3.7 MODELLING WAVES USING SIN t AND COS t
3.7.2 Wavelength, wave number and horizontal shift
The sine and cosine waves described earlier in this section hadt as their independent variable because the waves commonly met in engineering vary with time. There are occasions where the independent variable is distance,xsay, and in this case some of the terminology changes. Consider the wave
y=Asin(kx+φ )
As before,Ais the amplitude of the wave. The quantitykis called thewave number. It plays the same role as did the angular frequency,ω, whentwas the independent variable.
The length of one cycle of the wave, that is thewavelength, commonly denotedλ, is related tokby the formulaλ= 2π
k . The phase angle isφand its introduction has the effect of shifting the graph horizontally.
Example 3.13 Figure 3.17 shows a graph ofy=sin 2x.
(a) State the wave number for this wave.
(b) Find the wavelength of the wave.
(c) State the phase angle.
Solution (a) Comparingy=sin 2xwithy=sinkxwe see that the wave number,k, is 2.
(b) The wavelength, λ = 2π
k = π. Note by observing the graph that this result is consistent in that the distance required for one cycle of the wave isπunits.
(c) Comparingy=sin 2xwith sin(kx+φ )we see that the phase angle,φ, is 0.
3.7 Modelling waves using sintand cost 139
0 –2p
–1 1
2p x
y
–p p
Figure 3.17
A graph of the wavey=sin 2x.
Example 3.14 Figure 3.18 shows a graph of sin
2x+π 3
. (a) State the phase angle.
(b) By comparing Figures 3.17 and 3.18 we see that the introduction of the phase angle has caused a horizontal shift of the graph (to the left). Calculate this shift.
–2p 0
–1 1
2p
–p p x
y
p– 6
Figure 3.18 A graph of the wave y=sin 2x+ π
3
! .
Solution (a) By comparing sin
2x+ π 3
with sin(kx+φ )we see that the phase angle is π 3. (b) By writingy = sin
2x + π
3
as sin 2
x + π 6
we note that this isy= sin 2x shifted to the left by a horizontal distanceπ
6 units.
The results of this example can be generalized. The wavey = Asin(kx+φ )can be writteny=Asink(x+φ/k)so that a phase angle ofφintroduces a horizontal shift of lengthφ/k. (Compare this with the expression for time displacement in Section 3.7.)
Noting thatλ = 2π
k , then k = 2π
λ and we may writeAsin(kx+φ )equivalently asAsin
2πx
λ +φ
. Again, usingk = 2π
λ , the horizontal shift, φ
k, may similarly be written as
horizontal shift= φ k = φ
2π/λ = φλ 2π
from which
phase angle=φ= 2π×horizontal shift λ
This result is important in the engineering application that follows because, more gen- erally, when any two waves arrive at a receiver it enables the difference in their phases, φ, to be calculated from knowledge of the horizontal shift between them.
The presence ofφiny=Asin(kx+φ )causes a horizontal (left) shift ofφ k =φλ
2π. Note that adding any multiple of 2π onto the phase angleφ will result in the same graph because of the periodicity of the sine function. Consequently, a phase angle could be quoted as φ +2nπ. For example, the wave sin
2x+π 3
is the same wave as sin
2x+π
3 +2π
, sin
2x+π 3 +4π
and so on. Normally, we would quote a value of the phase that was less than 2πby subtracting multiples of 2πas necessary.
Engineering application 3.5
Two-ray propagation model
It is very useful to be able to model how an electromagnetic wave emitted by a trans- mitter 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 thetwo-ray propagationmodel. This model assumes that the signal re- ceived 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 re- flected off the ground.
Figure 3.19 shows a transmitter with height above the groundhttogether with a receiver with height above the groundhr. The distance between the transmitter and the receiver along the ground isd.
dd
dr d
hr ht
S T
Q
R Receiver Transmitter
Ground
A P O
Figure 3.19
A transmitter and receiver at different heights above the ground.
Note that the quantitiesht,hranddare all known.
