2 Engineering Example 5

2.1 Amplitude modulation

Introduction

Amplitude Modulation (the AM in AM radio) is a method of sending electromagnetic signals of a certain frequency (signal frequency) at another frequency (carrier frequency) which may be better for transmission. Modulation can be represented by the multiplication of the carrier and modulating signals. To demodulate the signal the carrier frequency must be removed from the modulated signal.

Problem in words

  1. A single frequency of 200 Hz (message signal) is amplitude modulated with a carrier frequency of 2 MHz. Show that the modulated signal can be represented by the sum of two frequencies at 2 × 1 0 6 ± 200 Hz
  2. Show that the modulated signal can be demodulated by using a locally generated carrier and applying a low-pass filter.

Mathematical statement of problem

  1. Express the message signal as m = a cos ( ω m t ) and the carrier as c = b cos ( u c t ) .

    Assume that the modulation gives the product m c = a b cos ( u c t ) cos ( ω m t ) .

    Use trigonometric identities to show that

    m c = a b cos ( ω c t ) cos ( u m t ) = k 1 cos ( ( ω c u m ) t ) + k 2 cos ( ( ω c + u m ) t )

    where k 1 and k 2 are constants.

    Then substitute ω c = 2 × 1 0 6 × 2 π and ω m = 200 × 2 π to calculate the two resulting frequencies.

  2. Use trigonometric identities to show that multiplying the modulated signal by b cos ( u c t ) results in the lowest frequency component of the output having a frequency equal to the original message signal.

Mathematical analysis

  1. The message signal has a frequency of f m = 200 Hz so ω m = 2 π f c = 2 π × 200 = 400 π radians per second.

    The carrier signal has a frequency of f c = 2 × 1 0 6 Hz.

    Hence ω c = 2 π f c = 2 π × 2 × 1 0 6 = 4 × 1 0 6 π radians per second.

    So m c = a b cos ( 4 × 1 0 6 π t ) cos ( 400 π t ) .

    Key Point 13 includes the identity:

    cos ( A + B ) + cos ( A B ) 2 cos ( A ) cos ( B )

    Rearranging gives the identity:

    cos ( A ) cos ( B ) 1 2 ( cos ( A + B ) + cos ( A B ) ) ( 1 )

    Using (1) with A = 4 × 1 0 6 π t and B = 400 π t gives

    m c = a b ( cos ( 4 × 1 0 6 π t ) cos ( 400 π t ) ) = a b ( cos ( 4 × 1 0 6 π t + 400 π t ) + cos ( 4 × 1 0 6 π t 400 π t ) ) = a b ( cos ( 4000400 π t ) + cos ( 3999600 π t ) )

    So the modulated signal is the sum of two waves with angular frequency of 4000400 π and 3999600 π radians per second corresponding to frequencies of 4000400 π ( 2 π ) and 39996000 π ( 2 π ) , that is 2000200 Hz and 1999800 Hz i.e. 2 × 1 0 6 ± 200 Hz.

  2. Taking identity (1) and multiplying through by cos ( A ) gives

    cos ( A ) cos ( A ) cos ( B ) 1 2 cos ( A ) ( cos ( A + B ) + cos ( A B ) )

    so

    cos ( A ) cos ( A ) cos ( B ) 1 2 ( cos ( A ) cos ( A + B ) + cos ( A ) cos ( A B ) ) ( 2 )

    Identity (1) can be applied to both expressions in the right-hand side of (2). In the first expression, using A + B instead of ‘ B ’, gives

    cos ( A ) cos ( A + B ) 1 2 ( cos ( A + A + B ) + cos ( A A B ) ) 1 2 ( cos ( 2 A + B ) + cos ( B ) )
    where we have used cos ( B ) cos ( B ) .

    Similarly, in the second expression, using A B instead of ‘ B ’, gives

    cos ( A ) cos ( A B ) 1 2 ( cos ( 2 A B ) + cos ( B ) )

    Together these give:

    cos ( A ) cos ( A ) cos ( B ) 1 2 ( cos ( 2 A + B ) + cos ( B ) + cos ( 2 A B ) + cos ( B ) ) cos ( B ) + 1 2 ( cos ( 2 A + B ) + cos ( 2 A B ) )

    With A = 4 × 1 0 6 π t and B = 400 π t and substituting for the given frequencies, the modulated signal multiplied by the original carrier signal gives

    a b 2 cos ( 4 × 1 0 6 π t ) cos ( 4 × 1 0 6 π t ) cos ( 400 π t ) =

    a b 2 cos ( 2 π × 200 t ) + 1 2 a b 2 ( cos ( 2 × 4 × 1 0 6 π t + 400 π t ) + cos ( 2 × 4 × 1 0 6 π t 400 π t ) )

    The last two terms have frequencies of 4 × 1 0 6 ± 200 Hz which are sufficiently high that a low-pass filter would remove them and leave only the term

    a b 2 cos ( 2 π × 200 t )

    which is the original message signal multiplied by a constant term.

