Signals and Systems

Fourier Series Representation of Continuous-Time Periodic Signals

Imron Rosyadi

2.3 Fourier Series Representation of Continuous-Time Periodic Signals

Unlocking the Frequency Domain

1. Introduction to Periodic Signals

A signal \(x(t)\) is periodic if, for some positive value of \(T\),

\[ x(t)=x(t+T) \quad \text { for all } t \tag{3.21} \]

The fundamental period \(T\) is the minimum positive, nonzero value satisfying this condition. The fundamental frequency \(\omega_{0}\) is related by \(\omega_{0} = 2\pi/T\).

Basic Periodic Signals

The building blocks of Fourier Series are complex exponentials.

Sinusoidal Signal:

\[ x(t)=\cos \omega_{0} t \tag{3.22} \]

Complex Exponential:

\[ x(t)=e^{j \omega_{0} t} \tag{3.23} \]

Both are periodic with fundamental frequency \(\omega_{0}\) and period \(T=2\pi/\omega_{0}\).

Harmonically Related Complex Exponentials:

\[ \phi_{k}(t)=e^{j k \omega_{0} t}=e^{j k(2 \pi / T) t}, \quad k=0, \pm 1, \pm 2, \ldots \tag{3.24} \] Each \(\phi_k(t)\) is periodic with period \(T\).

2. The Fourier Series Representation

A periodic signal \(x(t)\) can be represented as a linear combination of harmonically related complex exponentials:

\[ x(t)=\sum_{k=-\infty}^{+\infty} a_{k} e^{j k \omega_{0} t}=\sum_{k=-\infty}^{+\infty} a_{k} e^{j k(2 \pi / T) t} \tag{3.25} \]

  • The coefficients \(a_k\) are called Fourier Series Coefficients.
  • \(k=0\): DC component (constant term).
  • \(k=\pm 1\): Fundamental components (first harmonic).
  • \(k=\pm N\): \(N\)th harmonic components.

Example 3.2: Constructing a Signal (Part 1)

Consider a periodic signal \(x(t)\) with fundamental frequency \(2\pi\) (so \(T=1\)). It is expressed as:

\[ x(t)=\sum_{k=-3}^{+3} a_{k} e^{j k 2 \pi t} \tag{3.26} \]

With coefficients:

  • \(a_{0}=1\)
  • \(a_{1}=a_{-1}=\frac{1}{4}\)
  • \(a_{2}=a_{-2}=\frac{1}{2}\)
  • \(a_{3}=a_{-3}=\frac{1}{3}\)

Example 3.2: Constructing a Signal (Part 2)

Rewriting the sum and grouping terms:

\[ \begin{align*} x(t)= & 1+\frac{1}{4}\left(e^{j 2 \pi t}+e^{-j 2 \pi t}\right)+\frac{1}{2}\left(e^{j 4 \pi t}+e^{-j 4 \pi t}\right) \tag{3.27} \\ & +\frac{1}{3}\left(e^{j 6 \pi t}+e^{-j 6 \pi t}\right) \end{align*} \]

Using Euler’s relation (\(e^{j\theta} + e^{-j\theta} = 2\cos\theta\)), this simplifies to:

\[ x(t)=1+\frac{1}{2} \cos 2 \pi t+\cos 4 \pi t+\frac{2}{3} \cos 6 \pi t \tag{3.28} \]

Example 3.2: Interactive Signal Synthesis

See how a signal is built from its harmonics.

This interactive plot shows the individual harmonic components and their sum. Adjust the checkboxes to see how each harmonic contributes to the overall signal.

3. Properties for Real Periodic Signals

If \(x(t)\) is a real periodic signal, its Fourier series coefficients \(a_k\) have a specific property:

\[ a_{k}^{*}=a_{-k} \tag{3.29} \] This means the coefficients for negative frequencies are the complex conjugates of the coefficients for positive frequencies.

This property leads to alternative forms for the Fourier Series.

Alternative Forms for Real Signals

If \(x(t)\) is real, we can express its Fourier series in terms of real sinusoids:

Form 1 (Amplitude-Phase): If \(a_{k}=A_{k} e^{j \theta_{k}}\), then:

\[ x(t)=a_{0}+2 \sum_{k=1}^{\infty} A_{k} \cos \left(k \omega_{0} t+\theta_{k}\right) \tag{3.31} \]

Form 2 (Cosine-Sine): If \(a_{k}=B_{k}+j C_{k}\), then:

\[ x(t)=a_{0}+2 \sum_{k=1}^{\infty}\left[B_{k} \cos k \omega_{0} t-C_{k} \sin k \omega_{0} t\right] \tag{3.32} \]

Note

The complex exponential form (Eq. 3.25) is generally more convenient for analysis and manipulation in ECE, despite these real-valued alternatives.

