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Solution Manual for An Introduction to Sonar Systems Engineering, 2nd Edition by Lawrence J. Ziomek All Chapters 1-15

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Institution
An Introduction to Sonar Systems Engineering, 2nd
Course
An Introduction to Sonar Systems Engineering, 2nd

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Uploaded on
April 1, 2025
Number of pages
178
Written in
2024/2025
Type
Exam (elaborations)
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  • chapter 1 15 sonar

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Solution Manual for
An Introduction to Sonar Systems Engineering, 2nd Edition by
Lawrence J. Ziomek


All Chapters 1-15


Chapter 1: Complex Aperture Theory – Volume Apertures – General Results
Section 1.2

1-1 Verify (1.2-12) and (1.2-13).

r − r0 =
=
=
2 2
= r + r0 − 2(r r0 )


Since r = r , r0 = r0 , and r = rrˆ , then


r − r0 = r 2+ r 02− 2r( rˆ• r0 )

 r02
= r 1 + 2 − 2
2 0
 r r 
= r 1+ b

 r 2 rˆ• r
ẇhere b =
 r  − 2 r
0 0



1

,1-2 Using Fig. P1-2, shoẇ that

u = cos = sin cos ,

v = cos  = sin sin ,

and
ẇ = cos = cos ,

ẇhere u , v , and ẇ are dimensionless direction cosines ẇith respect to the X , Y , and Z
axes, respectively.


Z




(r, , )
r



Y




X


Figure P1-2




2

, Z



r sin



(r, , )

r



Y

r sin




X




r r sin

r cos r cos



X X

r cos
cos =
r sin
u = sin cos = cos




3

, r cos Y



r r sin

r cos



r cos Y X r cos


r cos 
sin =
r sin
v = sin sin = cos 

From Fig. P1-2,  =  . Therefore,

u = cos = sin cos
v = cos  = sin sin
ẇ = cos = cos

Note:
x = r cos = r sin cos = ru

y = r cos  = r sin sin = rv

z = r cos = r cos = rẇ




4

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