Signals and Systems

Properties of Continuous-Time Fourier Series

Imron Rosyadi

Properties of Continuous-Time Fourier Series

Introduction to Fourier Series Properties

Fourier series representations possess important properties useful for conceptual insights and simplifying calculations.

These properties help us understand how operations on a signal in the time domain affect its Fourier series coefficients.

Today, we’ll dive into the fundamental properties of Continuous-Time Fourier Series. Understanding these properties is crucial because they not only provide deeper insights into signal behavior but also offer powerful shortcuts for analyzing complex signals, often avoiding direct, tedious calculations.

Shorthand Notation for Fourier Series

To simplify discussion, we use a shorthand notation.

If \(x(t)\) is a periodic signal with period \(T\) and fundamental frequency \(\omega_0 = 2\pi/T\), and its Fourier series coefficients are \(a_k\), we write:

\[ x(t) \stackrel{\mathcal{F S}}{\longleftrightarrow} a_{k} \]

This notation, \(x(t)\) double-arrow FS \(a_k\), simply means that \(a_k\) are the complex Fourier series coefficients corresponding to the periodic signal \(x(t)\). It’s a convenient way to express the Fourier series pair without writing out the full synthesis or analysis equations every time.

Linearity

If \(x(t)\) and \(y(t)\) are periodic with period \(T\), with coefficients \(a_k\) and \(b_k\) respectively:

\[ \begin{aligned} & x(t) \stackrel{\mathcal{F S}}{\longleftrightarrow} a_{k}, \\ & y(t) \stackrel{\mathcal{F S}}{\longleftrightarrow} b_{k} . \end{aligned} \]

Then, a linear combination of these signals, \(z(t) = A x(t) + B y(t)\), will have Fourier series coefficients \(c_k\):

\[ z(t)=A x(t)+B y(t) \stackrel{\mathcal{F S}}{\longleftrightarrow} c_{k}=A a_{k}+B b_{k} \]

Linearity is one of the most fundamental properties. It tells us that if we combine signals in a linear way, their Fourier coefficients combine in the exact same linear way. This is incredibly useful because it allows us to break down complex signals into simpler components, analyze each component separately, and then combine the results. The proof follows directly from the definition of the Fourier series analysis equation.

Linearity: Interactive Demonstration

Let’s see linearity in action with two simple periodic signals.

In this Python example, we simulate the Fourier coefficients for two simple signals, a sine wave and a cosine wave. We then apply arbitrary scaling factors, A and B, to combine these signals. The output clearly shows that the coefficients of the combined signal are simply the scaled sum of the individual coefficients, confirming the linearity property. While this uses hypothetical coefficients for simplicity, it captures the essence.

Time Shifting

When a periodic signal \(x(t)\) is shifted in time by \(t_0\) to become \(y(t) = x(t-t_0)\), its period \(T\) remains unchanged.

The new Fourier series coefficients \(b_k\) are related to the original coefficients \(a_k\) by:

\[ x\left(t-t_{0}\right) \stackrel{\mathcal{F S}}{\longleftrightarrow} e^{-j k \omega_{0} t_{0}} a_{k}=e^{-j k(2 \pi / T) t_{0}} a_{k} \]

Tip

Key Insight: Time shifting a signal only changes the phase of its Fourier coefficients, not their magnitudes.

Time shifting is a very intuitive operation. If you take a periodic signal and just slide it along the time axis, its fundamental frequency and the ‘strength’ of its harmonics don’t change. What does change is the starting phase of each harmonic component. The exponential term, \(e^{-j k \omega_{0} t_{0}}\), represents this phase shift, which is linear with the harmonic index \(k\). This property is crucial for understanding delays and propagation effects in systems.

