The circuit shown in Figure 6.57 has R1 = R2 = 2 k and C =

Step 1 of 3

Apply voltage division to write the expression for output voltage, \({V_{out}}\).

\({V_{out}} = \frac{{{R_2}||\frac{1}{{j2\pi fC}}}}{{{R_1} + {R_2}||\frac{1}{{j2\pi fC}}}}{V_{in}}\)

\({V_{out}} = \frac{{\frac{{{R_2}}}{{{R_1} + {R_2}}}}}{{\frac{{j2\pi fC{R_1}{R_2}}}{{{R_1} + {R_2}}} + 1}}{V_{in}}\)

Substitute \(2\;{\rm{k}}\Omega\) for \({R_1}\),  \(2\;{\rm{k}}\Omega\)  for \({R_2}\), \(\frac{1}{\pi }\mu {\rm{F}}\)  for  C, and we get,

\({V_{out}} = \frac{{\frac{{2 \times {{10}^3}}}{{2 \times {{10}^3} + 2 \times {{10}^3}}}}}{{\frac{{j2\pi f\frac{1}{\pi } \times {{10}^{ - 6}}\left( {2 \times {{10}^3}} \right)\left( {2 \times {{10}^3}} \right)}}{{2 \times {{10}^3} + 2 \times {{10}^3}}} + 1}}{V_{in}}\)

\({V_{out}} = \frac{{0.5}}{{j\frac{f}{{100}} + 1}}{V_{in}}\)

Calculate the transfer function of the system, H(f).

\(H\left( f \right) = \frac{{{V_{out}}}}{{{V_{in}}}}\)

\(H\left( f \right) = \frac{{0.5}}{{j\frac{f}{{100}} + 1}}\)

Consider the 1000 Hz to be the break frequency \({f_B}\).

\(H\left( f \right) = \frac{{0.5}}{{\frac{{jf}}{{{f_B}}} + 1}}\)

Calculate the magnitude of the transfer function, |H(f)|.

\(\left| {H\left( f \right)} \right| = \frac{{0.5}}{{\sqrt {{{\left( {\frac{f}{{{f_B}}}} \right)}^2} + 1} }}\)

Convert this expression into decibels.

\({\left| {H\left( f \right)} \right|_{dB}} = 20\log \left( {\left| {H\left( f \right)} \right|} \right)\)

\({\left| {H\left( f \right)} \right|_{dB}} = 20\log \left( {\frac{{0.5}}{{\sqrt {{{\left( {\frac{f}{{{f_B}}}} \right)}^2} + 1} }}} \right)\)

\({\left| {H\left( f \right)} \right|_{dB}} =  - 6 - 10\log \left( {{{\left( {\frac{f}{{1000}}} \right)}^2} + 1} \right)\)