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{
}
{
}
F(
pxt
,
F(
pxt
,
r1
r 2
Rx f
(,)
=
,
Rx f
( ,)
=
,
(5.6)
1
{
}
2
{
}
F(
pxt
,
F(
pxt
,
i
1
i
2
and a transfer function H 12 for the total pressure in these two positions:
{
}
{
}
F
px t
( ,)
F
p x t
( ,)
+
p x t
( ,)
2
i
2
r
2
Hf
()
=
=
(5.7)
.
12
{
}
{
}
F(, )
px t
F (, )
p x t
+
p x t
(, )
1
i
1
r
1
p i ( x 1 , t )
p r ( x 1 , t )
p i ( x 2 , t )
p r ( x 2 , t )
x
Figure 5.5 Wave components in a standing wave tube.
Correspondingly, for the pressure in the incident and reflected wave, respectively,
we may define transfer functions
{
}
{
}
F(
pxt
, )
F(
pxt
, )
[
]
[
]
i
2
r
2
Hf
()
=
,
Hf
()
=
(5.8)
.
12
{
}
12
{
}
i
r
F(
pxt
, )
F(
pxt
, )
i
1
r1
From Equations (5.6) to (5.8) we get by eliminating R ( x 2 , f ):
[ ]
HH
12
12
i
Rx f
(,)
=
(5.9)
.
1
[ ]
HH
12
12
r
We have assumed plane wave propagation only and we may then write
[ ]
[ ]
−⋅
j
kxx
(
)
j
kxx
(
)
He
=
and
He
=
,
(5.10)
2 21
121
12
12
i
r
where k 12 and k 21 are wave numbers for the incident and reflected wave, respectively.
Furthermore, assuming no energy losses in the tube between these two positions, we may
write k 12 = k 21 = k 0 and Equation (5.9) will become
j
kd
He
0
12
Rx f
(,)
=
,
(5.11)
1
j
kd
e
H
0
12
 
 
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