Wave Propagation in Linear Media | re-examined

3.9 A Gaussian pulse in plasma
Like always, we introduce a normalised integration variable
and normalised parameters
= !
!p
T = !pt ; T0=!pt0; =!p ; X= !px
c
; = !0
!p
(3.101)
: (3.102)
With these new variables, we can write the voltage at any position along the line as
U(X; T )=U0 2 p
Z 1
,1
( , )
, 2 e
2
e ,jT0( , ) e j( T,K( )X) d ; (3.103)
with the normalised dispersion relation K( )= p 2 ,1. As for the sign of the square root,
we must keep in mind that since we regard only right-going waves, the negative branch applies
to negative frequencies. In the evanescent region, we must also choose the negative solution
of the root in order to obtain attenuation along the line. So in the three parts of the range of
integration, the dispersion relation is given by
K( )=
8
><
>:
, p 2 ,1 if 1
: (3.104)
Remark (Transfer function of the plasma) Looking at the wave integral from a
system theoretical point of view, we obtain still another reasoning for the choice of the
sign of the dispersion relation. We can think of the wave integral (3.99) as the response
of a system to a signal u(0;t),
u(x; t) =
Z 1
where the transfer function of the system is
A(!)H(!) e
,1
j!t d! ; (3.105)
H(!) =e ,jk(!)x : (3.106)
It is well known that a physically meaningful system must give a purely real response
when excited with a purely real signal. According to the basic properties of the Fourier
transform, this can only be guaranteed if the real and the imaginary parts of the transfer
function are even and odd functions in !, respectively. It is easy to see that the de nition
of the dispersion relation given above satis es these requirements. Within the pass
band, the imaginary part of H(!) is essentially the sine of an odd argument function
and thus also odd, whereas the real part is the cosine of an odd function and thus even.
In the evanescent region, the imaginary part of the transfer function vanishes, and the
exponentially decaying real part remains even.
The current in the transmission line could be calculated by inserting the voltage into the
di erential equations (3.75) and solving for I, which gives
i(x; t) =L 0
Z 1
A(!)
,1
k(!)
! ej(!t,k(!)x) d! : (3.107)
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