082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

3.7 Modelling waves using sint and cost 141
Waves can be considered to propagate between the transmitter and the receiver
in two ways. There is the direct route between transmitter and receiver. The direct
distance between transmitter and receiver is d d
. We obtain an expression for d d
in
terms of the known quantitiesh t
,h r
andd by considering the triangle RST. In this
triangle, RS =dand ST = TO − SO =h t
−h r
. Hence by Pythagoras’s theorem in
RST we have
and so
TR 2 = RS 2 +ST 2
d 2 d =d2 +(h t
−h r
) 2
d d
=
√
d 2 +(h t
−h r
) 2
Note thatd d
isexpressed intermsof the known quantitiesh t
,h r
andd.
ThereisalsoaroutewherebyawaveisreflectedoffthegroundatpointAbefore
arrivingatthereceiver.ThepointAonthegroundissuchthat ̸ TAOequals ̸ RAP.
The distance travelled in this case isd r
= TA +AR. We wish to find an expression
ford r
intermsoftheknownquantitiesh t
,h r
andd.Inordertosimplifythecalculation
of this distance we construct an isosceles triangle, TAQ, in which TA = QA and
̸ TAO = ̸ QAO. Note thatinthistriangle,TO = QO =h t
.
Then the distance travelled by thisreflected wave,d r
, is
distance travelled =d r
= TA +AR
=QA+AR
= QR
Consider now QSR. QR is the hypotenuse of this triangle. So by Pythagoras’s
theorem wehave
d 2 r
= QR 2 = SR 2 +SQ 2
WehaveSR=dandSQ=QO+SO=h t
+h r
.So
d 2 r
=d 2 +(h t
+h r
) 2
from which
√
d r
= d 2 +(h t
+h r
) 2
Now if the wavelength of the transmitted wave is λ then we can calculate the phase
differencebetweenthedirectwaveandthereflectedwave, φ,bynotingthedifference
in the distance travelled,d r
−d d
. Using the result for phase difference from Section
3.7.2 wehave
φ = 2π λ ×horizontal shift = 2π λ (d r −d d )
φ = 2π (√
)
d
λ
2 +(h t
+h r
) 2 −
√d 2 +(h t
−h r
) 2
⎛ √
φ = 2π ( )
√ ⎞
ht +h 2 ( )
⎝d 1 + r ht −h 2
−d 1 + r ⎠
λ d d
➔