## Properties of Convolution and LTI Systems

Linear Time Invariant (LTI) systems have properties that arise from the properties of convolution.

Property 1: Convolution is Commutative

$x_1(t)*x_2(t) = x_2(t)*x_1(t)\$

System Example: Convolving the input to a system with its impulse response is the same as convolving the impulse response with the input.

Property 2: Convolution is Distributive

$\displaystyle x_1(t)* \left ( x_2(t)+x_3(t) \right ) = x_1(t)*x_2(t)+x_1(t)*x_3(t)$

System Example: Convolving a single input with two impulse responses then adding the output is the same as convolving the input with the sum of the impulse responses.

Property 2: Convolution is Associative

$x_1(t)*x_2(t)*x_3(t) = x_2(t)*x_3(t)*x_1(t)\$

System Example: Convolving an input with an impulse response and convolving that with the impulse response of another system is the same as convolving the two impulse responses and then the input to the system.

## Convolution Simplification

Convolution of Unit Step Function:

To take a convolution, first determine whether the system is CT or DT and use the correct formula. Next it's time to simplify. Originally the bounds are set to negative and positive infinity. The unit step function will determine the new set of bounds. Consider the following unit step function as an example: $u(2t-1)\$. This function will be a zero as long as $(2t-1)$ is less than 0. Solve for t and apply the new bounds. Next its time for the real work!

Convolution of Delta Function:

Consider $\delta (ax+b)$. Simplify this convolution by solving for when the delta function is set to one. (This is when the $(ax+b)\$ is equal to zero). That is the only value of the integration or sum, so replace t accordingly and solve.

# Framework for computing the CT Convolution of two unit step exponentials

Let's take the convolution of the two most general unit-step exponentials in CT.

This solution can be very helpful in checking your work for convolutions of this form. Just plug in your numbers for the capital letters.

(I know this is kinda long, but it is very detailed to show the process of how to get to the general simplified solution.)

$x_1(t)=Ae^{Bt+C}u(Dt+E) \qquad x_2(t)=Fe^{Gt+H}u(It+J)$

\begin{align} x_1(t)*x_2(t) &= \int_{-\infty}^{\infty}x_1(\tau)x_2(t-\tau)d\tau \\ &=\int_{-\infty}^{\infty}Ae^{B\tau+C}u(D\tau+E)Fe^{G(t-\tau)+H}u(I(t-\tau)+J)d\tau \\ &=AF\int_{-\infty}^{\infty}e^{B\tau+C+G(t-\tau)+H}u(D\tau+E)u(It-I\tau+J)d\tau; \;(u(D\tau+E)=0\;,for\;D\tau+E<0\;\rightarrow\;\tau<\frac{-E}{D}) \\ &=AF\int_{\frac{-E}{D}}^{\infty}e^{\tau(B-G)+Gt+C+H}u(It-I\tau+J)d\tau; \;(u(It-I\tau+J)=0\;,for\;It-I\tau+J<0\;\rightarrow\;\tau>t+\frac{J}{I}) \\ &=AF\int_{\frac{-E}{D}}^{t+\frac{J}{I}}e^{\tau(B-G)+Gt+C+H}d\tau\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\ &=AFe^{Gt+C+H}\int_{\frac{-E}{D}}^{t+\frac{J}{I}}e^{\tau(B-G)}d\tau\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\ &=AFe^{Gt+C+H}\frac{1}{B-G}\left[e^{\tau(B-G)}\right]_{\frac{-E}{D}}^{t+\frac{J}{I}}\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\ &=AFe^{Gt+C+H}\frac{1}{B-G}(e^{(t+\frac{J}{I})\cdot(B-G)}-e^{\frac{-E}{D}\cdot(B-G)})\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\ &=\frac{AF}{B-G}(e^{Gt+CH+(t+\frac{J}{I})\cdot(B-G)}-e^{Gt+C+H-\frac{E}{D}(B-G)})\cdot u(t+\frac{J}{I}+\frac{E}{D}) \\ &=\frac{AF}{B-G}(e^{Bt+C+H+\frac{J}{I}(B-G)}-e^{Gt+C+H+\frac{E}{D}(G-B)})\cdot u(t+\frac{J}{I}+\frac{E}{D}) \end{align}

Example: Problem 2 on Fall 06 Midterm 1:

$Let:\;x_1(t)=x(t)=e^{-2t}u(t) \qquad x_2(t)=h(t)=u(t)$

$Thus:\;A=1,\;B=-2,\;C=0,\;D=1,\;E=0,\;F=1,\;G=0,\;H=0,\;I=1,\;J=0$

\begin{align} x(t)*h(t)&=x_1(t)*x_2(t) \\ &=\frac{1\cdot1}{-2-0}(e^{-2t+0+0+\frac{0}{1}(-2-0)}-e^{0t+0+0+\frac{0}{1}(0--2)})\cdot u(t+\frac{0}{1}+\frac{0}{1}) \\ &=\frac{-1}{2}(e^{-2t}-1)\cdot u(t) \\ &=\frac{1}{2}(1-e^{-2t})\cdot u(t) \end{align}

## Definition of Sampling Theorem

A band-limited signal can be recovered by sampling if the sampling frequency $\omega_s$ is greater than $2\omega_m$, where $\omega_m$ is the cut-off frequency. $T = \frac{2\pi}{\omega_s}$

Received $\frac{7}{10}$ because didn't specify cut-off frequency of what and should have used "recovered from sampling" instead of "recovered by sampling."

## Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett