You can check back here next Wednesday or Thursday and the solutions will be posted along with the assignment.

In class we showed that

by arguing that the only value of the function f() that matters in the integral is when the argument of the impulse is zero. Using the same idea, evaluate the following integrals.

In problems a-d, find the value of "t" that make the argument of the impulse=0. Replace the value of "t" in f() with this value.

For problem e, do the same for the value of "λ".

The function below shows x(λ) vs λ, and the location of t. Draw δ(t-λ) and x(λ)δ(t-λ) and then give a simple one or two sentence explanation of why the sifting property is obvious (in retrospect). The sifting property states:

Note: δ(t-λ)=δ(λ-t).

Remember for this problem that the independent variable is λ, and t is a constant.

The image below shows the point x(t), and the functions δ(t-λ) and x(λ)δ(t-λ). Note that f(λ)=x(λ)δ(t-λ)=x(t)δ(t-λ), so clearly f(λ) is simply an impulse of area x(t). If we integrate this impulse function over all λ, we get its area, x(t). This proves the sifting theorem.

In class we showed that if the input, f(t), to a linear time-invariant system has y(t) as output, then for an input f'(t) (i.e., the derivative of f(t)) the output is given by y'(t).

Similarly show that if the input to a system is , then the output is . You can use a trapezoidal approximation for the integral. Note the variable λ is used in the integration as a dummy variable (since t is in the limits of integration, it can't be in the integral).

We can represent the input as the sum of two function f(t) and f(t+h). The output due to f(t) is y(t) (this is given). The output due to f(t+h) will be y(t+h) because the system is time invariant. So the output is given as shown.