Chapter 8, Problem 42E

(a)

To determine

Sketch the functions over the range $-3\le t\le 3$.

Explanation of Solution

Given data:

The given function is:

$v\left(t\right)=3-u\left(2-t\right)-2u\left(t\right)\text{ V}$ (1)

The range of $t$ is $-3\le t\le 3$.

Calculation:

The unit-step forcing function as a function of time which is zero for all values of its argument less than zero and which is unity for all positive values of its argument.

$u\left(t-t_0\right)=\{\begin{array}{l}0t<t_0\\ 1t>t_0\end{array}$

Here,

$t_0$ is the time at which argument of the unit step function becomes zero.

Substitute $-3$ for $t$ in equation (1).

$\begin{array}{c}v\left(-3\right)=3-u\left(2-\left(-3\right)\right)-2u\left(-3\right)\text{ V}\\ =3-u\left(2+3\right)-2u\left(-3\right)\text{ V}\\ =3-u\left(5\right)-2u\left(-3\right)\text{ V}\\ =3-\left(1\right)-2\left(0\right)\text{ V}\end{array}$

$\begin{array}{c}v\left(-3\right)=3-1-0\text{ V}\\ =2\text{ V}\end{array}$

Substitute $-2$ for $t$ in equation (1).

$\begin{array}{c}v\left(-2\right)=3-u\left(2-\left(-2\right)\right)-2u\left(-2\right)\text{ V}\\ =3-u\left(2+2\right)-2u\left(-2\right)\text{ V}\\ =3-u\left(4\right)-2u\left(-2\right)\text{ V}\\ =3-\left(1\right)-2\left(0\right)\text{ V}\end{array}$

$\begin{array}{c}v\left(-2\right)=3-1-0\text{ V}\\ =2\text{ V}\end{array}$

Substitute $-1$ for $t$ in equation (1).

$\begin{array}{c}v\left(-1\right)=3-u\left(2-\left(-1\right)\right)-2u\left(-1\right)\text{ V}\\ =3-u\left(2+1\right)-2u\left(-1\right)\text{ V}\\ =3-u\left(3\right)-2u\left(-1\right)\text{ V}\\ =3-\left(1\right)-2\left(0\right)\text{ V}\end{array}$

$\begin{array}{c}v\left(-1\right)=3-1-0\text{ V}\\ =2\text{ V}\end{array}$

Substitute $0$ for $t$ in equation (1).

$\begin{array}{c}v\left(0\right)=3-u\left(2-0\right)-2u\left(0\right)\text{ V}\\ =3-u\left(2\right)-2u\left(0\right)\text{ V}\\ =3-\left(1\right)-2\left(0\right)\text{ V}\\ =\left(3-1-0\right)\text{ V}\end{array}$

$v\left(0\right)=2\text{ V}$

Substitute $1$ for $t$ in equation (1).

$\begin{array}{c}v\left(1\right)=3-u\left(2-1\right)-2u\left(1\right)\text{ V}\\ =3-u\left(1\right)-2u\left(1\right)\text{ V}\\ =3-\left(1\right)-2\left(1\right)\text{ V}\\ =\left(3-1-2\right)\text{ V}\end{array}$

$v\left(1\right)=0$

Substitute $2$ for $t$ in equation (1).

$\begin{array}{c}v\left(2\right)=3-u\left(2-2\right)-2u\left(2\right)\text{ V}\\ =3-u\left(0\right)-2u\left(2\right)\text{ V}\\ =3-\left(0\right)-2\left(1\right)\text{ V}\\ =\left(3-0-2\right)\text{ V}\end{array}$

$v\left(2\right)=1\text{ V}$

Substitute $3$ for $t$ in equation (1).

$\begin{array}{c}v\left(3\right)=3-u\left(2-3\right)-2u\left(3\right)\text{ V}\\ =3-u\left(-1\right)-2u\left(3\right)\text{ V}\\ =3-\left(0\right)-2\left(1\right)\text{ V}\\ =\left(3-0-2\right)\text{ V}\end{array}$

$v\left(3\right)=1\text{ V}$

The different value of the function $v\left(t\right)$ over the range $-3\le t\le 3$ is shown in the Table 1,

$\begin{array}{l}\begin{array}{cc}t& v\left(t\right)\text{ (V)}\\ -3& 2\\ -2& 2\\ -1& 2\\ 0& 2\\ 1& 0\\ 2& 1\\ 3& 1\end{array}\\ \text{ Table 1}\end{array}$

The sketch of the function over the range $-3\le t\le 3$ is shown in the Figure 1:

Conclusion:

Thus, the sketch for the function over the range $-3\le t\le 3$ is plotted and shown in Figure 1.

(b)

To determine

Sketch the functions over the range $-3\le t\le 3$.

Explanation of Solution

Given Data:

The function is

$i\left(t\right)=u\left(t\right)-u\left(t-0.5\right)+u\left(t-1\right)-u\left(t-1.5\right)+u\left(t-2\right)-u\left(t-2.5\right)\text{ A}$ (2)

The range of $t$ is $-3\le t\le 3$.