4. Determination of Fourier Series Coefficients

The Analysis Equation

To find the coefficients \(a_k\) for a given \(x(t)\), we use the following derivation:

  1. Multiply \(x(t)\) by \(e^{-j n \omega_{0} t}\): \[ x(t) e^{-j n \omega_{0} t}=\sum_{k=-\infty}^{+\infty} a_{k} e^{j k \omega_{0} t} e^{-j n \omega_{0} t} \tag{3.33} \]

  2. Integrate both sides over one period \(T\): \[ \int_{0}^{T} x(t) e^{-j n \omega_{0} t} d t=\int_{0}^{T} \sum_{k=-\infty}^{+\infty} a_{k} e^{j(k-n) \omega_{0} t} d t \tag{3.34} \]

  3. Interchange summation and integration: \[ \int_{0}^{T} x(t) e^{-j n \omega_{0} t} d t=\sum_{k=-\infty}^{+\infty} a_{k}\left[\int_{0}^{T} e^{j(k-n) \omega_{0} t} d t\right] \]

The Orthogonality of Complex Exponentials

The integral \(\int_{0}^{T} e^{j(k-n) \omega_{0} t} d t\) evaluates as:

\[ \int_{0}^{T} e^{j(k-n) \omega_{0} t} d t=\left\{\begin{array}{ll} T, & k=n \\ 0, & k \neq n \end{array}\right. \]

This property is called orthogonality. Applying this to the summation yields \(T a_n\).

Therefore, the formula for the Fourier Series coefficients is:

\[ a_{n}=\frac{1}{T} \int_{T} x(t) e^{-j n \omega_{0} t} d t \tag{3.37} \] The integration can be over any interval of length \(T\).

The Fourier Series Pair

The Fourier Series is defined by a pair of equations:

Synthesis Equation (Time Domain to Frequency Domain):

Reconstructs \(x(t)\) from its coefficients.

\[ x(t)=\sum_{k=-\infty}^{+\infty} a_{k} e^{j k \omega_{0} t} \tag{3.38} \]

Analysis Equation (Frequency Domain to Time Domain):

Determines the coefficients \(a_k\) from \(x(t)\).

\[ a_{k}=\frac{1}{T} \int_{T} x(t) e^{-j k \omega_{0} t} d t \tag{3.39} \]

Important

These two equations are fundamental to understanding and applying Fourier Series. They represent the bridge between the time domain and the frequency domain.

Example 3.3: Fourier Series of a Simple Sinusoid

Consider the signal \(x(t)=\sin \omega_{0} t\).

We can determine its Fourier series coefficients by inspection using Euler’s formula:

\[ \sin \omega_{0} t=\frac{1}{2 j} e^{j \omega_{0} t}-\frac{1}{2 j} e^{-j \omega_{0} t} \]

Comparing this to the synthesis equation (Eq. 3.38), we find:

  • \(a_{1}=\frac{1}{2 j}\)
  • \(a_{-1}=-\frac{1}{2 j}\)
  • \(a_{k}=0\), for \(k \neq +1\) or \(-1\).

Example 3.4: More Complex Sum of Sinusoids (Part 1)

Let \(x(t)=1+\sin \omega_{0} t+2 \cos \omega_{0} t+\cos \left(2 \omega_{0} t+\frac{\pi}{4}\right)\).

Expanding into complex exponentials and collecting terms:

\[ x(t)=1+\left(1-\frac{1}{2} j\right) e^{j \omega_{0} t}+\left(1+\frac{1}{2} j\right) e^{-j \omega_{0} t}+\left(\frac{1}{2} e^{j(\pi / 4)}\right) e^{j 2 \omega_{0} t}+\left(\frac{1}{2} e^{-j(\pi / 4)}\right) e^{-j 2 \omega_{0} t} \]

The Fourier series coefficients are:

  • \(a_{0}=1\)
  • \(a_{1}=1-\frac{1}{2} j\)
  • \(a_{-1}=1+\frac{1}{2} j\)
  • \(a_{2}=\frac{1}{2} e^{j(\pi / 4)}=\frac{\sqrt{2}}{4}(1+j)\)
  • \(a_{-2}=\frac{1}{2} e^{-j(\pi / 4)}=\frac{\sqrt{2}}{4}(1-j)\)
  • \(a_{k}=0\), for \(|k|>2\).

