Fourier Series (Periodic functions of Period 2)
The Fourier series provides a method of analysing periodic functions by considering their constituent components. Alternating currents, voltages and acoustic waves are typical of practical examples occurring in engineering and science where periodic functions have to be analysed.
A period function is periodic because , e.g.
This is true for all values of x and T is some positive constant. T is the interval between two successive repetitions and is called the period of the function x.
e.g. is a periodic function of period 2, because
Similarly,
In general if and these are periodic functions of period .
Note that since is given by two different expressions in the two halves of the range, the integration is performed in two parts i.e. for - to 0 and 0 to . The wave shown in the diagram has finite discontinuities at x = 0, , 2, etc. The advantage of the Fourier type series is that it can be applied to functions, which are discontinuous as well as to thosethat are continuous. The basis of the Fourier series is that all functions which can be defined in the interval - to + can be expressed as a convergent trig series of the form.
i)
where a0, a1, a2, ……b1, b2, ….. are real constants
and
[Note: this is the mean value of the function]
[Where n = 1, 2, 3, 4, …………]
a0, an and bn are called the Fourier coefficients. If these can be determined, the equation i)above then becomes the Fourier series for . In most practical cases only the first few terms of the series are required.
The sum of a Fourier series at a discontinuity is given by the mean of the 2 limiting values of the function.
The mean values at is
Example
Obtain the Fourier series for the periodic function determined by
The function is periodic outside of this range with the period 2
(no cosine terms appear as a1, a2, a3, … are all zero)
And therefore:
Fourier series for a non-periodic function over the range of 2
If a function of x ,, is not periodic then it cannot be expanded as a Fourier series for all values of x. However it is possible to determine a Fourier series to represent the function over any range provided the width is 2. This is done by constructing a new function so that is extends symmetrically every 2. The new function by construction is symmetrical and has a period of 2; it can therefore be expanded as a Fourier series for all x. Note that for a non-periodic function, such as the sum of the Fourier series is equal to at all points in the given range but not equal to outside the range.
Even Functions and Odd Functions
A function is said to be even if for all values of x
e.g.
Graphs of even functions are always symmetrical about the y-axis.
A function is said to be odd if for all values of x
e.g.
The graphs of odd functions are symmetrical about the origin.
There are many functions which are neither even or odd
e.g.
The Fourier series of an even periodic function having period 2 contains only the cosine terms and may contain a constant.
The Fourier series of an odd periodic function of period 2 contain only the Sine terms and no constant.
An alternative method for Fourier analysis
An alternative tabular method can be used to determine Fourier coefficients. This method is based on an application of the trapezoidal rule and is particularly useful when a number of voltage readings are available at regular time intervals over the period of one complete cycle of a waveform. See example 4.3.3 in Higher National Engineering by Tooley & Dingle.
The values of voltage over a complete cycle of a waveform (see figure above) are as follows:
Angle, / 30 / 60 / 90 / 120 / 150 / 180 / 210 / 240 / 270 / 300 / 330 / 360Voltage, v / 8.9 / 8.1 / 7.5 / 6.5 / 5.5 / 5.0 / 4.5 / 3.5 / 2.5 / 1.9 / 1.1 / 5
First we need to determine the value of a0. We can do this by calculating the mean value of voltage over the complete cycle. Applying the trapezoidal rule (i.e. adding together the values of voltage and dividing by the number of intervals) gives:
a0= (8.9+8.1+7.5+6.5+5.5+5.0+4.5+3.5+2.5+1.9_1.1+5)/12 = 5.
Similarly, to determine the vale of a1, we multiply each value of voltage by cos, before adding them together and dividing by half the number of intervals (a1 is twice the mean value of vcost over one complete cycle). Thus
a1 = (7.07.7+4.049-0.002-3.252+………+5)/6 = 0
The same method can be used to determine the remaining Fourier coefficients. The use of a spreadsheet for calculation is highly recommended.
Create a spreadsheet to determine the Fourier coefficients for the example above and show that the Fourier series can be expressed as:
v = 5 + 2.894 sin + 1.444 sin2 + 0.635 sin3 +0.522 sin4 + …..
(note that there are no terms in cos in the Fourier series for this waveform).
The Root mean square value of a waveform.
The root mean square (rms) value of a waveform is the effective value of the current or voltage concerned. It is defined as the value of DC voltage that would produce the same power in a pure resistive load.
Assume that we are dealing with a voltage given by:
If this voltage is applied to a pure resistance, R, the current flowing will be given by:
The instantaneous power dissipated in the resistor, p, will be given by the product of i and v. Thus:
This is an important result. From equation ii), we can infer that the power waveform (i.e. the waveform of p plotted against t) will comprise a cosine waveform at twice the frequency of the voltage (see figure below). Furthermore, if we apply Fourier series analysis, we can infer that the mean value of the power waveform (over a complete cycle of voltage or current) will be the same as its amplitude. The value of the Fourier coefficients being:
(i.e. the mean value of the power waveform over one cycle of voltage)
(i.e. the amplitude of the term in cos 2t)
(note that no other components are present).
The mean power, over one cycle, is thus given by a0.
Thus
And
Now, let VRMS be the equivalent DC voltage that will produce the same power.
Thus
And
Hence
In other words, a waveform given byV will produce the same power in a load as a direct current of 1 V. A similar relationship can be obtained for current:
where I is the amplitude of the current wave.
When a wave is complex, it follows that its rms value can be found by adding together the values of each individual harmonic component present. Thus
where Vo is the value of any DC component that may be present, and V1, V2, V3, ….., Vm are the amplitudes of the harmonic components present in the wave.
Power factor
When harmonic components are present, the power factor can be determined by adding together the power supplied by each harmonic before dividing by the product of RMS voltage and current. Hence:
Where PTOT is found by adding the power due to each harmonic component present. Thus:
See examples 4.3.4 and 4.3.5 in Higher National Engineering by Tooley & Dingle
U5: Electrical & Electronic PrinciplesPage 1 of 9
Fourier Series.docx