Chapter 8, Problem 41E

Evaluate the following functions at t = −1, 0, and +3 (assume u(0) = 1): (a) f (t) = tu(1 − t); (b) g(t) = 8 + 2u(2 − t); (c) h(t) = u(t + 1) − u(t − 1) + u(t + 2) − u(t − 4); (d) z(t) = 1 + u(3 − t) + u(t − 2).

To determine

Find the value of function $f\left(t\right)$ for time $t=-1,0,\text{and}+3$.

Answer to Problem 41E

The value of $f\left(t\right)$ for time $t=-1$ is $-1$, the value of $f\left(t\right)$ for time $t=0$ is $0$, value of $f\left(t\right)$ for time $t=3$ is $0$.

Explanation of Solution

Given data:

$f\left(t\right)=tu\left(1-t\right)$        (1)

Formula used:

Write a general expression to calculate the unit step function.

$u\left(t\right)=\{\begin{array}{l}0fort<0\\ 1fort\ge 0\end{array}$

Calculation:

Substitute $-1$ for $t$ in equation (1) to find $f\left(-1\right)$.

$\begin{array}{c}f\left(-1\right)=\left(-1\right)u\left(1-\left(-1\right)\right)\\ =\left(-1\right)u\left(2\right)\\ =-1\left(1\right)\left\{\because u\left(t\right)=1\text{for}t>0\right\}\\ =-1\end{array}$

Substitute $0$ for $t$ in equation (1) to find $f\left(0\right)$.

$\begin{array}{c}f\left(0\right)=\left(0\right)u\left(1-0\right)\\ =\left(0\right)u\left(1\right)\\ =\left(0\right)\left(1\right)\left\{\because u\left(t\right)=1\text{for}t>0\right\}\\ =0\end{array}$

Substitute $+3$ for $t$ in equation (1) to find $f\left(+3\right)$.

$\begin{array}{c}f\left(+3\right)=\left(3\right)u\left(1-3\right)\\ =\left(3\right)u\left(-2\right)\\ =\left(3\right)\left(0\right)\left\{\because u\left(t\right)=0\text{for}t<0\right\}\\ =0\end{array}$

The value of $f\left(t\right)$ for different time is tabulated as follows:

$t$	$-1$	$0$	$+3$
$f\left(t\right)$	$-1$	$0$	$0$

Conclusion:

Thus, the value of $f\left(t\right)$ for time $t=-1$ is $-1$, the value of $f\left(t\right)$ for time $t=0$ is $0$, value of $f\left(t\right)$ for time $t=3$ is $0$.

To determine

Find the value of function $g\left(t\right)$ for time $t=-1,0,\text{and}+3$.

Answer to Problem 41E

The value of function $g\left(t\right)$ or time $t=-1$ is $10$, the value of $g\left(t\right)$ for time $t=0$ is $10$, value of $g\left(t\right)$ for time $t=3$ is $8$.

Explanation of Solution

Given data:

$g\left(t\right)=8+2u\left(2-t\right)$        (2)

Calculation:

Substitute $-1$ for $t$ in equation (2) to find $g\left(-1\right)$.

$\begin{array}{c}g\left(-1\right)=8+2u\left(2-\left(-1\right)\right)\\ =8+2u\left(3\right)\\ =8+2\left(1\right)\left\{\because u\left(t\right)=1\text{for}t>0\right\}\\ =10\end{array}$

Substitute $0$ for $t$ in equation (2) to find $g\left(0\right)$.

$\begin{array}{c}g\left(0\right)=8+2u\left(2-0\right)\\ =8+2u\left(2\right)\\ =8+2\left(1\right)\left\{\because u\left(t\right)=1\text{for}t>0\right\}\\ =10\end{array}$

Substitute $+3$ for $t$ in equation (2) to find $g\left(+3\right)$.

$\begin{array}{c}g\left(+3\right)=8+2u\left(2-3\right)\\ =8+2u\left(-1\right)\\ =8+2\left(0\right)\left\{\because u\left(t\right)=0\text{for}t<0\right\}\\ =8\end{array}$

The value of $g\left(t\right)$ for different time is tabulated as follows:

$t$	$-1$	$0$	$+3$
$g\left(t\right)$	$10$	$10$	$8$

Conclusion:

Thus, the value of function $g\left(t\right)$ or time $t=-1$ is $10$, the value of $g\left(t\right)$ for time $t=0$ is $10$, value of $g\left(t\right)$ for time $t=3$ is $8$.

To determine

Find the value of function $h\left(t\right)$ for time $t=-1,0,\text{and}+3$.

