Question 6.5: Frequency Response of RC High-Pass Filter Compute the respon...
Frequency Response of RC High-Pass Filter
Compute the response of the R C filter in the circuit of Figure 6.16. Evaluate the response of the filter at \omega=2 \pi \times 100 and 2 \pi \times 10,000 \mathrm{rad} / \mathrm{s}.
Known Quantities: R=200 \Omega; C=0.199 \mu \mathrm{F}.
Find: The frequency response, H_{V}(j \omega).
Assumptions: None.
Analysis: We first recognize that the cutoff frequency of the high-pass filter is
\omega_{0}=1 / R C=2 \pi \times 4,000 \mathrm{rad} / \mathrm{s}. Next, we write the frequency response as in equation 6.25 :
\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}(j \omega)=\frac{j \omega C R}{1+j \omega C R} (6.25)
H_{V}(j \omega)=\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}(j \omega)=\frac{j \omega C R}{1+j \omega C R}=\frac{\frac{\omega}{\omega_{0}}}{\sqrt{1+\left(\frac{\omega}{\omega_{0}}\right)^{2}}} \angle\left[\frac{\pi}{2}-\arctan \left(\frac{\omega}{\omega_{0}}\right)\right]
We can now evaluate the response at the two frequencies:
H_{V}(\omega=2 \pi \times 100)=\frac{\frac{100}{4000}}{\sqrt{1+\left(\frac{100}{4000}\right)^{2}}} \angle\left[\frac{\pi}{2}-\arctan \left(\frac{100}{4000}\right)\right]=0.025 \angle 1.546
\begin{aligned}H_{V}(\omega&=2 \pi \times 10,000)=\frac{\frac{10,000}{4000}}{\sqrt{1+\left(\frac{10,000}{4000}\right)^{2}}}<\left[\frac{\pi}{2}-\arctan \left(\frac{10,000}{4000}\right)\right] \\&=0.929 \angle 0.38\end{aligned}
The frequency response plots are shown in Figure 6.18.
Comments: The effect of this high-pass filter is to preserve the amplitude of the input signal at frequencies substantially greater than \omega_{0}, while signals at frequencies below \omega_{0} would be strongly attenuated. With \omega_{0}=2 \pi \times 4,000 (i.e., 4,000 Hz), this filter would pass only the treble range of the audio frequency spectrum.
