Use the convolutional integral to find the response of an LTI system with impulse response ℎ() and input x(). Sketch the output signal. a) x() = () and ℎ() = () b) x() = c ( 2 ) and ℎ() = exp(−) () (You may write c ( 2 ) as ( + ) − ( − ).)

Use the convolution integral to determine the output of an LTI system with impulse response h(t) to input x(t) specified below.

h(t) = 4e^-6t-1 u(t+1) x(t)=2u(t – 2)

Explanation:

y(t)=∫0th(t−λ)f(λ)dλ

Start with a linear time-invariant (LTI) system (in box).  There is an input to the system, x(t), and an output from the system, y(t).

Now consider the same system with the input as an impulse, δ(t).  The output is then, by definition, the impulse response, h(t).

If we use a delayed impulse into the system, the output is just the impulse response with the same delay.

Because of linearity, if we scale the input by any factor, the output will be scaled by the same factor.  In particular, we can scale the input (and hence the output) by the factor x(λ)dλ.

If we integrate the input, the output is also integrated.  In this case we'll take the upper limit of the integration to be t+; but it could just as well have been positive infinity.  We could also take the lower lower limit to be negative infinity.

By the sifting property of the impulse function we know that (if λ>0, t>0):

∫0tδ(t−λ)x(λ)dλ=x(t)

Therefore the input to the system is x(t)

If the input is x(t), then the output must be y(t) (by definition, from the first diagram).  

Equating the outputs of the last two diagrams leads directly to the convolution theorem:

y(t)=∫0th(t−λ)f(λ)dλ

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