082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017
140 Chapter 3 The trigonometric functionsfrom whichphase angle = φ =2π ×horizontal shiftλThis result is important in the engineering application that follows because, more generally,when any two waves arrive at a receiver it enables the difference in their phases,φ, tobe calculated from knowledge of the horizontal shift between them.Thepresenceof φ iny =Asin(kx + φ)causesahorizontal(left)shiftof φ k = φλ2π .Note that adding any multiple of 2π onto the phase angle φ will result in the samegraphbecauseoftheperiodicityofthesinefunction.Consequently,aphaseanglecould(be quoted as φ + 2nπ. For example, the wave sin 2x + π )is the same wave as3sin(2x + π )3 +2π , sin(2x + π )3 +4π and so on. Normally, we would quote a valueof the phase thatwas lessthan 2πby subtracting multiples of 2πasnecessary.Engineeringapplication3.5Two-raypropagationmodelItisveryusefultobeabletomodelhowanelectromagneticwaveemittedbyatransmitterpropagates through space, in order to predict what signal is collected by thereceiver. This can be quite a complicated modelling exercise. One of the simplestmodels is the two-ray propagation model. This model assumes that the signal receivedconsists of two main components. There is the signal that is sent direct fromthe transmitter to the receiver and there is the signal that is received after being reflectedoffthe ground.Figure 3.19 shows a transmitter with height above the groundh ttogether with areceiver with height above the groundh r. The distance between the transmitter andthe receiver alongthe ground isd.TSh tTransmitterOGroundd dd rdARReceiverPh rQFigure3.19Atransmitterandreceiver atdifferent heightsabove theground.Note thatthe quantitiesh t,h randd areall known.
3.7 Modelling waves using sint and cost 141Waves can be considered to propagate between the transmitter and the receiverin two ways. There is the direct route between transmitter and receiver. The directdistance between transmitter and receiver is d d. We obtain an expression for d dinterms of the known quantitiesh t,h randd by considering the triangle RST. In thistriangle, RS =dand ST = TO − SO =h t−h r. Hence by Pythagoras’s theorem inRST we haveand soTR 2 = RS 2 +ST 2d 2 d =d2 +(h t−h r) 2d d=√d 2 +(h t−h r) 2Note thatd disexpressed intermsof the known quantitiesh t,h randd.ThereisalsoaroutewherebyawaveisreflectedoffthegroundatpointAbeforearrivingatthereceiver.ThepointAonthegroundissuchthat ̸ TAOequals ̸ RAP.The distance travelled in this case isd r= TA +AR. We wish to find an expressionford rintermsoftheknownquantitiesh t,h randd.Inordertosimplifythecalculationof this distance we construct an isosceles triangle, TAQ, in which TA = QA and̸ TAO = ̸ QAO. Note thatinthistriangle,TO = QO =h t.Then the distance travelled by thisreflected wave,d r, isdistance travelled =d r= TA +AR=QA+AR= QRConsider now QSR. QR is the hypotenuse of this triangle. So by Pythagoras’stheorem wehaved 2 r= QR 2 = SR 2 +SQ 2WehaveSR=dandSQ=QO+SO=h t+h r.Sod 2 r=d 2 +(h t+h r) 2from which√d r= d 2 +(h t+h r) 2Now if the wavelength of the transmitted wave is λ then we can calculate the phasedifferencebetweenthedirectwaveandthereflectedwave, φ,bynotingthedifferencein the distance travelled,d r−d d. Using the result for phase difference from Section3.7.2 wehaveφ = 2π λ ×horizontal shift = 2π λ (d r −d d )φ = 2π (√)dλ2 +(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➔
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140 Chapter 3 The trigonometric functions
from which
phase angle = φ =
2π ×horizontal shift
λ
This result is important in the engineering application that follows because, more generally,
when any two waves arrive at a receiver it enables the difference in their phases,
φ, tobe calculated from knowledge of the horizontal shift between them.
Thepresenceof φ iny =Asin(kx + φ)causesahorizontal(left)shiftof φ k = φλ
2π .
Note that adding any multiple of 2π onto the phase angle φ will result in the same
graphbecauseoftheperiodicityofthesinefunction.Consequently,aphaseanglecould
(
be quoted as φ + 2nπ. For example, the wave sin 2x + π )
is the same wave as
3
sin
(2x + π )
3 +2π , sin
(2x + π )
3 +4π and so on. Normally, we would quote a value
of the phase thatwas lessthan 2πby subtracting multiples of 2πasnecessary.
Engineeringapplication3.5
Two-raypropagationmodel
Itisveryusefultobeabletomodelhowanelectromagneticwaveemittedbyatransmitter
propagates through space, in order to predict what signal is collected by the
receiver. This can be quite a complicated modelling exercise. One of the simplest
models is the two-ray propagation model. This model assumes that the signal received
consists of two main components. There is the signal that is sent direct from
the transmitter to the receiver and there is the signal that is received after being reflectedoff
the ground.
Figure 3.19 shows a transmitter with height above the groundh t
together with a
receiver with height above the groundh r
. The distance between the transmitter and
the receiver alongthe ground isd.
T
S
h t
Transmitter
O
Ground
d d
d r
d
A
R
Receiver
P
h r
Q
Figure3.19
Atransmitterandreceiver at
different heightsabove the
ground.
Note thatthe quantitiesh t
,h r
andd areall known.