# Properties of Continuous-Time Fourier Series

This is a continuation from the previous tutorial - ** Fourier series representation of continuous-time periodic signals**.

As mentioned in the *Fourier series representation of continuous-time periodic signals tutorial*, Fourier series representations possess a number of important properties that are useful for developing conceptual insights into such representations, and they can also help to reduce the complexity of the evaluation of the Fourier series of many signals.

In Table 3.1 we have summarized these properties. In later tutorials, in which we develop the Fourier transform, we will see that most of these properties can be deduced from corresponding properties of the continuous-time Fourier transform.

Consequently we limit ourselves here to the discussion of several of these properties to illustrate how they may be derived, interpreted, and used.

Throughout the following discussion of selected properties from Table 3.1, we will find it convenient to use a shorthand notation to indicate the relationship between a periodic signal and its Fourier series coefficients.

Specifically, suppose that \(x(t)\) is a periodic signal with period \(T\) and fundamental frequency \(\omega_0=2\pi/T\). Then if the Fourier series coefficients of \(x(t)\) are denoted by \(a_k\), we will use the notation

\[x(t)\stackrel{\mathcal{FS}}{\longleftrightarrow}a_k\]

to signify the pairing of a periodic signal with its Fourier series coefficients.

## 1. Linearity

Let \(x(t)\) and \(y(t)\) denote two periodic signals with period \(T\) and which have Fourier series coefficients denoted by \(a_k\) and \(b_k\), respectively. That is,

\[\begin{align}x(t)\stackrel{\mathcal{FS}}{\longleftrightarrow}a_k\\y(t)\stackrel{\mathcal{FS}}{\longleftrightarrow}b_k\end{align}\]

Since \(x(t)\) and \(y(t)\) have the same period \(T\), it easily follows that any linear combination of the two signals will also be periodic with period \(T\).

Furthermore, the Fourier series coefficients \(c_k\) of the linear combination of \(x(t)\) and \(y(t)\), \(z(t)=Ax(t)+By(t)\), are given by the same linear combination of the Fourier series coefficients for \(x(t)\) and \(y(t)\). That is,

\[\tag{3.58}z(t)=Ax(t)+By(t)\stackrel{\mathcal{FS}}{\longleftrightarrow}c_k=Aa_k+Bb_k\]

The proof of this follows directly from the application of eq. (3.39) [refer to the *Fourier series representation of continuous-time periodic signals tutorial*]. We also note that the linearity property is easily extended to a linear combination of an arbitrary number of signals with period \(T\).

## 2. Time Shifting

When a time shift is applied to a periodic signal \(x(t)\), the period \(T\) of the signal is preserved. The Fourier series coefficients \(b_k\) of the resulting signal \(y(t)=x(t-t_0)\) may be expressed as

\[\tag{3.59}b_k=\frac{1}{T}\int_Tx(t-t_0)e^{-jk\omega_0t}\text{d}t\]

Letting \(\tau=t-t_0\) in the integral, and noting that the new variable \(\tau\) will also range over an interval of duration \(T\), we obtain

\[\tag{3.60}\begin{align}\frac{1}{T}\int_Tx(\tau)e^{-jk\omega_0(\tau+t_0)}\text{d}\tau&=e^{-jk\omega_0t_0}\frac{1}{T}\int_Tx(\tau)e^{-jk\omega_0\tau}\text{d}\tau\\=e^{-jk\omega_0t_0}a_k&=e^{-jk(2\pi/T)t_0}a_k\end{align}\]

where \(a_k\) is the \(k\)th Fourier series coefficient of \(x(t)\). That is, if

\[x(t)\stackrel{\mathcal{FS}}{\longleftrightarrow}a_k\]

then

\[x(t-t_0)\stackrel{\mathcal{FS}}{\longleftrightarrow}e^{-jk\omega_0t_0}a_k=e^{-jk(2\pi/T)t_0}a_k\]

One consequence of this property is that, when a periodic signal is shifted in time, the magnitudes of its Fourier series coefficients remain unaltered. That is, \(|b_k|=|a_k|\).

## 3. Time Reversal

The period \(T\) of a periodic signal \(x(t)\) also remains unchanged when the signal undergoes time reversal.

