Harmonic Analysis: Elementary Pipe and Collar Models - Measurement While Drilling Signal Analysis, Optimization and Design - page 89

x > 0 and the dipole source is again located at x = x s . For the domain 0 < x < x s ,
we take u 1 with linear combinations of sin Zx/c and cos Zx/c to model standing
waves, but for x > x s , our u 2 is proportional to e i Z(-x/c + t) to represent a non-
reflecting propagating wave traveling to the right. Since an open-ended reflector
satisfies wu/wx = 0 at x = 0, and Equations 3.A.8 and 3.A.9 apply at x = x s , we
find the solution
p 2 (x,t) = {2 cos (Zx s /c)} ½ p s e i Z(-x/c + t) , x > x s (3.A.13)

3.1.4 Case (d), “finite-finite” waveguide of length 2L.

The illustration in Figure 3.A.1d shows a dipole source centered at x = 0 in
a waveguide of length 2L. We will assume open-ended reflectors satisfying
wu/wx = 0 at x = r L and discuss its physical significance later. Since standing
waves are found at both sides of the source, linear combinations of sin Zx/c and
cos Zx/c are chosen on each side to represent the displacement u(x,t). Use of
our Equations 3.A.8 and 3.A.9 at the source x = 0 leads to the solutions
p 1 (x,t) = - {p s /(2 tan ZL/c)} [sin Zx/c + (tan ZL/c) cos Zx/c] e i Zt
on - L < x < 0 (3.A.14)
p 2 (x,t) = - {p s /(2 tan ZL/c)} [sin Zx/c - (tan ZL/c) cos Zx/c] e i Zt
on 0 < x < + L

(3.A.15)

3.1.5 Physical Interpretation.

Here we address the physical meanings and implications of the solutions
obtained in Cases (a) - (d). These solutions may not represent the detailed
telemetry channel discussed in Chapter 2, but they facilitate physical
understanding and enable us to build acoustic test fixtures whose data may be
interpreted unambiguously and accurately. More on testing and evaluation
methods appears later in Chapter 9.
Case (a) . Again, our exact solution states that a 'p pulse of strength p s
splits into two waves that travel in opposite directions having equal and opposite
strengths - ½ p s and + ½ p s . That is, we have p 1 (x,t) = - ½ p s e i Z(x/c + t) , x < 0
and p 2 (x,t) = + ½ p s e i Z(-x/c + t) , x > 0. These solutions are the result of not having
reflections - the waves on both sides travel away from the source and do not
return to the point of origin.
Case (b) . When the drillbit is modeled as a solid reflector, we have p 2 (x,t)
= + i {2 sin (Zx s /c)} ½ p s e i Z(-x/c + t) , x > x s . This important solution indicates
that the pressure is the product of an interference factor “2 sin (Zx s /c)” and the
no-reflection “½ p s e i Z(-x/c + t) ” solution of Case (a). This factor is the result of
constructive or destructive interference. It states that we can expect constructive
interference if Zx s /c = S/2, 3S/2, 5S/2 and so on, and destructive interference if
Zx s /c = S, 2S, 3S and so on.