Question 6.4: Frequency Response of RC Low-Pass Filter in a More Realistic...
Frequency Response of RC Low-Pass Filter in a More Realistic Circuit
Compute the response of the RC filter in the circuit of Figure 6.12.
Known Quantities: R_{S}=50 \Omega ; R_{1}=200 \Omega ; R_{L}=500 \Omega ; C=10 \mu \mathrm{F}.
Find: The output voltage, v_{o}(t), at each frequency.
Assumptions: None.
Analysis: The circuit shown in this problem is a more realistic representation of a filtering problem, in that we have inserted the R C filter circuit between source and load circuits (where the source and load are simply represented in equivalent form). To determine the response of the circuit, we compute the Thévenin equivalent representation of the circuit with respect to the load, as shown in Figure 6.13. Let R^{\prime}=R_{S}+R_{1} and
Z^{\prime}=R_{L} \| \frac{1}{j \omega C}=\frac{R_{L}}{1+j \omega C R_{L}}
Then the circuit response may be computed as follows:
\begin{aligned}\frac{\mathbf{V}_{L}}{\mathbf{V}_{S}}(j \omega) &=\frac{Z^{\prime}}{R^{\prime}+Z^{\prime}}=\frac{\frac{R_{L}}{1+j \omega C R_{L}}}{R_{S}+R_{1}+\frac{R_{L}}{1+j \omega C R_{L}}} \\&=\frac{R_{L}}{R_{L}+R_{S}+R_{1}+j \omega C R_{L}\left(R_{S}+R_{1}\right)}=\frac{\frac{R_{L}}{R_{L}+R^{\prime}}}{1+j \omega C R_{L} \| R^{\prime}}\end{aligned}
The above expression can be written as follows:
H(j \omega)=\frac{\frac{R_{L}}{R_{L}+R^{\prime}}}{1+j \omega C R_{L} \| R^{\prime}}=\frac{K}{1+j \omega C R_{\mathrm{EQ}}}=\frac{0.667}{1+j \frac{\omega}{600}}
Comments: Note the similarity and difference between the above expression and equation 6.16
\frac{\mathbf{V}_{o}}{\mathbf{V}_{i}}(j \omega)=\frac{1}{1+j \omega C R} (6.16)
: The numerator is different than 1 , because of the voltage divider effect resulting from the source and load resistances, and the cutoff frequency is given by the expression
\omega_{0}=\frac{1}{C R_{\mathrm{EQ}}}
