Question

Using the heaviside function write down the piecewise function that is 0 for t < 0 , t2 for t in [0,1] and t for t > 1 .

Answer 1

The piecewise function, using the Heaviside function, is defined as 0 for t < 0, t^2 for t in [0, 1], and t for t > 1.

Step-by-step explanation:

The piecewise function can be defined using the Heaviside function.

We know that the Heaviside function, H(x), is defined as:

H(x) = 0 for x < 0

H(x) = 1 for x ≥ 0

Using the Heaviside function, the piecewise function can be written as:

f(t) = 0 for t < 0

f(t) = t2 for t ∈ [0, 1]

f(t) = t for t ≥ 1

Answer 2

Recall that


`\Theta(t)=\begin{cases}0&\text{for }t<0\\1&\text{for }t\ge0\end{cases}`

We're looking for an equivalent form of


`f(t)=\begin{cases}0&\text{for }t<0\\t^2&\text{for }t\in[0,1]\\t&\text{for }t>1\end{cases}`

in terms of
`\Theta(t)`. For starters, we have


`t^2\Theta(t)=\begin{cases}0&\text{for }t<0\\t^2&\text{for }t\ge0\end{cases}`

which takes care of the negative domain, but we have "excess function" for
`t>1`. To get rid of it, note that


`t^2\Theta(t-1)=\begin{cases}0&\text{for }t<1\\t^2&\text{for }t\ge1\end{cases}`

so we can subtract this to get


`t^2\bigg(\Theta(t)-\Theta(t-1)\bigg)=\begin{cases}0&\text{for }t<0\text{ and }t>1\\t^2&\text{for }t\in[0,1]\end{cases}`

Finally, we just add another shifted Heaviside function scaled by
`t`:


`f(t)\equiv t^2\bigg(\Theta(t)-\Theta(t-1)\bigg)+t\Theta(t-1)`

which can also be expressed as


`f(t)\equiv t^2\Theta(t)+(t-t^2)\Theta(t-1)`