Time Shifting: Interactive Visualization

Observe a square wave and its time-shifted version. Notice how the magnitude of coefficients remains the same, but the phase changes.

viewof t0_slider = Inputs.range([-2, 2], {
  label: "Shift Amount (t0)",
  step: 0.1,
  value: 0
})

This interactive plot demonstrates the time-shifting property. The top panel shows the original square wave and its shifted version. As you adjust the Shift Amount (t0) slider, you’ll see the red dashed line (the shifted signal) move. Crucially, look at the middle panel: the magnitudes of the Fourier coefficients for both signals are identical. However, in the bottom panel, the phases of the shifted signal’s coefficients exhibit a linear change with respect to the harmonic index \(k\), which is exactly what the formula \(e^{-j k \omega_{0} t_{0}}\) predicts.

Time Reversal

When a periodic signal \(x(t)\) undergoes time reversal to become \(y(t) = x(-t)\), its period \(T\) remains unchanged.

The new Fourier series coefficients \(b_k\) are related to the original coefficients \(a_k\) by:

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

Note

Symmetry Implication:
If \(x(t)\) is even (\(x(-t) = x(t)\)), then \(a_k = a_{-k}\).
If \(x(t)\) is odd (\(x(-t) = -x(t)\)), then \(a_k = -a_{-k}\).

Time reversal flips the signal about the vertical axis at \(t=0\). The property states that the Fourier coefficients are also ‘flipped’ in their index. So the \(k\)-th coefficient of \(x(-t)\) is the negative \(k\)-th coefficient of \(x(t)\). This has direct implications for symmetric signals. For an even signal, the coefficients are also even; for an odd signal, the coefficients are odd. This can save a lot of calculation if you recognize the symmetry.

Time Reversal: Interactive Visualization

Observe an asymmetric periodic signal and its time-reversed version.

Here, we visualize time reversal using a sawtooth wave, which is an asymmetric signal. The top plot shows the original signal and its flipped version. In the coefficient plots, you’ll notice that the magnitudes are symmetric (\(|a_k| = |a_{-k}|\)), but the phases are anti-symmetric (\(\angle a_k = -\angle a_{-k}\)), reflecting the \(a_{-k}\) relationship. This clearly illustrates that reversing a signal in time corresponds to ‘reversing’ the indices of its Fourier coefficients.

Time Scaling

If \(x(t)\) is periodic with period \(T\) and fundamental frequency \(\omega_0\), then \(x(\alpha t)\) (for \(\alpha > 0\)) is periodic with period \(T/\alpha\) and fundamental frequency \(\alpha \omega_0\).

The Fourier coefficients for \(x(\alpha t)\) are the same as for \(x(t)\), i.e., \(a_k\).

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

Caution

Important: While the coefficients \(a_k\) remain the same, the series representation changes because the fundamental frequency is now \(\alpha \omega_0\).

Time scaling stretches or compresses the signal along the time axis. If you compress it (alpha > 1), it speeds up, and its period shortens. If you stretch it (alpha < 1), it slows down, and its period lengthens. The key here is that the weights or amplitudes of the harmonic components (the \(a_k\)’s) don’t change. What changes is the frequency at which each harmonic occurs, effectively scaling the fundamental frequency by \(\alpha\).

Time Scaling: Interactive Visualization

Observe how scaling affects the signal’s period and fundamental frequency, while the coefficients themselves remain invariant.

viewof alpha_slider = Inputs.range([0.5, 2], {
  label: "Scaling Factor (α)",
  step: 0.1,
  value: 1
})

This interactive plot helps illustrate the effect of time scaling. As you adjust the Scaling Factor (α), you’ll see the dashed red line (the scaled signal) either compress or expand. Notice that the period of the scaled signal changes to \(T/\alpha\). However, if you look at the bottom panel, the Fourier coefficients a_k themselves (their magnitudes) remain exactly the same. What changes is the mapping of these coefficients to the actual frequencies in the synthesis equation, as the fundamental frequency becomes \(\alpha \omega_0\).