To determine the Fourier series coefficients of \(y(t)=x(-t)\), let us consider the effect of time reversal on the synthesis equation (3.38) [refer to the *Fourier series representation of continuous-time periodic signals tutorial*]:

\[\tag{3.61}x(-t)=\sum_{k=-\infty}^{+\infty}a_ke^{-jk2\pi{t}/T}\]

Making the substitution \(k=-m\), we obtain

\[\tag{3.62}y(t)=x(-t)=\sum_{m=-\infty}^{+\infty}a_{-m}e^{jm2\pi{t}/T}\]

We observe that the right-hand side of this equation has the form of a Fourier series synthesis equation for \(x(-t)\), where the Fourier series coefficients \(b_k\) are

\[\tag{3.63}b_k=a_{-k}\]

That is, if

\[x(t)\stackrel{\mathcal{FS}}{\longleftrightarrow}a_k\]

then

\[x(-t)\stackrel{\mathcal{FS}}{\longleftrightarrow}a_{-k}\]

In other words time reversal applied to a continuous-time signal results in a time reversal of the corresponding sequence of Fourier series coefficients.

An interesting consequence of the time-reversal property is that if \(x(t)\) is even—that is, if \(x(-t)=x(t)\)—then its Fourier series coefficients are also even—i.e., \(a_{-k}=a_k\). Similarly, if \(x(t)\) is odd, so that \(x(-t)=-x(t)\), then so are its Fourier series coefficients—i.e., \(a_{-k}=-a_k\).

## 4. Time Scaling

Time scaling is an operation that in general changes the period of the underlying signal. Specifically, if \(x(t)\) is periodic with period \(T\) and fundamental frequency \(\omega_0=2\pi/T\), then \(x(\alpha{t})\), where \(\alpha\) is a positive real number, is periodic with period \(T/\alpha\) and fundamental frequency \(\alpha\omega_0\).

Since the time-scaling operation applies directly to each of the harmonic components of \(x(t)\), we may easily conclude that the Fourier coefficients for each of those components remain the same. That is, if \(x(t)\) has the Fourier series representation in eq. (3.38) [refer to the *Fourier series representation of continuous-time periodic signals tutorial*], then

\[x(\alpha{t})=\sum_{k=-\infty}^{+\infty}a_ke^{jk(\alpha\omega_0)t}\]

is the Fourier series representation of \(x(\alpha{t})\).

We emphasize that, while the Fourier coefficients have not changed, the Fourier series representation ** has** changed because of the change in the fundamental frequency.

## 5. Multiplication

Suppose that \(x(t)\) and \(y(t)\) are both periodic with period \(T\) and that

\[\begin{align}x(t)\stackrel{\mathcal{FS}}{\longleftrightarrow}a_k\\y(t)\stackrel{\mathcal{FS}}{\longleftrightarrow}b_k\end{align}\]

Since the product \(x(t)y(t)\) is also periodic with period \(T\), we can expand it in a Fourier series with Fourier series coefficients \(h_k\) expressed in terms of those for \(x(t)\) and \(y(t)\). The result is

\[\tag{3.64}x(t)y(t)\stackrel{\mathcal{FS}}{\longleftrightarrow}h_k=\sum_{l=-\infty}^{\infty}a_lb_{k-l}\]

One way to derive this relationship is to multiply the Fourier series representations of \(x(t)\) and \(y(t)\) and to note that the \(k\)th harmonic component in the product will have a coefficient which is the sum of terms of the form \(a_lb_{k-l}\).

Observe that the sum on the right-hand side of eq. (3.64) may be interpreted as the discrete-time convolution of the sequence representing the Fourier coefficients of \(x(t)\) and the sequence representing the Fourier coefficients of \(y(t)\).

## 6. Conjugation and Conjugate Symmetry

Taking the complex conjugate of a periodic signal \(x(t)\) has the effect of complex conjugation and time reversal on the corresponding Fourier series coefficients.

That is, if

\[x(t)\stackrel{\mathcal{FS}}{\longleftrightarrow}a_k\]

then

\[\tag{3.65}x^*(t)\stackrel{\mathcal{FS}}{\longleftrightarrow}a^*_{-k}\]

This property is easily proved by applying complex conjugation to both sides of eq. (3.38) [refer to the *Fourier series representation of continuous-time periodic signals tutorial*] and replacing the summation variable \(k\) by its negative.

