Electromagnetic Waves and Antennas combined - Chapter 2 potx
Electromagnetic Waves and Antennas combined - Chapter 2 potx
... expressions:
A
2
= A
2
+τ
s
AB cos φ =
A
2
−B
2
τ
2
s
1 −τ
2
s
=
1
2
A
2
+B
2
+sD
= A
2
B
2
λ
s
B
2
= B
2
−τ
s
AB cos φ =
B
2
−A
2
τ
2
s
1 −τ
2
s
=
1
2
A
2
+B
2
−sD
= A
2
B
2
λ
−s
(2. 13.4)
2. 13. Problems ... OD, are given by:
A
=
1
2
(A
2
+B
2
)+
s
2
(A
2
−B
2
)
2
+4A
2
B
2
cos
2
φ
B
=
1
2
(A
2
+B
2
)−...
Electromagnetic Waves and Antennas combined - Chapter 6 potx
... +ρ
2
2
ρ
2
3
+2
2
ρ
3
cos 2k
2
l
2
= ρ
2
1
This can be solved for cos 2k
2
l
2
:
cos 2
k
2
l
2
=
ρ
2
1
(1 +ρ
2
2
ρ
2
3
)−(ρ
2
2
+ρ
2
3
)
2
2
ρ
3
(1 −ρ
2
1
)
Using the identity, cos 2
k
2
l
2
= 2 cos
2
k
2
l
2
−1, ... condition
|Γ
2
|=
|ρ
1
|, or, |Γ
2
|
2
= ρ
2
1
, which is written as:
ρ
2
+ρ
3
e
−2jk
2
l
2
1 +ρ
2
ρ...
Electromagnetic Waves and Antennas combined - Chapter 7 potx
... (7.1.7):
k
2
z
+k
2
x
= k
2
= n
2
k
2
0
k
z
2
+k
x
2
= k
2
= n
2
k
2
0
Because, k
x
= k
x
= k sin θ = nk
0
sin θ, we may solve for k
z
to get:
k
2
z
= n
2
k
2
0
−k
2
x
= n
2
k
2
0
−k
2
x
= ... requires
that the numerator of (7. 12. 13) be positive, or,
(1 +β
2
)cos θ 2 ≥ 0 cos θ ≥
2
1 +β
2
θ ≤ acos
2
1 +β
2
(7. 12. 22)
Because 2
β...
Electromagnetic Waves and Antennas combined - Chapter 9 potx
... k
2
0
n
2
1
−β
2
−α
2
c
= k
2
0
n
2
2
−β
2
⇒
k
2
c
= k
2
0
n
2
1
−β
2
α
2
c
= β
2
−k
2
0
n
2
2
(9.11.3)
Similarly, Eqs. (9.11 .2) read:
∂
2
x
H
z
(x)+k
2
c
H
z
(x)= 0 for |x|≤a
∂
2
x
H
z
(x)−α
2
c
H
z
(x)= ... R
s
a
2
|H
0
|
2
+|H
1
|
2
+R
s
b|H
0
|
2
=
R
s
a
2
|H
0
|
2
+|H
1
|
2
+
2b
a
|H
0
|
2
Using |H
0
|
2
+|H
1
|
2
=|E
0
|...
Electromagnetic Waves and Antennas combined - Chapter 13 potx
... constants:
B
2
1
−4|C
1
|
2
= B
2
2
−4|C
2
|
2
= 4|S
12
S
21
|
2
(K
2
−1)
|C
1
|
2
=|S
12
S
21
|
2
+
1 −|S
22
|
2
D
1
|C
2
|
2
=|S
12
S
21
|
2
+
1 −|S
11
|
2
D
2
(13.5 .2)
534 13. S-Parameters
For ... −|S
11
|
2
|S
22
−ΔS
∗
11
|+|S
12
S
21
|
(Edwards-Sinsky stability parameter)
μ
2
=
1 −|S
22
|
2
|S
11
−ΔS
∗
22
|+|S
12
S...