Example 3.4: Interactive Spectrum Visualization

Magnitude and Phase Spectrum

This plot shows the magnitude and phase of the Fourier coefficients \(a_k\) for Example 3.4. Interact with the plot to examine the spectral content.

Example 3.5: Periodic Square Wave (Part 1)

The periodic square wave is a canonical signal in ECE.

Defined over one period as:

\[ x(t)= \begin{cases}1, & |t|<T_{1} \\ 0, & T_{1}<|t|<T / 2\end{cases} \tag{3.41} \]

This signal is periodic with fundamental period \(T\) and fundamental frequency \(\omega_{0}=2 \pi / T\).

Example 3.5: Periodic Square Wave (Part 2)

Calculating \(a_0\) and \(a_k\)

For \(k=0\) (DC component): \[ a_{0}=\frac{1}{T} \int_{-T_{1}}^{T_{1}} 1 \, dt=\frac{2 T_{1}}{T} \tag{3.42} \] \(a_0\) is the average value of \(x(t)\) over one period.

For \(k \neq 0\): \[ a_{k}=\frac{1}{T} \int_{-T_{1}}^{T_{1}} e^{-j k \omega_{0} t} d t = \frac{1}{T} \left[ \frac{e^{-j k \omega_{0} t}}{-j k \omega_{0}} \right]_{-T_{1}}^{T_{1}} \] \[ a_{k}=\frac{2}{k \omega_{0} T}\left[\frac{e^{j k \omega_{0} T_{1}}-e^{-j k \omega_{0} T_{1}}}{2 j}\right] \tag{3.43} \] Recognizing the term in brackets as \(\sin(k\omega_0 T_1)\), and using \(\omega_0 T = 2\pi\):

\[ a_{k}=\frac{2 \sin \left(k \omega_{0} T_{1}\right)}{k \omega_{0} T}=\frac{\sin \left(k \omega_{0} T_{1}\right)}{k \pi}, \quad k \neq 0 \tag{3.44} \]

Example 3.5: Periodic Square Wave (Part 3)

Spectrum for a 50% Duty Cycle

Consider the case where \(T=4T_1\) (a 50% duty cycle, i.e., \(x(t)=1\) for half the period). In this case, \(\omega_0 T_1 = \frac{2\pi}{T} T_1 = \frac{2\pi}{4T_1} T_1 = \frac{\pi}{2}\).

  • \(a_{0}=\frac{1}{2}\)
  • \(a_{k}=\frac{\sin (k \pi / 2)}{k \pi}, \quad k \neq 0\)

This implies:

  • \(a_k = 0\) for \(k\) even and non-zero.
  • \(a_k\) alternates sign for odd \(k\):
    • \(a_1 = a_{-1} = \frac{1}{\pi}\)
    • \(a_3 = a_{-3} = -\frac{1}{3\pi}\)
    • \(a_5 = a_{-5} = \frac{1}{5\pi}\)

Example 3.5: Interactive Square Wave Spectrum

Observe the effect of pulse width on the spectrum.

Adjust the duty cycle (ratio of pulse width \(2T_1\) to period \(T\)) to see how the frequency spectrum changes. The envelope of the spectrum is a \(\text{sinc}\) function.

Summary & Applications

Key Takeaways:

  • Periodic signals can be uniquely represented by their Fourier series coefficients.
  • The Fourier Series provides a powerful tool to move between time and frequency domains.
  • Complex exponentials are the fundamental building blocks.

Real-World Engineering Applications:

  • Audio Processing: Equalization, compression (MP3).
  • Image Processing: JPEG compression, filtering.
  • Communication Systems: Modulation, multiplexing, channel analysis.
  • Filter Design: Understanding how circuits respond to different frequencies.
  • Vibration Analysis: Identifying resonant frequencies in mechanical systems.

Tip

Practice deriving Fourier Series for different signals and interpreting their spectra. Understanding the relationship between time-domain features and frequency-domain characteristics is crucial!