Some interesting consequences of this property may be derived for \(x(t)\) real—that is, when \(x(t)=x^*(t)\). In particular, in this case, we see from eq. (3.65) that the Fourier series coefficients will be ** conjugate symmetric**, i.e.,

\[\tag{3.66}a_{-k}=a^*_k\]

as we previously saw in eq. (3.29) [refer to the *Fourier series representation of continuous-time periodic signals tutorial*].

This in turn implies various symmetry properties (listed in Table 3.1) for the magnitudes, phases, real parts, and imaginary parts of the Fourier series coefficients of real signals.

For example, from eq. (3.66), we see that if \(x(t)\) is real, then \(a_0\) is real and

\[|a_k|=|a_{-k}|\]

Also, if \(x(t)\) is real and even, then, from the ** Time Reversal** property above, \(a_k=a_{-k}\). However, from eq. (3.66) we see that \(a^*_k=a_{-k}\), so that \(a_k=a^*_k\). That is, if \(x(t)\) is real and even, then so are its Fourier series coefficients.

Similarly, it can be shown that if \(x(t)\) is real and odd, then its Fourier series coefficients are purely imaginary and odd. Thus, for example, \(a_0=0\) if \(x(t)\) is real and odd.

## 7. Parseval's Relation for Continuous-Time Periodic Signals

Parseval's relation for continuous-time periodic signals is

\[\tag{3.67}\frac{1}{T}\int_T|x(t)|^2\text{d}t=\sum_{k=-\infty}^{+\infty}|a_k|^2\]

where the \(a_k\) are the Fourier series coefficients of \(x(t)\) and \(T\) is the period of the signal.

Note that the left-hand side of eq. (3.67) is the average power (i.e., energy per unit time) in one period of the periodic signal \(x(t)\). Also,

\[\tag{3.68}\frac{1}{T}\int_T\left|a_ke^{jk\omega_0t}\right|^2\text{d}t=\frac{1}{T}\int_T|a_k|^2\text{d}t=|a_k|^2\]

so that \(|a_k|^2\) is the average power in the \(k\)th harmonic component of \(x(t)\). Thus, what Parseval's relation states is that the total average power in a periodic signal equals the sum of the average powers in all of its harmonic components.

## 8. Examples

Fourier series properties, such as those listed in Table 3.1, may be used to circumvent some of the algebra involved in determining the Fourier coefficients of a given signal. In the next three examples, we illustrate this.

The last example in this tutorial then demonstrates how properties of a signal can be used to characterize the signal in great detail.

**Example 3.6**

Consider the signal \(g(t)\) with a fundamental period of 4, shown in Figure 3.10.

We could determine the Fourier series representation of \(g(t)\) directly from the analysis equation (3.39) [refer to the *Fourier series representation of continuous-time periodic signals tutorial*].

Instead, we will use the relationship of \(g(t)\) to the symmetric periodic square wave \(x(t)\) in Example 3.5 [refer to the *Fourier series representation of continuous-time periodic signals tutorial*].

Referring to that example, we see that, with \(T=4\) and \(T_1=1\),

\[\tag{3.69}g(t)=x(t-1)-1/2\]

The time-shift property in Table 3.1 indicates that, if the Fourier Series coefficients of \(x(t)\) are denoted by \(a_k\), the Fourier coefficients of \(x(t-1)\) may be expressed as

\[\tag{3.70}b_k=a_ke^{-jk\pi/2}\]

The Fourier coefficients of the ** dc offset** in \(g(t)\)—i.e., the term \(-1/2\) on the right-hand side of eq. (3.69)—are given by

\[\tag{3.71}c_k=\begin{cases}0,\qquad\quad\text{for }k\ne0\\-\frac{1}{2},\qquad\text{for }k=0\end{cases}\]

Applying the linearity property in Table 3.1, we conclude that the coefficients for \(g(t)\) may be expressed as

\[d_k=\begin{cases}a_ke^{-jk\pi/2},\qquad\text{for }k\ne0\\a_0-\frac{1}{2},\quad\qquad\text{for }k=0\end{cases}\]

where each \(a_k\) may now be replaced by the corresponding expression from eqs. (3.45) and (3.46) [refer to the *Fourier series representation of continuous-time periodic signals tutorial*], yielding

\[\tag{3.72}d_k=\begin{cases}\frac{\sin(\pi{k}/2)}{k\pi}e^{-jk\pi/2},\qquad\text{for }k\ne0\\0,\quad\quad\qquad\qquad\qquad\text{for }k=0\end{cases}\]

**Example 3.7**

Consider the triangular wave signal \(x(t)\) with period \(T=4\) and fundamental frequency \(\omega_0=\pi/2\) shown in Figure 3.11.

