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Inside Collection (Textbook):

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

# Complex Form of the Fourier Series

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

The trigonometric form of the Fourier Series, shown in (Reference) can be converted into a more convenient form by doing the following substitutions:

cos ( n Ω 0 t ) = e j n Ω 0 t + e - j n Ω 0 t 2 cos ( n Ω 0 t ) = e j n Ω 0 t + e - j n Ω 0 t 2
(1)
sin ( n Ω 0 t ) = e j n Ω 0 t - e - j n Ω 0 t j 2 sin ( n Ω 0 t ) = e j n Ω 0 t - e - j n Ω 0 t j 2
(2)

After some straight-forward rearranging, we obtain

x ( t ) = a 0 + n = 1 a n - j b n 2 e j n Ω 0 t + n = 1 a n + j b n 2 e - j n Ω 0 t x ( t ) = a 0 + n = 1 a n - j b n 2 e j n Ω 0 t + n = 1 a n + j b n 2 e - j n Ω 0 t
(3)

Keeping in mind that anan and bnbn are only defined for positive values of nn, lets sum over the negative integers in the second summation:

x ( t ) = a 0 + n = 1 a n - j b n 2 e j n Ω 0 t + n = - 1 - a - n + j b - n 2 e j n Ω 0 t x ( t ) = a 0 + n = 1 a n - j b n 2 e j n Ω 0 t + n = - 1 - a - n + j b - n 2 e j n Ω 0 t
(4)

Next, let's assume that anan and bnbn are defined for both positive and negative nn. In this case, we find that an=a-nan=a-n and bn=-b-nbn=-b-n, since

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 x ( t ) cos ( n Ω 0 t ) d t = a n 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 x ( t ) cos ( n Ω 0 t ) d t = a n
(5)

and

b - n = 2 T t 0 t 0 + T x ( t ) sin ( - n Ω 0 t ) d t = - 2 T t 0 t 0 + T x ( t ) sin ( n Ω 0 t ) d t = - b n b - n = 2 T t 0 t 0 + T x ( t ) sin ( - n Ω 0 t ) d t = - 2 T t 0 t 0 + T x ( t ) sin ( n Ω 0 t ) d t = - b n
(6)

Using this fact, we can rewrite Equation 4 as

x ( t ) = a 0 + n = 1 a n - j b n 2 e j n Ω 0 t + n = - 1 - a n - j b n 2 e j n Ω 0 t x ( t ) = a 0 + n = 1 a n - j b n 2 e j n Ω 0 t + n = - 1 - a n - j b n 2 e j n Ω 0 t
(7)

If we define1c0a0c0a0, and

c n a n - j b n 2 c n a n - j b n 2
(8)

then we can rewrite Equation 7 as

x ( t ) = n = - c n e j n Ω 0 t x ( t ) = n = - c n e j n Ω 0 t
(9)

which is called the complex form of the Fourier Series. Note that since a-n=ana-n=an and b-n=-bnb-n=-bn, we have

c - n = c n * c - n = c n *
(10)

This means that

c - n = c n c - n = c n
(11)

and

c - n = - c n c - n = - c n
(12)

Next, we must find formulas for finding the cncn given x(t)x(t). We first look at a property of complex exponentials:

t 0 t 0 + T e j k Ω 0 t d t = T , k = 0 0 , otherwise t 0 t 0 + T e j k Ω 0 t d t = T , k = 0 0 , otherwise
(13)

To see this, we note that

t 0 t 0 + T e j k Ω 0 k t d t = t 0 t 0 + T cos ( k Ω 0 t ) + j t 0 t 0 + T sin ( k Ω 0 t ) d t t 0 t 0 + T e j k Ω 0 k t d t = t 0 t 0 + T cos ( k Ω 0 t ) + j t 0 t 0 + T sin ( k Ω 0 t ) d t
(14)

It's easy to see that kΩ0kΩ0 also has period TT, hence the integral is over kk periods of cos(kΩ0t)cos(kΩ0t) and sin(kΩ0t)sin(kΩ0t). Therefore, if k0k0, then

t 0 t 0 + T e j k Ω 0 t d t = 0 t 0 t 0 + T e j k Ω 0 t d t = 0
(15)

otherwise

t 0 t 0 + T e j k Ω 0 t d t = t 0 t 0 + T d t = T t 0 t 0 + T e j k Ω 0 t d t = t 0 t 0 + T d t = T
(16)

We use Equation 13 to derive an equation for cncn as follows. Consider the integral

t 0 t 0 + T x ( t ) e - j n Ω 0 t d t t 0 t 0 + T x ( t ) e - j n Ω 0 t d t
(17)

Substituting the complex form of the Fourier Series of x(t)x(t) in Equation 17, (using kk as the index of summation) we obtain

t 0 t 0 + T k = - c n e j k Ω 0 t e - j n Ω 0 t d t t 0 t 0 + T k = - c n e j k Ω 0 t e - j n Ω 0 t d t
(18)

Rearranging the order of integration and summation, combining the exponents, and using Equation 13 gives

k = - c n t 0 t 0 + T e j ( k - n ) Ω 0 t d t = T c n k = - c n t 0 t 0 + T e j ( k - n ) Ω 0 t d t = T c n
(19)

Using this result, we find that

c n = 1 T t 0 t 0 + T x ( t ) e - j n Ω 0 t d t c n = 1 T t 0 t 0 + T x ( t ) e - j n Ω 0 t d t
(20)

Example 2.1 Let's now find the complex form of the Fourier Series for the signal in Example (Reference). The integral to be evaluated is

c n = 2 3 - 0 . 5 0 . 5 2 t e - j 4 π 3 n t d t c n = 2 3 - 0 . 5 0 . 5 2 t e - j 4 π 3 n t d t
(21)

Integrating by parts yields

c n = j n π cos ( 2 π n / 3 ) - j 3 ( n π ) 2 sin ( 2 π n / 3 ) c n = j n π cos ( 2 π n / 3 ) - j 3 ( n π ) 2 sin ( 2 π n / 3 )
(22)

Figure 1 shows the magnitude of the coefficients, cncn. Note that the complex Fourier Series coefficients have even symmetry as was mentioned earlier.