Multiplication

If \(x(t)\) and \(y(t)\) are periodic with period \(T\), with coefficients \(a_k\) and \(b_k\) respectively:

\[ \begin{aligned} & x(t) \stackrel{\mathcal{F S}}{\longleftrightarrow} a_{k}, \\ & y(t) \stackrel{\mathcal{F S}}{\longleftrightarrow} b_{k} . \end{aligned} \]

The Fourier series coefficients \(h_k\) of their product \(x(t)y(t)\) are given by the discrete-time convolution of \(a_k\) and \(b_k\):

\[ x(t) y(t) \stackrel{\mathcal{F S}}{\longleftrightarrow} h_{k}=\sum_{l=-\infty}^{\infty} a_{l} b_{k-l} \]

The multiplication property is often called the convolution property in the frequency domain. Just as convolution in the time domain corresponds to multiplication in the frequency domain for Fourier transforms, here, multiplication of two periodic signals in the time domain corresponds to the convolution of their discrete Fourier series coefficients. This is a powerful concept, especially when analyzing systems where signals are modulated or mixed.

Conjugation and Conjugate Symmetry

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

\[ x^{*}(t) \stackrel{\mathcal{F S}}{\longleftrightarrow} a_{-k}^{*} \]

A significant consequence for real signals (\(x(t) = x^*(t)\)) is conjugate symmetry:

\[ a_{-k}=a_{k}^{*} \]

This implies various symmetry properties for magnitudes, phases, real, and imaginary parts of coefficients for real signals.

This property is fundamental when dealing with real-world signals, which are always real-valued. If a signal is real, its Fourier coefficients must satisfy conjugate symmetry. This means \(a_0\) must be real, the magnitudes \(|a_k|\) are even (\(|a_k| = |a_{-k}|\)), and the phases \(\angle a_k\) are odd (\(\angle a_k = -\angle a_{-k}\)). These symmetries simplify calculations and provide a way to check for errors.

Parseval’s Relation for Continuous-Time Periodic Signals

Parseval’s relation states that the average power of a periodic signal in the time domain equals the sum of the average powers of its harmonic components in the frequency domain.

\[ \frac{1}{T} \int_{T}|x(t)|^{2} d t=\sum_{k=-\infty}^{+\infty}\left|a_{k}\right|^{2} \]

The left side is the average power in one period of \(x(t)\). The term \(|a_k|^2\) represents the average power in the \(k\)-th harmonic component.

Important

Conservation of Energy: This relation highlights the conservation of power between the time and frequency domains.

Parseval’s relation is a cornerstone of signal analysis. It tells us that we can quantify the total energy or power in a signal by either integrating its squared magnitude over time or by summing the squared magnitudes of its Fourier coefficients. This is incredibly useful for power calculations, noise analysis, and understanding how power is distributed across different frequency components.

Parseval’s Relation: Interactive Demonstration

Let’s verify Parseval’s relation for a simple square wave.

In this example, we calculate the average power of a square wave in two ways. First, directly from its time-domain definition by integrating its squared magnitude over one period. Second, by summing the squared magnitudes of its Fourier series coefficients. Notice how closely these two values match, especially as we increase the number of harmonics considered. This vividly demonstrates Parseval’s relation, confirming that power is conserved across domains.

Summary of Properties

Many important properties of continuous-time Fourier series are summarized in the table below.

Property	Periodic Signal	Fourier Series Coeffs
Linearity	\(Ax(t) + By(t)\)	\(Aa_k + Bb_k\)
Time Shift	\(x(t-t_0)\)	\(a_k e^{-jk\omega_0 t_0}\)
Freq Shift	\(e^{jM\omega_0 t}x(t)\)	\(a_{k-M}\)
Conjugation	\(x^*(t)\)	\(a_{-k}^*\)
Time Reversal	\(x(-t)\)	\(a_{-k}\)
Time Scaling	\(x(\alpha t)\)	\(a_k\) (with new \(\omega_0\))

