Both Part A and Part B ClosedForm Convolution Find the convo

(Closed-Form Convolution) Find the convolution y[n] = x[n] * h[n] for the following: x[n] = u[n] h[n) = u[n] x[n] = (0.8)^n u[n] h[n] = (0.4)^n u[n]

Solution

Consider a DT LTI system, L. x(n) L y(n)

DT convolution is based on an earlier result where we showed that any signal x(n) can be expressed as a sum of impulses.

x(n) = X k= x(k)(n k)

So let us consider x(n) written in this form to be our input to the LTI system.

y(n) = L [x(n)] = L * X k= x(k)(n k)

This looks like our general linear form with a scalar x(k) and a signal in n, (n k). Recall that for an LTI system:

• Linearity (L): ax1(n) + bx2(n) L ay1(n) + by2(n)

• Time Invariance (TI): x(n no) L y(n no)

We can use the property of linearity to distribute the system L over our input.

y(n) = L \" X k= x(k)(n k) # = X k= x(k)L [(n k)]

So now we wonder, what is L [(n k)]? Well, we can figure it out. Suppose we know how L acts on one impulse (n), and we call it

h(n) = L [(n)]

then by time invariance we get our answer

h(n k) = L [(n k)]

(n k) L h(n k)

This means that if we know one input-output pair for this system, namely

(n) L h(n)

then we can infer

x(n) L y(n)

which gives us the following.

y(n) = X k= x(k)h(n k)

A system’s impulse response, h(n), completely characterizes the behaviour of the system.

The impulse response h(n) can be generated directly from (n) through L, since (n) is an actual signal in DT. Thus, we can actually find the impulse response experimentally.

• DT convolution has an output variable n and a dummy variable k which causes a shift and flip.

DT y(n) = x(n) h(n) = X k= x(k)h(n k)

Convolution is a commutative operation, meaning signals can be convolved in any order.

x(n) h(n) = h(n) x(n)

This quite naturally is true of the convolution sums themselves, as well.

X k= x(k)h(n k) = X k= h(k)x(n – k).