    Interpretation

    Amplitude modulation of a single frequency message signal ( f m ) with a single frequency carrier signal ( f c ) can be shown to be equal to the sum of two cosines with frequencies f c ± f m . Multiplying the modulated signal by a locally generated carrier signal and applying a low-pass filter can reproduce the frequency, f m , of the message signal.

    This is known as double side band amplitude modulation .

Example 2

Obtain expressions for cos θ in terms of the sine function and for sin θ in terms of the cosine function.

Solution

Using (9) with A = θ , B = π 2 we obtain

cos θ π 2 cos θ cos π 2 + sin θ sin π 2 cos θ ( 0 ) + sin θ ( 1 )

i.e. sin θ cos θ π 2 cos π 2 θ

This result explains why the graph of sin θ has exactly the same shape as the graph of cos θ but it is shifted to the right by π 2 . (See Figure 29 on page 28). A similar calculation using (6) yields the result

cos θ sin θ + π 2 .

2.2 Double angle formulae

If we put B = A in the identity given in (6) we obtain Key Point 15:

Key Point 15

sin 2 A sin A cos A + cos A sin A so sin 2 A 2 sin A cos A (12)

Task!

Substitute B = A in identity (7) in Key Point 13 on page 38 to obtain an identity for cos 2 A . Using sin 2 A + cos 2 A 1 obtain two alternative forms of the identity.

Using (7) with B A

cos ( 2 A ) ( cos A ) ( cos A ) ( sin A ) ( sin A ) cos ( 2 A ) cos 2 A sin 2 A

(13)

Substituting for sin 2 A in (13) we obtain

cos 2 A cos 2 A ( 1 cos 2 A ) 2 cos 2 A 1

(14)

Alternatively substituting for cos 2 A in (13)

cos 2 A ( 1 sin 2 A ) sin 2 A cos 2 A 1 2 sin 2 A

(15)

Task!

Use (14) and (15) to obtain, respectively, cos 2 A and sin 2 A in terms of cos 2 A .

From (14) cos 2 A 1 2 ( 1 + cos 2 A ) . From (15) sin 2 A 1 2 ( 1 cos 2 A ) .

Task!

Use (12) and (13) to obtain an identity for tan 2 A in terms of tan A .

tan 2 A sin 2 A cos 2 A 2 sin A cos A cos 2 A sin 2 A

Dividing numerator and denominator by cos 2 A we obtain

tan 2 A 2 sin A cos A 1 sin 2 A cos 2 A 2 tan A 1 tan 2 A ( 16 )

2.3 Half-angle formulae

If we replace A by A 2 and, consequently 2 A by A , in (12) we obtain

sin A 2 sin A 2 cos A 2 ( 17 )

Similarly from (13)

cos A 2 cos 2 A 2 1 . ( 18 )

These are examples of half-angle formulae . We can obtain a half-angle formula for tan A using (16). Replacing A by A 2 and 2 A by A in (16) we obtain

tan A 2 tan A 2 1 tan 2 A 2 ( 19 )

Other formulae, useful for integration when trigonometric functions are present, can be obtained using (17), (18) and (19) shown in the Key Point 16.

Key Point 16

If t = tan A 2 then

sin A = 2 t 1 + t 2 ( 20 )
cos A = 1 t 2 1 + t 2 ( 21 )
tan A = 2 t 1 t 2 ( 22 )

2.4 Sum of two sines and sum of two cosines

Finally, in this Section, we obtain results that are widely used in areas of science and engineering such as vibration theory, wave theory and electric circuit theory.

We return to the identities (6) and (9)

sin ( A + B ) sin A cos B + cos A sin B sin ( A B ) sin A cos B cos A sin B

Adding these identities gives

sin ( A + B ) + sin ( A B ) 2 sin A cos B ( 23 )

Subtracting the identities produces

sin ( A + B ) sin ( A B ) 2 cos A sin B ( 24 )

It is now convenient to let C = A + B and D = A B  so that

A = C + D 2 and B = C D 2

Hence (23) becomes

sin C + sin D 2 sin C + D 2 cos C D 2 ( 25 )

Similarly (24) becomes

sin C sin D 2 cos C + D 2 sin C D 2 ( 26 )

Task!

