

                             H = H h(x)e(+jωc −α)t ,
                                 ˜                            E = E e(x)e(+jωc −α)t .
                                                                  ˜                         (26)

If the conductivity, σ, is large then the boundary conditions are hardly disturbed by their lossy
nature; and so, in the interior of the cavity, we can use the same field shapes h and e as before
(equation 2). However, we must realize that there are now small electric fields, ∆Ee(+jωc −α)t ,
parallel to the bounding surfaces (and just inside the conducting material).

    Applying Stokes’ theorem to Maxwell’s equation for the ‘curl’ of magnetic field, we take a
circular path close to the circumference of the cavity, that is just inside the dielectric.

                                             ˜   ˜
                                             H = E(jωc − α)/k .                             (27)
    2
        We assume that the current carrying conductor is almost the length of the cavity.

                                                         5
    Applying Stokes’ theorem to Maxwell’s equation for the ‘curl’ of electric field, we take a
rectangular path along the centre line, and along the cavity walls.
                                ˜                  ˜    1
                           −µH(jωc − α)/k = E +              ∆E · dl .                     (28)
                                                        d
Evidently, we must first find an expression for the losses term appearing on the far right of
equation (28). Now ∆E = Jc /σ and the conduction current density can be found by employing
the relation ∇ ∧ H = Jc at the conducting surface. Let δ indicate the Dirac delta function. Let
Jc = nr Jr + nz Jz and recall that h = nφ J1 (kr). One finds:
  Jr            ˜
             = −HJ1 (kr)δ(z) ,    Jr            ˜
                                             = +HJ1 (kr)δ(z − d) ,   Jz            ˜
                                                                                = −HJ1 (ka)δ(r − a) . (29)
       z=0                             z=d                                r=a
We substitute and find
                                        ˜
                                        H δ(z) δ(z − d)
                         ∆E · dl =            +         + d J1 (ka)δ(r − a) .                         (30)
                                        σ  k       k
This last equation (30) is not much use to us. However, we recall that most of the electric
field is contained within a thickness δs , the skin-depth, of the surface. Hence, let us integrate
perpendicular to the path l to obtain:
                                                 ˜
                                                 H 2
                                δs ∆E · dl =            + d J1 (ka) .                        (31)
                                                 σ k
Hence, we can write (28) in the form
                                                       ˜
                          −µH ˜ (jωc − α) = E + H 2 + J1 (ka) .
                                               ˜                                             (32)
                                    k                 σδs kd
                                                                                           ˜ ˜
We have two equations (27, 32) for two unknowns: the decay rate, α, and the field ratio H/E .
It is usual to replace 1/(σδs ) by the surface resistivity Rs (ohms).

    We find that
                         µ                      ˜ 2 + J1 (P01 ) = 0 .
                           2
                             (jωc − α)2 E + Rs H
                             1+         ˜                                               (33)
                         k                         kd
              ˜
We eliminate H by using equation (27), and divide through out by (jωc − α)/ωc so as to leave
the eigenvalue equation:
                          (jωc − α)      ωc                 Rs 2
                                    +                  +         + J1 (P01 ) = 0 .                    (34)
                             ωc       (jωc − α)             η kd
                            2α        α        Rs 2a
                        or      1+          =            + J1 (P01 ) .                    (35)
                            ωc       j2ωc       η P01 d
Now, we assume the losses are small, so that α/(2ωc)     1. Further, we note that 2α/ωc = 1/Q
where Q is the quality factor. Hence
                       1      Rs 2a                  Rs a 2         P01
                          =            + J1 (P01 ) ≈            +       .                 (36)
                      Q        η P01 d               η d P01         2
This may be compared with the exact value from text books[2],
                                    1     Rs a         2
                                       =          +1       .                              (37)
                                   Q       η d        P01
Note that P01 ≈ 2.405, and that the agreement is remarkable.

                                                        6
4    Driven, lossy

H, E and Jb have the time dependencies assumed in section 2 – except that we may remove
the restriction ω = ωc .

