8 CT Convolution

8.1 Review CT LTI systems and superposition property

Recall the superposition property of LTI systems. If a CT system is LTI then the superposition property holds. Given a system where \[x_i(t) \mapsto y_i(t) \; \forall\; i\] then \[\sum\limits_{i} a_i x_i(t) \mapsto \sum\limits_{i} a_i y_i(t)\]

Superposition enables a powerful problem reduction strategy. The overall idea for is that if:

we can write an arbitrary signal as a sum of simple signals, and
we can determine the response to the simple signals, then
we can easily express the output due to the input using superposition

This will be a recurring pattern in this course. In this lecture, the simple signals are weighted, time shifts of one signal, the delta function, \(\delta(t)\).

8.2 Convolution Integral

To derive this we start with the sifting property of the CT impulse function (from chapter 2) \[\int\limits_{a}^{b} x(t)\delta(t-t_0) \; dt = x(t_0)\] for any \(a < t_0 < b\). A slight change of variables (\(t_0 \rightarrow \tau\)) and limits (\(a \rightarrow -\infty\) and \(b \rightarrow \infty\)) gives: \[x(t) = \int\limits_{-\infty}^{\infty} x(\tau)\delta(t-\tau) \; d\tau\] showing that we can write any CT signal as an infinite sum (integral) of weighted and time-shifted impluse functions.

Let \(h(t)\) be the CT impulse response, the output due to the input \(\delta(t)\), i.e. \(\delta(t) \mapsto h(t)\). Then if the system is time-invariant: \(\delta(t-\tau) \mapsto h(t-\tau)\) and by superposition if the input is writen as \[x(t) = \int\limits_{-\infty}^{\infty} x(\tau)\delta(t-\tau) \; d\tau\] then the output is given by \[y(t) = \int\limits_{-\infty}^{\infty} x(\tau)h(t-\tau) \; d\tau = x(t) * h(t)\] This is called the convolution integral .

It is worth pausing here to see the signifigance. For a LTI CT system, if I know its impulse response \(h(t)\), I can find the response due to any input using convolution. For this reason the impulse response is another way to represent an LTI system.

8.3 Graphical View of the Convolution Integral.

Let us break the convolution expression down into pieces. In its general form the convolution of two signals \(x_1(t)\) and \(x_2(t)\) is \[x_1(t) * x_2(t) = \int\limits_{-\infty}^{\infty} x_1(\tau)x_2(t-\tau) \; d\tau\]

Suppose \(x_1(t)\) and \(x_2(t)\) are signals that look like

diagram showing the signals being used for convolution — The two signals being convolved.

Then \(x_1(\tau)\) and \(x_2(-\tau)\) look like

diagram showing the signals being used for convolution, see caption — The second signal reflected.

The signal \(x_2(t-\tau)\) is \(x_2(-\tau)\) shifted by \(t\) (since \(x_2(-\tau+t)= x_2(t-\tau)\)) and then looks like

Then the integrand of convolution is the product \(x_1(\tau)x_2(t-\tau)\) whose plot depends of the value of \(t\). Some examples, where the individual signals are dashed and their product is in bold:

Then convolution is the total integral of the product (bold curves above) for that value of \(t\). For the example above we see the integral will be zero for \(t\) less than \(t_0\) since the two signals do not overlap and their product is zero. For \(t_0 < t < t_1\) the signals overap and the product is non-zero, and the effective bounds of integration are \([t_0,t]\). For \(t > t_1\) the signals again overap and the product is non-zero, but the effective bounds of integration are \([t_0,t_1]\).

8.4 Examples of CT Convolution

Example

Consider the convolution of two unit step functions. \[u(t) * u(t) = \int\limits_{-\infty}^{\infty} u(\tau)u(t-\tau) \; d\tau\] The product \(u(\tau) u(t-\tau)\) is non-zero only when \(t\geq 0\) as illustrated here

The convolution integral is then the shaded area \[u(t) * u(t) = \left\{ \begin{array}{lc} 0 & t< 0\\ \int\limits_{0}^{t} d\tau = t & t \geq 0\\ \end{array}\right.\] Combining this back into a single expression gives: \[u(t) * u(t) = tu(t)\] Thus the convolution of two step signals is a ramp signal.

Example

Let \(x_1(t) = u(t)\) and \(x_2(t) = e^{-at}u(t)\) for constant \(a\in\mathbb{C}\), then \[u(t) * e^{-at}u(t) = \int\limits_{-\infty}^{\infty} u(\tau)e^{-a(t-\tau)}u(t-\tau) \; d\tau\] Similar to the previous example, the product \(u(\tau) e^{-a(t-\tau)} u(t-\tau)\) is non-zero only when \(t\geq 0\)

The convolution integral is then the shaded area \[u(t) * e^{-at}u(t) = \left\{ \begin{array}{lc} 0 & t< 0\\ \int\limits_{0}^{t} e^{-a(t-\tau)} d\tau = \frac{1-e^{-at}}{a} & t \geq 0\\ \end{array}\right.\] Combining this back into a single expression gives: \[u(t) * e^{-at}u(t) = \frac{1-e^{-at}}{a}u(t)\]

Example

Let \(x_1(t) = \delta(t)\) and \(x_2(t)\) be an arbitrary signal. Then \[\delta(t) * x_2(t) = \int\limits_{-\infty}^{\infty} \delta(\tau)x_2(t-\tau) \; d\tau\] By the sifting property of the delta function this evaluates to \[\delta(t) * x_2(t) = x_2(t)\] or in other words convolution with a delta function just results in the signal it was convolved with. That is it acts like the identity function, with respect to convolution.

