Signals and Systems

3.6 Fourier Series Representation of Discrete-Time Periodic Signals

Imron Rosyadi

Fourier Series Representation of Discrete-Time Periodic Signals

Welcome to today’s lecture on Discrete-Time Fourier Series. This section builds upon our understanding of continuous-time Fourier series, but introduces some crucial differences unique to discrete-time signals. We’ll cover the fundamental concepts, derive the key equations, and explore several examples with interactive demonstrations.

Introduction: Discrete-Time Periodic Signals

Discrete-time (DT) signals \(x[n]\) are periodic with period \(N\) if \(x[n] = x[n+N]\).

The fundamental period \(N\) is the smallest positive integer for which this holds.

The fundamental frequency is \(\omega_0 = 2\pi/N\).

Note

Key Difference from Continuous-Time (CT): The Fourier series representation for discrete-time signals is a finite series, unlike the infinite series required for continuous-time signals. This simplifies convergence issues significantly.

Today, we’re diving into the discrete-time Fourier series. Recall from Chapter 1 that a discrete-time signal is periodic if it repeats itself after a certain number of samples, denoted by \(N\). The fundamental frequency \(\omega_0\) is directly related to this period \(N\). A crucial point to remember from the outset is that the DTFS is a finite sum, which has profound implications, especially regarding convergence, as we’ll see later.

Harmonically Related Complex Exponentials

The set of all discrete-time complex exponential signals periodic with period \(N\) is given by:

\[ \phi_k[n] = e^{j k \omega_0 n} = e^{j k(2\pi/N)n}, \quad k = 0, \pm 1, \pm 2, \ldots \]

There are only \(N\) distinct signals in this set. This is because discrete-time complex exponentials differing by a multiple of \(2\pi\) in frequency are identical.

\[ \phi_k[n] = \phi_{k+rN}[n] \]

This means \(e^{j k(2\pi/N)n}\) and \(e^{j (k+N)(2\pi/N)n}\) are the same sequence.

Just like in continuous time, we build our periodic signals from harmonically related complex exponentials. However, discrete-time is special. While \(k\) can theoretically go from negative to positive infinity, only \(N\) of these complex exponentials are unique. After \(N\) terms, the sequence of complex exponentials repeats itself. This is a direct consequence of the periodic nature of discrete-time exponentials. Let’s see an example.

Demonstrating Distinct Exponentials

Let’s visualize how \(\phi_k[n]\) repeats for \(k\) and \(k+N\). Consider \(N=5\). We expect \(\phi_0[n] = \phi_5[n]\), \(\phi_1[n] = \phi_6[n]\), etc.

Demonstrating Distinct Exponentials (cont.)

The plots show the real and imaginary parts of \(\phi_k[n]\) for \(k=1\) and \(k=1+N\) (with \(N=5\)). Observe that the two plots are identical, confirming that \(\phi_k[n] = \phi_{k+N}[n]\).

Tip

Interactive Element: Feel free to change \(N_{val}\) and \(k\) (by modifying the get_phi_k calls) in the Python code to explore this property further! For example, try \(k=0\) and \(k=N_{val}\).

Here, we use Python to plot two complex exponentials. The top plot shows \(\phi_1[n]\) for \(N=5\). The bottom plot shows \(\phi_{1+5}[n]\), which is \(\phi_6[n]\). As you can clearly see, both plots produce the exact same sequence of real and imaginary values. This visually demonstrates the periodicity of the discrete-time complex exponentials in the index \(k\). Try changing the value of N_val or the k value in the get_phi_k calls in the code block to see how this relationship holds. For instance, set k_val=0 and observe k_val=N_val for any N_val.

Discrete-Time Fourier Series (DTFS) Synthesis

A periodic discrete-time signal \(x[n]\) can be represented as a linear combination of these harmonically related complex exponentials.

\[ x[n] = \sum_{k=\langle N\rangle} a_k \phi_k[n] = \sum_{k=\langle N\rangle} a_k e^{j k \omega_0 n} = \sum_{k=\langle N\rangle} a_k e^{j k(2\pi/N)n} \]

The notation \(\sum_{k=\langle N\rangle}\) indicates summation over any set of \(N\) successive integers for \(k\). Common choices include \(k=0, 1, \ldots, N-1\) or \(k=- (N-1)/2, \ldots, (N-1)/2\) (for odd \(N\)).

