Math / Algebra

QuestionThe function $f(t)=\left\{\begin{array}{cc} 3 & \text { if } 0<t \leq 1 \\ 0 & \text { if } 1<t \leq 4 \\ 2 & \text { it } t>4 \end{array}\right.$ can bewritten as a sum of unit step functions $f(t)=a+b u(t-c)+d u(t-e)$ wherea= $\square$ , $b=$ $\square$ , $c=$ $\square$ , d= $\square$ , ande = $\square$

Studdy Solution

STEP 1

What is this asking? We need to rewrite a piecewise function $f(t)$ using unit step functions $u(t-c)$ , which basically act like on/off switches at specific times $c$ . Watch out! Be careful with the signs of the coefficients when switching between pieces of the function!

STEP 2

1. Define the function initially
2. Adjust the function at $t=1$
3. Adjust the function at $t=4$

STEP 3

Alright, let's start with the first piece!
For $0 < t \leq 1$ , $f(t)$ is 3.
This is our initial value, so we set $a = \textbf{3}$ .
So far, we have $f(t) = 3$ .
Easy peasy!

STEP 4

Now, at $t=1$ , the function jumps down to 0.
This means we need to subtract 3 from our current function when $t$ becomes greater than 1.
We do this using a unit step function: $f(t) = 3 - 3u(t-1)$ .

STEP 5

The unit step function $u(t-1)$ is like a switch that turns on at $t=1$ .
Before $t=1$ , it's 0, so our function is just $3$ .
After $t=1$ , it's 1, so our function becomes $3 - 3 \cdot 1 = 0$ , exactly what we want!
So, we have $b = \textbf{-3}$ and $c = \textbf{1}$ .

STEP 6

Finally, at $t=4$ , the function jumps up to 2.
Since we're currently at 0, we need to add 2 to our function when $t$ becomes greater than 4.
We use another unit step function: $f(t) = 3 - 3u(t-1) + 2u(t-4)$ .

STEP 7

Similar to before, $u(t-4)$ is off before $t=4$ and on after $t=4$ .
So, before $t=4$ , our function remains at $0$ .
After $t=4$ , it becomes $3 - 3 \cdot 1 + 2 \cdot 1 = 2$ , perfect!
Thus, $d = \textbf{2}$ and $e = \textbf{4}$ .

STEP 8

We found $a = \textbf{3}$ , $b = \textbf{-3}$ , $c = \textbf{1}$ , $d = \textbf{2}$ , and $e = \textbf{4}$ .
So, our final representation of $f(t)$ using unit step functions is: $f(t) = 3 - 3u(t-1) + 2u(t-4)$

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Studdy Solution

STEP 1

What is this asking? We need to rewrite a piecewise function $f(t)$ using unit step functions $u(t-c)$ , which basically act like on/off switches at specific times $c$ . Watch out! Be careful with the signs of the coefficients when switching between pieces of the function!

STEP 2

1. Define the function initially
2. Adjust the function at $t=1$
3. Adjust the function at $t=4$

STEP 3

Alright, let's start with the first piece!
For $0 < t \leq 1$ , $f(t)$ is 3.
This is our initial value, so we set $a = \textbf{3}$ .
So far, we have $f(t) = 3$ .
Easy peasy!

STEP 4

Now, at $t=1$ , the function jumps down to 0.
This means we need to subtract 3 from our current function when $t$ becomes greater than 1.
We do this using a unit step function: $f(t) = 3 - 3u(t-1)$ .

STEP 5

The unit step function $u(t-1)$ is like a switch that turns on at $t=1$ .
Before $t=1$ , it's 0, so our function is just $3$ .
After $t=1$ , it's 1, so our function becomes $3 - 3 \cdot 1 = 0$ , exactly what we want!
So, we have $b = \textbf{-3}$ and $c = \textbf{1}$ .

STEP 6

Finally, at $t=4$ , the function jumps up to 2.
Since we're currently at 0, we need to add 2 to our function when $t$ becomes greater than 4.
We use another unit step function: $f(t) = 3 - 3u(t-1) + 2u(t-4)$ .

STEP 7

Similar to before, $u(t-4)$ is off before $t=4$ and on after $t=4$ .
So, before $t=4$ , our function remains at $0$ .
After $t=4$ , it becomes $3 - 3 \cdot 1 + 2 \cdot 1 = 2$ , perfect!
Thus, $d = \textbf{2}$ and $e = \textbf{4}$ .

