Question

Figure (39-E1) shows a typical circuit for a low-pass filter. An AC input V_i = 10 mV is applied at the left end and the output V₀ is received at the right end. Find the output voltage for ν = 10 k Hz, 1.0 MHz and 10.0 MHz. Note that as the frequency is increased the output decreases and, hence, the name low-pass filter.

Answer 1

Here,
Input voltage to the filter, Vi = 10 × 10⁻³ V
Resistance of the circuit, R = 1 × 10³ Ω
Capacitance of the circuit, C = 10 × 10⁻⁹ F
(a) When frequency, f = 10 kHz
A low pass filter consists of resistance and capacitance. Voltage across the capacitor is taken as the output.
Capacitive reactance $\left(X_C\right)$ is given by,
$X_{\mathrm{C}}=\frac{\mathit{1}}{\mathit{\omega }\mathit{C}}=\frac{1}{2\mathrm{\pi }fC}$
$$\begin{gathered} \Rightarrow X_C=\frac{1}{2\mathrm{\pi }\times 10\times {10}^3\times 10\times {10}^{-9}} \\ \Rightarrow X_C=\frac{1}{2\mathrm{\pi }\times {10}^{-4}} \\ \Rightarrow X_C=\frac{{10}^4}{2\mathrm{\pi }}=\frac{5000}{\mathrm{\pi }}\mathrm{\Omega } \end{gathered}$$
Net impedence of the resistance-capacitance circuit (Z) is given by,
$$\begin{gathered} Z=\sqrt{R^2+X_{\mathrm{C}}^2} \\ \Rightarrow Z=\sqrt{\left(1+{10}^3\right)+{\left(5000/\mathrm{\pi }\right)}^2} \\ \Rightarrow Z=\sqrt{{10}^6+{\left(5000/\mathrm{\pi }\right)}^2} \end{gathered}$$
Current (I₀) is given by,
$$\begin{gathered} I_0=\frac{V_i}{Z} \\ \Rightarrow I_0=\frac{10\times {10}^{-3}}{\sqrt{{10}^6+{\left(5000/\mathrm{\pi }\right)}^2}} \end{gathered}$$
Output across the capacitor $\left(V_0\right)$ is given by,
$$\begin{gathered} V_0=\frac{{10}^2}{\sqrt{{10}^5+{\left(50/\mathrm{\pi }\right)}^2}}\times \frac{500}{\mathrm{\pi }} \\ \Rightarrow V_0=1.6124\mathrm{V}=1.6\mathrm{mV} \end{gathered}$$
(b)When frequency, f = 1 MHz = $1\times$10⁶ Hz
Capacitive reactance $\left(X_C\right)$ is given by,
$$\begin{gathered} X_c=\frac{1}{\omega C} \\ \Rightarrow X_C=\frac{\mathit{1}}{\mathit{2}\pi fC} \\ \Rightarrow X_C=\frac{1}{2\mathrm{\pi }\times {10}^6\times {10}^{-9}\times 10} \\ \Rightarrow X_C=\frac{1}{2\mathrm{\pi }\times {10}^{-2}} \\ \Rightarrow X_C=\frac{100}{2\mathrm{\pi }} \\ \Rightarrow X_C=\frac{500}{\mathrm{\pi }}\mathrm{\Omega } \\ \mathrm{Total}\mathrm{impedence}\left(Z\right)=\sqrt{R^2+{{\mathrm{X}}_{\mathrm{C}}}^2} \\ \Rightarrow Z=\sqrt{{\left({10}^3\right)}^2+{\left(50/\mathrm{\pi }\right)}^2} \\ \mathrm{Current}\left(I_0\right)=\frac{V_{\mathit{1}}}{Z} \\ \Rightarrow I_0=\frac{10\times {10}^{-3}}{\sqrt{{10}^6+{\left(50/\mathrm{\pi }\right)}^2}} \\ \mathrm{Output}\mathrm{voltage}\left(V_0\right)\mathit{=}{I}_{\mathit{0}}{X}_C \\ \Rightarrow V_0=\frac{{10}^{-2}}{\sqrt{{10}^5+{\left(50/\mathrm{\pi }\right)}^2}}\times \frac{50}{\mathrm{\pi }} \\ \Rightarrow V_0=0.16\mathrm{mV} \end{gathered}$$

(c) When frequency, f = 10 MHz = 10 × 10⁶ Hz = 10⁷ Hz
Capacitive reactance $\left(X_C\right)$ is given by,
$$\begin{gathered} X_c=\frac{1}{\omega C} \\ \Rightarrow X_C=\frac{1}{2\pi fC} \\ \Rightarrow X_C=\frac{1}{2\mathrm{\pi }\times {10}^7\times 10\times {10}^{-9}} \\ \Rightarrow X_C=\frac{5}{\mathrm{\pi }}\mathrm{\Omega } \\ \mathrm{Impedence}\left(Z\right)=\sqrt{R^2+X_c^2} \\ \Rightarrow Z=\sqrt{{\left({10}^3\right)}^2+{\left(5/\mathrm{\pi }\right)}^2} \\ \mathrm{Current}\left(I_0\right)=\frac{V_1}{Z} \\ \Rightarrow I_0=\frac{10\times {10}^{-3}}{\sqrt{{10}^6+{\left(5/\mathrm{\pi }\right)}^2}} \\ {V}_0=I_0{X}_{\mathrm{C}} \\ \Rightarrow V_0=\frac{{10}^{-2}}{\sqrt{{10}^6+{\left(5/\mathrm{\pi }\right)}^2}}\times \frac{5}{\mathrm{\pi }} \\ \Rightarrow V_o=16\mathrm{\mu V} \end{gathered}$$