2.4 Complex form of the fourier series

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The trigonometric form of the Fourier Series, shown in [link] can be converted into a more convenient form by doing the following substitutions:

cos (n Ω_{0} t) = \frac{e^{j n Ω_{0} t} + e^{- j n Ω_{0} t}}{2}

sin (n Ω_{0} t) = \frac{e^{j n Ω_{0} t} - e^{- j n Ω_{0} t}}{j 2}

After some straight-forward rearranging, we obtain

x (t) = a_{0} + \sum_{n = 1}^{\infty} [\frac{a_{n} - j b_{n}}{2}] e^{j n Ω_{0} t} + \sum_{n = 1}^{\infty} [\frac{a_{n} + j b_{n}}{2}] e^{- j n Ω_{0} t}

Keeping in mind that $a_{n}$ and $b_{n}$ are only defined for positive values of $n$ , lets sum over the negative integers in the second summation:

x (t) = a_{0} + \sum_{n = 1}^{\infty} [\frac{a_{n} - j b_{n}}{2}] e^{j n Ω_{0} t} + \sum_{n = - 1}^{- \infty} [\frac{a_{- n} + j b_{- n}}{2}] e^{j n Ω_{0} t}

Next, let's assume that $a_{n}$ and $b_{n}$ are defined for both positive and negative $n$ . In this case, we find that $a_{n} = a_{- n}$ and $b_{n} = - b_{- n}$ , since

\begin{matrix} a_{- n} & = & \frac{2}{T} \int_{t_{0}}^{t_{0} + T} x (t) cos (- n Ω_{0} t) d t \\ = & \frac{2}{T} \int_{t_{0}}^{t_{0} + T} x (t) cos (n Ω_{0} t) d t \\ = & a_{n} \end{matrix}

and

\begin{matrix} b_{- n} & = & \frac{2}{T} \int_{t_{0}}^{t_{0} + T} x (t) sin (- n Ω_{0} t) d t \\ = & - \frac{2}{T} \int_{t_{0}}^{t_{0} + T} x (t) sin (n Ω_{0} t) d t \\ = & - b_{n} \end{matrix}

Using this fact, we can rewrite [link] as

x (t) = a_{0} + \sum_{n = 1}^{\infty} [\frac{a_{n} - j b_{n}}{2}] e^{j n Ω_{0} t} + \sum_{n = - 1}^{- \infty} [\frac{a_{n} - j b_{n}}{2}] e^{j n Ω_{0} t}

If we define The notation “ $\equiv$ " is often used instead of the “ $=$ " sign when defining a new variable. $c_{0} \equiv a_{0}$ , and

c_{n} \equiv \frac{a_{n} - j b_{n}}{2}

then we can rewrite [link] as

x (t) = \sum_{n = - \infty}^{\infty} c_{n} e^{j n Ω_{0} t}

which is called the complex form of the Fourier Series. Note that since $a_{- n} = a_{n}$ and $b_{- n} = - b_{n}$ , we have

c_{- n} = c_{n}^{*}

This means that

|c_{- n}| = |c_{n}|

and

∠ c_{- n} = - ∠ c_{n}

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

\int_{t_{0}}^{t_{0} + T} e^{j k Ω_{0} t} d t = \{\begin{matrix} T, & k = 0 \\ 0, & otherwise \end{matrix})

To see this, we note that

\int_{t_{0}}^{t_{0} + T} e^{j k Ω_{0} k t} d t = \int_{t_{0}}^{t_{0} + T} cos (k Ω_{0} t) + j \int_{t_{0}}^{t_{0} + T} sin (k Ω_{0} t) d t

It's easy to see that $k Ω_{0}$ also has period $T$ , hence the integral is over $k$ periods of $cos (k Ω_{0} t)$ and $sin (k Ω_{0} t)$ . Therefore, if $k \neq 0$ , then

\int_{t_{0}}^{t_{0} + T} e^{j k Ω_{0} t} d t = 0

otherwise

\int_{t_{0}}^{t_{0} + T} e^{j k Ω_{0} t} d t = \int_{t_{0}}^{t_{0} + T} d t = T

We use [link] to derive an equation for $c_{n}$ as follows. Consider the integral

\int_{t_{0}}^{t_{0} + T} x (t) e^{- j n Ω_{0} t} d t

Substituting the complex form of the Fourier Series of $x (t)$ in [link] , (using $k$ as the index of summation) we obtain

\int_{t_{0}}^{t_{0} + T} [\sum_{k = - \infty}^{\infty}, c_{n}, e^{j k Ω_{0} t}] e^{- j n Ω_{0} t} d t

Rearranging the order of integration and summation, combining the exponents, and using [link] gives

\sum_{k = - \infty}^{\infty} c_{n} \int_{t_{0}}^{t_{0} + T} e^{j (k - n) Ω_{0} t} d t = T c_{n}

Using this result, we find that

c_{n} = \frac{1}{T} \int_{t_{0}}^{t_{0} + T} x (t) e^{- j n Ω_{0} t} d t

Example 2.1 Let's now find the complex form of the Fourier Series for the signal in Example [link] . The integral to be evaluated is

c_{n} = \frac{2}{3} \int_{- 0.5}^{0.5} 2 t e^{- j \frac{4 π}{3} n t} d t

Integrating by parts yields

c_{n} = \frac{j}{n π} cos (2 π n / 3) - \frac{j 3}{{(n π)}^{2}} sin (2 π n / 3)

[link] shows the magnitude of the coefficients, $|c_{n}|$ . Note that the complex Fourier Series coefficients have even symmetry as was mentioned earlier.

