Question 10.3: Figure 10.15 shows a high pass filter. Find the cut-off freq......

The cut-off frequency is,

f_{1}={\frac{1}{2\pi R C}}={\frac{1}{2\pi\times100\times0.03\times10^{-6}}}=53.05\,\mathrm{kHz}     (10.73)

The output voltage is calculated as,

|V_{0}|={\frac{|V_{i}|}{\sqrt{\left({\frac{f_{i}}{f}}\right)^{2}+1}}}={\frac{80}{\sqrt{\left({\frac{53.05}{15}}\right)^{2}+1}}}=21.80\,\mathrm{V}     (10.74)

At 30 kHz, the phase angle of the output voltage is determined as,

\theta=\left\lfloor H(\omega)=90^{\circ}-\tan^{-1}(2\pi\times30\times10^{3}\times100\times0.03\times10^{-6}\right)=60.53^{\circ}     (10.75)