Signals and Systems

4.4 The Convolution Property

Imron Rosyadi

The Convolution Property

4.4 The Convolution Property: Frequency Domain Perspective

Understanding how LTI systems interact with signals is fundamental in Electrical and Computer Engineering. The convolution property establishes a powerful link between the time and frequency domains.

This lecture introduces one of the most important properties in Signals and Systems: the convolution property. It’s a cornerstone for analyzing LTI systems, allowing us to simplify complex time-domain operations by moving to the frequency domain. We’ll start by building intuition from Fourier series and then move to a formal derivation.

From Fourier Series to Fourier Transform

Recall that for periodic signals, the output of an LTI system is simply the Fourier series coefficients of the input multiplied by the system’s frequency response.

\[ \begin{equation*} x(t)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty} X(j \omega) e^{j \omega t} d \omega=\lim _{\omega_{0} \rightarrow 0} \frac{1}{2 \pi} \sum_{k=-\infty}^{+\infty} X\left(j k \omega_{0}\right) e^{j k \omega_{0} t} \omega_{0} \tag{4.47} \end{equation*} \]

The response of an LTI system to a complex exponential \(e^{j k \omega_{0} t}\) is \(H\left(j k \omega_{0}\right) e^{j k \omega_{0} t}\). Where \(H(j \omega)\) is the Fourier Transform of the impulse response \(h(t)\).

We’re building on the intuition we developed for periodic signals in Chapter 3. Remember that complex exponentials are eigenfunctions of LTI systems. This means if you put \(e^{j\omega t}\) into an LTI system, the output is simply \(H(j\omega)e^{j\omega t}\), where \(H(j\omega)\) is the system’s frequency response. Equation 4.47 shows how an aperiodic signal \(x(t)\) can be represented as a continuous sum (an integral) of these complex exponentials. This is essentially the Fourier Transform synthesis equation. The key idea here is that because an LTI system scales each individual exponential component, we can extend this concept to the continuous spectrum of aperiodic signals.

Intuitive Derivation of the Convolution Property

Applying the superposition principle, each exponential component in the input \(x(t)\) is scaled by the frequency response.

\[ \frac{1}{2 \pi} \sum_{k=-\infty}^{+\infty} X\left(j k \omega_{0}\right) e^{j k \omega_{0} t} \omega_{0} \longrightarrow \frac{1}{2 \pi} \sum_{k=-\infty}^{+\infty} X\left(j k \omega_{0}\right) H\left(j k \omega_{0}\right) e^{j k \omega_{0} t} \omega_{0} \]

Taking the limit as \(\omega_0 \rightarrow 0\), the output \(y(t)\) is:

\[ \begin{align*} y(t) & =\lim _{\omega_{0} \rightarrow 0} \frac{1}{2 \pi} \sum_{k=-\infty}^{+\infty} X\left(j k \omega_{0}\right) H\left(j k \omega_{0}\right) e^{j k \omega_{0} t} \omega_{0} \tag{4.49}\\ & =\frac{1}{2 \pi} \int_{-\infty}^{+\infty} X(j \omega) H(j \omega) e^{j \omega t} d \omega . \end{align*} \]

By definition, the inverse Fourier Transform of \(Y(j\omega)\) is \(y(t)\):

\[ \begin{equation*} y(t)=\frac{1}{2 \pi} \int_{-\infty}^{+\infty} Y(j \omega) e^{j \omega t} d \omega \tag{4.50} \end{equation*} \]

Comparing these, we arrive at the convolution property:

\[ \begin{equation*} Y(j \omega)=X(j \omega) H(j \omega) \tag{4.51} \end{equation*} \]

This slide shows the intuitive leap from discrete Fourier series to the continuous Fourier Transform. The first equation represents the input signal as a sum of scaled exponentials. The arrow indicates that an LTI system scales each of these components by its frequency response \(H(j k \omega_0)\). As the spacing between frequencies \(\omega_0\) approaches zero, the sum becomes an integral. By comparing the resulting expression for \(y(t)\) (Equation 4.49) with the general definition of an inverse Fourier Transform (Equation 4.50), we can directly infer that the Fourier Transform of the output, \(Y(j\omega)\), must be the product of the input’s Fourier Transform \(X(j\omega)\) and the system’s frequency response \(H(j\omega)\).

