contestada

(1 point) consider the function f(t)=⎧⎩⎨⎪⎪⎪⎪0,−5,−6,6,t<00≤t<11≤t<7t≥7;f(t)={0,t<0−5,0≤t<1−6,1≤t<76,t≥7; 1. write the function in terms of unit step function

Respuesta :

LammettHash
[tex]f(t)=\begin{cases}0&\text{for }t<0\\-5&\text{for }0\le t<1\\-6&\text{for }1\le t<7\\6&\text{for }t\ge7\end{cases}[/tex]

Recall that

[tex]u(t)=\begin{cases}0&\text{for }t<0\\1&\text{for }t\ge0\end{cases}[/tex]

Take it one piece at a time. For [tex]t\ge0[/tex], we can scale [tex]u(t)[/tex] by -5:

[tex]-5u(t)=\begin{cases}0&\text{for }t<0\\-5&\text{for }t\ge0\end{cases}[/tex]

If we shift the argument by 1 and scale by -5, we have

[tex]-5u(t-1)=\begin{cases}0&\text{for }t<1\\-5&\text{for }t\ge1\end{cases}[/tex]

so if we subtract this from [tex]-5u(t)[/tex], we'll end up with

[tex]-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t<0\text{ and }t\ge1\\-5&\text{for }0\le t<1\end{cases}[/tex]

For the next piece, we can add another scaled and shifted step like

[tex]-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t<1\text{ and }t\ge7\\-6&\text{for }1\le t<7\end{cases}[/tex]

so that

[tex]-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t<0\text{ and }t\ge7\\-5&\text{for }0\le t<1\\-6&\text{for }1\le t<7\end{cases}[/tex]

For the last piece, we add one more term:

[tex]6u(t-7)=\begin{cases}0&\text{for }t<7\\6&\text{for }t\ge7\end{cases}[/tex]

and so putting everything together, we get [tex]f(t)[/tex]:

[tex]f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)[/tex]
[tex]f(t)\equiv-5u(t)-u(t-1)+12u(t-7)[/tex]

Otras preguntas