Page 470 - Electromagnetics

Here a is a positive, finite real quantity, and λ and µ are finite complex quantities.
Outside the range |t| < a the time-domain function is zero.
Letting a = z/v, µ = 0, and λ = 2 in the above expression, we find
v − t
2 2
2
Q(x, y, z, t) = e J 0 z − v t [U(t + z/v) − U(t − z/v)] (A.38)
2 v
where U(x) is the unit step function (A.5). From (A.37) we see that
∞
z/v
ψ 2 (x, y, z, t) = g(x, y, t − τ)Q(x, y, z,τ) dτ = g(x, y, t − τ)Q(x, y, z,τ) dτ.
−∞ −z/v
Using the change of variables u = t − τ and substituting (A.38), we then have
z

v − t t+ v u
2 2
2
ψ 2 (x, y, z, t) = e g(x, y, u)e J 0 z − (t − u) v du. (A.39)
2 t− z v
v
To find ψ 1 we must compute ∂ Q/∂z. Using the product rule we have
∂ Q(x, y, z, t) v − t 2 2 2 ∂
= e J 0 z − v t [U(t + z/v) − U(t − z/v)] +
∂z 2 v ∂z
v − t ∂
2 2
2
+ e [U(t + z/v) − U(t − z/v)] J 0 z − v t .
2 ∂z v
Next, using dU(x)/dx = δ(x) and remembering that J (x) =−J 1 (x) and J 0 (0) = 1,we

0
can write
∂ Q(x, y, z, t) 1 − t
= e [δ(t + z/v) + δ(t − z/v)] −
∂z 2
√
2 2 2
z 2 − t J 1 v z − v t
− e √ [U(t + z/v) − U(t − z/v)].
2v z − v t
2 2
2
v
Convolving this expression with f (x, y, t) we obtain
1 − z z 1 z z

ψ 1 (x, y, z, t) = e v f x, y, t − + e v f x, y, t + −
2 v 2 v
2 2 2
z 2 − t t+ z v u J 1 v z − (t − u) v
− e f (x, y, u)e du. (A.40)
2 2
2
2v z z − (t − u) v
t−
v v
Finally, adding (A.40) and (A.39), we obtain
1 − z z 1 z z

ψ(x, y, z, t) = e v f x, y, t − + e v f x, y, t + −
2 v 2 v

2 2
2
z z − (t − u) v
2
z − t t+ v u J 1 v

− e f (x, y, u)e du +
2v z z − (t − u) v
2
2 2
t−
v v
z
v − t t+ v u

2
2 2
+ e g(x, y, u)e J 0 z − (t − u) v du. (A.41)
2 t− z v
v
Note that when = 0 this reduces to
z
1 z 1 z v t+ v

ψ(x, y, z, t) = f x, y, t − + f x, y, t + + g(x, y, u) du,
2 v 2 v 2 t− z
v
which matches (A.32) for the homogeneous case where S = 0.
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