Formal Derivation

Starting from the convolution integral, we can formally derive the property.

\[ \begin{equation*} y(t)=\int_{-\infty}^{+\infty} x(\tau) h(t-\tau) d \tau \tag{4.52} \end{equation*} \]

Taking the Fourier Transform of \(y(t)\):

\[ \begin{equation*} Y(j \omega)=\mathcal{F}\{y(t)\}=\int_{-\infty}^{+\infty}\left[\int_{-\infty}^{+\infty} x(\tau) h(t-\tau) d \tau\right] e^{-j \omega t} d t \tag{4.53} \end{equation*} \]

Interchanging integration order and using the time-shift property \(\mathcal{F}\{h(t-\tau)\} = H(j\omega)e^{-j\omega\tau}\):

\[ \begin{equation*} Y(j \omega)=\int_{-\infty}^{+\infty} x(\tau)\left[\int_{-\infty}^{+\infty} h(t-\tau) e^{-j \omega t} d t\right] d \tau = \int_{-\infty}^{+\infty} x(\tau) e^{-j \omega \tau} H(j \omega) d \tau \tag{4.54} \end{equation*} \]

Formal Derivation

Factoring out \(H(j\omega)\):

\[ \begin{equation*} Y(j \omega)=H(j \omega) \int_{-\infty}^{+\infty} x(\tau) e^{-j \omega \tau} d \tau = H(j \omega) X(j \omega) \tag{4.55} \end{equation*} \]

Thus, the convolution property is:

\[ \begin{equation*} y(t)=h(t) * x(t) \stackrel{\mathfrak{F}}{\longleftrightarrow} Y(j \omega)=H(j \omega) X(j \omega) . \tag{4.56} \end{equation*} \]

Important

This property is fundamental! It transforms complex convolution operations in the time domain into simple multiplication in the frequency domain.

The formal derivation provides a rigorous proof, starting directly from the definition of convolution.
First, we apply the Fourier Transform definition to the output \(y(t)\), substituting the convolution integral.
Next, we interchange the order of integration, which is a common technique in Fourier analysis.
The crucial step is recognizing that the inner integral, \(\int_{-\infty}^{+\infty} h(t-\tau) e^{-j \omega t} d t\), is the Fourier Transform of a time-shifted impulse response, \(h(t-\tau)\). By the time-shift property of the Fourier Transform, this equals \(H(j\omega)e^{-j\omega\tau}\).
Finally, we factor out \(H(j\omega)\), leaving the Fourier Transform of \(x(\tau)\), which is \(X(j\omega)\). This directly leads to \(Y(j\omega) = H(j\omega)X(j\omega)\). This derivation solidifies the intuitive understanding and provides a powerful tool for analyzing LTI systems.

Significance of the Convolution Property

The convolution property is a cornerstone of LTI system analysis. \(H(j \omega)\), the Fourier transform of the impulse response \(h(t)\), is the frequency response of the system.

• It completely characterizes an LTI system, just like \(h(t)\).
• It describes how the system scales and shifts the phase of different frequency components.
• Crucial for frequency-selective filtering, where certain frequency bands are passed and others are attenuated.

The overall frequency response of cascaded LTI systems is the product of their individual frequency responses, regardless of order.

graph LR
    subgraph System 1
        A["Input X(jω)"] --> B("H1(jω)")
    end
    subgraph System 2
        C["Output Y1(jω)"] --> D("H2(jω)")
    end
    B --> C
    D --> E["Output Y(jω)"]

The frequency response \(H(j\omega)\) is a complex-valued function. Its magnitude, \(|H(j\omega)|\), tells us how much the amplitude of a sinusoidal component at frequency \(\omega\) is scaled by the system. Its phase, \(\angle H(j\omega)\), tells us how much the phase of that component is shifted.
Understanding \(H(j\omega)\) is essential for designing filters to pass or block specific frequency ranges.
The Mermaid diagram illustrates two LTI systems connected in cascade. If the first system has frequency response \(H_1(j\omega)\) and the second has \(H_2(j\omega)\), then the output of the first is \(Y_1(j\omega) = X(j\omega)H_1(j\omega)\). This becomes the input to the second system, so \(Y(j\omega) = Y_1(j\omega)H_2(j\omega) = X(j\omega)H_1(j\omega)H_2(j\omega)\).
This means the overall frequency response of the cascade is \(H_1(j\omega)H_2(j\omega)\). Since multiplication is commutative (\(H_1H_2 = H_2H_1\)), the order in which LTI systems are cascaded does not affect the overall frequency response. This is a powerful consequence of the convolution property.