Summary of Properties

Property	Periodic Signal	Fourier Series Coeffs
Periodic Convolution	\(\int_T x(\tau)y(t-\tau)d\tau\)	\(T a_k b_k\)
Multiplication	\(x(t)y(t)\)	\(\sum_{l=-\infty}^\infty a_l b_{k-l}\)
Differentiation	\(dx(t)/dt\)	\(jk\omega_0 a_k\)
Integration	\(\int x(\tau)d\tau\)	\((1/(jk\omega_0)) a_k\)
Real Signal	\(x(t)\) real	\(a_k = a_{-k}^*\)
Real & Even	\(x(t)\) real & even	\(a_k\) real & even
Real & Odd	\(x(t)\) real & odd	\(a_k\) purely imag. & odd

\[ \frac{1}{T} \int_{T}|x(t)|^{2} d t=\sum_{k=-\infty}^{+\infty}\left|a_{k}\right|^{2} \quad \text{(Parseval's Relation)} \]

This table provides a comprehensive overview of all the key properties we’ve discussed and some additional ones. It’s an invaluable reference for any Fourier series problem. Notice how operations in one domain (time) often have a dual or inverse effect in the other domain (frequency). For instance, differentiation in time corresponds to multiplication by \(jk\omega_0\) in the frequency domain. These dualities are a recurring theme in signal processing.

Example 3.6: Using Linearity and Time Shifting

Consider the signal \(g(t)\) with period 4.

We know \(g(t) = x(t-1) - 1/2\), where \(x(t)\) is a symmetric square wave with \(T=4, T_1=1\).

The coefficients \(a_k\) for \(x(t)\) are \(\frac{\sin(\pi k/2)}{k\pi}\) (for \(k \neq 0\)) and \(a_0 = 1/2\).

graph LR
    A["x(t) - Square Wave T=4, T1=1"] --> B{"Shift by 1: x(t-1)"}
    C["Constant -1/2"] --> D{Add to B}
    B --> E["g(t) = x(t-1) - 1/2"]
    D --> E

This example shows how to leverage properties to find Fourier coefficients without direct integration. We’re given a signal \(g(t)\) that looks like a shifted and DC-offset version of a standard square wave \(x(t)\). The square wave \(x(t)\) has easily derivable coefficients. By recognizing \(g(t)\) as a transformation of \(x(t)\), we can use the linearity and time-shifting properties.

Example 3.6: Deriving Coefficients

1. Time Shifting: Coefficients \(b_k\) for \(x(t-1)\) are \(a_k e^{-j k \omega_0 t_0}\).

   Here, \(\omega_0 = 2\pi/T = 2\pi/4 = \pi/2\) and \(t_0 = 1\).

   So, \(b_k = a_k e^{-j k (\pi/2)(1)} = a_k e^{-j k \pi/2}\).

2. DC Offset: The term \(-1/2\) has coefficients \(c_k = -1/2\) for \(k=0\) and \(0\) for \(k \neq 0\).

3. Linearity: Coefficients \(d_k\) for \(g(t) = x(t-1) - 1/2\) are \(b_k + c_k\).

   \[ d_{k}=\left\{\begin{array}{ll} \frac{\sin (\pi k / 2)}{k \pi} e^{-j k \pi / 2}, & \text { for } k \neq 0 \\ 0, & \text { for } k=0 \end{array}\right. \]

First, we apply the time-shifting property. We know the coefficients of the unshifted square wave, \(x(t)\). Shifting it by 1 in time simply multiplies each coefficient \(a_k\) by a phase factor \(e^{-j k \pi/2}\). Next, we consider the DC offset. A constant value in the time domain only contributes to the \(a_0\) coefficient in the frequency domain. Finally, we use linearity: the coefficients of the sum are the sum of the coefficients. We combine the shifted square wave’s coefficients with the DC offset’s coefficients to get the coefficients for \(g(t)\). Notice how the \(a_0\) term for \(x(t)\) (\(1/2\)) combined with the DC offset (\(-1/2\)) to yield \(d_0=0\).

Example 3.7: Using Differentiation

Consider the triangular wave signal \(x(t)\) with period \(T=4\).

Its derivative, \(dx(t)/dt\), is the square wave \(g(t)\) from Example 3.6.