Calculation:

The unit-step forcing function as a function of time which is zero for all values of its argument less than zero and which is unity for all positive values of its argument.

$u\left(t-t_0\right)=\{\begin{array}{l}0t<t_0\\ 1t>t_0\end{array}$

Here,

$t_0$ is the time at which argument of the unit step function becomes zero.

Substitute $-3$ for $t$ in equation (2).

$\begin{array}{c}i\left(-3\right)=u\left(-3\right)-u\left(-3-0.5\right)+u\left(-3-1\right)-u\left(-3-1.5\right)+u\left(-3-2\right)-u\left(-3-2.5\right)\text{ A}\\ =u\left(-3\right)-u\left(-3.5\right)+u\left(-4\right)-u\left(-4.5\right)+u\left(-5\right)-u\left(-5.5\right)\text{ A}\\ =\left(0-0-0-0-0-0\right)\text{ A}\\ =0\text{ A}\end{array}$

Substitute $-2$ for $t$ in the given function in equation (2).

$\begin{array}{c}i\left(-2\right)=u\left(-2\right)-u\left(-2-0.5\right)+u\left(-2-1\right)-u\left(-2-1.5\right)+u\left(-2-2\right)-u\left(-2-2.5\right)\text{ A}\\ =u\left(-2\right)-u\left(-2.5\right)+u\left(-3\right)-u\left(-3.5\right)+u\left(-4\right)-u\left(-4.5\right)\text{ A}\\ =\left(0-0-0-0-0-0\right)\text{ A}\\ =0\text{ A}\end{array}$

Substitute $-1$ for $t$ in equation (2).

$\begin{array}{c}i\left(-1\right)=u\left(-1\right)-u\left(-1-0.5\right)+u\left(-1-1\right)-u\left(-1-1.5\right)+u\left(-1-2\right)-u\left(-1-2.5\right)\text{ A}\\ =u\left(-1\right)-u\left(-1.5\right)+u\left(-2\right)-u\left(-2.5\right)+u\left(-3\right)-u\left(-3.5\right)\text{ A}\\ =\left(0-0+0-0+0-0\right)\text{ A}\\ =0\text{ A}\end{array}$

Substitute $0$ for $t$ in equation (2).

$\begin{array}{c}i\left(0\right)=u\left(0\right)-u\left(0-0.5\right)+u\left(0-1\right)-u\left(0-1.5\right)+u\left(0-2\right)-u\left(0-2.5\right)\text{ A}\\ =u\left(0\right)-u\left(-0.5\right)+u\left(-1\right)-u\left(-1.5\right)+u\left(-2\right)-u\left(-2.5\right)\text{ A}\\ =\left(1-0+0-0+0-0\right)\text{ A}\\ =1\text{ A}\end{array}$

Substitute $0.5$ for $t$ in equation (2).

$\begin{array}{c}i\left(0.5\right)=u\left(0.5\right)-u\left(0.5-0.5\right)+u\left(0.5-1\right)-u\left(0.5-1.5\right)+u\left(0.5-2\right)-u\left(0.5-2.5\right)\text{ A}\\ =u\left(0.5\right)-u\left(0\right)+u\left(-0.5\right)-u\left(-1\right)+u\left(-1.5\right)-u\left(-2\right)\text{ A}\\ =\left(1-1+0-0+0-0\right)\text{ A}\\ =0\text{ A}\end{array}$

Substitute $1$ for $t$ in equation (2).

$\begin{array}{c}i\left(1\right)=u\left(1\right)-u\left(1-0.5\right)+u\left(1-1\right)-u\left(1-1.5\right)+u\left(1-2\right)-u\left(1-2.5\right)\text{ A}\\ =u\left(1\right)-u\left(0.5\right)+u\left(0\right)-u\left(-0.5\right)+u\left(-1\right)-u\left(-1.5\right)\text{ A}\\ =\left(1-1+1-0+0-0\right)\text{ A}\\ =1\text{ A}\end{array}$

Substitute $1.5$ for $t$ in equation (2).

$\begin{array}{c}i\left(1.5\right)=u\left(1.5\right)-u\left(1.5-0.5\right)+u\left(1.5-1\right)-u\left(1.5-1.5\right)+u\left(1.5-2\right)-u\left(1.5-2.5\right)\text{ A}\\ =u\left(1.5\right)-u\left(1\right)+u\left(0.5\right)-u\left(0\right)+u\left(-0.5\right)-u\left(-1\right)\text{ A}\\ =\left(1-1+1-1+0-0\right)\text{ A}\\ =0\text{ A}\end{array}$

Substitute $2$ for $t$ in equation (2).

$\begin{array}{c}i\left(2\right)=u\left(2\right)-u\left(2-0.5\right)+u\left(2-1\right)-u\left(2-1.5\right)+u\left(2-2\right)-u\left(2-2.5\right)\text{ A}\\ =u\left(2\right)-u\left(1.5\right)+u\left(1\right)-u\left(0.5\right)+u\left(0\right)-u\left(-0.5\right)\text{ A}\\ =\left(1-1+1-1+1-0\right)\text{ A}\\ =1\text{ A}\end{array}$

Substitute $2.5$ for $t$ in equation (2).

