2.3: Convolution

Learning Objectives: CT Convolution

1. Understand how the convolution integral is a consequence of the linearity and time-invariance properties
2. Interpret the convolution integral as three steps: flip and shift, then multiply with \(x(\tau)\) to form a product signal, then compute the area under the product signal
3. Practice identifying the different regions for the convolution integral expression.

YouTube Videos

CT Convolution Integral

CT Convolution Example

(Optional) CT Convolution with Barker Code Sequences. Credit to Prof. Kathleen E. Wage

Summary

Any arbitrary continuous-time signal can be represented as idealized pulses with vanishingly small duration. Let’s consider \(\hat{x}(t)\) the pulse or staircase approximation of the continuous-time signal \(x(t)\), see Figure 2.3.1.

Figure 2.3.1: Illustration of the compensation of an arbitrary continuous signal following the staircase approximation.

Thus \[ \hat{x}(t)=\sum_{k=-\infty}^{\infty} x(k\Delta)\,\delta_{\Delta}(t-k\Delta)\,\Delta \] where \[ \delta_{\Delta}(t)= \begin{cases} \frac{1}{\Delta}, & 0\le t \le \Delta\\ 0, & \text{otherwise} \end{cases} \] If \(\Delta \to 0,\) \[ x(t)=\int_{-\infty}^{+\infty} x(\tau)\,\delta(t-\tau)\,d\tau \] shifting property of a CT impulse

Example:

\[ u(t)=\int_{-\infty}^{+\infty} u(\tau)\,\delta(t-\tau)\,d\tau \]

\(u(\tau)=0\) for \(\tau<0\)
\(u(\tau)=1\) for \(\tau\ge 0\)

Note that

\[ \int_{-\infty}^{+\infty} x(\tau)\,\delta(t-\tau)\,d\tau = \int_{-\infty}^{+\infty} x(t)\,\delta(t-\tau)\,d\tau = x(t)\int_{-\infty}^{+\infty}\delta(t-\tau)\,d\tau = x(t) \]

\(x(\tau)\,\delta(t-\tau)=x(t)\,\delta(t-\tau)\)

The response/output \(\hat{y}(t)\) of a linear system is the superposition of the responses to the scaled and shifted versions of \(\delta_{\Delta}(t)\)\[ \hat{y}(t)=\sum_{k=-\infty}^{\infty} x(k\Delta)\,\hat{h}_{k\Delta}(t)\,\Delta \] when \(\Delta\to 0\) and the summation becomes an integral

CONVOLUTION INTEGRAL

\[ y(t)=\int_{-\infty}^{+\infty} x(\tau)\,h_{\tau}(t)\,d\tau \]

Response of a linear-time invariant system in continuous time

\(h_{\tau}(t)=h_0(t-\tau)\) response of the LTI system to a shifted unit impulse \(\delta(t-\tau)\)

Unit response \(h(t)=h_0(t)\) “centered” unit impulse

\[ y(t)=x(t)*h(t) \]

Figure 2.3.2: Animation showing the convolution between two rectangular functions.

A continuous-time LTI system is completely characterized by its unit response (i.e., the system’s response to a single elementary signal \(\delta(t)\))

Example

\(x(t)=e^{-at}u(t)\) for \(a>0\) and \(h(t)=u(t)\)

Figure 2.3.3: Animation showing the convolution (i.e., y(t)) between a decaying exponential signal (i.e., x(t)) and a unit step signal (i.e., h(t)).

For \(t<0\), \(x(\tau)\,h(t-\tau)=0\) since there is no an overlapping region \(\Rightarrow\; y(t)=0\)

For \(t>0\), \[ x(\tau)\,h(t-\tau)= \begin{cases} e^{-a\tau}, & 0<\tau<t\\ 0, & \text{otherwise} \end{cases} \]

\[ y(t)=\int_{0}^{t} e^{-a\tau}\,d\tau =\left.-\frac{1}{a}e^{-a\tau}\right|_{0}^{t} =\frac{1}{a}-\frac{1}{a}e^{-at} =\frac{1}{a}(1-e^{-at}) \]

In general: \[ y(t)=\frac{1}{a}(1-e^{-at})\,u(t) \]

Example

\[ x(t)= \begin{cases} 1, & 0<t<T\\ 0, & \text{otherwise} \end{cases} \]

\[ h(t)= \begin{cases} t, & 0<t<2T\\ 0, & \text{otherwise} \end{cases} \]

Figure 2.3.4: Animation showing the convolution (i.e., y(t)) between a rectangular signal (i.e., x(t)) and a linear ramp (i.e., h(t))).

for \(t<0\), \(y(t)=0\)

for \(t>3T\), \(y(t)=0\)

\[ y(t)= \begin{cases} 0, & t<0\\[4pt] \frac{1}{2}t^2, & 0<t<T \;\;\text{I}\\[4pt] Tt-\frac{1}{2}T^2, & T<t<2T \;\;\text{II}\\[4pt] -\frac{1}{2}t^2+Tt+\frac{3}{2}T^2, & 2T<t<3T\\[4pt] 0, & t>3T \end{cases} \]

Example

\(x(t)=e^{2t}u(-t)\) and \(h(t)=u(t-3)\)

Figure 2.3.5: Animation showing the convolution (i.e., y(t)) between a growing exponential signal (i.e., x(t)) and a laterally-shifted unit step signal (i.e., h(t))).

\(x(t)\) and \(h(t)\) always overlap regardless the \(t\) value

When \(t-3\le 0\) (i.e., \(t\le 3\)), \[ y(t)=\int_{-\infty}^{t-3} e^{2\tau}\,d\tau=\frac{1}{2}e^{2(t-3)} \]

For \(t-3\ge 0\) (i.e., \(t\ge 3\)), \[ y(t)=\int_{-\infty}^{0} e^{2\tau}\,d\tau=\frac{1}{2} \] total overlap between \(x(t)\) and \(h(t)\)

Unit impulse

\(x(t)=x(t)*\delta(t)=\displaystyle \int_{-\infty}^{+\infty} x(t-\tau)\,\delta(\tau)\,d\tau\)

\(\delta(t)=\delta(t)*\delta(t)\)

Properties of the unit impulse

1) \(x(t)=x(t)*\delta(t)=\delta(t)*x(t)=\displaystyle \int_{-\infty}^{+\infty}\delta(\tau)\,x(t-\tau)\,d\tau\)

\[ 1=\int_{-\infty}^{+\infty}\delta(\tau)\,d\tau \] -> the unit impulse has unit area, i.e., area of 1.

2) \(g(-t)=g(-t)*\delta(t)=\delta(t)*g(-t)=\displaystyle \int_{-\infty}^{+\infty}\delta(\tau)\,g(\tau-t)\,d\tau\)

@ \(t=0\) \(g(0)=\displaystyle \int_{-\infty}^{+\infty}\delta(\tau)\,g(\tau)\,d\tau\)

3) \(f(t)\cdot \delta(t)=f(0)\cdot \delta(t)\)