Question 6.FOM.2: AC Line Interference Filter One application of narrow-band f...

AC Line Interference Filter

One application of narrow-band filters is in rejecting interference due to AC line power. Any undesired 60-\mathrm{Hz} signal originating in the AC line power can cause serious interference in sensitive instruments. In medical instruments such as the electrocardiograph, 60-Hz notch filters are often provided to reduce the effect of this interference { }^{2} on cardiac measurements. Figure 6.23 depicts a circuit in which the effect of 60-\mathrm{Hz} noise is represented by way of a 60-Hz sinusoidal generator connected in series with a signal source \left(\mathbf{V}_{S}\right), representing the desired signal. In this example we design a 60-Hz narrow-band (or notch) filter to remove the unwanted 60-Hz noise.

{ }^{2} See Example 13.3 and Section 15.2 for further information on electrocardiograms and line noise, respectively.

6.23
The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

Known QuantitiesR_{S}=50 \Omega.

Find– Appropriate values of L and C for the notch filter.

Assumptions– None.

Analysis– To determine the appropriate capacitor and inductor values, we write the expression for the notch filter impedance:

Z_{\|}=Z_{L} \| Z_{C}=\frac{\frac{j \omega L}{j \omega C}}{j \omega L+\frac{1}{j \omega C}}=\frac{j \omega L}{1-\omega^{2} L C} .

Note that when \omega^{2} L C=1, the impedance of the circuit is infinite! The frequency

\omega_{0}=\frac{1}{\sqrt{L C}}

is the resonant frequency of the L C circuit. If this resonant frequency were selected to be equal to 60 \mathrm{~Hz}, then the series circuit would show an infinite impedance to 60-Hz currents, and would therefore block the interference signal, while passing most of the other frequency components. We thus select values of L and C that result in \omega_{0}=2 \pi \times 60. Let L=100  \mathrm{mH}. Then

C=\frac{1}{\omega_{0}^{2} L}=70.36  \mu \mathrm{F}

The frequency response of the complete circuit is given below:

H_{V}(j \omega)=\frac{\mathbf{V}_{o}(j \omega)}{\mathbf{V}_{i}(j \omega)}=\frac{R_{L}}{R_{S}+R_{L}+Z_{\|}}=\frac{R_{L}}{R_{S}+R_{L}+\frac{j \omega L}{1-\omega^{2} L C}}

and is plotted in Figure 6.24.

Comments– It would be instructive for you to calculate the response of the notch filter at frequencies in the immediate neighborhood of 60 \mathrm{~Hz}, to verify the attenuation effect of the notch filter.

6.24

Related Answered Questions

Question: 6.10

Verified Answer:

Known Quantities: Function to be inverse Laplace-t...
Question: 6.11

Verified Answer:

Known Quantities: Function to be inverse Laplace-t...
Question: 6.9

Verified Answer:

Known Quantities: Function to be Laplace-transform...
Question: 6.8

Verified Answer:

Known Quantities: Function to be Laplace-transform...
Question: 6.12

Verified Answer:

Known Quantities: Values of resistor, inductor, an...