- #1

roam

- 1,271

- 12

## Homework Statement

For each of these sketch and provide a formula for the function (i.e. in terms of ##u(t)##, ##\delta(t)##) and its derivative and anti-derivative. Denote the ##\delta## function with a vertical arrow of length 1.

**(a)**##f(t)=\frac{|t|}{t}##

**(b)**##f(t)=u(t) exp(-t)##

## Homework Equations

$$u'(t)=\delta(t)$$

$$\int^t_{-\infty} \delta (\tau) d \tau = u(t)$$

## The Attempt at a Solution

**(a)**##f(t)=\frac{|t|}{t}## in terms of the Heaviside unit step can be written as:

$$f(t)=2 u(t)-1$$

Using the property stated above the derivative becomes:

$$f'(t)=2 \delta(t)$$

But shouldn't it be zero? Because when I take the "normal" derivative of ##f(t)=\frac{|t|}{t}## it becomes 0 (derivative of -1 for t<0 an 1 for t>0).

Here is the sketch of f(t) and the two possible versions of f'(t):

For the antiderivative, if we integrate the original function we will have ##\int f(t) dt=|x| + c##. But I am not sure how to represent it in terms of ##\delta## because we do not know the indefinite integral of u(t).

**(b)**So here I made a graph of this function and its derivative:

From the graph for t<0 the derivative of zero is just 0. For t>0 derivative of e

^{-t}is -e

^{-t}(this is also equal to this function's anti-derivative).

But now if we differentiate this in terms of delta using the product rule, we get ##f'(t)=-u(t)e^{-t}+e^{-t} \delta (t)##. This is very different from the previous answer. And how can I represent ##e^{-t} \delta (t)## on the graph?

I have noticed that in some websites they draw the delta function as just being scaled by the size of the jump discontinuity, so in this case:

So I am not sure which one of these solutions is correct. Any help is greatly appreciated.