14.5 Series Resonance

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14.5 Series Resonance

The most prominent feature of the frequenc y response of a circuit may be the sharp peak (or resonant peak ) exhibited in its amplitude char acteristic. The concept of resonance applies in se veral areas of science and engineering. Resonance occurs in an y system that has a comple x conjugate pair of poles; it is the cause of oscillations of stored ener gy from one form to another . It is the phenomenon that allo ws frequency

ω

‒20 dB/decade

discrimination in communications netw orks. Resonance occurs in an y circuit that has at least one inductor and one capacitor.

Resonance is a condition in an RLC circuit in which the capacitive and inductive reactances are equal in magnitude, thereby resulting in a purely resistive impedance.

Resonant circuits (series or parallel) are useful for constructing filters, as their transfer functions can be highly frequency selective. They are used in many applications such as selecting the desired stations in radio and TV receivers.

Consider the series RLC circuit shown in Fig. 14.21 in the frequency domain. The input impedance is

$$\mathbf{Z} = \mathbf{H}(\omega) = \frac{\mathbf{V}_s}{\mathbf{I}} = R + j\omega L + \frac{1}{j\omega C}$$

(14.22)

or

$$\mathbf{Z} = R + j \left( \omega L - \frac{1}{\omega C} \right) \tag{14.23}$$

Resonance results when the imaginary part of the transfer function is zero, or

$$\operatorname{Im}(\mathbf{Z}) = \omega L - \frac{1}{\omega C} = 0 \tag{14.24}$$

The value of ω that satisfies this condition is called the resonant frequency ω0. Thus, the resonance condition is

$$\omega_0 L = \frac{1}{\omega_0 C} \tag{14.25}$$

or

$$\omega_0 = \frac{1}{\sqrt{LC}} \text{rad/s} \tag{14.26}$$

Since ω0 = 2 π f0,

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

(14.27)

Note that at resonance:

• 1. The impedance is purely resistive, thus, Z = R. In other words, the LC series combination acts like a short circuit, and the entire voltage is across R.
• 2. The voltage Vs and the current I are in phase, so that the po wer factor is unity.
• 3. The magnitude of the transfer function H(ω) = Z(ω) is minimum.
• 4. The inductor v oltage and capacitor v oltage can be much more than the source voltage.

The frequency response of the circuit’s current magnitude

response of the circuit’s current magnitude
\n

$$I = |\mathbf{I}| = \frac{V_m}{\sqrt{R^2 + (\omega L - 1/\omega C)^2}}$$

\n(14.28)

$$|\mathbf{V}_L| = \frac{V_m}{R} \omega_0 L = Q V_m$$ $$|\mathbf{V}_C| = \frac{V_m}{R} \frac{1}{\omega_0 C} = Q V_m$$

where Q is the quality factor, defined in Eq. (14.38).

The series resonant circuit.

is shown in Fig. 14.22; the plot only sho ws the symmetry illustrated in this graph when the frequenc y axis is a log arithm. The average power dissipated by the RLC circuit is

$$P(\omega) = \frac{1}{2} \hat{I}^2 R \tag{14.29}$$

The highest po wer dissipated occurs at resonance, when I = Vm∕R, so that

$$P(\omega_0) = \frac{1}{2} \frac{V_m^2}{R}$$

(14.30)

At certain frequencies ω = ω1, ω2, the dissipated power is half the maximum value; that is,

$$P(\omega_1) = P(\omega_2) = \frac{(V_m/\sqrt{2})^2}{2R} = \frac{V_m^2}{4R}$$

(14.31)

Hence, ω1 and ω2 are called the half-power frequencies.

The half-power frequencies are obtained by setting Z equal to √ __ 2 R, and writing

equences are obtained by setting Z equal to V2R,
\n

$$\sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} = \sqrt{2}R$$

\n(14.32)

Solving for ω, we obtain

$$\omega_1 = -\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \frac{1}{LC}}$$

\n

$$\omega_2 = \frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \frac{1}{LC}}$$

\n(14.33)

We can relate the half-po wer frequencies with the resonant frequenc y. From Eqs. (14.26) and (14.33),

$$\omega_0 = \sqrt{\omega_1 \omega_2} \tag{14.34}$$

showing that the resonant frequenc y is the geometric mean of the halfpower frequencies. Notice that ω1 and ω2 are in general not symmetrical around the resonant frequency ω0, because the frequency response is not generally symmetrical. However, as will be explained shortly, symmetry of the half-power frequencies around the resonant frequenc y is often a reasonable approximation.