The coefficients \(a_k\) are called the Fourier series coefficients.

Now that we understand the building blocks, we can construct any periodic discrete-time signal using a sum of these exponentials. This equation is the synthesis equation. Notice the summation is over \(N\) terms. This is because, as we just saw, there are only \(N\) distinct complex exponentials. The specific range for \(k\) doesn’t change the set of exponentials used, only which \(N\) coefficients we’re solving for.

Determination of DTFS Coefficients (1/2)

Given \(x[n]\) periodic with fundamental period \(N\), we want to find \(a_k\).

We can solve a system of \(N\) linear equations, but a more direct method exists.

The key identity (similar to orthogonality in CT):

\[ \sum_{n=\langle N\rangle} e^{j k(2\pi/N)n} = \begin{cases} N, & k = 0, \pm N, \pm 2N, \ldots \\ 0, & \text{otherwise} \end{cases} \]

This identity is crucial for isolating each \(a_k\).

How do we find these \(a_k\) coefficients? Similar to continuous time, we use an orthogonality property. This specific identity states that if you sum a complex exponential over one period, the result is zero unless the exponential is a constant (i.e., its frequency is a multiple of \(2\pi\)). When it’s a constant, the sum is simply \(N\) times the constant. Let’s verify this interactively.

Interactive Identity Verification

Let’s verify the summation identity: \(\sum_{n=\langle N\rangle} e^{j k(2\pi/N)n}\).

Tip

Experiment:

Change N_period and k_val in the code above and run it. Observe the sum. What happens when k_val is a multiple of N_period? What happens otherwise?

This interactive code block allows you to directly test the identity. When k_val is 0, the exponential is \(e^0 = 1\), and summing \(N\) ones gives \(N\). When k_val is a multiple of \(N\), for example \(k=N\), the exponential is \(e^{jN(2\pi/N)n} = e^{j2\pi n} = 1\) for all integer \(n\), so the sum is again \(N\). For any other \(k\), the terms \(e^{j k(2\pi/N)n}\) will trace a full circle (or multiple full circles) in the complex plane, summing to zero over one period. This is a fundamental property.

Determination of DTFS Coefficients (2/2)

To derive \(a_k\):

1. Multiply the synthesis equation by \(e^{-j r(2\pi/N)n}\).
2. Sum over one period \(n=\langle N\rangle\).

\[ \sum_{n=\langle N\rangle} x[n] e^{-j r(2\pi/N)n} = \sum_{n=\langle N\rangle} \sum_{k=\langle N\rangle} a_k e^{j(k-r)(2\pi/N)n} \]

Interchanging summation order and applying the identity:

\[ \sum_{n=\langle N\rangle} x[n] e^{-j r(2\pi/N)n} = \sum_{k=\langle N\rangle} a_k \left( \sum_{n=\langle N\rangle} e^{j(k-r)(2\pi/N)n} \right) \]

The inner sum is \(N\) if \(k-r\) is a multiple of \(N\) (i.e., \(k=r\) within the \(\langle N \rangle\) range), and \(0\) otherwise. This simplifies to \(N a_r\), leading to:

\[ a_k = \frac{1}{N} \sum_{n=\langle N\rangle} x[n] e^{-j k(2\pi/N)n} \]

This is the Discrete-Time Fourier Series Analysis Equation.

With that identity established, the derivation of the analysis equation is straightforward. We multiply the synthesis equation by a specific complex exponential, \(e^{-j r(2\pi/N)n}\), and sum over one period. This process effectively “filters out” all terms except for the one corresponding to \(a_r\). The result is a direct formula to compute each \(a_k\) from the signal \(x[n]\).

DTFS Synthesis and Analysis Pair

These two equations form the Discrete-Time Fourier Series pair:

Synthesis Equation:

(How to build \(x[n]\) from coefficients \(a_k\))

\[ x[n] = \sum_{k=\langle N\rangle} a_k e^{j k \omega_0 n} = \sum_{k=\langle N\rangle} a_k e^{j k(2\pi/N)n} \]

Analysis Equation:

(How to find coefficients \(a_k\) from \(x[n]\))

\[ a_k = \frac{1}{N} \sum_{n=\langle N\rangle} x[n] e^{-j k \omega_0 n} = \frac{1}{N} \sum_{n=\langle N\rangle} x[n] e^{-j k(2\pi/N)n} \]

Important

The Fourier series coefficients \(a_k\) are periodic with period \(N\): \(a_k = a_{k+N}\). This is a direct consequence of the periodicity of \(\phi_k[n]\).

