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s2008m [1.1K]
3 years ago
7

(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. wr

ite the function in terms of unit step function
Mathematics
1 answer:
sergij07 [2.7K]3 years ago
5 0
f(t)=\begin{cases}0&\text{for }t

Recall that

u(t)=\begin{cases}0&\text{for }t

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

-5u(t)=\begin{cases}0&\text{for }t

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

-5u(t-1)=\begin{cases}0&\text{for }t

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

-5u(t)+5u(t-1)=\begin{cases}0&\text{for }t

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

-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

so that

-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)=\begin{cases}0&\text{for }t

For the last piece, we add one more term:

6u(t-7)=\begin{cases}0&\text{for }t

and so putting everything together, we get f(t):

f(t)\equiv-5u(t)+5u(t-1)-6u(t-1)+6u(t-7)+6u(t-7)
f(t)\equiv-5u(t)-u(t-1)+12u(t-7)
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