Stability and Existence of \(H(j\omega)\)

The Fourier Transform \(H(j\omega)\) does not exist for every LTI system.
For a stable LTI system, its impulse response \(h(t)\) must be absolutely integrable:

\[ \begin{equation*} \int_{-\infty}^{+\infty}|h(t)| d t<\infty \tag{4.57} \end{equation*} \]

This condition is one of the Dirichlet conditions required for the existence of the Fourier Transform.

Note

A stable LTI system always has a frequency response \(H(j\omega)\).
This is a crucial link between time-domain stability and frequency-domain analysis.

For unstable LTI systems, a generalization called the Laplace Transform is used.

This slide addresses an important theoretical point regarding the applicability of Fourier Transforms to LTI systems.
The absolute integrability condition (Equation 4.57) ensures that the integral defining the Fourier Transform converges. If an LTI system is stable, its impulse response will decay over time, making it absolutely integrable.
Therefore, for any stable LTI system, we are guaranteed to be able to find its frequency response \(H(j\omega)\).
This means that Fourier analysis is a suitable tool for analyzing the behavior of stable LTI systems. For systems that are not stable, we need a more general transform, such as the Laplace Transform, which allows for poles in the right-half plane.

Example 4.15: Time Shift

Consider an LTI system with impulse response \(h(t)=\delta\left(t-t_{0}\right)\).

\[ \begin{equation*} h(t)=\delta\left(t-t_{0}\right) \tag{4.58} \end{equation*} \]

The frequency response \(H(j\omega)\) is the Fourier Transform of \(h(t)\):

\[ \begin{equation*} H(j \omega)=e^{-j \omega t_{0}} . \tag{4.59} \end{equation*} \]

Thus, the Fourier Transform of the output \(y(t)\) is:

\[ \begin{align*} Y(j \omega) & =H(j \omega) X(j \omega) \tag{4.60}\\ & =e^{-j \omega t_{0}} X(j \omega) . \end{align*} \]

This is consistent with the time-shift property: \(y(t) = x(t-t_0)\).
A pure time shift system has unity magnitude response \(|H(j\omega)|=1\) and a linear phase \(-\omega t_0\).

This example is a perfect illustration of the convolution property in action.
A system that simply shifts its input by \(t_0\) has an impulse response that is a shifted Dirac delta function.
When we take the Fourier Transform of this impulse response, we get \(e^{-j\omega t_0}\).
Applying the convolution property, \(Y(j\omega) = X(j\omega)e^{-j\omega t_0}\), which perfectly matches the time-shift property we learned earlier.
Notice that the magnitude of \(H(j\omega)\) is always 1, meaning all frequency components are passed without attenuation. The phase, however, is linear with frequency, \(-\omega t_0\). This linear phase ensures that all frequency components are delayed by the same amount, preserving the signal’s shape without distortion.

Interactive Time Shift

Adjust the time shift \(t_0\) and observe its effect on the signal and its frequency response.

viewof t0_shift = Inputs.range([-2, 2], {step: 0.1, value: 0.5, label: "Time Shift (t0)"})

Example 4.16: Differentiator

For a differentiator, \(y(t) = \frac{d x(t)}{d t}\).
From the differentiation property:

\[ \begin{equation*} Y(j \omega)=j \omega X(j \omega) \tag{4.61} \end{equation*} \]

Therefore, the frequency response of a differentiator is:

\[ \begin{equation*} H(j \omega)=j \omega \tag{4.62} \end{equation*} \]

This example shows that the differentiation operation in the time domain corresponds to a simple multiplication by \(j\omega\) in the frequency domain.
The magnitude response \(|H(j\omega)| = |\omega|\) indicates that a differentiator amplifies higher frequencies linearly. This is why differentiation can make a signal “noisier” if high-frequency noise is present, as it enhances those components.
The phase response \(\angle H(j\omega)\) is \(\pi/2\) for \(\omega > 0\) and \(-\pi/2\) for \(\omega < 0\). This constant phase shift means that all frequency components are phase-shifted by 90 degrees, which is characteristic of differentiation.