Use (7) and (10) to obtain results for the sum and difference of two cosines.

cos ( A + B ) cos A cos B sin A sin B and cos ( A B ) cos A cos B + sin A sin B cos ( A + B ) + cos ( A B ) 2 cos A cos B cos ( A + B ) cos ( A B ) 2 sin A sin B

Hence with C = A + B  and  D = A B

cos C + cos D 2 cos C + D 2 cos C D 2 ( 27 )

cos C cos D 2 sin C + D 2 sin C D 2 ( 28 )

2.5 Summary

In this Section we have covered a large number of trigonometric identities. The most important of them and probably the ones most worth memorising are given in the following Key Point.

Key Point 17

cos 2 θ + sin 2 θ 1 sin 2 θ 2 sin θ cos θ cos 2 θ cos 2 θ sin 2 θ 2 cos 2 θ 1 1 2 sin 2 θ sin ( A ± B ) sin A cos B ± cos A sin B cos ( A ± B ) cos A cos B ∓ sin A sin B
Task!

A projectile is fired from the ground with an initial speed u m s 1 at an angle of elevation α ∘ . If air resistance is neglected, the vertical height, y m, is related to the horizontal distance, x m, by the equation

y = x tan α g x 2 s e c 2 α 2 u 2 where g m s 2 is the gravitational constant.

[This equation is derived in HELM booklet  34 Modelling Motion pages 16-17.]

  1. Confirm that y = 0 when x = 0 :

    When y = 0 , the left-hand side of the equation is zero. Since x appears in both of the terms on the right-hand side, when x = 0 , the right-hand side is zero.

  2. Find an expression for the value of x other than x = 0 at which y = 0 and state how this value is related to the maximum range of the projectile:

    When y = 0 , the equation can be written g x 2 s e c 2 α 2 u 2 x tan α = 0

    If x = 0 is excluded from consideration, we can divide through by x and rearrange to give

    g x s e c 2 α 2 u 2 = tan α

    To make x the subject of the equation we need to multiply both sides by 2 u 2 g s e c 2 α .

    Given that 1 s e c 2 α cos 2 α , tan α sin α cos α and sin 2 α 2 sin α cos α , this results in

    x = 2 u 2 sin α cos α g = u 2 sin 2 α g

    This represents the maximum range.

  3. Find the value of x for which the value of y would be a maximum and thereby obtain an expression for the maximum height:

    If air resistance is neglected, we can assume that the parabolic path of the projectile is symmetrical about its highest point. So the highest point will occur at half the maximum range i.e. where

    x = u 2 sin 2 α 2 g

    Substituting this expression for x in the equation for y gives

    y = u 2 sin 2 α 2 g tan α u 2 sin 2 α 2 g 2 g s e c 2 α 2 u 2

    Using the same trigonometric identities as before,

    y = u 2 sin 2 α g u 2 sin 2 α 2 g = u 2 sin 2 α 2 g This represents the maximum height.

  4. Assuming u = 20 m s 1 , α = 6 0 ∘ and g = 10 m s 2 , find the maximum value of the range and the horizontal distances travelled when the height is 10 m:

    Substitution of u = 20 , α = 60 , g = 10 and y = 10 in the original equation gives a quadratic for x :

    10 = 1.732 x 0.05 x 2 or 0.05 x 2 1.732 x + 10 = 0

    Solution of this quadratic yields x = 7.33 or x = 27.32 as the two horizontal ranges at which y = 10 . These values are illustrated in the diagram below which shows the complete trajectory of the projectile.

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Exercises
  1. Show that sin t s e c t tan t .
  2. Show that ( 1 + sin t ) ( 1 + sin ( t ) ) cos 2 t .
  3. Show that 1 tan θ + cot θ 1 2 sin 2 θ .
  4. Show that sin 2 ( A + B ) sin 2 ( A B ) sin 2 A sin 2 B .

    (Hint: the left-hand side is the difference of two squared quantities.)

