Solution:

← Back to Fundamentals of Electric Circuits Overview From Example 17.1,

$$v_s(t) = \frac{1}{2} + \frac{2}{\pi} \sum_{k=1}^{\infty} \frac{1}{n} \sin n\pi t$$

, $n = 2k - 1$

where ωn = nω0 = nπrad/s. Using phasors, we obtain the response Vo in the circuit of Fig. 17.20 by voltage division:

$$\mathbf{V}_o = \frac{j\omega_n L}{R + j\omega_n L} \mathbf{V}_s = \frac{j2n\pi}{5 + j2n\pi} \mathbf{V}_s$$

For the dc component (ωn = 0 or n = 0)

$$\mathbf{V}_s = \frac{1}{2} \qquad \Rightarrow \qquad \mathbf{V}_o = 0$$

This is expected, given that the inductor is a short circuit to dc. For the nth harmonic,

$$V_s = \frac{2}{n\pi} \sqrt{-90^\circ}$$

(17.6.1)

and the corresponding response is

$$\mathbf{v}_s = \frac{2n\pi}{n\pi} \frac{7 - 90^{\circ}}{(17.6.1)}$$

\nThe corresponding response is

\n

$$\mathbf{V}_o = \frac{2n\pi/90^{\circ}}{\sqrt{25 + 4n^2\pi^2} \left(\frac{\tan^{-1}2n\pi/5}{\tan^{-1}2n\pi/5}\right)} \left(\frac{2}{n\pi} \left(\frac{7 - 90^{\circ}}{\tan^{-1}2n\pi/5}\right)\right)$$

\n(17.6.2)

\n

$$= \frac{4(-\tan^{-1}2n\pi/5)}{\sqrt{25 + 4n^2\pi^2}}$$

In the time domain,

$$v_o(t) = \sum_{k=1}^{\infty} \frac{4}{\sqrt{25 + 4n^2 \pi^2}} \cos\left(n\pi t - \tan^{-1}\frac{2n\pi}{5}\right), \qquad n = 2k - 1$$

The first three terms (k = 1, 2, 3 or n = 1, 3, 5) of the odd harmonics in the summation give us

$$v_o(t) = 0.4981 \cos(\pi t - 51.49^\circ) + 0.2051 \cos(3\pi t - 75.14^\circ)$$

• 0.1257 \cos(5\pi t - 80.96^\circ) + … V

Figure 17.21 shows the amplitude spectrum for output v oltage vo(t), while that of the input voltage vs(t) is in Fig. 17.4(a). Notice that the two spectra are close. Why? We observe that the circuit in Fig. 17.20 is a high-pass filter with the corner frequency ωc = R∕L = 2.5 rad/s, which is less than the fundamental frequenc y ω0 = πrad/s. The dc component is not passed and the first harmonic is slightly attenuated, but higher harmonics are passed. In fa ct, from Eqs. (17.6.1) and (17.6.2), Vo is identical to Vs for large n, which is characteristic of a high-pass filter.

Figure 17.22 For Practice Prob. 17.6.

the output voltage.

Example 17.7

If the sawtooth waveform in Fig. 17.9 (see Practice Prob. 17.2) is the voltage source vs(t) in the circuit of Fig. 17.22, find the response vo(t).

6CUCE PTODIEIII 17.0 If the sawtooth waveform in Fig. 17.9 (see Practice
voltage source

$$v_s(t)$$

in the circuit of Fig. 17.22, find
$v_s(t)$
$v_s(t)$

$$1F = \frac{1}{v_s(t)}$$

Answer: $v_o(t) = \frac{3}{2} - \frac{3}{\pi} \sum_{n=1}^{\infty} \frac{\sin(2\pi nt - \tan^{-1} 4n\pi)}{n\sqrt{1 + 16n^2 \pi^2}} V.$

Find the response io(t) of the circuit of Fig. 17.23 if the input voltage v(t) has the Fourier series expansion

$$v(t) = 1 + \sum_{n=1}^{\infty} \frac{2(-1)^n}{1 + n^2} (\cos nt - n \sin nt)$$

For Example 17.6: Amplitude spectrum of

Solution:

Using Eq. (17.13), we can express the input voltage as

$$v(t) = 1 + \sum_{n=1}^{\infty} \frac{2(-1)^n}{\sqrt{1 + n^2}} \cos(nt + \tan^{-1} n)$$

= 1 - 1.414 \cos(t + 45^\circ) + 0.8944 \cos(2t + 63.45^\circ)
-0.6345 \cos(3t + 71.56^\circ) - 0.4851 \cos(4t + 78.7^\circ) + …

We notice that ω0 = 1, ωn = n rad/s. The impedance at the source is

$$\mathbf{Z} = 4 + j\omega_n 2 \mid 4 = 4 + \frac{j\omega_n 8}{4 + j\omega_n 2} = \frac{8 + j\omega_n 8}{2 + j\omega_n}$$