Here we have the complete Discrete-Time Fourier Series pair. The synthesis equation tells us how to construct the signal from its frequency components, and the analysis equation tells us how to decompose the signal into those components. Crucially, remember the coefficients \(a_k\) themselves are periodic with period \(N\). This means if you calculate \(a_0, a_1, \ldots, a_{N-1}\), you automatically know \(a_N, a_{N+1}\), etc., because they simply repeat.

Example 3.10: Discrete-Time Sine Wave

Consider the signal \(x[n] = \sin(\omega_0 n)\). This signal is periodic only if \(2\pi/\omega_0\) is an integer (or ratio of integers). If \(\omega_0 = 2\pi/N\), then \(x[n]\) is periodic with fundamental period \(N\).

We can expand \(x[n]\) using Euler’s formula:

\[ x[n] = \frac{1}{2j} e^{j(2\pi/N)n} - \frac{1}{2j} e^{-j(2\pi/N)n} \]

By comparing this with the synthesis equation \(x[n] = \sum_{k=\langle N\rangle} a_k e^{j k(2\pi/N)n}\), we can identify the coefficients:

\[ a_1 = \frac{1}{2j}, \quad a_{-1} = -\frac{1}{2j} \]

All other \(a_k\) (for \(k\) within one period) are zero.

Let’s apply our new tools to an example. A discrete-time sine wave is only periodic under specific conditions related to its frequency. If \(\omega_0\) is a multiple of \(2\pi/N\), it’s periodic. By expressing the sine wave as a sum of two complex exponentials, we can directly compare it to the Fourier series synthesis equation and pick out the coefficients by inspection. This is a very common technique when the signal is already in the form of complex exponentials.

Example 3.10: Sine Wave DTFS Visualization

If \(\omega_0 = (2\pi M)/N\) (where \(M, N\) are coprime), \(x[n]\) has fundamental period \(N\).

Then, \(a_M = \frac{1}{2j}\), \(a_{-M} = -\frac{1}{2j}\), and other \(a_k=0\) within one period.

viewof N_val = Inputs.range([5, 20], {
  label: "N:",
  step: 1,
  value: 5
})

viewof M_val = Inputs.range([1, 4], {
  label: "M:",
  step: 1,
  value: 1
})

This interactive plot visualizes the Fourier series coefficients for a discrete-time sine wave. You can adjust \(N\) (the period) and \(M\) (the multiplier for the fundamental frequency) using the sliders. Observe how only two coefficients are non-zero within any given period of \(N\). When \(M=1\), you see coefficients at \(k=1\) and \(k=N-1\) (which is equivalent to \(k=-1\)). When \(M\) changes, the locations of these non-zero coefficients shift to \(k=M\) and \(k=N-M\). Notice the magnitudes are always 0.5, and the phases are \(\pm 90^\circ\) (or \(\pm \pi/2\) radians), which corresponds to \(\pm j/2\). This clearly illustrates the spectral content of a discrete-time sine wave.

Example 3.11: More Complex Signal

Consider \(x[n]=1+\sin\left(\frac{2\pi}{N}\right)n+3\cos\left(\frac{2\pi}{N}\right)n+\cos\left(\frac{4\pi}{N}n+\frac{\pi}{2}\right)\).

This signal is periodic with period \(N\).

We expand each term into complex exponentials:

\(1 \implies a_0 = 1\)

\(\sin\left(\frac{2\pi}{N}\right)n \implies a_1 = \frac{1}{2j}, a_{-1} = -\frac{1}{2j}\)

\(3\cos\left(\frac{2\pi}{N}\right)n \implies a_1 = \frac{3}{2}, a_{-1} = \frac{3}{2}\)

\(\cos\left(\frac{4\pi}{N}n+\frac{\pi}{2}\right) \implies \frac{1}{2}e^{j\pi/2}e^{j2(2\pi/N)n} + \frac{1}{2}e^{-j\pi/2}e^{-j2(2\pi/N)n}\)

\(\implies a_2 = \frac{1}{2}e^{j\pi/2} = \frac{1}{2}j, a_{-2} = \frac{1}{2}e^{-j\pi/2} = -\frac{1}{2}j\)

Example 3.11: More Complex Signal

Combining terms for each \(k\):

\(a_0 = 1\)
\(a_1 = \frac{1}{2j} + \frac{3}{2} = \frac{3}{2} - \frac{1}{2}j\)
\(a_{-1} = -\frac{1}{2j} + \frac{3}{2} = \frac{3}{2} + \frac{1}{2}j\)
\(a_2 = \frac{1}{2}j\)
\(a_{-2} = -\frac{1}{2}j\)

All other \(a_k=0\) within the period.

