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Half-Wave Symmetry

Module by: Carlos E. Davila. E-mail the author

Half-Wave Symmetry

Periodic signals having half-wave symmetry have the property

x ( t ) = - x ( t - T / 2 ) x ( t ) = - x ( t + T / 2 ) x ( t ) = - x ( t - T / 2 ) x ( t ) = - x ( t + T / 2 )
(1)

It turns out that signals with this type of symmetry only have odd-numbered harmonics, the even harmonics are zero. To see this, lets look at the formula for the coefficients anan:

a n = 2 T t 0 t 0 + T x ( t ) cos ( n Ω 0 t ) d t = 2 T t 0 t 0 + T / 2 x ( t ) cos ( n Ω 0 t ) d t + t 0 + T / 2 t 0 + T x ( t ) cos ( n Ω 0 t ) d t = 2 T I 1 + I 2 a n = 2 T t 0 t 0 + T x ( t ) cos ( n Ω 0 t ) d t = 2 T t 0 t 0 + T / 2 x ( t ) cos ( n Ω 0 t ) d t + t 0 + T / 2 t 0 + T x ( t ) cos ( n Ω 0 t ) d t = 2 T I 1 + I 2
(2)

Making the substitution τ=t-T/2τ=t-T/2 in I2I2 gives

I 2 = t 0 t 0 + T / 2 x ( τ + T / 2 ) cos ( n Ω 0 ( τ + T / 2 ) ) d τ = - t 0 t 0 + T / 2 x ( τ ) cos ( n Ω 0 ( τ + T / 2 ) ) d τ I 2 = t 0 t 0 + T / 2 x ( τ + T / 2 ) cos ( n Ω 0 ( τ + T / 2 ) ) d τ = - t 0 t 0 + T / 2 x ( τ ) cos ( n Ω 0 ( τ + T / 2 ) ) d τ
(3)

The quantity cos(nΩ0(τ+T/2))=cos(nΩτ+nπ)cos(nΩ0(τ+T/2))=cos(nΩτ+nπ) can be simplified using the trigonometric identity

cos ( u ± v ) = cos ( u ) cos ( v ) sin ( u ) sin ( v ) cos ( u ± v ) = cos ( u ) cos ( v ) sin ( u ) sin ( v )
(4)

We have

cos ( n Ω τ + n π ) = cos ( n Ω τ ) cos ( n π ) - sin ( n Ω τ ) sin ( n π ) = ( - 1 ) n cos ( n Ω τ ) - 0 cos ( n Ω τ + n π ) = cos ( n Ω τ ) cos ( n π ) - sin ( n Ω τ ) sin ( n π ) = ( - 1 ) n cos ( n Ω τ ) - 0
(5)

Therefore

I 2 = - ( - 1 ) n t 0 t 0 + T / 2 x ( τ ) cos ( n Ω 0 τ ) d τ I 2 = - ( - 1 ) n t 0 t 0 + T / 2 x ( τ ) cos ( n Ω 0 τ ) d τ
(6)

and we can write:

a n = 2 T ( 1 - ( - 1 ) n ) t 0 t 0 + T / 2 x ( t ) cos ( n Ω 0 t ) d t a n = 2 T ( 1 - ( - 1 ) n ) t 0 t 0 + T / 2 x ( t ) cos ( n Ω 0 t ) d t
(7)

From this expression we find that an=0an=0 whenever nn is even. In fact, we have

a n = 4 T t 0 t 0 + T / 2 x ( t ) cos ( n Ω 0 t ) d t , n , odd 0 , n , even a n = 4 T t 0 t 0 + T / 2 x ( t ) cos ( n Ω 0 t ) d t , n , odd 0 , n , even
(8)

A similar derivation leads to

b n = 4 T t 0 t 0 + T / 2 x ( t ) sin ( n Ω 0 t ) d t , n , odd 0 , n , even b n = 4 T t 0 t 0 + T / 2 x ( t ) sin ( n Ω 0 t ) d t , n , odd 0 , n , even
(9)

A good choice of t0t0 can lead to a considerable savings in time when calculating the Fourier Series of half-wave symmetric signals. Note that half-wave symmetric signals need not have odd or even symmetry for the above formulae to apply. If a signal has half-wave symmetry and in addition has odd or even symmetry, then some additional simplification is possible. Consider the case when a half-wave symmetric signal also has even symmetry. Then clearly bn=0bn=0, and Equation 8 applies. However since the integrand in Equation 8 is the product of two even signals, x(t)x(t) and cos(nΩ0t)cos(nΩ0t), it too has even symmetry. Therefore, instead of integrating from, say, -T/4-T/4 to T/4T/4, we need only integrate from 0 to T/4T/4 and multiply the result by 2. Therefore the formula for anan for an even, half-wave symmetric signal becomes:

a n = 8 T 0 T / 4 x ( t ) cos ( n Ω 0 t ) d t , n , odd 0 , n , even a n = 8 T 0 T / 4 x ( t ) cos ( n Ω 0 t ) d t , n , odd 0 , n , even
(10)
b n = 0 b n = 0
(11)

For an odd half-wave symmetric signals, a similar argument leads to

a n = 0 a n = 0
(12)
b n = 8 T 0 T / 4 x ( t ) sin ( n Ω 0 t ) d t , n , odd 0 , n , even b n = 8 T 0 T / 4 x ( t ) sin ( n Ω 0 t ) d t , n , odd 0 , n , even
(13)

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