Can the basic formula for computing the cncn in Equation 20 be simplified when x(t)x(t) has either even, odd, or half-wave symmetry? The answer is yes. We simply use the fact that

c n = a n - j b n 2 c n = a n - j b n 2
(23)

and solve for anan and bnbn using the formulae given above for even, odd, or half-wave symmetric signals. This avoids having to integrate complex quantities. This can also be seen by noting that (setting t0=T/2t0=T/2 in Equation 20):

c n = 1 T - T / 2 T / 2 x ( t ) e - j n Ω 0 t d t = 1 T - T / 2 T / 2 x ( t ) cos ( n Ω 0 t ) d t - j T - T / 2 T / 2 x ( t ) sin ( n Ω 0 t ) d t = 1 2 a n - j 2 b n c n = 1 T - T / 2 T / 2 x ( t ) e - j n Ω 0 t d t = 1 T - T / 2 T / 2 x ( t ) cos ( n Ω 0 t ) d t - j T - T / 2 T / 2 x ( t ) sin ( n Ω 0 t ) d t = 1 2 a n - j 2 b n
(24)

Alternately, if x(t)x(t) has half-wave symmetry, we can use Equation 24, (Reference), and (Reference) to get

c n = 2 T - T / 4 T / 4 x ( t ) e - j n Ω 0 t d t , n odd 0 , n even c n = 2 T - T / 4 T / 4 x ( t ) e - j n Ω 0 t d t , n odd 0 , n even
(25)

Unlike the trigonometric form, we cannot simplify this further if x(t)x(t) is even or odd symmetric since e-jnΩ0te-jnΩ0t has neither even nor odd symmetry.

Example 2.2 In this example we will look at the effect of adjusting the period of a pulse train signal. Consider the signal depicted in Figure 2.

The Fourier Series coefficients for this signal are given by

c n = 1 T - τ / 2 τ / 2 e - j n Ω 0 t d t = - 1 j n Ω 0 T e - j n Ω 0 τ / 2 - e j n Ω 0 τ / 2 = τ T sin ( n Ω 0 τ / 2 ) n Ω 0 τ / 2 τ T sinc ( n Ω 0 τ / 2 ) c n = 1 T - τ / 2 τ / 2 e - j n Ω 0 t d t = - 1 j n Ω 0 T e - j n Ω 0 τ / 2 - e j n Ω 0 τ / 2 = τ T sin ( n Ω 0 τ / 2 ) n Ω 0 τ / 2 τ T sinc ( n Ω 0 τ / 2 )
(26)

Figure 3 shows the magnitude of |cn||cn|, the amplitude spectrum, for T=1T=1 and τ=1/2τ=1/2 as well as the Fourier Series for the signal based on the first 30 coefficients

x ^ ( t ) = n = - 30 30 c n e n Ω 0 t x ^ ( t ) = n = - 30 30 c n e n Ω 0 t
(27)

Similar plots are shown in Figures Figure 4, and Figure 5, for T=4T=4, and T=8T=8, respectively.

This example illustrates several important points about the Fourier Series: As the period TT increases, Ω0Ω0 decreases in magnitude (this is obvious since Ω0=2π/TΩ0=2π/T). Therefore, as the period increases, successive Fourier Series coefficients represent more closely spaced frequencies. The frequencies corresponding to each nn are given by the following table:

 n n Ω Ω 0 0 ± 1 ± 1 ± Ω 0 ± Ω 0 ± 2 ± 2 ± 2 Ω 0 ± 2 Ω 0 ⋮ ⋮ ⋮ ⋮ ± n ± n ± n Ω 0 ± n Ω 0

This table establishes a relation between nn and the frequency variable ΩΩ. In particular, if T=1T=1, we have Ω0=2πΩ0=2π and

 n n Ω Ω 0 0 ± 1 ± 1 ± 2 π ± 2 π ± 2 ± 2 ± 4 π ± 4 π ⋮ ⋮ ⋮ ⋮ ± n ± n ± 2 n π ± 2 n π

If T=TT=T, then Ω0=π/2Ω0=π/2 and

 n n Ω Ω 0 0 ± 1 ± 1 ± π / 2 ± π / 2 ± 2 ± 2 ± π ± π ⋮ ⋮ ⋮ ⋮ ± n ± n ± n π / 2 ± n π / 2

and if T=8T=8, we have Ω0=π/4Ω0=π/4 and

 n n Ω Ω 0 0 ± 1 ± 1 ± π / 4 ± π / 4 ± 2 ± 2 ± π / 2 ± π / 2 ⋮ ⋮ ⋮ ⋮ ± n ± n ± n π / 4 ± n π / 4

Note that in all three cases, the first zero coefficient corresponds to the value of nn for which Ω=4πΩ=4π. Also, as TT gets bigger, the cncn appear to resemble more closely spaced samples of a continuous function of frequency (since the nΩnΩ are more closely spaced). Can you determine what this function is?

As we shall see, by letting the period TT get large (infinitely large), we will derive the Fourier Transform in the next chapter.

References

## Footnotes

1. The notation “" is often used instead of the “==" sign when defining a new variable.

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