Fourier Series coefficients for Example "Complex Form of the Fourier Series" .

Can the basic formula for computing the $c_{n}$ in [link] be simplified when $x (t)$ has either even, odd, or half-wave symmetry? The answer is yes. We simply use the fact that

c_{n} = \frac{a_{n} - j b_{n}}{2}

and solve for $a_{n}$ and $b_{n}$ 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 $t_{0} = T / 2$ in [link] ):

\begin{matrix} c_{n} & = & \frac{1}{T} \int_{- T / 2}^{T / 2} x (t) e^{- j n Ω_{0} t} d t \\ = & \frac{1}{T} \int_{- T / 2}^{T / 2} x (t) cos (n Ω_{0} t) d t - \frac{j}{T} \int_{- T / 2}^{T / 2} x (t) sin (n Ω_{0} t) d t \\ = & \frac{1}{2} a_{n} - \frac{j}{2} b_{n} \end{matrix}

Alternately, if $x (t)$ has half-wave symmetry, we can use [link] , [link] , and [link] to get

c_{n} = \{\begin{matrix} \frac{2}{T} \int_{- T / 4}^{T / 4} x (t) e^{- j n Ω_{0} t} d t, & n odd \\ 0, & n even \end{matrix})

Unlike the trigonometric form, we cannot simplify this further if $x (t)$ is even or odd symmetric since $e^{- j n Ω_{0} t}$ 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 [link] .

Pulse train having period $T$ used in Example "Complex Form of the Fourier Series" .

The Fourier Series coefficients for this signal are given by

\begin{matrix} c_{n} & = & \frac{1}{T} \int_{- τ / 2}^{τ / 2} e^{- j n Ω_{0} t} d t \\ = & \frac{- 1}{j n Ω_{0} T} (e^{- j n Ω_{0} τ / 2} - e^{j n Ω_{0} τ / 2}) \\ = & \frac{τ}{T} \frac{sin (n Ω_{0} τ / 2)}{n Ω_{0} τ / 2} \\ \equiv & \frac{τ}{T} sinc (n Ω_{0} τ / 2) \end{matrix}

[link] shows the magnitude of $| c_{n} |$ , the amplitude spectrum , for $T = 1$ and $τ = 1 / 2$ as well as the Fourier Series for the signal based on the first 30 coefficients

\hat{x} (t) = \sum_{n = - 30}^{30} c_{n} e^{n Ω_{0} t}

Similar plots are shown in Figures [link] , and [link] , for $T = 4$ , and $T = 8$ , respectively.

Example "Complex Form of the Fourier Series" , $T = 1, τ = 1 / 2$ : (top) Fourier Series coefficient magnitudes, (b) $\hat{x} (t)$ .

Example "Complex Form of the Fourier Series" , $T = 4, τ = 1 / 2$ : (top) Fourier Series coefficient magnitudes, (b) $\hat{x} (t)$ .

Example "Complex Form of the Fourier Series" , $T = 8, τ = 1 / 2$ : (top) Fourier Series coefficient magnitudes, (b) $\hat{x} (t)$ .

This example illustrates several important points about the Fourier Series: As the period $T$ increases, $Ω_{0}$ decreases in magnitude (this is obvious since $Ω_{0} = 2 π / T$ ). Therefore, as the period increases, successive Fourier Series coefficients represent more closely spaced frequencies. The frequencies corresponding to each $n$ are given by the following table:

$n$	$Ω$
0	0
$\pm 1$	$\pm Ω_{0}$
$\pm 2$	$\pm 2 Ω_{0}$
$⋮$	$⋮$
$\pm n$	$\pm n Ω_{0}$

This table establishes a relation between $n$ and the frequency variable $Ω$ . In particular, if $T = 1$ , we have $Ω_{0} = 2 π$ and

$n$	$Ω$
0	0
$\pm 1$	$\pm 2 π$
$\pm 2$	$\pm 4 π$
$⋮$	$⋮$
$\pm n$	$\pm 2 n π$

If $T = T$ , then $Ω_{0} = π / 2$ and

$n$	$Ω$
0	0
$\pm 1$	$\pm π / 2$
$\pm 2$	$\pm π$
$⋮$	$⋮$
$\pm n$	$\pm n π / 2$

and if $T = 8$ , we have $Ω_{0} = π / 4$ and

$n$	$Ω$
0	0
$\pm 1$	$\pm π / 4$
$\pm 2$	$\pm π / 2$
$⋮$	$⋮$
$\pm n$	$\pm n π / 4$

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

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

References

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Source: OpenStax, Signals, systems, and society. OpenStax CNX. Oct 07, 2012 Download for free at http://cnx.org/content/col10965/1.15

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