  5. Show that sin 4 θ + sin 2 θ cos 4 θ + cos 2 θ tan 3 θ .
  6. Show that cos 4 A sin 4 A cos 2 A
  7. Express each of the following as the sum (or difference) of 2 sines (or cosines)
    1. sin 5 x cos 2 x
    2. 8 cos 6 x cos 4 x
    3. 1 3 sin 1 2 x cos 3 2 x
  8. Express
    1. sin 3 θ in terms of cos θ .
    2. cos 3 θ in terms of cos θ .
  9. By writing cos 4 x as cos 2 ( 2 x ) , or otherwise, express cos 4 x in terms of cos x .
  10. Show that tan 2 t 2 tan t 2 s e c 2 t .
  11. Show that cos 10 t cos 12 t sin 10 t + sin 12 t tan t .
  12. Show that the area of an isosceles triangle with equal sides of length x is x 2 2 sin θ

    where θ is the angle between the two equal sides. Hint: use the following diagram:

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  1. sin t . s e c t sin t . 1 cos t sin t cos t tan t .
  2. ( 1 + sin t ) ( 1 + sin ( t ) ) ( 1 + sin t ) ( 1 sin t ) 1 sin 2 t cos 2 t
  3. 1 tan θ + cos θ 1 sin θ cos θ + cos θ sin θ 1 sin 2 θ + cos 2 θ sin θ cos θ sin θ cos θ sin 2 θ + cos 2 θ sin θ cos θ 1 2 sin 2 θ
  4. Using the hint and the identity x 2 y 2 ( x y ) ( x + y ) we have

    sin 2 ( A + B ) sin 2 ( A B ) ( sin ( A + B ) sin ( A B ) ) ( sin ( A + B ) + sin ( A B ) )

    The first bracket gives

    sin A cos B + cos A sin B ( sin A cos B cos A sin B ) 2 cos A sin B

    Similarly the second bracket gives 2 sin A cos B .

    Multiplying we obtain   ( 2 cos A sin A ) ( 2 cos B sin B ) sin 2 A . sin 2 B

  5. sin 4 θ + sin 2 θ cos 4 θ + cos 2 θ 2 sin 3 θ cos θ 2 cos 3 θ cos θ sin 3 θ cos 3 θ tan 3 θ
  6. cos 4 A sin 4 A ( cos A ) 4 ( sin A ) 4 ( cos 2 A ) 2 ( sin 2 A ) 2 ( cos 2 A sin 2 A ) ( cos 2 A + sin 2 A ) cos 2 A sin 2 A cos 2 A
  7. (a) Using sin A + sin B 2 sin A + B 2 cos A B 2

    Clearly here A + B 2 = 5 x A B 2 = 2 x giving A = 7 x B = 3 x

    sin 5 x cos 2 x 1 2 ( sin 7 x + sin 3 x )

    (b) Using cos A + cos B 2 cos A + B 2 cos A B 2 .

    With A + B 2 = 6 x A B 2 = 4 x giving A = 10 x B = 2 x

    8 cos 6 x cos 4 x 4 ( cos 6 x + cos 2 x )

    (c) 1 3 sin 1 2 x cos 3 x 2 1 6 ( sin 2 x sin x )

  8. (a) sin 3 θ sin ( 2 θ + θ ) = sin 2 θ cos θ + cos 2 θ sin θ 2 sin θ cos 2 θ + ( cos 2 θ sin 2 θ ) sin θ 3 sin θ cos 2 θ sin 3 θ 3 sin θ ( 1 sin 2 θ ) sin 3 θ 3 sin θ 4 sin 3 θ

    (b) cos 3 θ cos ( 2 θ + θ ) cos 2 θ cos θ sin 2 θ sin θ ( cos 2 θ sin 2 θ ) cos θ 2 sin θ cos θ sin θ cos 3 θ 3 sin 2 θ cos θ cos 3 θ 3 ( 1 cos 2 θ ) cos θ 4 cos 3 θ 3 cos θ
  9.    cos 4 x = cos 2 ( 2 x ) 2 cos 2 ( 2 x ) 1 2 ( cos 2 x ) 2 1 2 ( 2 cos 2 x 1 ) 2 1 2 ( 4 cos 4 x 4 cos 2 x + 1 ) 1 8 cos 4 x 8 cos 2 x + 1 .
  10.    tan 2 t 2 tan t 1 tan 2 t 2 tan t 1 ( s e c 2 t 1 ) 2 tan t 2 s e c 2 t
  11.    cos 10 t cos 12 t 2 sin 11 t sin t sin 10 t + sin 12 t 2 sin 11 t cos ( t )

    cos 10 t cos 12 t sin 10 t + sin 12 t sin t cos ( t ) sin t cos t tan t

  12.   The right-angled triangle A C D has area    1 2 ( C D ) ( A D ) But sin θ 2 = C D x C D = x sin θ 2 cos θ 2 = A D x A D = x cos θ 2

    area of Δ A C D = 1 2 x 2 sin θ 2 cos θ 2 = 1 4 x 2 sin θ area of Δ A B C = 2 × area of Δ A C D = 1 2 x 2 sin θ