Interactive Differentiator Frequency Response

Observe the magnitude and phase response of an ideal differentiator.

Example 4.17: Integrator

An integrator is an LTI system where \(y(t)=\int_{-\infty}^{t} x(\tau) d \tau\).
Its impulse response is the unit step \(u(t)\).
The frequency response \(H(j\omega)\) is:

\[ H(j \omega)=\frac{1}{j \omega}+\pi \delta(\omega) \]

Using the convolution property, the output transform is:

\[ \begin{aligned} Y(j \omega) & =H(j \omega) X(j \omega) \\ & =\frac{1}{j \omega} X(j \omega)+\pi X(j \omega) \delta(\omega) \\ & =\frac{1}{j \omega} X(j \omega)+\pi X(0) \delta(\omega) \end{aligned} \]

This is consistent with the integration property.

In contrast to differentiation, integration in the time domain corresponds to division by \(j\omega\) in the frequency domain, along with a term related to the input’s DC component.
The magnitude response \(|H(j\omega)| = 1/|\omega|\) shows that an integrator attenuates higher frequencies and amplifies lower frequencies. This makes it a lowpass-like filter, often used for smoothing signals.
The phase response \(\angle H(j\omega)\) is \(-\pi/2\) for \(\omega > 0\) and \(\pi/2\) for \(\omega < 0\), a constant -90 degree phase shift.
The \(\pi X(0) \delta(\omega)\) term is important because the integral of a signal can have a DC component even if the original signal has none. This term ensures that the DC value of the output is correctly handled, especially when \(X(0)\) is non-zero.

Interactive Integrator Frequency Response

Observe the magnitude and phase response of an ideal integrator.

Example 4.18: Ideal Lowpass Filter

An ideal lowpass filter passes frequencies below a cutoff \(\omega_c\) and blocks others.

\[ H(j \omega)= \begin{cases}1 & |\omega|<\omega_{c} \tag{4.63}\\ 0 & |\omega|>\omega_{c}\end{cases} \]

The impulse response \(h(t)\) of this filter is the inverse Fourier Transform of \(H(j\omega)\):

\[ \begin{equation*} h(t)=\frac{\sin \omega_{c} t}{\pi t} \tag{4.64} \end{equation*} \]

Warning

The ideal lowpass filter is non-causal (\(h(t)\) is not zero for \(t<0\)) and its impulse response has oscillatory behavior that extends infinitely in time.
This makes it impossible to realize perfectly in practice.

The ideal lowpass filter is a theoretical concept that is highly desired for its perfect frequency selectivity—it completely passes frequencies within a band and completely blocks those outside.
However, when we find its impulse response, \(h(t)\), it turns out to be a sinc function. This sinc function has two major practical drawbacks:

1. Non-causality: It is non-zero for \(t < 0\), meaning the filter output would depend on future input values, which is impossible for real-time physical systems.
2. Infinite Duration and Oscillations: The impulse response extends infinitely in time and exhibits oscillatory behavior (ringing). This can cause undesirable artifacts in the time domain, such as overshoot and undershoot, especially when processing signals with sharp transitions. This example highlights a fundamental trade-off in filter design between ideal frequency-domain characteristics and practical time-domain constraints.

Interactive Ideal Lowpass Filter

Adjust the cutoff frequency \(\omega_c\) and observe the frequency and impulse responses.

viewof wc_ideal = Inputs.range([1, 10], {step: 0.1, value: 3, label: "Cutoff Frequency (ωc)"})

Example 4.18: Practical Lowpass Filter

Consider a more practical LTI system with impulse response \(h(t)=e^{-at}u(t)\).

\[ \begin{equation*} h(t)=e^{-t} u(t) \tag{4.65} \end{equation*} \]

The frequency response of this system (an RC circuit) is:

\[ \begin{equation*} H(j \omega)=\frac{1}{j \omega+1} \tag{4.66} \end{equation*} \]

Characteristics:

• Causal: \(h(t)=0\) for \(t<0\).
• No Oscillations: Impulse response decays monotonically.
• Easier to implement: Simple RC circuit (first-order system).
• Trade-off: Less sharp frequency selectivity compared to ideal.