If \(e_k\) are the coefficients for \(x(t)\) and \(d_k\) for \(g(t)\), the differentiation property states:

\[ d_{k}=j k \omega_{0} e_{k} \]

Since \(\omega_0 = \pi/2\):

\[ d_{k}=j k (\pi/2) e_{k} \]

This example showcases the differentiation property. We have a triangular wave, and its derivative is a square wave, which we just analyzed in Example 3.6. Instead of directly calculating the Fourier series for the triangular wave (which involves integration by parts), we can use the differentiation property. This property states that differentiating in the time domain corresponds to multiplying the Fourier coefficients by \(j k \omega_0\).

Example 3.7: Deriving Coefficients

From \(d_k = j k (\pi/2) e_k\), we can find \(e_k\):

\[ e_{k}=\frac{2 d_{k}}{j k \pi}, \quad k \neq 0 \]

Substituting \(d_k\) from Example 3.6:

\[ e_{k}=\frac{2 \sin (\pi k / 2)}{j(k \pi)^{2}} e^{-j k \pi / 2}, \quad k \neq 0 \]

For \(k=0\), \(e_0\) is the average value of \(x(t)\) over one period:

\[ e_{0}=\frac{1}{T} \int_T x(t) dt = \frac{1}{4} \times (\text{Area of triangle}) = \frac{1}{4} \times (2) = \frac{1}{2} \]

We rearrange the differentiation formula to solve for \(e_k\) in terms of \(d_k\). We already know \(d_k\) from the previous example. For the DC component, \(e_0\), we cannot use the differentiation formula because it would involve division by zero. Instead, \(e_0\) is simply the average value of the signal, which we can find by calculating the area under one period of the triangular wave and dividing by the period. This demonstrates how properties can be combined with basic definitions to solve problems efficiently.

Example 3.8: Impulse Train Coefficients

The impulse train \(x(t) = \sum_{k=-\infty}^{\infty} \delta(t-kT)\) with period \(T\).

The Fourier series coefficients \(a_k\) are found by integrating over one period (e.g., \([-T/2, T/2]\)):

\[ a_{k}=\frac{1}{T} \int_{-T / 2}^{T / 2} \delta(t) e^{-j k 2 \pi / T} d t=\frac{1}{T} \]

All coefficients are equal: \(a_k = 1/T\).

graph LR
    A[…] --- B{…}
    B -- T --> C{"δ(t)"}
    C -- T --> D{"δ(t-T)"}
    D -- T --> E{"δ(t-2T)"}
    E -- T --> B
    style C fill:#f9f,stroke:#333,stroke-width:2px

The impulse train is a very important signal in signal processing. It consists of a series of Dirac delta functions equally spaced in time. To find its Fourier coefficients, we use the analysis equation and integrate over a single period that contains one impulse, typically at \(t=0\). Due to the sifting property of the delta function, the integral simply evaluates to the value of the exponential at \(t=0\), which is 1. Thus, all Fourier coefficients for an impulse train are constant, \(1/T\). This also aligns with the property that a real and even signal (like the impulse train) has real and even coefficients.

Example 3.8: Relating Signals using Properties

The derivative of a square wave \(g(t)\) is \(q(t)\), which can be expressed as a difference of two shifted impulse trains:

\[ q(t)=x\left(t+T_{1}\right)-x\left(t-T_{1}\right) \]

• Coefficients for \(x(t)\) are \(a_k = 1/T\).

• By time-shifting and linearity, coefficients \(b_k\) for \(q(t)\) are:

  \(b_k = a_k e^{j k \omega_0 T_1} - a_k e^{-j k \omega_0 T_1}\)

  \(b_k = \frac{1}{T} (e^{j k \omega_0 T_1} - e^{-j k \omega_0 T_1}) = \frac{2j \sin(k \omega_0 T_1)}{T}\)

graph TD
    A["Square Wave g(t)"]
    B["Derivative d/dt"]
    C["Impulse Train x(t)"]
    D["Shifted x(t+T1)"]
    E["Shifted x(t-T1)"]
    F["q(t) = D - E"]

    A -- B --> F
    C -- Shift --> D
    C -- Shift --> E
    D -- Subtract --> F
    E -- Subtract --> F

This part of the example cleverly connects several properties. We observe that the derivative of a square wave is a series of positive and negative impulses. This q(t) can be seen as the difference between two time-shifted impulse trains. We already know the coefficients for a basic impulse train. By applying the time-shifting property to get the coefficients for \(x(t+T_1)\) and \(x(t-T_1)\), and then the linearity property for their difference, we can quickly derive the Fourier coefficients \(b_k\) for \(q(t)\).