3.7 Modelling waves using sintand 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 isdd. We obtain an expression fordd in terms of the known quantitiesht,hranddby considering the triangle1RST. In this triangle, RS=dand ST=TO−SO=ht−hr. Hence by Pythagoras’s theorem in 1RST we have
TR2=RS2+ST2 dd2=d2+(ht−hr)2 and so
dd= q
d2+(ht−hr)2
Note thatddis expressed in terms of the known quantitiesht,hrandd.
There is also a route whereby a wave is reflected off the ground at point A before arriving at the receiver. The point A on the ground is such that6 TAO equals6 RAP.
The distance travelled in this case isdr =TA+AR. We wish to find an expression fordrin terms of the known quantitiesht,hrandd. In order to simplify the calculation of this distance we construct an isosceles triangle,1TAQ, in which TA=QA and
6 TAO=6 QAO. Note that in this triangle, TO=QO=ht. Then the distance travelled by this reflected wave,dr, is
distance travelled=dr =TA+AR
=QA+AR
=QR
Consider now1QSR. QR is the hypotenuse of this triangle. So by Pythagoras’s theorem we have
dr2=QR2=SR2+SQ2
We have SR=dand SQ=QO+SO=ht+hr. So dr2=d2+(ht+hr)2
from which dr=
q
d2+(ht+hr)2
Now if the wavelength of the transmitted wave isλthen we can calculate the phase difference between the direct wave and the reflected wave,φ,by noting the difference in the distance travelled,dr−dd. Using the result for phase difference from Section 3.7.2 we have
φ= 2π
λ ×horizontal shift= 2π
λ (dr−dd) φ= 2π
λ q
d2+(ht+hr)2− q
d2+(ht−hr)2
φ= 2π λ
d s
1+
ht+hr d
2
−d s
1+
ht−hr d
2
➔
Now the binomial expansion for√
(1+x)is (see Section 6.4) p(1+x)=(1+x)1/2=1+ x
2 −x2 8 + x3
16− ã ã ã ≈1+x
2, if|x|<1 Using this expansion in the expression forφand noting that the moduli of both (ht+ hr)/dand (ht−hr)/dare less than 1, we have
φ≈ 2πd λ
1+(ht+hr)2
2d2 −1−(ht−hr)2 2d2
Expanding the bracketed terms gives φ≈ 2πd
λ
h2t +2hthr+h2r −h2t +2hthr−h2r 2d2
So
φ≈ 4hthrπ λd
This is a simplified approximation for the phase difference between the direct wave and the reflected wave. Note that it depends on the height of the transmitter, the height of the receiver and the distance between the transmitter and the receiver.
This calculation is important because under some conditions the phase difference between the two paths means that the directed and reflected wavesdestructively interfere. In severe cases this causes the signal to decrease at the receiver enough so that the communications link is lost. The effect is often termedmultipath-induced fading.
EXERCISES 3.7
1 State the amplitude, angular frequency, frequency, phase angle and time displacement of the following waves:
(a) 3 sin 2t (b) 1
2sin 4t (c) sin(t+1) (d) 4 cos 3t (e) 2 sin(t−3) (f) 5 cos(0.4t) (g) sin(100πt) (h) 6 cos(5t+2) (i) 2
3sin(0.5t) (j) 4 cos(πt−20)
2 State the period of
(a) 2 sin 7t (b) 7 sin(2t+3) (c) tant
2 (d) sec 3t
(e) cosec(2t−1) (f) cot 2t 3 +2
!
3 A voltage source produces a time-varying voltage, v(t), given by
v(t)=15 sin(20πt+4) t>0 (a) State the amplitude ofv(t).
(b) State the angular frequency ofv(t).
(c) State the period ofv(t).
(d) State the phase ofv(t).
(e) State the time displacement ofv(t).
(f) State the minimum value ofv(t).
