Solution Found!
The input to a low-pass filter is vi t 10 cos t 10 cos 10t 10 cos 100t V The output of
Chapter 15, Problem P15.6-3(choose chapter or problem)
The input to a low-pass filter is
\(v_{\mathrm{i}}(t)=10 \cos t+10 \cos 10 t+10 \cos 100 t \mathrm{~V}\)
The output of the filter is the voltage \(v_{\mathrm{o}}(t)\). The network function of the low-pass filter is
\(\mathbf{H}(\omega)=\frac{\mathbf{V}_{\mathrm{o}}(\omega)}{\mathbf{V}_{\mathrm{i}}(\omega)}=\frac{2}{\left(1+j \frac{\omega}{5}\right)^2}\)
Plot the Fourier spectrum of the input and the output of the low pass filter.
Questions & Answers
QUESTION:
The input to a low-pass filter is
\(v_{\mathrm{i}}(t)=10 \cos t+10 \cos 10 t+10 \cos 100 t \mathrm{~V}\)
The output of the filter is the voltage \(v_{\mathrm{o}}(t)\). The network function of the low-pass filter is
\(\mathbf{H}(\omega)=\frac{\mathbf{V}_{\mathrm{o}}(\omega)}{\mathbf{V}_{\mathrm{i}}(\omega)}=\frac{2}{\left(1+j \frac{\omega}{5}\right)^2}\)
Plot the Fourier spectrum of the input and the output of the low pass filter.
ANSWER:Step 1 of 8
The frequency response of the input signal .