    Applying Stokes’ theorem to Maxwell’s equation for the ‘curl’ of magnetic field, we take a
circular path close to the circumference of the cavity, that is just inside the dielectric.
                                            ˜
                                           Ib         ω˜
                                 ˜
                                 H =               + j E.                                (38)
                                        2πaJ1 (ka)    k

    Applying Stokes’ theorem to Maxwell’s equation for the ‘curl’ of electric field, we take a
rectangular path along the centre line, and along the cavity walls.
                                        ω˜      ˜     η ˜
                                   −µj H = E + H .                                       (39)
                                        k             Q
The last term in equation (39) represents the losses due to finite conductivity. We have two
                                     ˜      ˜                              ˜
equations (38, 39) for two unknowns: E and H in terms of the total current Ib .
                 ˜
    We eliminate E to find:
                                  ω2 ˜        ˜
                                             Ib          ωη ˜
                           1− µ     2
                                      H =            − j    H.                           (40)
                                  k       2πaJ1 (ka)     kQ
Now ck = ωc and so
                               2
                             (ωc − ω 2 )   j ω ˜        ˜
                                                       Ib
                                  2
                                         +      H =            .                         (41)
                                 ωc        Q ωc     2πaJ1 (ka)
We now perform some mathematical manipulation. We multiply throughout by Qωc /(jω). We
define the detuning angle by
                                                2
                                             Q(ωc − ω 2 )
                                   tan Ψ =                .                        (42)
                                                ωωc
Hence, the magnetic field phasor is given by:
                                                   Q ωc    ˜
                                                          Ib
                                           ˜
                             [1 − j tan Ψ] H =                    .                      (43)
                                                   j ω 2πaJ1 (ka)
This may be simplified, using trigonometric identities, to give:
                                                          ˜
                                                         Ib
                              H = −jQ cos Ψe+jΨ
                              ˜                                  .                       (44)
                                                      2πaJ1 (ka)
Notice how the amplification factor Q cos Ψ replaces the resonant denominator appearing in
equation (21).

    We may go on to solve for the electric field phasor:

                                   ˜     1    ω   ˜
                                   E = −   +j    ηH .                                    (45)
                                         Q    ωc

                                               7
Now, ω ≈ ωc and Q        1, and so to good approximation:

                                η˜          ˜
                                           Ib
                            ˜
                            E = H = −ηQ            cos Ψe+jΨ .                           (46)
                                j       2πaJ1 (ka)


4.1    gap voltage and shunt resistance

Finally, we may find the peak, instantaneous gap voltage

                           ˜
                                        d                 ηdQ cos Ψe+jΨ ˜
                           V =−             E · nz dz = +               Ib .             (47)
                                    0                      2πaJ1 (P01 )
If we compare with the conventional representation

                                     V = Rshunt cos Ψe+jΨ Ib ,
                                     ˜                    ˜                              (48)

then it is clear, the parallel (or shunt) resistance is given by:
                                        Rshunt      ηd
                                               =              .                          (49)
                                         Q       πa2J1 (P01 )

This may be compared with the value in text books[3]:
             Rshunt        ηd                      ηd                     ηd
                    =        2
                                      ≈                            ≈                .    (50)
              Q       πaP01 J1 (P01 )   πaP01 J1 (P01 ) × (P01 /2)   πa2.9J1 (P01 )
Given the roughness of our derivation, the agreement is remarkable.


4.2    power flow

We may calculate the power W exchanged with the cavity:

                            2W = R[I V ∗ ] = R[I ∗ V ]
                                   ˜˜           ˜ ˜                                      (51)
                               = +Rshunt cos2 Ψ|Ib |2
                                                 ˜                 (watt) .              (52)

The electrical power is positive, indicating that energy flows into the cavity and out of the
beam.