The derivative of this signal is the signal \(g(t)\) in Example 3.6. Denoting the Fourier coefficients of \(g(t)\) by \(d_k\) and those of \(x(t)\) by \(e_k\), we see that the differentiation property in Table 3.1 indicates that

\[\tag{3.73}d_k=jk(\pi/2)e_k\]

This equation can be used to express \(e_k\) in terms of \(d_k\) except when \(k=0\). Specifically, from eq. (3.72),

\[\tag{3.74}e_k=\frac{2d_k}{jk\pi}=\frac{2\sin(\pi{k}/2)}{j(k\pi)^2}e^{-jk\pi/2},\qquad{k\ne0}\]

For \(k=0\), \(e_0\) can be determined by finding the area under one period of \(x(t)\) and dividing by the length of the period:

\[e_0=\frac{1}{2}\]

**Example 3.8**

Let us examine some properties of the Fourier series representation of a periodic train of impulses, or ** impulse train**. This signal and its representation in terms of complex exponentials will play an important role when we discuss the topic of sampling in later tutorials.

The impulse train with period \(T\) may be expressed as

\[\tag{3.75}x(t)=\sum_{k=-\infty}^{\infty}\delta(t-kT)\]

it is illustrated in Figure 3.12(a).

To determine the Fourier series coefficients \(a_k\), we use eq. (3.39) [refer to the *Fourier series representation of continuous-time periodic signals tutorial*] and select the interval of integration to be \(-T/2\le{t}\le{T/2}\), avoiding the placement of impulses at the integration limits.

Within this interval, \(x(t)\) is the same as \(\delta(t)\), and it follows that

\[\tag{3.76}a_k=\frac{1}{T}\int_{-T/2}^{T/2}\delta(t)e^{-jk2\pi{t}/T}\text{d}t=\frac{1}{T}\]

In other words, all the Fourier series coefficients of the impulse train are identical. These coefficients are also real valued and even (with respect to the index \(k\)). This is to be expected, since, according to Table 3.1, any real and even signal (such as our impulse train) should have real and even Fourier coefficients.

The impulse train also has a straightforward relationship to square-wave signals such as \(g(t)\) in Figure 3.6 [refer to the *Fourier series representation of continuous-time periodic signals tutorial*], which we repeat in Figure 3.12(b).

The derivative of \(g(t)\) is the signal \(q(t)\) illustrated in Figure 3.12(c). We may interpret \(q(t)\) as the difference of two shifted versions of the impulse train \(x(t)\). That is,

\[\tag{3.77}q(t)=x(t+T_1)-x(t-T_1)\]

Using the properties of Fourier series, we can now compute the Fourier series coefficients of \(q(t)\) and \(g(t)\) without any further direct evaluation of the Fourier series analysis equation.

First, from the time-shifting and linearity properties, we see from eq. (3.77) that the Fourier series coefficients \(b_k\) of \(q(t)\) may be expressed in terms of the Fourier series coefficients \(a_k\) of \(x(t)\); that is,

\[b_k=e^{jk\omega_0T_1}a_k-e^{-jk\omega_0T_1}a_k\]

where \(\omega_0=2\pi/T\).

Using eq. (3.76), we then have

\[b_k=\frac{1}{T}[e^{jk\omega_0T_1}-e^{-jk\omega_0T_1}]=\frac{2j\sin(k\omega_0T_1)}{T}\]

Finally, since \(q(t)\) is the derivative of \(g(t)\), we can use the differentiation property in Table 3.1 to write

\[\tag{3.78}b_k=jk\omega_0c_k\]

where the \(c_k\) are the Fourier series coefficients of \(g(t)\). Thus

\[\tag{3.79}c_k=\frac{b_k}{jk\omega_0}=\frac{2j\sin(k\omega_0T_1)}{jk\omega_0T}=\frac{\sin(k\omega_0T_1)}{k\pi},\qquad{k\ne0}\]

where we have used the fact that \(\omega_0T=2\pi\).