The input current is

$$\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} = \frac{2 + j\omega_n}{8 + j\omega_n 8} \mathbf{V}$$

where V is the phasor form of the source voltage v(t). By current division,

**^I**o = ________ ⁴ 4 + jωn2 ^I = ________ ^V 4 + jωn4

Because ωn = n, Io can be expressed as

Because

$$\omega_n = n$$

, $\mathbf{I}_o$ can be expressed as
\n

$$\mathbf{I}_o = \frac{\mathbf{V}}{4\sqrt{1 + n^2 / \tan^{-1} n}}$$

\nFor the dc component ( $\omega_n = 0$ or $n = 0$ )

$$\mathbf{V} = 1 \qquad \Rightarrow \qquad \mathbf{I}_o = \frac{\mathbf{V}}{4} = \frac{1}{4}$$

For the nth harmonic,

$$\mathbf{V} = \frac{2(-1)^n}{\sqrt{1 + n^2}} \frac{1}{\tan^{-1} n}$$

so that

$$\mathbf{I}_o = \frac{1}{4\sqrt{1 + n^2}/\tan^{-1}n} \frac{2(-1)^n}{\sqrt{1 + n^2}} / \tan^{-1}n = \frac{(-1)^n}{2(1 + n^2)}$$

In the time domain,

$$i_o(t) = \frac{1}{4} + \sum_{n=1}^{\infty} \frac{(-1)^n}{2(1+n^2)} \cos nt \, \text{A}$$

If the input voltage in the circuit of Fig. 17.24 is

$$v(t) = \frac{7}{3} + \frac{1}{\pi^2} \sum_{n=1}^{\infty} \left( \frac{1}{n^2} \cos nt - \frac{\pi}{n} \sin nt \right) \text{ V}$$

determine the response io(t).

Practice Problem 17.7

Figure 17.24 For Practice Prob. 17.7.

17.5 Average Power and RMS Values

Recall the concepts of average power and rms value of a periodic signal that we discussed in Chapter 11. To find the average power absorbed by a circuit due to a periodic excitation, we write the voltage and current in amplitude-phase form [see Eq. (17.10)] as

$$v(t) = V_{\text{dc}} + \sum_{n=1}^{\infty} V_n \cos(n\omega_0 t - \theta_n)$$

(17.42)

$$i(t) = I_{\text{dc}} + \sum_{m=1}^{\infty} I_m \cos(m\omega_0 t - \phi_m)$$

(17.43)

Following the passi ve sign con vention (Fig. 17.25), the a verage power is

$$P = \frac{1}{T} \int_0^T v i \, dt \tag{17.44}$$

Substituting Eqs. (17.42) and (17.43) into Eq. (17.44) gives

$$P = \frac{1}{T} \int_0^T V_{dc} I_{dc} dt + \sum_{m=1}^{\infty} \frac{I_m V_{dc}}{T} \int_0^T \cos(m\omega_0 t - \phi_m) dt$$

$$\sum_{n=1}^{\infty} \frac{V_n I_{dc}}{T} \int_0^T \cos(n\omega_0 t - \theta_n) dt$$

(17.45)
+

$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{V_n I_m}{T} \int_0^T \cos(n\omega_0 t - \theta_n) \cos(m\omega_0 t - \phi_m) dt$$

The second and third integrals vanish, since we are integrating the cosine over its period. According to Eq. (17.4e), all terms in the fourth inte gral are zero when m ≠ n. By evaluating the first integral and applying Eq. (17.4g) to the fourth integral for the case m = n, we obtain

$$P = V_{\rm dc} I_{\rm dc} + \frac{1}{2} \sum_{n=1}^{\infty} V_n I_n \cos(\theta_n - \phi_n)$$

(17.46)

This shows that in average-power calculation involving periodic voltage and current, the total average power is the sum of the average powers in each harmonically related voltage and current.

Given a periodic function f(t), its rms value (or the effective value) is given by

$$F_{\rm rms} = \sqrt{\frac{1}{T} \int_0^T f^2(t) \, dt} \tag{17.47}$$

Figure 17.25

The voltage polarity reference and current reference direction.

Substituting f(t) in Eq. (17.10) into Eq. (17.47) and noting that (a + b) 2 = a² + 2ab + b² , we obtain

$$F_{\text{rms}}^2 = \frac{1}{T} \int_0^T \left[ a_0^2 + 2 \sum_{n=1}^\infty a_0 A_n \cos (n\omega_0 t + \phi_n) \right. \\ \left. + \sum_{n=1}^\infty \sum_{m=1}^\infty A_n A_m \cos(n\omega_0 t + \phi_n) \cos(m\omega_0 t + \phi_m) \right] dt$$

\n

$$= \frac{1}{T} \int_0^T a_0^2 dt + 2 \sum_{n=1}^\infty a_0 A_n \frac{1}{T} \int_0^T \cos(n\omega_0 t + \phi_n) dt$$

\n

$$+ \sum_{n=1}^\infty \sum_{m=1}^\infty A_n A_m \frac{1}{T} \int_0^T \cos(n\omega_0 t + \phi_n) \cos(m\omega_0 t + \phi_m) dt$$