Chapter 25, Problem 1P

Determine the following for the pulse waveforms of Fig. 25.51:

a. whether it is positive-or negative-going

b. base-line voltage

c. pulse width

d. amplitude

e. % tilt

To determine

(a)

Whether the waveform is positive going or negative going.

Answer to Problem 1P

The pulse waveform is positive going for case 1 and pulse waveform is positive going for case 2.

Explanation of Solution

Calculation:

Case 1:

Pulse waveform increases in positive direction from the base line therefore it is positive going pulse waveform.

Case 2:

Pulse waveform increases in positive direction from the base line therefore it is positive going pulse waveform.

Conclusion:

Thus, pulse waveform is positive going for case 1 and pulse waveform is positive going for case 2.

To determine

(b)

The value of base line voltage.

Answer to Problem 1P

The base line voltage is $0\text{V}$ for case 1 and base line voltage is $-6\text{V}$ for case 2.

Explanation of Solution

Concept used:

Base line voltage is the value of starting point voltage for pulse waveform.

Calculation:

Case 1:

In the waveform starting value of voltage is $0\text{V}$ for pulse waveform therefore value of base line voltage is $0\text{V}\text{.}$

Case 2:

In the waveform starting value of voltage is $-6\text{V}$ for pulse waveform therefore value of base line voltage is $-6\text{V}$.

Conclusion:

Thus, base line voltage is $0\text{V}$ for case 1 and base line voltage is $-6\text{V}$ for case 2.

To determine

(c)

The value of pulse width.

Answer to Problem 1P

The value of pulse width is $2.5\mu \text{s}$ for case 1 and value of pulse width is $1\text{ms}$ for case 2.

Explanation of Solution

Concept used:

Write the expression for pulse width.

  $t_p=t_2-t_1$ ...... (1)

Here, $t_p$ is pulse width, $t_2$ is final value of time for pulse and $t_1$ is initial value of time for pulse.

Calculation:

Case 1:

Substitute $2.5\mu \text{s}$ for $t_2$, $0\mu \text{s}$ for $t_1$ in equation (1).

  $\begin{array}{c}{t}_p=2.5\mu \text{s}-0\mu \text{s}\\ =2.5\mu \text{s}\end{array}$

Therefore, value of pulse width is $2.5\mu \text{s}$.

Case 2:

Substitute $1\text{ms}$ for $t_2$, $0\text{ms}$ for $t_1$ in equation (1).

  $\begin{array}{c}{t}_p=1\text{ms}-0\text{ms}\\ =1\text{ms}\end{array}$

Therefore, value of pulse width is $1\text{ms}$.

Conclusion:

Thus ,value of pulse width is $2.5\mu \text{s}$ for case 1 and value of pulse width is $1\text{ms}$ for case 2.

To determine

(d)

The value of amplitude.

Answer to Problem 1P

The value of amplitude is $12\text{V}$ for case 1 and value of amplitude is $8\text{V}$ for case 2.

Explanation of Solution

Concept used:

Write the expression for peak to peak value.

  $v_{p-p}=v_1-v_2$   ....... (2)

Here, $v_{p-p}$ is peak to peak value of voltage, $v_1$ is maximum value of voltage and $v_2$ is minimum value of voltage.

Calculation:

Amplitude of pulse waveform is equal to peak to peak value of the waveform.

Case 1:

Substitute $12\text{V}$ for $v_1$ and $0\text{V}$ for $v_2$ in equation (2).

  $\begin{array}{c}{v}_{p-p}=12\text{V}-0\text{V}\\ \text{=}12\text{V}\end{array}$

The value of amplitude is $12\text{V}$.

Case 2:

Substitute $2\text{V}$ for $v_1$ and $-6\text{V}$ for $v_2$ in equation (2).

  $\begin{array}{c}{v}_{p-p}=2\text{V}-\left(-6\text{V}\right)\\ =8\text{V}\end{array}$

The value of amplitude is $8\text{V}$.

Conclusion:

Thus, value of amplitude is $12\text{V}$ for case 1 and value of amplitude is $8\text{V}$ for case 2.

To determine

(e)

The $\%$ tilt.

Answer to Problem 1P

The $\%$ tilt of first pulse waveform is $0\%$ and the $\%$ tilt of second pulse waveform is $200\%$.

Explanation of Solution

Concept used:

Write the expression for $\%$ tilt

  $T=\frac{v_1-v_3}{\left(\frac{v_1+v_3}{2}\right)}\times 100\%$   ....... (3)

Here, $T$ is the $\%$ tilt and $v_3$ is the voltage where tilt end.

Calculation:

Case 1:

Substitute $12\text{V}$ for $v_1$ and $12\text{V}$ for $v_3$ in equation (3).

  $\begin{array}{c}T=\frac{12\text{V}-12\text{V}}{\left(\frac{12\text{V}+12\text{V}}{2}\right)}\times 100\%\\ =0\times 100\%\\ =0\%\end{array}$

Case 2:

Substitute $2\text{ V}$ for $v_1$ and $0\text{ V}$ for $v_2$ in equation (3).

  $\begin{array}{c}T=\frac{2\text{ V}-\left(0\text{ V}\right)}{\left(\frac{2\text{ V}+\left(0\text{ V}\right)}{2}\right)}\times 100\%\\ =\frac{2\text{ V}}{1\text{ V}}\times 100\%\\ =200\%\end{array}$

Conclusion:

Thus, the $\%$ tilt of first pulse waveform is $0\%$ and the $\%$ tilt of second pulse waveform is $200\%$.