Interactive RC Filter

Adjust the parameter ‘a’ and observe the impulse and magnitude responses.

viewof a_rc = Inputs.range([0.1, 5], {step: 0.1, value: 1, label: "Parameter 'a'"})

This example contrasts the ideal lowpass filter with a practical, realizable filter. The system with impulse response \(h(t) = e^{-at}u(t)\) is causal because it is zero for \(t < 0\). Its impulse response decays monotonically without oscillations, which is often desirable in applications to avoid ringing artifacts. The frequency response, \(H(j\omega) = 1/(a+j\omega)\), is that of a first-order lowpass filter, commonly implemented with an RC circuit. While it doesn’t have the “brick-wall” cutoff of an ideal filter, its practical advantages often outweigh the less-than-ideal frequency selectivity. The parameter ‘a’ controls the filter’s bandwidth: a larger ‘a’ means a faster decay in the time domain and a wider bandwidth (higher cutoff frequency) in the frequency domain. This demonstrates the trade-offs engineers often face in filter design.

Example 4.19: Convolution of Exponentials

Let’s find the response of an LTI system with \(h(t)=e^{-at}u(t)\) to an input \(x(t)=e^{-bt}u(t)\).

From previous examples, their Fourier Transforms are:
\(X(j \omega)=\frac{1}{b+j \omega}\)
\(H(j \omega)=\frac{1}{a+j \omega}\)

Using the convolution property \(Y(j\omega) = X(j\omega)H(j\omega)\):

\[ \begin{equation*} Y(j \omega)=\frac{1}{(a+j \omega)(b+j \omega)} \tag{4.67} \end{equation*} \]

To find \(y(t)\), we use partial fraction expansion (assuming \(a \neq b\)):

\[ \begin{equation*} Y(j \omega)=\frac{A}{a+j \omega}+\frac{B}{b+j \omega}, \tag{4.68} \end{equation*} \]

where \(A=\frac{1}{b-a}\) and \(B=\frac{1}{a-b} = -A\).

\[ \begin{equation*} Y(j \omega)=\frac{1}{b-a}\left[\frac{1}{a+j \omega}-\frac{1}{b+j \omega}\right] \tag{4.69} \end{equation*} \]

Taking the inverse Fourier Transform (using linearity): \(y(t)=\frac{1}{b-a}\left[e^{-a t} u(t)-e^{-b t} u(t)\right]\).

This example highlights the practical utility of the convolution property for solving convolution integrals that would be tedious in the time domain.
Instead of direct time-domain convolution, we transform \(x(t)\) and \(h(t)\) into their respective Fourier Transforms, multiply them in the frequency domain to get \(Y(j\omega)\), and then find the inverse Fourier Transform of \(Y(j\omega)\).
The key step in finding the inverse transform of complex rational functions like \(Y(j\omega)\) is often partial fraction expansion. This technique decomposes the complex fraction into simpler terms whose inverse transforms are known (e.g., \(e^{-at}u(t)\)).
The resulting \(y(t)\) shows how the system’s exponential decay and the input’s exponential decay combine to form the output.

Example 4.19: Convolution of Exponentials

Symbolic Partial Fraction Expansion

Verify the partial fraction expansion using SymPy.

Example 4.19: Special Case (\(a=b\))

When \(b=a\), the partial fraction expansion is different:

\[ Y(j \omega)=\frac{1}{(a+j \omega)^{2}} . \]

We can recognize this form using the dual of the differentiation property:
\(\frac{1}{(a+j \omega)^{2}}=j \frac{d}{d \omega}\left[\frac{1}{a+j \omega}\right]\).

Recall the Fourier Transform pair:
\(e^{-a t} u(t) \stackrel{\mathcal{F}}{\longleftrightarrow} \frac{1}{a+j \omega}\).

Using the differentiation in frequency property (\(\mathcal{F}\{t x(t)\} = j \frac{d}{d\omega} X(j\omega)\)):
\(t e^{-a t} u(t) \stackrel{\mathcal{F}}{\longleftrightarrow} j \frac{d}{d \omega}\left[\frac{1}{a+j \omega}\right]=\frac{1}{(a+j \omega)^{2}}\).

