A telephone transmission system suffers from 60-Hz interference caused by nearby power utility lines. Let us use the network in Fig. 12.45 to design a simple notch filter to eliminate the 60-Hz interference.

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The resistor [latex]R_{ eq }[/latex] represents the equivalent resistance of the telephone system to the right of the LC combination. The LC parallel combination has an equivalent impedance of
[latex]Z =(j \omega L) / /(1 / j \omega C)=\frac{(L / C)}{j \omega L+1 /(j \omega C)}[/latex]
Now the voltage transfer function is
[latex]\frac{ V _{o}}{ V _{ in }}=\frac{R_{ eq }}{R_{ eq }+ Z }=\frac{R_{ eq }}{R_{ eq }+\frac{(L / C)}{j \omega L+(1 / j \omega C)}}[/latex]
which can be written
[latex]\frac{ V _{o}}{ V _{ in }}=\frac{(j \omega)^{2}+\frac{1}{L C}}{(j \omega)^{2}+\left(\frac{j \omega}{R_{ eq } C} ight)+\frac{1}{L C}}[/latex]
Note that at resonance, the numerator and thus [latex]V _{o}[/latex] go to zero. We want resonance to occur at 60 Hz. Thus,
[latex]\omega_{0}=\frac{1}{\sqrt{L C}}=2 \pi(60)=120 \pi[/latex]
If we select C = 100 μF, then the required value for L is 70.3 mH—both are reasonable values. To demonstrate the effectiveness of the filter, let the input voltage consist of a 60-Hz sinusoid and a 1000-Hz sinusoid of the form
[latex]v_{ ext {in }}(t)=1 \sin [(2 \pi) 60 t]+0.2 \sin [(2 \pi) 1000 t] V[/latex]
The input and output waveforms are both shown in Fig. 12.46. Note that the output voltage, as desired, contains none of the 60-Hz interference.

Question 12.19: A telephone transmission system suffers from 60-Hz interfere...

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