Tip

Property for Real Signals: For a real signal \(x[n]\), the Fourier coefficients exhibit conjugate symmetry: \(a_{-k} = a_k^*\). Verify this for the calculated coefficients!

This example shows how to find Fourier coefficients for a more complex signal by breaking it down into its constituent complex exponentials. Each sinusoidal component contributes two complex exponential terms, and the constant term contributes to \(a_0\). We then sum the coefficients for each harmonic index \(k\). Notice that \(a_{-1}\) is the complex conjugate of \(a_1\), and \(a_{-2}\) is the complex conjugate of \(a_2\). This is a general property for real-valued signals, known as conjugate symmetry.

Example 3.11: Coefficients Visualization

Visualizing the real, imaginary, magnitude, and phase of the coefficients for \(N=10\).

Here we see a visual representation of the coefficients calculated in the previous slide. The top-left plot shows the real parts, and the bottom-left shows the imaginary parts. Notice the symmetry: Real parts are even (\(Re\{a_k\} = Re\{a_{-k}\}\)), and imaginary parts are odd (\(Im\{a_k\} = -Im\{a_{-k}\}\)). On the right, we have the magnitude and phase. Magnitudes are even (\(|a_k| = |a_{-k}|\)), and phases are odd (\(Arg\{a_k\} = -Arg\{a_{-k}\}\)). This visual confirms the conjugate symmetry property for real signals: \(a_{-k} = a_k^*\).

Example 3.12: Discrete-Time Periodic Square Wave

Consider a discrete-time periodic square wave \(x[n]\) with period \(N\).

It is defined as \(x[n]=1\) for \(-N_1 \leq n \leq N_1\), and \(x[n]=0\) otherwise within one period.

The analysis equation is: \[ a_k = \frac{1}{N} \sum_{n=\langle N\rangle} x[n] e^{-j k(2\pi/N)n} \]

Choosing the summation range from \(-N_1\) to \(N_1\):

\[ a_k = \frac{1}{N} \sum_{n=-N_1}^{N_1} e^{-j k(2\pi/N)n} \]

This sum is a geometric series.

Let’s analyze a common signal: the discrete-time periodic square wave. This signal is 1 for a certain duration around \(n=0\) and 0 otherwise within its period. To find its Fourier coefficients, we plug its definition into the analysis equation. The summation simplifies because \(x[n]\) is only non-zero for a specific range. The sum then becomes a geometric series, which we can solve using a known formula.

Example 3.12: Square Wave Coefficients

Applying the geometric series sum formula and simplifying, we get:

\[ a_k = \frac{1}{N} \frac{\sin\left[2\pi k(N_1+1/2)/N\right]}{\sin(\pi k/N)}, \quad k \neq 0, \pm N, \pm 2N, \ldots \]

And for \(k=0, \pm N, \pm 2N, \ldots\):

\[ a_k = \frac{2N_1+1}{N} \]

This formula resembles the continuous-time sinc function, but uses \(\sin(\cdot)/\sin(\cdot)\) due to discrete nature.

After some algebraic manipulation, which involves using the geometric series sum formula and Euler’s identity, we arrive at this closed-form expression for the Fourier coefficients. It looks similar to the sinc function we encountered in continuous-time, but it’s a ratio of sines. This is typical for discrete-time Fourier transforms and series, as the discrete nature often leads to these periodic sinc-like functions. Note the special case for \(k=0\) (and its multiples), where the denominator would be zero if not handled separately. At \(k=0\), \(a_0\) represents the average value of the signal over one period, which is simply the number of non-zero samples (\(2N_1+1\)) divided by the total period (\(N\)).

