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
- 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 Hz
- Show that the modulated signal can be demodulated by using a locally generated carrier and applying a low-pass filter.
Mathematical statement of problem
-
Express the message signal as
and the carrier as
.
Assume that the modulation gives the product .
Use trigonometric identities to show that
where and are constants.
Then substitute and to calculate the two resulting frequencies.
- Use trigonometric identities to show that multiplying the modulated signal by results in the lowest frequency component of the output having a frequency equal to the original message signal.
Mathematical analysis
-
The message signal has a frequency of
Hz so
radians per second.
The carrier signal has a frequency of Hz.
Hence radians per second.
So .
Key Point 13 includes the identity:
Rearranging gives the identity:
Using (1) with and gives
So the modulated signal is the sum of two waves with angular frequency of 4000400 and 3999600 radians per second corresponding to frequencies of 4000400 and 39996000 , that is 2000200 Hz and 1999800 Hz i.e. Hz.
-
Taking identity (1) and multiplying through by
gives
so
Identity (1) can be applied to both expressions in the right-hand side of (2). In the first expression, using instead of ‘ ’, gives
Similarly, in the second expression, using instead of ‘ ’, gives
Together these give:
With and and substituting for the given frequencies, the modulated signal multiplied by the original carrier signal gives
The last two terms have frequencies of Hz which are sufficiently high that a low-pass filter would remove them and leave only the term
which is the original message signal multiplied by a constant term.
Interpretation
Amplitude modulation of a single frequency message signal with a single frequency carrier signal can be shown to be equal to the sum of two cosines with frequencies . Multiplying the modulated signal by a locally generated carrier signal and applying a low-pass filter can reproduce the frequency, , of the message signal.
This is known as double side band amplitude modulation .
Example 2
Obtain expressions for in terms of the sine function and for in terms of the cosine function.
Solution
Using (9) with we obtain
i.e.
This result explains why the graph of has exactly the same shape as the graph of but it is shifted to the right by . (See Figure 29 on page 28). A similar calculation using (6) yields the result
2.2 Double angle formulae
If we put in the identity given in (6) we obtain Key Point 15:
Key Point 15
so (12)
Task!
Substitute in identity (7) in Key Point 13 on page 38 to obtain an identity for . Using obtain two alternative forms of the identity.
Using (7) with
(13)
Substituting for in (13) we obtain
(14)
Alternatively substituting for in (13)
(15)
Task!
Use (14) and (15) to obtain, respectively, and in terms of .
From (14) . From (15) .
Task!
Use (12) and (13) to obtain an identity for in terms of .
Dividing numerator and denominator by we obtain
2.3 Half-angle formulae
If we replace by and, consequently by , in (12) we obtain
Similarly from (13)
These are examples of half-angle formulae . We can obtain a half-angle formula for using (16). Replacing by and by in (16) we obtain
Other formulae, useful for integration when trigonometric functions are present, can be obtained using (17), (18) and (19) shown in the Key Point 16.
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)
Adding these identities gives
Subtracting the identities produces
It is now convenient to let and so that
Hence (23) becomes
Similarly (24) becomes
Task!
Use (7) and (10) to obtain results for the sum and difference of two cosines.
Hence with and
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.
Task!
A projectile is fired from the ground with an initial speed at an angle of elevation . If air resistance is neglected, the vertical height, m, is related to the horizontal distance, m, by the equation
where is the gravitational constant.
[This equation is derived in HELM booklet 34 Modelling Motion pages 16-17.]
-
Confirm that
when
:
When , the left-hand side of the equation is zero. Since appears in both of the terms on the right-hand side, when , the right-hand side is zero.
-
Find an expression for the value of
other than
at which
and state how this value is related to the maximum range of the projectile:
When , the equation can be written
If is excluded from consideration, we can divide through by and rearrange to give
To make the subject of the equation we need to multiply both sides by .
Given that and , this results in
This represents the maximum range.
-
Find the value of
for which the value of
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
Substituting this expression for in the equation for gives
Using the same trigonometric identities as before,
This represents the maximum height.
-
Assuming
,
and
, find the maximum value of the range and the horizontal distances travelled when the height is 10 m:
Substitution of and in the original equation gives a quadratic for :
or
Solution of this quadratic yields or as the two horizontal ranges at which . These values are illustrated in the diagram below which shows the complete trajectory of the projectile.
Exercises
- Show that .
- Show that
- Show that .
-
Show that
.
(Hint: the left-hand side is the difference of two squared quantities.)
- Show that
- Show that
-
Express each of the following as the sum (or difference) of 2 sines (or cosines)
-
Express
- in terms of .
- in terms of .
- By writing as , or otherwise, express in terms of .
- Show that .
- Show that .
-
Show that the area of an isosceles triangle with equal sides of length
is
where is the angle between the two equal sides. Hint: use the following diagram:
- .
-
Using the hint and the identity
we have
The first bracket gives
Similarly the second bracket gives
Multiplying we obtain
-
(a) Using
Clearly here giving
(b) Using .
With giving
(c)
-
-
-
The right-angled triangle
has area