Chapter 12, Problem 1P

(a)

To determine

Write the expression for the function given in Figure P12.1 (a) using step functions.

Answer to Problem 1P

The expression for the given function is $\left(t+10\right)u\left(t+10\right)-2tu\left(t\right)+\left(t-10\right)u\left(t-10\right)$.

Explanation of Solution

Given data:

Refer to Figure P12.1 (a) in the textbook for the given function.

Calculation:

The function in Figure P12.1 (a) is a piecewise linear function

Write the piecewise-linear definition for the function as follows:

$f\left(t\right)=\{\begin{array}{l}0,t\le -10s;\\ 10+t,-10s\ge t\ge 0s\\ 10-t,0s\le t\le 10s\\ 0,t\ge 10s\end{array}$

Form a single continuous definition for the function using step function as follows:

$\begin{array}{l}10+t,\text{on}\text{at}t=-10,\text{off}\text{at}t=0;\\ 10-t,\text{on}\text{at}t=0,\text{off}\text{at}t=10.\end{array}$

Use the values and write the expression for $f\left(t\right)$ as follows:

$f\left(t\right)=\left(10+t\right)\left[u\left(t+10\right)-u\left(t\right)\right]+\left(10-t\right)\left[u\left(t\right)-u\left(t-10\right)\right]$

Rearrange the expression as follows:

$\begin{array}{c}f\left(t\right)=\left\{\begin{array}{l}10u\left(t+10\right)+tu\left(t+10\right)-10u\left(t\right)-tu\left(t\right)\\ +10u\left(t\right)-tu\left(t\right)-10u\left(t-10\right)+tu\left(t-10\right)\end{array}\right\}\\ =\left(t+10\right)u\left(t+10\right)-2tu\left(t\right)+\left(t-10\right)u\left(t-10\right)\end{array}$

Conclusion:

The expression for the given function is $\left(t+10\right)u\left(t+10\right)-2tu\left(t\right)+\left(t-10\right)u\left(t-10\right)$

(b)

To determine

Write the expression for the function given in Figure P12.1 (b) using step functions.

Answer to Problem 1P

The expression for the given function is $\left\{\begin{array}{l}-8\left(t+3\right)u\left(t+3\right)+8\left(t+2\right)u\left(t+2\right)+8\left(t+1\right)u\left(t+1\right)\\ -8\left(t-1\right)u\left(t-1\right)-8\left(t-2\right)u\left(t-2\right)+8\left(t-3\right)u\left(t-3\right)\end{array}\right\}$.

Explanation of Solution

Given data:

Refer to Figure P12.1 (b).in the textbook for the given function.

Calculation:

The function in Figure P12.1 (b) is a piecewise linear function

Write the piecewise-linear definition for the function as follows:

$f\left(t\right)=\{\begin{array}{l}0,t\le -3s;\\ -24-8t,-3s\le t\le -2s\\ -8,-2s\le t\le -1s\\ 8t,-1s\le t\le 1s\\ 8,1s\le t\le 2s\\ -8t+24,2s\le t\le 3s\\ 0,t\ge 3s\end{array}$

Form a single continuous definition for the function using step function as follows:

$\begin{array}{l}-24-8t,\text{on}\text{at}t=-3,\text{off}\text{at}t=-2;\\ -8,\text{on}\text{at}t=-2,\text{off}\text{at}t=-1;\\ 8t,\text{on}\text{at}t=-1,\text{off}\text{at}t=1;\\ 8,\text{on}\text{at}t=1,\text{off}\text{at}t=2;\\ -8t+24,\text{on}\text{at}t=2,\text{off}\text{at}t=3;\end{array}$

Use the values and write the expression for $f\left(t\right)$ as follows:

$f\left(t\right)=\left\{\begin{array}{l}\left(-24-8t\right)\left[u\left(t+3\right)-u\left(t+2\right)\right]-8\left[u\left(t+2\right)-u\left(t+1\right)\right]\\ +8t\left[u\left(t+1\right)-u\left(t-1\right)\right]+8\left[u\left(t-1\right)-u\left(t-2\right)\right]\\ \left(24-8t\right)\left[u\left(t-2\right)-u\left(t-3\right)\right]\end{array}\right\}$

Rearrange the expression as follows:

$\begin{array}{c}f\left(t\right)=\left\{\begin{array}{l}-24u\left(t+3\right)-8tu\left(t+3\right)+24u\left(t+2\right)+8tu\left(t+2\right)\\ -8u\left(t+2\right)+8u\left(t+1\right)\\ +8tu\left(t+1\right)-8tu\left(t-1\right)+8u\left(t-1\right)-8u\left(t-2\right)\\ +24u\left(t-2\right)-8tu\left(t-2\right)-24u\left(t-3\right)+8tu\left(t-3\right)\end{array}\right\}\\ =\left\{\begin{array}{l}-8\left(t+3\right)u\left(t+3\right)+8\left(t+2\right)u\left(t+2\right)+8\left(t+1\right)u\left(t+1\right)\\ -8\left(t-1\right)u\left(t-1\right)-8\left(t-2\right)u\left(t-2\right)+8\left(t-3\right)u\left(t-3\right)\end{array}\right\}\end{array}$

Conclusion:

The expression for the given function is $\left\{\begin{array}{l}-8\left(t+3\right)u\left(t+3\right)+8\left(t+2\right)u\left(t+2\right)+8\left(t+1\right)u\left(t+1\right)\\ -8\left(t-1\right)u\left(t-1\right)-8\left(t-2\right)u\left(t-2\right)+8\left(t-3\right)u\left(t-3\right)\end{array}\right\}$.