Example 3.12: Coefficients Visualization

Let’s visualize the coefficients for \(2N_1+1=5\) and varying \(N\).

viewof N_val_sq = Inputs.range([10, 40], {
  label: "N (Period):",
  step: 5,
  value: 10
})
viewof N1_val_sq = Inputs.range([1, 5], {
  label: "N1 (Pulse Half-Width):",
  step: 1,
  value: 2
})

This interactive plot lets you see how the Fourier coefficients of a discrete-time square wave change with the period \(N\) and the pulse width \(N_1\). The shape of the coefficients resembles a sampled sinc function. As \(N\) increases (making the signal sparser in time), the coefficients become more closely spaced in frequency. The main lobe width is inversely proportional to the pulse width. Experiment with the sliders to see these effects. For instance, notice how the \(a_k\) values are scaled by \(1/N\).

Convergence of Discrete-Time Fourier Series

Unlike continuous-time Fourier series, discrete-time Fourier series have no convergence issues.

The DTFS is a finite sum of \(N\) terms.

A discrete-time periodic sequence \(x[n]\) is completely specified by its \(N\) values over one period.

The DTFS analysis equation transforms these \(N\) values into \(N\) Fourier coefficients.

The DTFS synthesis equation perfectly reconstructs the original \(N\) values from these \(N\) coefficients.

Important

There is no Gibbs phenomenon in discrete-time Fourier series. The partial sum (if it includes all \(N\) distinct terms) will exactly equal \(x[n]\).

This is a critical distinction between continuous-time and discrete-time Fourier series. Because a discrete-time signal is defined by a finite number of samples within a period, its Fourier series also has a finite number of terms. This means that if you sum up all \(N\) terms of the DTFS, you will exactly reconstruct the original signal. There are no approximation errors, no ripples at discontinuities, and therefore, no Gibbs phenomenon. This makes DTFS mathematically much simpler in terms of convergence.

DTFS Reconstruction: No Gibbs Phenomenon

Let’s reconstruct the square wave using partial sums.

\(\hat{x}[n] = \sum_{k=-M}^{M} a_k e^{j k(2\pi/N)n}\) (for odd \(N\), \(M=(N-1)/2\) includes all terms).

viewof N_rec = Inputs.range([5, 15], {
  label: "N (Period, odd):",
  step: 2,
  value: 9
})

viewof N1_rec = Inputs.range([1, 3], {
  label: "N1 (Pulse Half-Width):",
  step: 1,
  value: 2
}) // Note: 2N1+1 should be <= N

viewof M_rec = Inputs.range([0, 4], {
  label: "M (Summation Terms):",
  step: 1,
  value: 0
})

This interactive demonstration highlights the key difference in convergence. We plot the original square wave and its reconstruction \(\hat{x}[n]\) using a partial sum of \(2M+1\) terms. Adjust the M slider. Notice that as M increases, the reconstructed signal gets closer to the original. When M reaches its maximum value, which corresponds to including all \(N\) unique terms (e.g., \(M=(N-1)/2\) for odd \(N\)), the reconstructed signal perfectly matches the original signal. There are no overshoots or undershoots, no Gibbs phenomenon, because the series is finite and exact. This is a powerful advantage of the discrete-time Fourier series.

Summary and Key Takeaways

Discrete-Time Fourier Series (DTFS):

• Represents periodic discrete-time signals \(x[n]\) with period \(N\).
• Uses a finite sum of \(N\) harmonically related complex exponentials.

Key Equations:

• Synthesis: \(x[n] = \sum_{k=\langle N\rangle} a_k e^{j k(2\pi/N)n}\)
• Analysis: \(a_k = \frac{1}{N} \sum_{n=\langle N\rangle} x[n] e^{-j k(2\pi/N)n}\)

Crucial Differences from Continuous-Time (CTFS):

• Only \(N\) distinct complex exponentials \(\phi_k[n]\).
• Fourier coefficients \(a_k\) are periodic with period \(N\) (\(a_k = a_{k+N}\)).
• No convergence issues or Gibbs phenomenon. The finite sum perfectly reconstructs \(x[n]\).

To summarize, the Discrete-Time Fourier Series is a powerful tool for analyzing periodic discrete-time signals. Its finite nature is its most distinguishing feature, leading to exact reconstruction and avoiding the convergence complexities of its continuous-time counterpart. Understanding these differences is crucial for working with sampled signals and discrete systems. Thank you.