The appendix lists several CT convolution results.

8.5 Properties of CT Convolution

There are several useful properties of convolution. We do not prove these here, but it is not terribly difficult to do so. Given signals \(x_1(t)\), \(x_2(t)\), and \(x_3(t)\):

Communative Property: The ordering of the signals does not matter. \[x_1(t) * x_2(t) = x_2(t) * x_1(t)\]
Distributive Property: Convolution is distributed over addition. \[x_1(t) * \left[x_2(t) + x_3(t)\right] = \left[x_1(t) * x_2(t) \right] + \left[x_1(t) * x_3(t) \right]\]
Associative Property: The order of convolution does not matter. \[x_1(t) * \left[x_2(t) * x_3(t)\right] = \left[x_1(t) * x_2(t) \right] * x_3(t)\]
Time Shift: Given \(x_3(t) = x_1(t) * x_2(t)\) then for time shifts \(\tau_1, \tau_2 \in \mathbb{R}\) \[x_1(t-\tau_1) * x_2(t-\tau_2) = x_3(t-\tau_1 - \tau_2)\]
Multiplicative Scaling: Given \(x_3(t) = x_1(t) * x_2(t)\) then for constants \(a,b \in \mathbb{C}\) \[\left[a\, x_1(t)\right] * \left[b\, x_2(t)\right] = a\, b\, x_3(t)\]

These properties can be used in combination with a table like that above to compute the convolution of a wide variety of signals without evaluating the integrals.

Example

Here is a simple example. Let \(x_1(t) = e^tu(t)\) and \(x_2(t) = 2\delta(t) + 5e^{-3t}u(t)\). \[x_1(t) * x_2(t) = e^tu(t) * \left[2\delta(t) + 5e^{-3t}u(t)\right]\] Using the distributive property \[x_1(t) * x_2(t) = 2\left[\delta(t) * e^tu(t)\right] + 5\left[e^tu(t) * e^{-3t}u(t)\right]\] Using previously derived results involving the delta function and the table row 3 \[x_1(t) * x_2(t) = 2 e^t\, u(t) + 5\left[ \frac{e^t-e^{-3t}}{4}\right]u(t)\] Doing some simplification gives the result \[x_1(t) * x_2(t) = \left[ \frac{13}{4}e^t-\frac{5}{4}e^{-3t}\right]u(t)\]

Example

Here is a more complicated example. Let \(x_1(t) = 2e^{-5t}u(t-1)\) and \(x_2(t) = \left(1-e^{-t}\right)u(t)\). \[x_1(t) * x_2(t) = \left[2e^{-5t}u(t-1)\right] * \left[\left(1-e^{-t}\right)u(t)\right]\] We first rewrite \(e^{-5t}u(t-1)=e^{-5}e^{-5(t-1)}u(t-1) = e^{-5}e^{-5t}u(t)\Big|_{t=t-1}\) so that we can remove the time shift \[x_1(t) * x_2(t) = 2e^{-5}\left[e^{-5t}u(t)\right] * \left[\left(1-e^{-t}\right)u(t)\right]\Big|_{t=t-1}\] We now apply the distributive property \[x_1(t) * x_2(t) = 2e^{-5}\left[\left(e^{-5t}u(t) * u(t)\right) - \left(e^{-5t}u(t)* e^{-t}u(t)\right)\right]\Big|_{t=t-1}\] Using the table rows 1 and 3 we get \[x_1(t) * x_2(t) = 2e^{-5}\left[\frac{1}{5}\left(1-e^{-5t}\right)u(t) + \frac{1}{4}\left(e^{-5t} - e^{-t}\right)u(t)\right]\Big|_{t=t-1}\] Combining terms we simplify to \[x_1(t) * x_2(t) = 2e^{-5}\left[\frac{1}{5} - \frac{1}{4}e^{-t} + \frac{1}{20}e^{-5t} \right]u(t)\Big|_{t=t-1}\] Replacing the time shift gives the final result \[x_1(t) * x_2(t) = 2e^{-5}\left[\frac{1}{5} - \frac{1}{4}e^{-(t-1)} + \frac{1}{20}e^{-5(t-1)} \right]u(t-1)\] which can be cleaned up a bit more by distributing the leading term \[x_1(t) * x_2(t) =\left[\frac{2}{5}e^{-5} -\frac{1}{2}e^{-(t+4)} +\frac{1}{10}e^{-5t}\right]u(t-1)\]