Consider the network shown in Fig. 15.24a. This network represents a simple low-pass filter, as shown in Chapter 12. We wish to illustrate the impact of this network on the input signal by examining the frequency characteristics of the output signal and the relationship between the 1-Ω or

Question

Consider the network shown in Fig. 15.24a. This network represents a simple low-pass filter, as shown in Chapter 12. We wish to illustrate the impact of this network on the input signal by examining the frequency characteristics of the output signal and the relationship between the 1-Ω or normalized energy at the input and output of the network.

Accepted Answer

The network transfer function is
[latex]H (\omega)=\frac{1 / R C}{1 / R C+j \omega}=\frac{5}{5+j \omega}=\frac{1}{1+0.2 j \omega}[/latex]
The Fourier transform of the input signal is
[latex]V _{i}(\omega)=\frac{20}{20+j \omega}=\frac{1}{1+0.05 j \omega}[/latex]
Then, using Eq. (15.45)[latex]V _{o}(\omega)= H (\omega) V _{i}(\omega)[/latex], the Fourier transform of the output is
[latex]V _{o}(\omega)=\frac{1}{(1+0.2 j \omega)(1+0.05 j \omega)}[/latex]
Using the techniques of Chapter 12, we note that the straight-line log-magnitude plot (frequency characteristic) for these functions is shown in Figs. 15.24b–d. Note that the lowpass filter passes the low frequencies of the input signal but attenuates the high frequencies.
The normalized energy at the filter input is
[latex]W_{i}=\int_{0}^{\infty}\left(20 e^{-20 t}\right)^{2} d t[/latex]
[latex]=\left.\frac{400}{-40} e^{-40 t}\right|_{0} ^{\infty}[/latex]
= 10 J
The normalized energy at the filter output can be computed using Parseval’s theorem. Since
[latex]V _{o}(\omega)=\frac{100}{(5+j \omega)(20+j \omega)}[/latex]
and
[latex]\left| V _{o}(\omega)\right|^{2}=\frac{10^{4}}{\left(\omega^{2}+25\right)\left(\omega^{2}+400\right)}[/latex]
[latex]\left| V _{o}(\omega)\right|^{2}[/latex] is an even function, and therefore
[latex]W_{o}=2\left(\frac{1}{2 \pi}\right) \int_{0}^{\infty} \frac{10^{4} d \omega}{\left(\omega^{2}+25\right)\left(\omega^{2}+400\right)}[/latex]
However, we can use the fact that
[latex]\frac{10^{4}}{\left(\omega^{2}+25\right)\left(\omega^{2}+400\right)}=\frac{10^{4} / 375}{\omega^{2}+25}-\frac{10^{4} / 375}{\omega^{2}+400}[/latex]
Then
[latex]W_{o}=\frac{1}{\pi}\left(\int_{0}^{\infty} \frac{10^{4} / 375}{\omega^{2}+25} d \omega-\int_{0}^{\infty} \frac{10^{4} / 375}{\omega^{2}+400} d \omega\right)[/latex]
[latex]=\frac{10^{4}}{375}\left(\frac{1}{\pi}\right)\left[\frac{1}{5}\left(\frac{\pi}{2}\right)-\frac{1}{20}\left(\frac{\pi}{2}\right)\right][/latex]
= 2.0 J

Question 15.16: Consider the network shown in Fig. 15.24a. This network repr...

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