Example 3.8: Connecting Back to Square Wave

Since \(q(t)\) is the derivative of \(g(t)\), we use the differentiation property:

\[ b_{k}=j k \omega_{0} c_{k} \]

Where \(c_k\) are the Fourier coefficients of \(g(t)\).

Solving for \(c_k\) (for \(k \neq 0\)):

\[ c_{k}=\frac{b_{k}}{j k \omega_{0}}=\frac{2 j \sin \left(k \omega_{0} T_{1}\right)}{j k \omega_{0} T}=\frac{\sin \left(k \omega_{0} T_{1}\right)}{k \pi}, \quad k \neq 0 \]

For \(c_0\), the average value of \(g(t)\) from inspection of a square wave with width \(2T_1\):

\[ c_{0}=\frac{2 T_{1}}{T} \]

These are the same coefficients derived directly for a square wave!

Finally, we tie it all together. Since \(q(t)\) is the derivative of \(g(t)\), we can use the differentiation property in reverse. We divide the coefficients \(b_k\) (which we just found for \(q(t)\)) by \(jk\omega_0\) to get the coefficients \(c_k\) for \(g(t)\). The \(k=0\) term (DC component) needs to be found separately by calculating the average value of the square wave. This confirms that the Fourier coefficients obtained using these properties match those found by direct integration, demonstrating the power and consistency of these properties.

Example 3.9: Characterizing a Signal

Given facts about \(x(t)\):

1. \(x(t)\) is a real signal.
2. \(x(t)\) is periodic with \(T=4\), coefficients \(a_k\).
3. \(a_k = 0\) for \(|k|>1\).
4. The signal with coefficients \(b_k = e^{-j\pi k/2} a_{-k}\) is odd.
5. \(\frac{1}{4} \int_{4}|x(t)|^{2} d t = 1/2\).

Let’s determine \(x(t)\) to within a sign factor.

This is a comprehensive problem that requires applying multiple properties to deduce the form of an unknown signal. We are given five distinct facts about a signal \(x(t)\) and its Fourier coefficients. Our goal is to use these facts to uniquely identify \(x(t)\) up to a sign. This is a great example of how understanding these properties allows for powerful deductive reasoning in signal analysis.

Example 3.9: Step-by-Step Deduction

Fact 3: \(a_k = 0\) for \(|k|>1\).

\(\implies x(t) = a_0 + a_1 e^{j\pi t/2} + a_{-1} e^{-j\pi t/2}\) (since \(\omega_0 = 2\pi/4 = \pi/2\)).

Fact 1: \(x(t)\) is real.

\(\implies a_0\) is real, and \(a_1 = a_{-1}^*\).

So, \(x(t) = a_0 + 2\operatorname{Re}\{a_1 e^{j\pi t/2}\}\).

Fact 4: Signal with \(b_k = e^{-j\pi k/2} a_{-k}\) is odd.

• \(a_{-k}\) corresponds to \(x(-t)\) (Time Reversal).
• \(e^{-j\pi k/2}\) corresponds to a time shift of \(t_0=1\) (since \(e^{-jk\omega_0 t_0}\) with \(\omega_0=\pi/2, t_0=1\) gives \(e^{-jk\pi/2}\)).

\(\implies b_k\) correspond to \(x(-(t-1)) = x(-t+1)\).

Since \(x(-t+1)\) is odd and real (Fact 1), its Fourier coefficients \(b_k\) must be purely imaginary and odd. \(\implies b_0 = 0\) and \(b_{-1} = -b_1\).

