Chapter 22, Problem 1P

1. Determine the frequencies (in kHz) at the points indicated on the plot in Fig. 22.104(a).
2. Determine the voltages (in mV) at the points indicated on the plot in Fig. 22.104(b).

To determine

(a)

The frequencies at the points indicated in the plot

Answer to Problem 1P

The frequencies at the points are $1.5kHz$ & $3.98kHz$

Explanation of Solution

Given:

Formula used:

The frequency at the point is calculated by

  $value={10}^3\times {10}^{\frac{d_1}{d_2}}$

Calculation:

The distance of the first point, $d_1=1.8$

The distance between two vertical axes is $d_2=10$

The value of the first point,

  $\begin{array}{l}value={10}^3\times {10}^{\frac{d_1}{d_2}}\hfill \\ ={10}^3\times {10}^{\frac{1.8}{10}}\hfill \\ =1513.56=1.5kHz.\hfill \end{array}$

The value of the first point is $1.5kHz$

The values of the second point is,

  $\begin{array}{l}value={10}^3\times {10}^{\frac{d_1}{d_2}}\hfill \\ ={10}^3\times {10}^{\frac{6}{10}}\hfill \\ =3981.1=3.98kHz\hfill \end{array}$

The value of the second point is $3.98kHz$

Conclusion:

Thus, the value of the first point is $1.5kHz$ and the value of the second point is $3.98kHz$

To determine

(b)

The voltages at the points indicated on the plot

Answer to Problem 1P

The voltage of the top point is 527.5mVand the bottom point is 181.7mV

Explanation of Solution

Given:

Formula used:

The voltage at the point is calculated by

  $v_r={10}^{(-1+y_r)}$

Calculation:

The total length of the vertical axis is,

  ${\log }_{10}{10}^8-{\log }_{10}{10}^{-1}=1$

Using a scale on the vertical axis, from the plot, we can find that the top and bottom points are at a fractionof total length of the vertical axis from the bottom.

  $\begin{array}{l}{y}_r=\frac{19.5}{27}\hfill \\ y_g=\frac{7}{27}\hfill \end{array}$

The voltage value of the top point is

  $\begin{array}{l}{v}_r={10}^{(-1+y_r)}\hfill \\ ={10}^{(-1+\frac{19.5}{27})}\hfill \\ =0.5275\hfill \end{array}$

The voltage of the top point is 527.5mV

The voltage value of the bottom point is

  $\begin{array}{l}{v}_r={10}^{(-1+y_b)}\hfill \\ ={10}^{(-1+\frac{7}{27})}\hfill \\ =0.1817\hfill \end{array}$

The voltage of the bottom point is 181.7mV

Conclusion:

Thus, the voltage of the top point is 527.5mVand the voltage of the bottom point is 181.7mV