STEP 8

We found $a = \textbf{3}$ , $b = \textbf{-3}$ , $c = \textbf{1}$ , $d = \textbf{2}$ , and $e = \textbf{4}$ .
So, our final representation of $f(t)$ using unit step functions is: $f(t) = 3 - 3u(t-1) + 2u(t-4)$

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Studdy Solution

STEP 1

What is this asking? We need to rewrite a piecewise function f(t)f(t)f(t) using unit step functions u(t−c)u(t-c)u(t−c), which basically act like on/off switches at specific times ccc. Watch out! Be careful with the signs of the coefficients when switching between pieces of the function!

STEP 2

1. Define the function initially2. Adjust the function at t=1t=1t=13. Adjust the function at t=4t=4t=4

STEP 3

Alright, let's **start** with the first piece!For 0<t≤10 < t \leq 10<t≤1, f(t)f(t)f(t) is **3**.This is our **initial value**, so we set a=3a = \textbf{3}a=3.So far, we have f(t)=3f(t) = 3f(t)=3.Easy peasy!

STEP 4

Now, at t=1t=1t=1, the function **jumps down** to **0**.This means we need to **subtract 3** from our current function when ttt becomes greater than 1.We do this using a unit step function: f(t)=3−3u(t−1)f(t) = 3 - 3u(t-1)f(t)=3−3u(t−1).

STEP 5

STEP 6

Finally, at t=4t=4t=4, the function **jumps up** to **2**.Since we're currently at **0**, we need to **add 2** to our function when ttt becomes greater than 4.We use another unit step function: f(t)=3−3u(t−1)+2u(t−4)f(t) = 3 - 3u(t-1) + 2u(t-4)f(t)=3−3u(t−1)+2u(t−4).

STEP 7

Similar to before, u(t−4)u(t-4)u(t−4) is off before t=4t=4t=4 and on after t=4t=4t=4.So, before t=4t=4t=4, our function remains at 000.After t=4t=4t=4, it becomes 3−3⋅1+2⋅1=23 - 3 \cdot 1 + 2 \cdot 1 = 23−3⋅1+2⋅1=2, perfect!Thus, d=2d = \textbf{2}d=2 and e=4e = \textbf{4}e=4.

STEP 8

We found a=3a = \textbf{3}a=3, b=-3b = \textbf{-3}b=-3, c=1c = \textbf{1}c=1, d=2d = \textbf{2}d=2, and e=4e = \textbf{4}e=4.So, our final representation of f(t)f(t)f(t) using unit step functions is: f(t)=3−3u(t−1)+2u(t−4) f(t) = 3 - 3u(t-1) + 2u(t-4) f(t)=3−3u(t−1)+2u(t−4)

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Studdy solves anything!

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We need to rewrite a piecewise function $f(t)$ using unit step functions $u(t-c)$ , which basically act like on/off switches at specific times $c$ . Watch out! Be careful with the signs of the coefficients when switching between pieces of the function!

1. Define the function initially
2. Adjust the function at $t=1$
3. Adjust the function at $t=4$

Alright, let's start with the first piece!
For $0 < t \leq 1$ , $f(t)$ is 3.
This is our initial value, so we set $a = \textbf{3}$ .
So far, we have $f(t) = 3$ .
Easy peasy!

Now, at $t=1$ , the function jumps down to 0.
This means we need to subtract 3 from our current function when $t$ becomes greater than 1.
We do this using a unit step function: $f(t) = 3 - 3u(t-1)$ .

Finally, at $t=4$ , the function jumps up to 2.
Since we're currently at 0, we need to add 2 to our function when $t$ becomes greater than 4.
We use another unit step function: $f(t) = 3 - 3u(t-1) + 2u(t-4)$ .

Similar to before, $u(t-4)$ is off before $t=4$ and on after $t=4$ .
So, before $t=4$ , our function remains at $0$ .
After $t=4$ , it becomes $3 - 3 \cdot 1 + 2 \cdot 1 = 2$ , perfect!
Thus, $d = \textbf{2}$ and $e = \textbf{4}$ .

We found $a = \textbf{3}$ , $b = \textbf{-3}$ , $c = \textbf{1}$ , $d = \textbf{2}$ , and $e = \textbf{4}$ .
So, our final representation of $f(t)$ using unit step functions is: $f(t) = 3 - 3u(t-1) + 2u(t-4)$