Question 6.3: Frequency Response of RC Filter Compute the response of the ...
Frequency Response of RC Filter
Compute the response of the R C filter of Figure 6.9 to sinusoidal inputs at the frequencies of 60 and 10,000 \mathrm{~Hz}.
Known Quantities: R=1 \mathrm{k} \Omega ; C=0.47 \mu \mathrm{F} ; v_{i}(t)=5 \cos (\omega t) \mathrm{V}.
Find: The output voltage, v_{o}(t), at each frequency.
Assumptions: None.
Analysis: In this problem, we know the input signal voltage and the frequency response of the circuit (equation 6.18), and we need to find the output voltage at two different frequencies. If we represent the voltages in phasor form, we can use the frequency response to calculate the desired quantities:
\begin{aligned}H(j \omega) &=\frac{\mathbf{V}_o}{\mathbf{V}_i}(j \omega)=\frac{1}{1+j \omega C R} \\&=\frac{1}{\sqrt{1+(\omega C R)^2}} \frac{e^{j 0^{\circ}}}{e^{j \arctan (\omega C R / 1)}} \\&=\frac{1}{\sqrt{1+(\omega C R)^2}} \cdot e^{-j \arctan (\omega C R)}\end{aligned} (6.18)
\begin{aligned}&\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}(j \omega)=H_{V}(j \omega)=\frac{1}{1+j \omega C R} \\&\mathbf{V}_{o}(j \omega)=H_{V}(j \omega) \mathbf{V}_{i}(j \omega)=\frac{1}{1+j \omega C R} \mathbf{V}_{i}(j \omega)\end{aligned}
If we recognize that the cutoff frequency of the filter is \omega_{0}=1 / R C=2,128 \mathrm{rad} / \mathrm{s}, we can write the expression for the frequency response in the form of equations 6.20 and 6.21 :
\begin{aligned}& |H(j \omega)|=\frac{1}{\sqrt{1+(\omega C R)^{2}}}=\frac{1}{\sqrt{1+\left(\omega / \omega_{0}\right)^{2}}} \qquad (6.20) \\& \phi_{H}(j \omega)=-\arctan (\omega C R)=-\arctan \left(\frac{\omega}{\omega_{0}}\right) \qquad (6.21)\end{aligned}
H_{V}(j \omega)=\frac{1}{1+\frac{j \omega}{\omega_{0}}} \quad\left|H_{V}(j \omega)\right|=\frac{1}{\sqrt{1+\left(\frac{\omega}{\omega_{0}}\right)^{2}}} \quad \phi_{H}(j \omega)=-\arctan \left(\frac{\omega}{\omega_{0}}\right)
Next, we recognize that at \omega=120 \pi \mathrm{rad} / \mathrm{s}, the ratio \omega / \omega_{0}=0.177, and at \omega=20,000 \pi, \omega / \omega_{0}=29.5. Thus we compute the output voltage at each frequency as follows:
\begin{aligned}&\mathbf{V}_{o}(\omega=2 \pi 60)=\frac{1}{1+j 0.177} \mathbf{V}_{i}(\omega=2 \pi 60)=0.985 \times 5 \angle-0.175 \mathrm{~V} \\&\mathbf{V}_{o}(\omega=2 \pi 10,000)=\frac{1}{1+j 29.5} \mathbf{V}_{i}(\omega=2 \pi 10,000)=0.0345 \times 5 \angle-1.537 \mathrm{~V}\end{aligned}
And finally write the time-domain response for each frequency:
\begin{array}{lll}v_{o}(t)=4.923 \cos (2 \pi 60 t-0.175) \mathrm{V} & \text { at } \omega=2 \pi 60 \mathrm{rad} / \mathrm{s} \\v_{o}(t)=0.169 \cos (2 \pi 10,000 t-1.537) \mathrm{V} & \text { at } \omega=2 \pi 10,000 \mathrm{rad} / \mathrm{s}\end{array}
The magnitude and phase responses of the filter are plotted in Figure 6.11. It should be evident from these plots that only the low-frequency components of the signal are passed by the filter. This low-pass filter would pass only the bass range of the audio spectrum.
Comments: Can you think of a very quick, approximate way of obtaining the answer to this problem from the magnitude and phase plots of Figure 6.11? Try to multiply the input voltage amplitude by the magnitude response at each frequency, and determine the phase shift at each frequency. Your answer should be pretty close to the one computed analytically.
Focus on Computer-Aided Tools: A computer-generated solution of this problem generated by MathCad may be found in the CD-ROM that accompanies this book.
