Calculation of radio electrical coverage in Medium‐Wave Frequencies
The Wave Function Heart beat Electrical Many wave shapes, whether occurring as sound, light, water...
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Transcript of The Wave Function Heart beat Electrical Many wave shapes, whether occurring as sound, light, water...
The Wave Function
Heart beat
Electrical
Many wave shapes, whether occurring as sound, light, water or electrical waves, can be described mathematically
as a combination of sine and cosine waves.
Spectrum Analysis
Expressing a cos x + b sin x in the form k cos(x ± ) or
k sin(x ± )
General shape for y = sin x + cos x
1. Like y = sin x shifted left
2. Like y = cos x shifted right
3. Vertical height (amplitude) different
y = sin x
y = cos x
y = sin x +cos x
Whenever a function is formed by adding cosine and sine functions the result can be expressed as a related
cosine or sine function. In general:
With these constants the expressions on the right hand sides = those on the left hand side
FOR ALL VALUES OF x
a cos x + b sin x = k cos(x ± ) or = k sin(x ± )
Where a, b, k and are constants
Given a and b, we can calculate k and .
Write 4 cos xo + 3 sin xo in the form k cos(x – )o, where 0 ≤ ≤ 360
cos(x – ) = cos x cos + sin x sin
k cos(x – )o = k cos xo cos + k sin xo sin 4 cos xo
Now equate with 4 cos xo + 3 sin xo+ 3 sin xo
It follows that : k cos = 4 and k sin = 3
cos2 x + sin2 x = 1
k2 = 42 + 32
k = √25k = 5
tan = ¾
= tan-1 0∙75 = 36∙9
4 cos xo + 3 sin xo = 5 cos(x – 36∙9)o
(kcos )2 + (ksin )2 = k2tan = sin
cos
Write cos x – √3 sin x in the form R cos(x + ), where 0 ≤ ≤ 2π
cos(x + ) = cos x cos – sin x sin
R cos(x + ) = R cos x cos – R sin x sin cos x – √3 sin x
It follows that : R cos = 1 and R sin = √3
cos2 x + sin2 x = 1
R2 = 12 + (√3)2
R = √4R = 2
tan = √3/1
= tan-1 √3
= π/3 (60)
cos x – √3 sin x = 2 cos(x + π/3)
(Rcos )2 + (Rsin )2 = R2 tan = sin cos
Write 5 cos 2x + 12 sin x in the form k sin(2x + ), where 0 ≤ ≤ 360
sin(2x + ) = sin 2x cos + cos 2x sin
k sin(2x + ) = k sin 2xo cos + k cos 2xo sin 12 sin 2xo + 5 cos 2xo
It follows that : k cos = 12 and k sin = 5
cos2 x + sin2 x = 1
k2 = 122 + 52
k = √169k = 13
tan = 5/12
= tan-1 0∙417 = 22∙6
5 cos 2x + 12 sin 2x = 13 sin(2x + 22∙6)
(kcos )2 + (ksin )2 = k2 tan = sin cos
Maximum and Minimum Values
MAX k cos (x ± ) is kk when (x ± ) = 0 or 360 (0 or 2π)
MIN k cos (x ± ) is – k when (x ± ) = 180 (π)
MAX k sin (x ± ) is kk when (x ± ) = 90 (π/2)
MIN k sin (x ± ) is – kk when (x ± ) = 270 (3π/2)
MAX cos x = 1
when x = 0o or 360o
MAX sin x = 1
when x = 90o
MIN cos x = –1
when x = 180o
MIN sin x = –1
when x = 270o
Write f(x) = sin x – cos x in the form k cos (x – ) and find the maximum of f(x) and the value of x at which occurs.
k cos(x – )o = k cos xo cos + k sin xo sin sin xo – cos xo
k cos = –1 k sin = 1
k2 = (–1)2 + 12
k = √2
AS
T C
cos –ve
sin +ve
tan = sin cos
tan = – 1
= (180 – 45)o angle = tan-1 1 = 45o
= 135o
f(x) = √2 cos (x – 135)o
MAX f(x) = √2
When angle = 0
x – 135 = 0
x = 135o
A synthesiser adds two sound waves together to make a new sound. The first wave is described by V = 75sin to and the second by V = 100cos to, where V is the amplitude in decibels and t is the time
in milliseconds.
sin( ) a) Express the resultant wave in the form ok t
Find the minimum value of the resultant wave and the value of t at which it occurs.
