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Fourier series: review

A function or signal x ( t ) is called periodic with period T if x ( t + T ) = x ( t ) . All “typical” periodic function x ( t ) with period T can be developed as follows

x ( t ) = a 0 2 + k = 1 a k cos ( k 2 π T t ) + b k sin ( k 2 π T t )

where the coefficients are computed as follows:

a k = 2 T - T / 2 T / 2 x ( t ) cos ( k 2 π T t ) d t ( k = 0 , 1 , 2 , . . . )
b k = 2 T - T / 2 T / 2 x ( t ) sin ( k 2 π T t ) d t ( k = 1 , 2 , 3 , . . . )

The natural interpretation of [link] is as a decomposition of the signal x ( t ) into individual oscillations where a k indicates the amplitude of the even oscillation cos ( k 2 π T t ) of frequency k / T (meaning its period is T / k ), and b k indicates the amplitude of the odd oscillation sin ( k 2 π T t ) of frequency k / T . For an audio signal x ( t ) , frequency corresponds to how high a sound is and amplitude to how loud it is. The oscillations appearing in the Fourier decomposition are oftenalso called harmonics (first, second, third harmonic etc).

Note: one can also integrate over [ 0 , T ] or any other interval of length T . Note also, that the average value of x ( t ) over one period is equal to a 0 / 2 .

Complex representation of Fourier series

Often it is more practical to work with complex numbers in the area of Fourier analysis. Using the famous formula

e j α = cos ( α ) + j sin ( α )

it is possible to simplify several formulas at the price of working with complex numbers.Towards this end we write

a cos ( α ) + b sin ( α ) = a 1 2 ( e j α + e - j α ) + b 1 2 j ( e j α - e - j α ) = 1 2 ( a - b j ) e j α + 1 2 ( a + b j ) e - j α

From this we observe that we may replace the cos and sin harmonics by a pair of exponential harmonicswith opposite frequencies and with complex amplitudes which are conjugate complex to each other.In fact, we arrive at the more simple complex Fourier series :

x ( t ) = k = - X k e j 2 π k t / T with X k = 1 T 0 T x ( t ) e - j 2 π k t / T d t

Note that X is complex, but x is real-valued (the imaginary parts of all the terms in X k e j 2 π k t / T add up to zero; in other words, they cancel each other out). The absolute value of X k gives the amplitude of the complex harmonic with frequency k / T (meaning its period is T / k ); the argument of X k provides the phase difference between the complex harmonics.If x is even, X k is real for all k and all harmonics are in phase.

To verify [link] note that by [link] and [link] we have for positive k

X k = 1 T 0 T x ( t ) cos ( 2 π k t / T ) - j sin ( 2 π k t / T ) d t = 1 2 ( a k - b k j ) .

For negative k we note that X - k = X k * by [link] , where ( ) * denotes the conjugate complex. By [link] , the X k are exactly as they are supposed to be.

Properties

  • Linearity: The Fourier coefficients of the signal z ( t ) = c x ( t ) + y ( t ) are simply
Z k = c X k + Y k
  • Change of frequency: The signal z ( t ) = x ( λ t ) has the period T / λ and has the same Fourier coefficients as x ( t ) — but they correspond to different frequencies f :
Z k | f = k T / λ = X k | f = k T since z ( t ) = x ( λ t ) = k = - X k e j 2 π λ k t / T = k = - Z k e j 2 π t k T / λ

The equation on the right allows to read off the Fourier coefficients and to establish Z k = X k . (For an alternative computation see  Comment 1 )

Comment 1

Z k = 1 T / λ 0 T / λ z ( t ) e - 2 π j k t / ( T / λ ) d t = λ T 0 T / λ x ( t λ ) e - 2 π j k t λ / T d t = 1 T 0 T x ( s ) e - 2 π j k s / T d s = X k

  • Shift: The Fourier coefficients of z ( t ) = x ( t + d ) are simply
Z k = X k e j 2 π k d / T

The modulation is much more simple in complex writing then it would be with real coefficients.For the special shift by half a period, i.e., d = T / 2 we have Z k = X k e j π k = ( - 1 ) k X k .

  • Derivative: The Fourier series of the derivative of x ( t ) with development [link] can be obtained simply by taking the derivative of [link] term by term:
x ' ( t ) = k = - X k · k 2 π j T · e j 2 π k t / T

Short: when taking the derivative of a signal, the complex Fourier coefficients get multiplied by k 2 π j T . Consequently, the coefficients of the derivative decay slower.

Examples

  • The pure oscillation (containing only one real but two complex frequencies)
    x ( t ) = sin ( 2 π t / T ) X 1 = - X - 1 = j 2
    or B 1 = 1 / 2 = - B - 1 , or b 1 = 1 , and all other coefficients are zero.This formula can be obtained without computing integrals by noting that sin ( α ) = ( e j α - e - j α ) / ( 2 j ) = ( j / 2 ) ( e - j α - e j α ) and setting α = 2 π t / T .
  • Functions which are time-limited , i.e., defined on a finite interval can be periodically extended. Example with T = 2 π :
    x ( t ) = 1 for 0 < t < π - 1 for - π < t < 0 c for t = 0 , π b k = 2 B k = 4 π k for odd k 1
    and all other coefficients zero. Note that c is any constant; the value of c does not affect the coefficients b k . We have for 0 < t < π
    1 = 4 π ( sin ( t ) + 1 3 sin ( 3 t ) + 1 5 sin ( 5 t ) + . . . )
    Note that for t = 0 the value of the series on the right is 0, which is equal to x ( 0 - ) + x ( 0 + ) , the middle of the jump of x ( t ) at 0, no matter what c is. Similar for t = π .

Questions & Answers

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Source:  OpenStax, Sampling rate conversion. OpenStax CNX. Sep 05, 2013 Download for free at http://legacy.cnx.org/content/col11529/1.2
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