$\begin{array}{c}i\left(2.5\right)=u\left(2.5\right)-u\left(2.5-0.5\right)+u\left(2.5-1\right)-u\left(2.5-1.5\right)+u\left(2.5-2\right)-u\left(2.5-2.5\right)\text{ A}\\ =u\left(2.5\right)-u\left(2\right)+u\left(1.5\right)-u\left(1\right)+u\left(0.5\right)-u\left(0\right)\text{ A}\\ =\left(1-1+1-1+1-1\right)\text{ A}\\ =0\text{ A}\end{array}$

Substitute $3$ for $t$ in equation (2).

$\begin{array}{c}i\left(3\right)=u\left(3\right)-u\left(3-0.5\right)+u\left(3-1\right)-u\left(3-1.5\right)+u\left(3-2\right)-u\left(3-2.5\right)\text{ A}\\ =u\left(3\right)-u\left(2.5\right)+u\left(2\right)-u\left(1.5\right)+u\left(1\right)-u\left(0.5\right)\text{ A}\\ =\left(1-1+1-1+1-1\right)\text{ A}\\ =0\text{ A}\end{array}$

The different value of the function $v\left(t\right)$ over the range $-3\le t\le 3$ is shown in Table 2:

$\begin{array}{l}\begin{array}{cc}t& i\left(t\right)\left(\text{A}\right)\\ -3& 0\\ -2& 0\\ -1& 0\\ 0& 1\\ 0.5& 0\\ 1& 1\\ 1.5& 0\\ 2& 1\\ 2.5& 0\\ 3& 0\end{array}\\ \text{ Table 1}\end{array}$

The sketch of the function over the range $-3\le t\le 3$ is shown in the Figure 2:

Conclusion:

Thus, the sketch for the function over the range $-3\le t\le 3$ is plotted and shown in Figure 2.

(c)

To determine

Sketch the functions over the range $-3\le t\le 3$.

Explanation of Solution

Given data:

The function is

$q\left(t\right)=8u\left(-t\right)C$ (3)

The range of $t$ is $-3\le t\le 3$.

Calculation:

The unit-step forcing function as a function of time which is zero for all values of its argument less than zero and which is unity for all positive values of its argument.

$u\left(t-t_0\right)=\{\begin{array}{l}0t<t_0\\ 1t>t_0\end{array}$

Here,

$t_0$ is the time at which argument of the unit step function becomes zero.

Substitute $-3$ for $t$ in equation (3).

$\begin{array}{c}q\left(-3\right)=8u\left(-\left(-3\right)\right)\text{ C}\\ =8u\left(+3\right)\text{ C}\\ =8\left(1\right)\text{ C}\\ =8\text{ C}\end{array}$

Substitute $-2$ for $t$ in equation (3).

$\begin{array}{c}q\left(-2\right)=8u\left(-\left(-2\right)\right)\text{ C}\\ =8u\left(+2\right)\text{ C}\\ =8\left(1\right)\text{ C}\\ =8\text{ C}\end{array}$

Substitute $-1$ for $t$ in equation (3).

$\begin{array}{c}q\left(-1\right)=8u\left(-\left(-1\right)\right)\text{ C}\\ =8u\left(+1\right)\text{ C}\\ =8\left(1\right)\text{ C}\\ =8\text{ C}\end{array}$

Substitute $0$ for $t$ in equation (3).

$\begin{array}{c}q\left(0\right)=8u\left(-\left(0\right)\right)\text{ C}\\ =8\left(0\right)\text{ C}\\ =0\text{ C}\end{array}$

Substitute $1$ for $t$ in equation (3).

$\begin{array}{c}q\left(1\right)=8u\left(-1\right)\text{ C}\\ =8\left(0\right)\text{ C}\\ =0\text{ C}\end{array}$

Substitute $2$ for $t$ in equation (3).

$\begin{array}{c}q\left(2\right)=8u\left(-2\right)\text{ C}\\ =8\left(0\right)\text{ C}\\ =0\text{ C}\end{array}$

Substitute $3$ for $t$ in equation (3).

$\begin{array}{c}q\left(3\right)=8u\left(-3\right)\text{ C}\\ =8\left(0\right)\text{ C}\\ =0\text{ C}\end{array}$

The different value of the function $v\left(t\right)$ over the range $-3\le t\le 3$ is shown in the Table 3:

$\begin{array}{l}\begin{array}{cc}t& q\left(t\right)\text{ C}\\ -3& 8\\ -2& 8\\ -1& 8\\ 0& 0\\ 1& 0\\ 2& 0\\ 3& 0\end{array}\\ \text{ Table 3}\end{array}$

The sketch the function over the range $-3\le t\le 3$ is shown in the Figure 3:

Conclusion:

Thus, the sketch for the function over the range $-3\le t\le 3$ is plotted and shown in Figure 3.