Electromagnetic Waves and Antennas combined - Chapter 23 potx
... length
dualband - two-section dual-band Chebyshev impedance transformer
dualbw - two-section dual-band transformer bandwidths
stub1 - single-stub matching
stub2 - double-stub matching
stub3 - triple-stub ... x
=
⎡
⎢
⎢
⎢
⎣
τ
x
y
z
⎤
⎥
⎥
⎥
⎦
,L=
⎡
⎢
⎢
⎢
⎣
γ
00−γβ
0100
0010
−γβ 00 γ
⎤
⎥
⎥
⎥
⎦
(H .2)
Such transformations leave the quadratic form
(c
2
t
2
−x
2...
Electromagnetic Waves and Antennas combined - Chapter 1 pps
... energies:
¯
w
mech
=
1
2
Re
N
1
2
m|v|
2
+
1
2
mω
2
0
|x|
2
=
1
4
Nm(ω
2
+ω
2
0
)|x|
2
=
1
4
Nm(ω
2
+ω
2
0
)e
2
|E|
2
/m
2
(ω
2
0
−ω
2
)
2
=
1
4
0
|E|
2
ω
2
p
(ω
2
+ω
2
0
)
(ω
2
0
−ω
2
)
2
where we used the definition (1.11.6) of the plasma frequency. It follows that Eq. (1.16 .2)
can ... integrals, for
which you may assume that 0
&...
Electromagnetic Waves and Antennas combined - Chapter 3 ppt
... gain case:
n = 1 +
fω
2
p
/2
ω
2
1
−ω
2
+jωγ
+
fω
2
p
/2
ω
2
2
−ω
2
+jωγ
n
g
= 1 +
fω
2
p
(ω
2
+ω
2
1
) /2
(ω
2
1
−ω
2
+jωγ)
2
+
fω
2
p
(ω
2
+ω
2
2
) /2
(ω
2
2
−ω
2
+jωγ)
2
(3.9.8)
The two peaks ... parts:
Re
(n
g
) = 1 +
fω
2
p
(ω
2
+ω
2
r
)
(ω
2
−ω
2
r
)
2
−ω
2
γ
2
(ω
2
−ω
2
r
)
2
+ω
2
γ
2
2
Im(n
g
) =
fω
2
p
γω...
Electromagnetic Waves and Antennas combined - Chapter 4 pdf
... =
n
2
1
+n
2
3
−N
2
N
2
(4.7.11)
sin
2
θ =
1 −
n
2
1
N
2
1 −
n
2
1
n
2
3
, cos
2
θ =
1 −
n
2
3
N
2
1 −
n
2
3
n
2
1
(4.7. 12)
cos
2
θ −
n
2
1
N
2
=−
n
2
1
n
2
3
sin
2
θ, sin
2
θ −
n
2
3
N
2
=−
n
2
3
n
2
1
cos
2
θ (4.7.13)
Using ... relationships:
sin
2
θ
1 −
n
2
1
N
2
+
cos
2
θ
1 −
n
2
3
N
2
= 1 (4.7.9)
n
3
n
1
cos
2
θ +
n
1
n...
Electromagnetic Waves and Antennas combined - Chapter 5 ppt
... have:
Z
1
=
η
2
1
Z
2
=
η
2
1
η
2
2
/Z
3
=
η
2
1
η
2
2
Z
3
=
η
2
1
η
2
2
η
b
Inserting this into Γ
1
= (Z
1
−η
a
)/(Z
1
+η
a
), we obtain:
Γ
1
=
η
2
1
η
b
−η
2
2
η
a
η
2
1
η
b
+η
2
2
η
a
The two ... η
b
.
If
η
1
is half- and η
2
quarter-wavelength, then, Z
1
= Z
2
= η
2
2
/Z
3
= η
2
2
/η
b
. And, if the
quarter-wavelength is first and the half-wavelength se...