14.12 Applications

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14.12 Applications

Resonant circuits and filters are widely used, particularly in electronics, power systems, and communications systems. For example, a Notch filter with a cutoff frequency at 60 Hz may be used to eliminate the 60-Hz power line noise in various communications electronics. Filtering of signals in communications systems is necessary in order to select the desired signal from a host of others in the same range (as in the case of radio receivers discussed next) and also to minimize the effects of noise and interference on the desired signal. In this section, we consider one practical application of resonant circuits and two applications of filters. The focus of each application is not to understand the details of how each device works but to see how the circuits considered in this chapter are applied in the practical devices.

14.12.1 Radio Receiver

Series and parallel resonant circuits are commonly used in radio and TV receivers to tune in stations and to separate the audio signal from the radio-frequency carrier wave. As an example, consider the block diagram of an AM radio receiver shown in Fig. 14.62. Incoming amplitudemodulated radio waves (thousands of them at different frequencies from different broadcasting stations) are received by the antenna. A resonant circuit (or a band-pass filter) is needed to select just one of the incoming waves. The selected signal is very weak and is amplified in stages in order to generate an audible audio-frequency wave. Thus, we have the radiofrequency (RF) amplifier to amplify the selected broadcast signal, the intermediate frequency (IF) amplifier to amplify an internally generated signal based on the RF signal, and the audio amplifier to amplify the audio signal just before it reaches the loudspeaker. It is much easier to amplify the signal at three stages than to build an amplifier to provide the same amplification for the entire band.

Figure 14.62

A simplified block diagram of a superheterodyne AM radio receiver.

The type of AM receiver shown in Fig. 14.62 is known as the superheterodyne receiver. In the early de velopment of radio, each amplification stage had to be tuned to the frequenc y of the incoming signal. This way, each stage must have several tuned circuits to cover the entire AM band (540 to 1600 kHz). To avoid the problem of having several resonant circuits, modern recei vers use a frequency mixer or heterodyne circuit, which always produces the same IF signal (445 kHz) but retains the audio frequencies carried on the incoming signal. To produce the constant IF frequency, the rotors of two separate variable capacitors are mechanically coupled with one another so that they can be rotated simultaneously with a single control; this is called ganged tuning. A local oscillator ganged with the RF amplifier produces an RF signal that is combined with the incoming wave by the frequenc y mixer to produce a n output signal that contains the sum and the difference frequencies of the two signals. For example, if the resonant circuit is tuned to recei ve an 800-kHz incoming signal, the local oscillator must produce a 1,255-kHz signal, so that the sum (1,255 + 800 = 2,055 kHz) and the difference (1,255 – 800 = 455 kHz) of frequencies are a vailable at the output of the mix er. However, only

the difference, 455 kHz, is used in practice. This is the only frequency to which all the IF amplifier stages are tuned, regardless of the station dialed. The original audio signal (containing the “intelligence”) is e xtracted in the detector stage. The detector basically removes the IF signal, leaving the audio signal. The audio signal is amplified to drive the loudspeaker, which acts as a transducer converting the electrical signal to sound.

Our major concern here is the tuning circuit for the AM radio re ceiver. The operation of the FM radio recei ver is different from that of the AM receiver discussed here, and in a much dif ferent range of fre quencies, but the tuning is similar.

The resonant or tuner circuit of an AM radio is portrayed in Fig. 14.63. Example 14.17 Given that L = 1μH, what must be the range of C to have the resonant frequency adjustable from one end of the AM band to another?

Solution:

The frequency range for AM broadcasting is 540 to 1,600 kHz. We consider the low and high ends of the band. Since the resonant circuit in Fig. 14.63 is a parallel type, we apply the ideas in Section 14.6. From Eq. (14.44),

$$\omega_0 = 2\pi f_0 = \frac{1}{\sqrt{LC}}$$

or

$$C = \frac{1}{4\pi^2 f_0^2 L}$$

For the high end of the AM band, f0 = 1,600 kHz, and the corresponding C is

$$4\pi J_0 L$$

d of the AM band, $f_0 = 1,600$ kHz, and the

$$C_1 = \frac{1}{4\pi^2 \times 1,600^2 \times 10^6 \times 10^{-6}} = 9.9$$

nF

For the low end of the AM band, f0 = 540 kHz, and the corresponding C is

of the AM band,

$$f_0 = 540$$

kHz, and the co-

$$C_2 = \frac{1}{4\pi^2 \times 540^2 \times 10^6 \times 10^{-6}} = 86.9$$

nF

Thus, C must be an adjustable (gang) capacitor varying from 9.9 to 86.9 nF.

For an FM radio receiver, the incoming wave is in the frequency range Practice Problem 14.17 from 88 to 108 MHz. The tuner circuit is a parallel RLC circuit with a 4-μH coil. Calculate the range of the variable capacitor necessary to cover the entire band.

Answer: From 0.543 to 0.818 pF.

Figure 14.63 The tuner circuit for Example 14.17.