Therefore, for \(a=b\):
\(y(t)=t e^{-a t} u(t)\).

The case where the parameters \(a\) and \(b\) are equal leads to a repeated pole in the frequency domain.
Direct partial fraction expansion for repeated poles is slightly different, but the Fourier Transform properties provide a very elegant way to find the inverse transform.
Specifically, the differentiation in frequency property states that multiplying a signal by \(t\) in the time domain corresponds to differentiating its Fourier Transform with respect to \(\omega\) and multiplying by \(j\).
Since we know the Fourier Transform of \(e^{-at}u(t)\) is \(1/(a+j\omega)\), applying this property directly gives us the inverse transform of \(1/(a+j\omega)^2\) as \(t e^{-at}u(t)\).
This highlights the power and interconnectedness of Fourier Transform properties in solving complex signal processing problems.

Interactive Convolution of Exponentials

Adjust parameters \(a\) and \(b\) to observe \(x(t)\), \(h(t)\), and \(y(t)\).

viewof param_a = Inputs.range([0.1, 5], {step: 0.1, value: 1, label: "Parameter 'a'"})
viewof param_b = Inputs.range([0.1, 5], {step: 0.1, value: 2, label: "Parameter 'b'"})

Example 4.20: Convolution of Sinc Functions

Consider the response of an ideal lowpass filter (sinc impulse response) to a sinc input signal.
Input: \(x(t)=\frac{\sin \omega_{i} t}{\pi t}\)
Impulse Response: \(h(t)=\frac{\sin \omega_{c} t}{\pi t}\)

Their Fourier Transforms are rectangular pulses:
\(X(j \omega)= \begin{cases}1 & |\omega| \leq \omega_{i} \\ 0 & \text { elsewhere }\end{cases}\)
\(H(j \omega)= \begin{cases}1 & |\omega| \leq \omega_{c} \\ 0 & \text { elsewhere }\end{cases}\)

The output in the frequency domain is \(Y(j\omega)=X(j\omega)H(j\omega)\):

\[ Y(j \omega)= \begin{cases}1 & |\omega| \leq \omega_{0} \\ 0 & \text { elsewhere }\end{cases} \]

where \(\omega_{0} = \min(\omega_i, \omega_c)\).

The inverse Fourier Transform gives the output \(y(t)\):

\[ y(t)=\left\{\begin{array}{ll} \frac{\sin \omega_{c} t}{\pi t} & \text { if } \omega_{c} \leq \omega_{i} \\ \frac{\sin \omega_{i} t}{\pi t} & \text { if } \omega_{i} \leq \omega_{c} \end{array} .\right. \]

That is, the output is a sinc function with the smaller of the two cutoff frequencies.

This final example beautifully illustrates the filtering action. Both the input signal \(x(t)\) and the filter’s impulse response \(h(t)\) are sinc functions, which means their Fourier Transforms \(X(j\omega)\) and \(H(j\omega)\) are rectangular pulses.
When these two rectangular pulses are multiplied in the frequency domain, the result \(Y(j\omega)\) is another rectangular pulse whose width is determined by the narrower of the two input pulses.
This means that if the input signal’s bandwidth (\(\omega_i\)) is wider than the filter’s bandwidth (\(\omega_c\)), the filter truncates the signal’s spectrum, and the output signal’s bandwidth is limited by the filter. If the input’s bandwidth is narrower, the signal passes through the filter’s passband largely unchanged in its spectral shape.
The output \(y(t)\) is therefore a sinc function corresponding to this narrower bandwidth. This is a fundamental concept in filtering, spectrum shaping, and is integral to understanding processes like ideal sampling and reconstruction.

Interactive Convolution of Sinc Functions

Adjust \(\omega_i\) (input bandwidth) and \(\omega_c\) (filter bandwidth) and observe the spectra and signals.

viewof wi_sinc = Inputs.range([1, 10], {step: 0.1, value: 5, label: "Input Bandwidth (ωi)"})
viewof wc_sinc = Inputs.range([1, 10], {step: 0.1, value: 3, label: "Filter Bandwidth (ωc)"})