Although the height of the curv e in Fig. 14.22 is determined by R, the width of the curv e depends on other f actors. The width of the re sponse curve depends on the bandwidth B, which is defined as the difference between the two half-power frequencies,

$$B = \omega_2 - \omega_1 \tag{14.35}$$

This definition of bandwidth is just one of several that are commonly used. Strictly speaking, B in Eq. (14.35) is a half-po wer bandwidth, because it is the width of the frequenc y band between the half-po wer frequencies.

The “sharpness” of the resonance in a resonant circuit is measured quantitatively by the quality factor Q. At resonance, the reactive energy

Figure 14.22

The current amplitude versus frequency for the series resonant circuit of Fig. 14.21.

Although the same symbol Q is used for the reactive power, the two are not equal and should not be confused. Q here is dimensionless, whereas reactive power Q is in VAR. This may help distinguish between the two.

in the circuit oscillates between the inductor and the capacitor. The quality factor relates the maximum or peak ener gy stored to the energy dissipated in the circuit per cycle of oscillation:

is the maximum or peak energy stored to the energy dis-
rcuit per cycle of oscillation:

\n

$$Q = 2\pi \frac{\text{Peak energy stored in the circuit}}{\text{Energy dissipated by the circuit}} \qquad (14.36)$$

\nin one period at resonance

It is also regarded as a measure of the energy storage property of a circuit in relation to its energy dissipation property. In the series RLC circuit, the peak energy stored is __1 ² LI² , while the energy dissipated in one period is __1 2 (I 2 R)(1∕f0). Hence,

$$Q = 2\pi \frac{\frac{1}{2}LI^2}{\frac{1}{2}I^2R(1/f_0)} = \frac{2\pi f_0L}{R}$$

(14.37)

B3 Q3 (greatest selectivity) Q2 (medium selectivity) Q1 (least selectivity) B2 B1 ω

Amplitude

The higher the circuit Q, the smaller the bandwidth.

The quality factor is a measure of the selectivity (or “sharpness” of resonance) of the circuit.

or

$$Q = \frac{\omega_0 L}{R} = \frac{1}{\omega_0 CR}$$

(14.38)

Notice that the quality factor is dimensionless. The relationship between the bandwidth B and the quality f actor Q is obtained by substituting Eq. (14.33) into Eq. (14.35) and utilizing Eq. (14.38).

$$B = \frac{R}{L} = \frac{\omega_0}{Q} \tag{14.39}$$

or B = ω⁰ 2 CR. Thus,

The quality factor of a resonant circuit is the ratio of its resonant frequency to its bandwidth.

Keep in mind that Eqs. (14.33), (14.38), and (14.39) only apply to a series RLC circuit.

As illustrated in Fig. 14.23, the higher the v alue of Q, the more selective the circuit is but the smaller the bandwidth. The selectivity of an RLC circuit is the ability of the circuit to respond to a certain frequenc y and discriminate against all other frequencies. If the band of frequencies to be selected or rejected is narrow, the quality f actor of the resonant circuit must be high. If the band of frequencies is wide, the quality factor must be low.

A resonant circuit is designed to operate at or near its resonant fre quency. It is said to be a high-Q circuit when its quality factor is equal to or greater than 10. F or high -Q circuits (Q ≥ 10), the half- power frequencies are, for all practical purposes, symmetrical around the resonant frequency and can be approximated as

$$\omega_1 \simeq \omega_0 - \frac{B}{2}, \qquad \omega_2 \simeq \omega_0 + \frac{B}{2} \qquad (14.40)$$

High-Q circuits are used often in communications networks.

We see that a resonant circuit is characterized by five related parameters: the two half-power frequencies ω1 and ω2, the resonant frequency ω0, the bandwidth B, and the quality factor Q.

In the circuit of Fig. 14.24, R = 2 Ω, L = 1 mH, and C = 0.4 μF. (a) Find Example 14.7 the resonant frequency and the half-power frequencies. (b) Calculate the quality factor and bandwidth. (c) Determine the amplitude of the current at ω0, ω1, and ω2.