Note that eq. (3.79) is valid for \(k\ne0\), since we cannot solve for \(c_0\) from eq. (3.78) with \(k=0\). However, since \(c_0\) is just the average value of \(g(t)\) over one period, we can determine it by inspection from Figure 3.12(b):

\[\tag{3.80}c_0=\frac{2T_1}{T}\]

Eqs. (3.80) and (3.79) are identical to eqs. (3.42) and (3.44) [refer to the *Fourier series representation of continuous-time periodic signals tutorial*], respectively, for the Fourier series coefficients of the square wave derived in Example 3.5 [refer to the *Fourier series representation of continuous-time periodic signals tutorial*].

The next example is chosen to illustrate the use of many of the properties in Table 3.1.

**Example 3.9**

Suppose we are given the following facts about a signal \(x(t)\):

- \(x(t)\) is a real signal.
- \(x(t)\) is periodic with period \(T=4\), and it has Fourier series coefficients \(a_k\).
- \(a_k=0\) for \(|k|\gt1\).
- The signal with Fourier coefficients \(b_k=e^{-j\pi{k}/2}a_{-k}\) is odd.
- \(\frac{1}{4}\int_4|x(t)|^2\text{d}t=1/2\).

Let us show that this information is sufficient to determine the signal \(x(t)\) to within a sign factor.

According to Fact 3, \(x(t)\) has at most three nonzero Fourier series coefficients \(a_k:a_0,a_1,\) and \(a_{-1}\). Then, since \(x(t)\) has fundamental frequency \(\omega_0=2\pi/4=\pi/2\), it follows that

\[x(t)=a_0+a_1e^{j\pi{t}/2}+a_{-1}e^{-j\pi{t}/2}\]

Since \(x(t)\) is real (Fact 1), we can use the symmetry properties in Table 3.1 to conclude that \(a_0\) is real and \(a_1=a^*_{-1}\). Consequently,

\[\tag{3.81}x(t)=a_0+a_1e^{j\pi{t}/2}+(a_1e^{j\pi{t}/2})^*=a_0+2\mathcal{Re}\{a_1e^{j\pi{t}/2}\}\]

Let us now determine the signal corresponding to the Fourier coefficients \(b_k\) given in Fact 4.

Using the time-reversal property from Table 3.1, we note that \(a_{-k}\) corresponds to the signal \(x(-t)\).

Also, the time-shift property in the table indicates that multiplication of the \(k\)th Fourier coefficient by \(e^{-jk\pi/2}=e^{-jk\omega_0}\) corresponds to the underlying signal being shifted by 1 to the right (i.e., having \(t\) replaced by \(t-1\)).

We conclude that the coefficients \(b_k\) correspond to the signal \(x(-(t-1))=x(-t+1)\), which, according to Fact 4, must be odd.

Since \(x(t)\) is real, \(x(-t+1)\) must also be real. From Table 3.1, it then follows that the Fourier coefficients of \(x(-t+1)\) must be purely imaginary and odd. Thus, \(b_0=0\) and \(b_{-1}=-b_1\).

Since time-reversal and time-shift operations cannot change the average power per period, Fact 5 holds even if \(x(t)\) is replaced by \(x(-t+1)\). That is,

\[\tag{3.82}\frac{1}{4}\int_4|x(-t+1)|^2\text{d}t=1/2\]

We can now use Parseval's relation to conclude that

\[\tag{3.83}|b_1|^2+|b_{-1}|^2=1/2\]

Substituting \(b_1=-b_{-1}\) in this equation, we obtain \(|b_1|=1/2\). Since \(b_1\) is also known to be purely imaginary, it must be either \(j/2\) or \(-j/2\).

Now we can translate these conditions on \(b_0\) and \(b_1\) into equivalent statements on \(a_0\) and \(a_1\).

First, since \(b_0=0\), Fact 4 implies that \(a_0=0\).

With \(k=-1\), this condition implies that \(a_1=e^{-j\pi/2}b_{-1}=-jb_{-1}=jb_1\). Thus, if we take \(b_1=j/2\), then \(a_1=-1/2\), and therefore, from eq. (3.81), \(x(t)=-\cos(\pi{t}/2)\). Alternatively, if we take \(b_1=-j/2\), then \(a_1=1/2\), and therefore, \(x(t)=\cos(\pi{t}/2)\).

The next tutorial discusses about ** Fourier series representation of discrete-time periodic signals**.