4 A sinusoidal function has an amplitude of23and a period of 2. State a possible form of the function.
5 State the phase angle and time displacement of (a) 2 sin(t+3)relative to 2 sint
(b) sin(2t−3)relative to sin 2t (c) cos t
2+0.2
!
relative to cost 2 (d) cos(2−t)relative to cost (e) sin 3t+4
5
!
relative to sin3t 5 (f) sin(4−3t)relative to sin 3t
3.7 Modelling waves using sintand cost 143 (g) sin(2πt+π)relative to sin 2πt
(h) 3 cos(5πt−3)relative to 3 cos 5πt (i) sin πt
3 +2
!
relative to sinπt 3 (j) cos(3π−t)relative to cost 6 Write each of the following in the form
Asin(3t+θ ),θ>0:
(a) 2 sin 3t+3 cos 3t (b) cos 3t−2 sin 3t (c) sin 3t−4 cos 3t (d) −cos 3t−4 sin 3t
7 Write each of the following in the form Acos(t−θ ),θ>0:
(a) 2 sint−3 cost (b) 9 sint+6 cost (c) 4 cost−sint (d) 3 sint
8 Write each of the following expressions in the form (i)Asin(ωt+θ ), (ii)Asin(ωt−θ ),
(iii)Acos(ωt+θ ), (iv)Acos(ωt−θ )whereθ>0:
(a) 5 sint+4 cost (b) −2 sin 3t+2 cos 3t (c) 4 sin 2t−6 cos 2t (d) −sin 5t−3 cos 5t
Solutions
1 (a) 3,2,1 π
,0,0 (b) 1
2,4,2 π
,0,0 (c) 1,1, 1
2π,1,1 (d) 4,3, 3 2π,0,0 (e) 2,1, 1
2π,−3,−3 (f) 5,0.4, 1 5π,0,0 (g) 1,100π,50,0,0 (h) 6,5, 5
2π,2,0.4 (i) 2
3,0.5, 1
4π,0,0 (j) 4,π,1
2,−20,−20 π
2 (a) 2π
7 (b) π (c) 2π
(d) 2π
3 (e) π (f) 3π
2
3 (a) 15 (b) 20π (c) 0.1
(d) 4 (e) 1
5π (f) −15
4 2
3sin(πt+k)or2
3cos(πt+k) 5 (a) 3, 3 (b) −3,−3
2 (c) 0.2, 0.4 (d) −2,−2 (e) 4
5,4
3 (f) −0.858,−0.286 (g) π,1
2 (h) −3,− 3
5π (i) 2,6 π (j) −3π,−3π
6 (a) √
13 sin(3t+0.9828) (b) √
5 sin(3t+2.6779) (c) √
17 sin(3t+4.9574) (d) √
17 sin(3t+3.3866)
7 (a) √
13 cos(t−2.5536) (b) √
117 cos(t−0.9828) (c) √
17 cos(t−6.0382) (d) 3 cos t−π
2
!
8 (a) (i)√
41 sin(t+0.675) (ii)√
41 sin(t−5.608) (iii)√
41 cos(t+5.387) (iv)√
41 cos(t−0.896) (b) (i) √
8 sin 3t+3π 4
!
(ii) √
8 sin 3t−5π 4
!
(iii)√
8 cos 3t+ π 4
!
(iv)√
8 cos 3t− 7π 4
!
(c) (i) √
52 sin(2t+5.300) (ii) √
52 sin(2t−0.983) (iii)√
52 cos(2t+3.730) (iv)√
52 cos(2t−2.554) (d) (i) √
10 sin(5t+4.391) (ii) √
10 sin(5t−1.893) (iii)√
10 cos(5t+2.820) (iv)√
10 cos(5t−3.463)
Technical Computing Exercises 3.7
1 Ploty=sin 2tfor 06t62π.
2 Ploty=cos 3tfor 06t63π.
3 Ploty=sin t 2
!
for 06t64π.
4 Ploty=cos 2t 3
!
for 06t66π.
5 Ploty=sint+3 costfor 06t63π. By reading from your graph, state the maximum value of sint+3 cost.
6 (a) Ploty=2 sin 3t−cos 3tfor 06t62π.
Use your graph to find the amplitude of 2 sin 3t−cos 3t.
(b) On the same axes ploty=sin 3t. Estimate the time displacement of 2 sin 3t−cos 3t.