Problem I.

-4

1) Speed of sound in PTZ - SH n 4560 Ms . L = Iz = ¥ = 452=9 .Ixion

2×2.5×106

⇒ L = 0.9 mm

2) element #4 : 2- y = ✓ ¥2 = ) 4-624 = 68 mm44 = Z=f§ymoI% =

4.42×155s

element # 3 .. Zz = ) ¥2 = JUST = 67.08 mm

es - ZI -- ffjf.omm.my,

=

4.35×55s

element # 2.

- Zz - -✓ ¥82 = ✓ 438-4 = 66 . 21 mm

ez - - ZZ - - bfj7.IT#m,

=4.30×155selement # 5 : Zs = J

¥42=) # = 68.96 mm

es = Z{- = g = 4 . 48×155 s

Dtc = ITs - Ey1=0.06 x toS

= o .6×156

S = 0.6 µs

µe , yes . ↳I = o . ,z× , £5 = , ,3× if s = , .rs µ ,DEZ = Its - Tz I = o . 18×155 = 1.8×156 S = 1.8 µs

3) FD -- DIG = = Ditz - - 41%143%3=6.49 mm

chime the transducer is not circularly symmetric, I will accept any dimension you chose LD .

Problem 2 .

• Cosine aperture

tdvantage : the first side lobe is at about - 23 dB

disadvantage : the main lobe is wider .

• Rect aperture

quantage i the main lobe is tighter at -3 dB .

disadvantage : the first side lobe appears at - 13 dB

Problem 3 .

I .

2. A common compression A use

is : calm ( imam, varying ⇐ will compress the

image .

3.

Come up aieh a function that will weigh the top rows heavily compared * the deeper

rows . An example would be let weigh each row as : flu ) = K x *÷ :

comparethis A the o . z - 0.8

image in port 2 . You can see

thetop rows in this one better .

4.

the point of this problem is to show that higher frequencies don't make is to the deeperlayers .

As you can see in the frequency spectrum , at deeper layers ( orange ) high frequency

components are vanished .