Solution:

(a) The resonant frequency is

ant frequency is
\n

$$\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{10^{-3} \times 0.4 \times 10^{-6}}} = 50 \text{ krad/s}$$

■ METHOD 1 The lower half-power frequency is

$$\omega_1 = -\frac{R}{2L} + \sqrt{\left(\frac{R}{2L}\right)^2 + \frac{1}{LC}}$$

= $-\frac{2}{2 \times 10^{-3}} + \sqrt{(10^3)^2 + (50 \times 10^3)^2}$
= $-1 + \sqrt{1 + 2500}$ krad/s = 49 krad/s

Similarly, the upper half-power frequency is

$$\omega_2 = 1 + \sqrt{1 + 2500} \text{ krad/s} = 51 \text{ krad/s}$$

(b) The bandwidth is

$$B = \omega_2 - \omega_1 = 2 \text{ krad/s}$$

or

$$B = \frac{R}{L} = \frac{2}{10^{-3}} = 2 \text{ krad/s}$$

The quality factor is

$$Q = \frac{\omega_0}{B} = \frac{50}{2} = 25$$

■ METHOD 2 Alternatively, we could find

Alternatively, we could find
\n

$$Q = \frac{\omega_0 L}{R} = \frac{50 \times 10^3 \times 10^{-3}}{2} = 25$$

From Q, we find

$$B = \frac{\omega_0}{Q} = \frac{50 \times 10^3}{25} = 2 \text{ krad/s}$$

Since Q > 10, this is a high-Q circuit and we can obtain the half-power frequencies as

$$\omega_1 = \omega_0 - \frac{B}{2} = 50 - 1 = 49 \text{ krad/s}$$ $$\omega_2 = \omega_0 + \frac{B}{2} = 50 + 1 = 51 \text{ krad/s}$$

as obtained earlier.

(c) At ω = ω0,

$$I = \frac{V_m}{R} = \frac{20}{2} = 10 \text{ A}$$

At ω = ω1, ω2,

$$I = \frac{V_m}{\sqrt{2} R} = \frac{10}{\sqrt{2}} = 7.071 \text{ A}$$

Practice Problem 14.7 A series-connected circuit has R = 4 Ω and L = 25 mH. (a) Calculate the value of C that will produce a quality factor of 50. (b) Find ω1, ω2, and B. (c) Determine the average power dissipated at ω = ω0, ω1, ω2. Take Vm = 100 V.

Answer: (a) 0.625 μF, (b) 7920 rad/s, 8080 rad/s, 160 rad/s, (c) 1.25 kW , 0.625 kW, 0.625 kW.

14.6 Parallel Resonance

The parallel RLC circuit in Fig. 14.25 is the dual of the series RLC circuit. So we will avoid needless repetition. The admittance is

$$Y = H(\omega) = \frac{I}{V} = \frac{1}{R} + j\omega C + \frac{1}{j\omega L}$$

(14.41)

or

or

$$\mathbf{Y} = \frac{1}{R} + j \left( \omega C - \frac{1}{\omega L} \right) \tag{14.42}$$

Resonance occurs when the imaginary part of Y is zero,

$$\omega C - \frac{1}{\omega L} = 0 \tag{14.43}$$ $$\omega_0 = \frac{1}{\sqrt{LC}} \text{ rad/s}$$

(14.44)

which is the same as Eq. (14.26) for the series resonant circuit. The voltage ∣V∣ is sketched in Fig. 14.26 as a function of frequenc y. Notice that at resonance, the parallel LC combination acts like an open circuit, so that the entire current flows through R. Also, the inductor and capacitor current can be much more than the source current at resonance.

We exploit the duality between Figs. 14.21 and 14.25 by comparing Eq. (14.42) with Eq. (14.23). By replacing R, L, and C in the expressions

Figure 14.25

The parallel resonant circuit.

The current amplitude versus frequency for the series resonant circuit of Fig. 14.25.

We can see this from the fact that

$$|\mathbf{I}_L| = \frac{I_m R}{\omega_0 L} = Q I_m$$ $$|\mathbf{I}_C| = \omega_0 C I_m R = Q I_m$$

where Q is the quality factor, defined in Eq. (14.47).