Fourier Series Representation of Continuous-Time Periodic Signals
Unlocking the Frequency Domain
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 \(\omega_{0} = 2\pi/T\).
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 \(T\).
A periodic signal \(x(t)\) can be represented as a linear combination of harmonically related complex exponentials:
\[ x(t)=\sum_{k=-\infty}^{+\infty} a_{k} e^{j k \omega_{0} t}=\sum_{k=-\infty}^{+\infty} a_{k} e^{j k(2 \pi / T) t} \tag{3.25} \]
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:
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:
\[ x(t)=1+\frac{1}{2} \cos 2 \pi t+\cos 4 \pi t+\frac{2}{3} \cos 6 \pi t \tag{3.28} \]
This interactive plot shows the individual harmonic components and their sum. Adjust the checkboxes to see how each harmonic contributes to the overall signal.
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.
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.
To find the coefficients \(a_k\) for a given \(x(t)\), we use the following derivation:
Multiply \(x(t)\) by \(e^{-j n \omega_{0} t}\): \[ x(t) e^{-j n \omega_{0} t}=\sum_{k=-\infty}^{+\infty} a_{k} e^{j k \omega_{0} t} e^{-j n \omega_{0} t} \tag{3.33} \]
Integrate both sides over one period \(T\): \[ \int_{0}^{T} x(t) e^{-j n \omega_{0} t} d t=\int_{0}^{T} \sum_{k=-\infty}^{+\infty} a_{k} e^{j(k-n) \omega_{0} t} d t \tag{3.34} \]
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 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 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.
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:
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:
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.
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 \(\omega_{0}=2 \pi / T\).
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 \(\omega_0 T = 2\pi\):
\[ a_{k}=\frac{2 \sin \left(k \omega_{0} T_{1}\right)}{k \omega_{0} T}=\frac{\sin \left(k \omega_{0} T_{1}\right)}{k \pi}, \quad k \neq 0 \tag{3.44} \]
Consider the case where \(T=4T_1\) (a 50% duty cycle, i.e., \(x(t)=1\) for half the period). In this case, \(\omega_0 T_1 = \frac{2\pi}{T} T_1 = \frac{2\pi}{4T_1} T_1 = \frac{\pi}{2}\).
This implies:
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.
Key Takeaways:
Real-World Engineering Applications:
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!