14.12.2 Touch-Tone Telephone

A typical application of filtering is the touch-tone telephone set shown in Fig. 14.64. The keypad has 12 buttons arranged in four rows and three columns. The arrangement provides 12 distinct signals by using seven tones divided into two groups: the low-frequency group (697 to 941 Hz) and the high-frequency group (1,209 to 1,477 Hz). Pressing a button generates a sum of two sinusoids corresponding to its unique pair of frequencies. For example, pressing the number 6 button generates sinusoidal tones with frequencies 770 and 1,477 Hz.

Figure 14.64 Frequency assignments for touch-tone dialing.

When a caller dials a telephone number, a set of signals is transmitted to the telephone of fice, where the touch-tone signals are de coded by detecting the frequencies they contain. Figure 14.65 shows the block diagram for the detection scheme. The signals are f irst amplified and separated into their respective groups by the low-pass (LP) and high-pass (HP) f ilters. The limiters (L) are used to con vert the separated tones into square w aves. The individual tones are identified using seven band-pass (BP) filters, each filter passing one tone and rejecting other tones. Each f ilter is follow ed by a detec tor (D), which is energized when its input voltage exceeds a certain level. The outputs of the detectors pro vide the required dc signals needed by the switching system to connect the caller to the party being called.

Block diagram of detection scheme.

Using the standard 600-Ω resistor used in telephone circuits and a series Example 14.18 RLC circuit, design the band-pass filter BP2 in Fig. 14.65.

Solution:

The band-pass filter is the series RLC circuit in Fig. 14.35. Inasmuch as BP2 passes frequencies 697 to 852 Hz and is centered at f0 = 770 Hz, its bandwidth is

$$B = 2\pi(f_2 - f_1) = 2\pi(852 - 697) = 973.89 \text{ rad/s}$$

From Eq. (14.39),

$$L = \frac{R}{B} = \frac{600}{973.89} = 0.616 \text{ H}$$

From Eq. (14.27) or (14.55),

(14.27) or (14.55),
\n

$$C = \frac{1}{\omega_0^2 L} = \frac{1}{4\pi^2 f_0^2 L} = \frac{1}{4\pi^2 \times 770^2 \times 0.616} = 69.36 \text{ nF}$$

Repeat Example 14.18 for band-pass filter BP6. Practice Problem 14.18

Answer: 356 mH, 39.83 nF.

14.12.3 Crossover Network

Another typical application of filters is the crossover network that couples an audio amplifier to woofer and tweeter speakers, as shown in Fig. 14.66(a). The network basically consists of one high-pass RC filter

Figure 14.66

(a) A crossover network for two loudspeakers, (b) equivalent model.

Figure 14.67 Frequency responses of the crossover network in Fig. 14.66.

and one low-pass RL filter. It routes frequencies higher than a prescribed crossover frequency fc to the tweeter (high-frequency l oudspeaker) and frequencies below fc into the woofer (low-frequency loudspeaker). These loudspeakers have been designed to accommodate certain frequency responses. A woofer is a low- frequency loudspeaker designed to repro duce the lower part of the frequency range, up to about 3 kHz. A tweeter can reproduce audio frequencies from about 3 kHz to about 20 kHz. The two speaker types can be combined to reproduce the entire audio range of interest and provide the optimum in frequency response.

By replacing the amplifier with a voltage source, the approximate equivalent circuit of the crosso ver network is sho wn in Fig. 14.66(b), where the loudspeak ers are modeled by resistors. As a high-pass filter, the transfer function V1∕Vs is given by

$$H_1(\omega) = \frac{V_1}{V_s} = \frac{j\omega R_1 C}{1 + j\omega R_1 C}$$

(14.87)

Similarly, the transfer function of the low-pass filter is given by

$$H_2(\omega) = \frac{V_2}{V_s} = \frac{R_2}{R_2 + j\omega L}$$

(14.88)

The values of R1, R2, L, and C may be selected such that the two filters have the same cutoff frequency, known as the crossover frequency, as shown in Fig. 14.67.

The principle behind the crossover network is also used in the resonant circuit for a TV receiver, where it is necessary to separate the video and audio bands of RF carrier frequencies. The lower-frequency band (picture information in the range from about 30 Hz to about 4 MHz) is channeled into the receiver’s video amplifier, while the high-frequency band (sound information around 4.5 MHz) is channeled to the receiver’s sound amplifier.

Example 14.19 In the crossover network of Fig. 14.66, suppose each speak er acts as a 6-Ω resistance. Find C and L if the crossover frequency is 2.5 kHz.

Solution:

For the high-pass filter,

$$\omega_c = 2\pi f_c = \frac{1}{R_1 C}$$

or

or

$$C = \frac{1}{2\pi f_c R_1} = \frac{1}{2\pi \times 2.5 \times 10^3 \times 6} = 10.61 \,\mu\text{F}$$

For the low-pass filter,

$$\omega_c = 2\pi f_c = \frac{R_2}{L}$$ $$L = \frac{R_2}{2\pi f_c} = \frac{6}{2\pi \times 2.5 \times 10^3} = 382 \,\mu\text{H}$$

If each speaker in Fig. 14.66 has an 8-Ω resistance and C = 10μF, find L Practice Problem 14.19 and the crossover frequency.

Answer: 0.64 mH, 1.989 kHz.