4.3    Effect of detuning

Let us consider the voltage and current relations for a parallel LCR resonator. Let us assume
e+jωt time dependence.
                                ˜     1               1    ˜
                                I=      + j ωC −           V .                            (53)
                                     R              ωL

                                                      8
                        2                                                   ˜    ˜
At resonance we find ωc = 1/(LC), and that voltage and current are in phase: V = RI. Off
(but close to) resonance we find

                            ˜  1             1             ˜
                            I≈   + j ωc C +      (ω − ωc ) V                               (54)
                               R            ωc L
Clearly if ω > ωc then current leads and voltage lags. Conversely, if ω < ωc then voltage leads
and current lags. Let us compare this behaviour with the pill-box resonator.

    From equation (42) we find that Ψ < 0 when ω > ωc , and so from (48) voltage lags the
current. Conversely, when ω < ωc then Ψ > 0, and so the voltage leads the current. Evidently,
with respect to detuning, the resonator displays the same behaviour as the parallel LCR circuit.



5    Driven, on resonance, lossless

This particular case is somewhat away from the main stream of our presentation, but is given
here for completeness. The fields are anticipated to increase linearly with time.

    We postulate magnetic and electric fields of the form:

                 H = [ωtH + h] h(x)e+jωt ,
                        ˜ ˜                          E = [ωtE + e] e(x)e+jωt .
                                                            ˜ ˜                            (55)
                   ˜     ˜
The ‘extra’ fields, h and e, account for the fact that the beam current will give rise to fields
even at time t = 0.

    Applying Stokes’ theorem to Maxwell’s equation for the ‘curl’ of magnetic field, we take a
circular path close to the circumference of the cavity, that is just inside the dielectric.

                             ˜   ˜    ˜    ω ˜
                          [ωtH + h] = Ib +                  e
                                             [E(1 + jωt) + j˜] .                           (56)
                                           k
Here, for brevity, we introduced the current per unit length of azimuth:
                                                 ˜
                                                 Ib
                                      ˜
                                      Ib =                .                                (57)
                                             2πaJ1 (P01 )

Applying Stokes’ theorem to Maxwell’s equation for the ‘curl’ of electric field, we take a rect-
angular path along the centre line, and along the cavity walls.
                             ω ˜              ˜       ˜
                           −µ [H(1 + jωt) + j h] = [ωtE + e] .
                                                          ˜                                (58)
                             k

    We compare the coefficients of t to obtain the simultaneous equations:
                                          ˜         ˜
                                          H = j(ω/k)E                                      (59)
                                          ˜   ˜
                                  −µj(ω/k)H = E .                                          (60)

                                               9
These relations are identical with those of section 1, and imply that ω = ωc = kc
     ˜         ˜
and E = −jη H. Note, it was the assumed form of the time dependence (i.e. fields linearly
proportional to t) that forces ω = ωc – it was not that we assumed, at the outset, that the drive
was ‘on resonance’.

    We compare the constant terms to give the simultaneous equations:
                                        ˜   ˜    ˜    e
                                       H = Ib + [E + j˜]/η                                  (61)
                                  ˜     ˜ = e,
                               −η[H + j h]  ˜                                               (62)

where we have used ωc /k = 1/η and µ ωc /k = η . These relations may be rearranged to obtain
a more symmetrical form:
                                   ˜          ˜    ˜
                                 η h − j˜ = η Ib + E
                                        e                                                   (63)
                                        ˜        ˜    ˜
                                 e + jη h = −η H = −j E .
                                 ˜                                                          (64)

We may solve equations (63) and (64) to obtain

                                 e      ˜      ˜
                                (˜ + jη h) = j Ib η /2                                      (65)
                                                η      ˜
                                                       Ib
                                         ˜
                                        E = −                  .                            (66)
                                                2 2πaJ1 (P01 )

Evidently, the electric field induced by the beam opposes its motion and acts to decelerate it.



References
        o
[1] G. Dˆme: RF systems: waveguides and cavities; CERN SPS/86-24 (ARF).

[2] Ramo, Whinnery, Van Duzer: Fields and waves in communication electronics; John Wiley
    & Sons, New York.

[3] S. Koscielniak: Introduction to RF systems in Accelerators; USPAS, San Diego, January
    1996. http://decu10.triumf.ca:8080/uspas/




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