Homework SourseThe 300N tension force F is exerted on the pipe assembly as
Solution
The fields E1,H1 are obtained by successively applying Eq. (5.4.9):
E1
H1
=
cos k1l1 j1 sin k1l1
j1
1 sin k1l1 cos k1l1
E2
H2
=
cos k1l1 j1 sin k1l1
j1
1 sin k1l1 cos k1l1
cos k2l2 j2 sin k2l2
j1
2 sin k2l2 cos k2l2
E3
H3
But at interface-3, E3 = E
3 = E
3+ and H3 = Z1
3 E3 = 1
b E
3+, because Z3 = b.
Therefore, we can obtain the fields E1,H1 by the matrix multiplication:
E1
H1
=
cos k1l1 j1 sin k1l1
j1
1 sin k1l1 cos k1l1
cos k2l2 j2 sin k2l2
j1
2 sin k2l2 cos k2l2
1
1
b
E
3+
Because Z1 is the ratio of E1 and H1, the factor E
3+ cancels out and can be set equal
to unity.
Example 5.7.1: Determine 1 if both slabs are quarter-wavelength slabs. Repeat if both slabs
are half-wavelength and when one is half- and the other quarter-wavelength.
Solution: Because l1 = 1/4 and l2 = 2/4, we have 2k1l1 = 2k2l2 = , and it follows that
z1 = z2 = 1. Then, Eq. (5.7.1) becomes:
1 = 1 2 123 + 3
1 12 23 + 13
A simpler approach is to work with wave impedances. Using Z3 = b, we have:
Z1 = 21
Z2
= 21
22
/Z3
= 21
22
Z3 = 21
22
b
Inserting this into 1 = (Z1 a)/(Z1 + a), we obtain:
1 = 21
b 22
a
21
b + 22
a
The two expressions for 1 are equivalent. The input impedance Z1 can also be obtained
by matrix multiplication. Because k1l1 = k2l2 = /2, we have cos k1l1 = 0 and sin k1l1 = 1
and the propagation matrices for E1,H1 take the simplified form:
E1
H1
=
0 j1
j1
1 0
0 j2
j1
2 0
1
1
b
E
3+ =
11
2
21
1 1
b
E
3+
The ratio E1/H1 gives the same answer for Z1 as above. When both slabs are half-wavelength,
the impedances propagate unchanged: Z1 = Z2 = Z3, but Z3 = b.
If 1 is half- and 2 quarter-wavelength, then, Z1 = Z2 = 22
/Z3 = 22
/b. And, if the
quarter-wavelength is first and the half-wavelength second, Z1 = 21
/Z2 = 21
/Z3 = 21
/b.
The corresponding reflection coefficient 1 is in the three cases:
1 = b a
b + a
, 1 = 22
ab
22
+ ab
, 1 = 21
ab
21
+ ab
These expressions can also be derived by Eq. (5.7.1), or by the matrix method