We start by using Fact 3 to limit the number of non-zero coefficients to \(a_0, a_1, a_{-1}\). Then, Fact 1 (x(t) is real) tells us about the symmetry of these coefficients: \(a_0\) is real and \(a_1\) is the complex conjugate of \(a_{-1}\). This simplifies the time-domain expression for \(x(t)\). Fact 4 is where it gets interesting. We recognize \(a_{-k}\) as the coefficients for \(x(-t)\) (time reversal) and the exponential term as a time shift. So, \(b_k\) are the coefficients for \(x(-t+1)\). Since this signal is explicitly stated to be odd AND we know \(x(t)\) is real (making \(x(-t+1)\) also real), its coefficients \(b_k\) must be purely imaginary and odd. This immediately gives us \(b_0=0\) and \(b_{-1}=-b_1\).

Example 3.9: Final Steps

From \(b_0 = 0\):

\(b_0 = e^{-j\pi (0)/2} a_{-0} = a_0 = 0\).

From \(b_{-1} = -b_1\):

We also know \(b_k = e^{-j\pi k/2} a_{-k}\).

For \(k=1\): \(b_1 = e^{-j\pi /2} a_{-1} = -j a_{-1}\).

Since \(a_{-1} = a_1^*\), we have \(b_1 = -j a_1^*\).

Since \(b_k\) are purely imaginary, \(b_1\) must be purely imaginary. This is consistent with \(-j a_1^*\).

Example 3.9: Final Steps

Fact 5: Parseval’s Relation.

The average power of \(x(-t+1)\) is also \(1/2\).

\(\frac{1}{4} \int_{4}|x(-t+1)|^{2} d t=1/2\).

Using Parseval’s: \(\sum_{k=-\infty}^\infty |b_k|^2 = |b_{-1}|^2 + |b_0|^2 + |b_1|^2 = 1/2\).

Since \(b_0=0\) and \(b_{-1}=-b_1\), we have \(2|b_1|^2 = 1/2 \implies |b_1|^2 = 1/4 \implies |b_1| = 1/2\).

Since \(b_1\) is purely imaginary, \(b_1 = j/2\) or \(b_1 = -j/2\).

We use the deduced properties of \(b_k\). From \(b_0 = 0\), we find that \(a_0\) must be zero. This significantly simplifies \(x(t)\) to just two exponential terms. Next, we use Parseval’s relation. Since time reversal and time shifting don’t change the average power, the average power of \(x(-t+1)\) is also \(1/2\). Applying Parseval’s to the \(b_k\) coefficients, and recalling that \(b_0=0\) and \(b_{-1}=-b_1\), we find that \(|b_1|\) must be \(1/2\). Coupled with the fact that \(b_1\) is purely imaginary, this means \(b_1\) can only be \(j/2\) or \(-j/2\).

Example 3.9: Determining \(x(t)\)

We have \(a_0=0\) and \(b_1 = -j a_1^*\).

Case 1: \(b_1 = j/2\).

\(j/2 = -j a_1^* \implies a_1^* = -1/2 \implies a_1 = -1/2\).

Then \(x(t) = 2\operatorname{Re}\{(-1/2) e^{j\pi t/2}\} = 2(-1/2 \cos(\pi t/2)) = -\cos(\pi t/2)\).

Case 2: \(b_1 = -j/2\).

\(-j/2 = -j a_1^* \implies a_1^* = 1/2 \implies a_1 = 1/2\).

Then \(x(t) = 2\operatorname{Re}\{(1/2) e^{j\pi t/2}\} = 2(1/2 \cos(\pi t/2)) = \cos(\pi t/2)\).

Thus, \(x(t) = \pm \cos(\pi t/2)\).

Finally, we substitute the possible values of \(b_1\) back into our relationship between \(a_1\) and \(b_1\). This allows us to determine the possible values for \(a_1\). With \(a_0=0\) and the determined \(a_1\), we can reconstruct \(x(t)\) using the synthesis equation. As the problem stated, we arrive at \(x(t)\) being \(\cos(\pi t/2)\) or \(-\cos(\pi t/2)\), demonstrating that the given facts are sufficient to determine the signal up to a sign factor. This is a powerful illustration of how properties can be used in reverse, from frequency domain characteristics to time domain signal reconstruction.