Answer to Problem 41E

The value of function $h\left(t\right)$ or time $t=-1$ is $2$, the value of $h\left(t\right)$ for time $t=0$ is $2$, value of $h\left(t\right)$ for time $t=3$ is $1$.

Explanation of Solution

Given data:

$h\left(t\right)=u\left(t+1\right)-u\left(t-1\right)+u\left(t+2\right)-u\left(t-4\right)$        (3)

Calculation:

Substitute $-1$ for $t$ in equation (3) to find $h\left(-1\right)$.

$\begin{array}{c}h\left(-1\right)=u\left(-1+1\right)-u\left(-1-1\right)+u\left(-1+2\right)-u\left(-1-4\right)\\ =u\left(0\right)-u\left(-2\right)+u\left(1\right)-u\left(-5\right)\\ =1-0+1-0\left\{\because u\left(t\right)=\{\begin{array}{l}0fort<0\\ 1fort\ge 0\end{array}\right\}\\ =2\end{array}$

Substitute $0$ for $t$ in equation (3) to find $h\left(0\right)$.

$\begin{array}{c}h\left(0\right)=u\left(0+1\right)-u\left(0-1\right)+u\left(0+2\right)-u\left(0-4\right)\\ =u\left(1\right)-u\left(-1\right)+u\left(2\right)-u\left(-4\right)\\ =1-0+1-0\left\{\because u\left(t\right)=\{\begin{array}{l}0fort<0\\ 1fort\ge 0\end{array}\right\}\\ =2\end{array}$

Substitute $+3$ for $t$ in equation (3) to find $h\left(+3\right)$.

$\begin{array}{c}h\left(+3\right)=u\left(3+1\right)-u\left(3-1\right)+u\left(3+2\right)-u\left(3-4\right)\\ =u\left(4\right)-u\left(2\right)+u\left(5\right)-u\left(-1\right)\\ =1-1+1-0\left\{\because u\left(t\right)=\{\begin{array}{l}0fort<0\\ 1fort\ge 0\end{array}\right\}\\ =1\end{array}$

The value of $h\left(t\right)$ for different time is tabulated as follows:

$t$	$-1$	$0$	$+3$
$h\left(t\right)$	$2$	$2$	$1$

Conclusion:

Thus, the value of function $h\left(t\right)$ or time $t=-1$ is $2$, the value of $h\left(t\right)$ for time $t=0$ is $2$, value of $h\left(t\right)$ for time $t=3$ is $1$.

To determine

Find the value of function $z\left(t\right)$ for time $t=-1,0,\text{and}+3$.

Answer to Problem 41E

The value of function $z\left(t\right)$ or time $t=-1$ is $2$, the value of $z\left(t\right)$ for time $t=0$ is $2$, value of $z\left(t\right)$ for time $t=3$ is $3$.

Explanation of Solution

Given data:

$z\left(t\right)=1+u\left(3-t\right)+u\left(t-2\right)$        (4)

Calculation:

Substitute $-1$ for $t$ in equation (4) to find $z\left(-1\right)$.

$\begin{array}{c}z\left(-1\right)=1+u\left(3-\left(-1\right)\right)+u\left(-1-2\right)\\ =1+u\left(4\right)+u\left(-3\right)\\ =1+1+0\left\{\because u\left(t\right)=\{\begin{array}{l}0fort<0\\ 1fort\ge 0\end{array}\right\}\\ =2\end{array}$

Substitute $0$ for $t$ in equation (4) to find $z\left(0\right)$.

$\begin{array}{c}z\left(0\right)=1+u\left(3-0\right)+u\left(0-2\right)\\ =1+u\left(3\right)+u\left(-2\right)\\ =1+1+0\left\{\because u\left(t\right)=\{\begin{array}{l}0fort<0\\ 1fort\ge 0\end{array}\right\}\\ =2\end{array}$

Substitute $+3$ for $t$ in equation (4) to find $z\left(+3\right)$.

$\begin{array}{c}z\left(+3\right)=1+u\left(3-3\right)+u\left(3-2\right)\\ =1+u\left(0\right)+u\left(1\right)\\ =1+1+1\left\{\because u\left(t\right)=\{\begin{array}{l}0fort<0\\ 1fort\ge 0\end{array}\right\}\\ =3\end{array}$

The value of $z\left(t\right)$ for different time is tabulated as follows:

$t$	$-1$	$0$	$+3$
$z\left(t\right)$	$2$	$2$	$3$

Conclusion:

Thus, the value of function $z\left(t\right)$ or time $t=-1$ is $2$, the value of $z\left(t\right)$ for time $t=0$ is $2$, value of $z\left(t\right)$ for time $t=3$ is $3$.