75sin 100cos sin( ) resultV t t K t
25 3sin 4cos 25 sin( ) resultV t t k t
3sin 4cos sin( ) t t k t For later, remember K = 25k
Maximum and Minimum Values
3sin 4cos sin( ) t t k t
3sin 4cos sin cos cos sin t t k t k t
cos 3
sin 4
k
k
2 2 2 2 2cos sin 3 4 k 5 k1sin 4
tan tan 53.1cos 3
o
th is in the 4 quadrant
360 53.1 306.9 o
Expand and equate
coefficients
C
AS
T0o180o
270o
90o
3
2
2
cos is +ve
sin is –ve
306.9 270 The minumum occurs where ot
125sin( 306.9)resultV t
The minimum value of sin is -1 and it occurs where the angle is 270o
Therefore, the minimum value of Vresult is – 125
270 306.9 576.9 ot
576.9 360 216.9 ot
216.9 ot
Adding or subtracting
360o leaves the sin unchanged
remember K = 25k =25 × 5 = 125
Minimum, we have:
125sin( 216.9) minimum of oy x
sin 1 270 the minimum of is - when o oy x x
125si 125n the minimum of is oy x
216.9occurrs when ox
75sin 100cos 125sin( 216.9 ) oresultV t t t
Solving Trig Equations
3 cos sin 2 0 2 Solve for x x x
3 cos sin cos( ).x x k x
cos 3
sin 1
k
k
2 2 2 2cos sin 3 1 k k2 4 k
2 k
3 cos sin cos cos sin sin x x k x k x
3 cos sin cos( ) Write in the form x x k x
1sin 1tan tan 30
cos 3
o
3 cos sin 2cos6
x x x
C
AS
T0o180
o
270o
90o
3
2
2
cos 3
sin 1
k
k
306
o
o
180 =
o
cos is +ve
sin is +ve
3 cos sin 2 x x
Re-write the trig. equation using your result from step 1, then solve.
2cos 26
x
1cos
6 2
x
C
AS
T0o180
o
270o
90o
3
2
2
1 1
cos6 2
x
6
o 0 45 and 315x7
6
and 4 4
x
cos is +ve
7
4 6 4 6
or x x
5 23
12 12
o o (75 ) or (345 ) x x
5 23
12 12
o o (75 ) or (345 ) x x
2cos2 3sin 2 sin(2 )
2cos2 3sin 2 1 0 360
a) Express in the form
b) Hence solve for ox x k x
x x x
sin 2
cos 3
k
k
2 13 k 13 k
1sin 2tan tan 33.7
cos 3
o
180 33.7 213.7o o o
C
AS
T0o180
o
270o
90o
3
2
2
sinxcoskcosxsink)xsin(k 222 xcosxsin)xsin(k 22232
cos is –ve
sin is –ve
02cos2 3sin 2 13sin(2 213.7)x x x b) We now have
2cos2 3sin 2 1
13sin(2 213.7) 1
x x
x
We solve
by solving
1sin(2 213.7)
13x 1 01
sin 16.113
st In the 1 quadrant
2x – 213.7 = 16.1o , (180-16.1o),(360+16.1o),(360+180-16.1o)
2x – 213.7 = 16.1o , 163.9o, 376.1o, 523.9o, ….
2x = 229∙8o , 310∙2o, 589∙9o, 670∙2o, ….
x = 114∙9o , 188∙8o, 294∙9o, 368∙8o, ….
Part of the graph of y = 2 sin x + 5 cos x is shown
a) Express y = 2 sin x + 5 cos x in the form
k sin (x + a) where k > 0 and 0 a 360
b) Find the coordinates of the minimum turning point P.
Expand ksin(x + a):
Equate coefficients:
Square and add
Dividing:
Put together:
Minimum when:
P has coords.
asinxcoskacosxsink)axsin(k xcosxsiny 52
2acosk 5asink222 52 k 29k
2
5atan 6852acute 1 tana
sin + , cos +
68a
)xsin(xcosxsin 682952
27068 x 202 x 1s inMin
),( 29202
2
2
Expand k sin(x - a): sin( ) sin cos cos sink x a k x a k x a
Equate coefficients: cos 1 sin 1k a k a
Square and add2 2 21 1 2k k
Dividing:
Put together:4 4
sin cos 2 sin( ) 2x x x k a
Sketch Graph
a) Write sin x - cos x in the form k sin (x - a) stating the values of k and a where k > 0 and 0 a 2
b) Sketch the graph of sin x - cos x for 0 a 2 showing clearly the graph’s
maximum and minimum values and where it cuts the x-axis and the y-axis.
max min2 2
3 7max at min at
4 4x x
tan 1a acute4
a
a is in 1st quadrant
(sin and cos are +)
4a