# Circuit Response to a Nonsinusoidal Input

The Fourier series representation of a nonsinusoidal input can be applied to a linear network using the principle of superposition. Recall that this theorem allowed us to consider the effects of each source of a circuit independently. If we replace the nonsinusoidal input with the terms of the Fourier series deemed necessary for practical considerations, we can use superposition to find the response of the network to each term (Fig. 1).
Fig. 1: Setting up the application of a Fourier series of terms to a linear network.
The total response of the system is then the algebraic sum of the values obtained for each term. The major change between using this theorem for nonsinusoidal circuits and using it for the circuits previously described is that the frequency will be different for each term in the nonsinusoidal application. Therefore, the reactances
$$X_{L}=2 \pi f L \quad \text { and } \quad X_{C}=\frac{1}{2 \pi f C}$$
will change for each term of the input voltage or current. In Chapter 12, we found that the rms value of any waveform was given by
$$\sqrt{\frac{1}{T} \int_{0}^{T} f^{2}(t) d t}$$
If we apply this equation to the following Fourier series:
$$V(\alpha)=V_{0}+V_{m_{1}} \sin \alpha+\cdots+V_{m_{n}} \sin n \alpha+V_{m_{1}}^{\prime} \cos \alpha+\cdots+V_{m_{n}}^{\prime} \cos n \sigma$$
then
$$\bbox[10px,border:1px solid grey]{V_{\mathrm{nms}}=\sqrt{V_{0}^{2}+\frac{V_{m_{1}}^{2}+\cdots+V_{m_{n}}^{2}+V_{m_{1}}^{\prime 2}+\cdots+V_{m_{n}}^{\prime 2}}{2}}} \tag{1}$$
However, since
$$\frac{V_{m_{1}}^{2}}{2}=\left(\frac{V_{m_{1}}}{\sqrt{2}}\right)\left(\frac{V_{m_{1}}}{\sqrt{2}}\right)=\left(V_{1_{\mathrm{ms}}}\right)\left(V_{1_{\mathrm{ms}}}\right)=V_{1_{\mathrm{ms}}}^{2}$$
then
$$V_{\mathrm{rms}}=\sqrt{V_{0}^{2}+V_{\mathrm{I}_{\mathrm{rms}}}^{2}+\cdots+V_{n_{\mathrm{mms}}}^{2}+V_{\mathrm{I}_{\mathrm{rms}}}^{\prime 2}+\cdots+V_{n_{\mathrm{ms}}}^{\prime 2}} \tag{2}$$
Similarly, for
$$i(\alpha)=I_{0}+I_{m_{1}} \sin \alpha+\cdots+I_{m_{n}} \sin n \alpha+I_{m_{1}}^{\prime} \cos \alpha+\cdots+I_{m_{n}}^{\prime} \cos n \alpha$$
we have
$$I_{\mathrm{mus}}=\sqrt{I_{0}^{2}+\frac{I_{m_{1}}^{2}+\cdots+I_{m_{n}}^{2}+I_{m_{1}}^{\prime 2}+\cdots+I_{m_{n}}^{\prime 2}}{2}} \tag{3}$$
and
$$I_{\mathrm{rms}}=\sqrt{I_{0}^{2}+I_{1_{\mathrm{ms}}}^{2}+\cdots+I_{n_{\mathrm{mus}}}^{2}+I_{1_{\mathrm{mus}}}^{2}+\cdots+I_{n_{\mathrm{mus}}}^{2}} \tag{4}$$
The total power delivered is the sum of that delivered by the corresponding terms of the voltage and current. In the following equations, all voltages and currents are rms values:
$$P_{T}=V_{0} I_{0}+V_{1} I_{1} \cos \theta_{1}+\cdots+V_{n} I_{n} \cos \theta_{n}+\cdots \tag{5}$$
or
$$P_{T}=I_{\text {mas }}^{2} R \tag{6}$$
with $I_{\mathrm{mms}}$ as defined by Eq. (3), and, similarly,
$$P_{T}=\frac{V_{\text {rms }}^{2}}{R} \tag{7}$$
with $V_{\mathrm{rms}}$ as defined by Eq. (1).
Example 1:
a. Sketch the input resulting from the combination of sources in Fig. 2.
b. Determine the rms value of the input of Fig. 2.
Fig. 2: For Example 1.
Solution:
a. Note Fig. 3.
Fig. 3: Wave pattern generated by the source of Fig. 2.
b. Eq. (2):
\begin{aligned} V_{\mathrm{rms}} & =\sqrt{V_{0}^{2}+\frac{V_{m}^{2}}{2}} \\ & =\sqrt{(4 \mathrm{~V})^{2}+\frac{(6 \mathrm{~V})^{2}}{2}}=\sqrt{16+\frac{36}{2}} \mathrm{~V}=\sqrt{34} \mathrm{~V} \\ & =\mathbf{5 . 8 3 1 \mathrm { V }}\end{aligned}
It is particularly interesting to note from Example 1 that the rms value of a waveform having both dc and ac components is not simply the sum of the effective values of each. In other words, there is a temptation in the absence of Eq. (2) to state that
$$Vrms = 4 V + 0.707 (6 V) = 8.242 V$$
which is incorrect and, in fact, exceeds the correct level by some 41%.
Example 2: Determine the rms value of the square wave of Fig. 4 with $V_{m}=20 \mathrm{~V}$ using the first six terms of the Fourier expansion, and compare the result to the actual rms value of $20 \mathrm{~V}$.
Fig. 4: For example 2.
Solution:
\begin{aligned}V=\frac{4}{\pi}(20 \mathrm{~V}) \sin \omega t+\frac{4}{\pi}\left(\frac{1}{3}\right)(20 \mathrm{~V}) \sin 3 \omega t & +\frac{4}{\pi}\left(\frac{1}{5}\right)(20 \mathrm{~V}) \sin 5 \omega t+\frac{4}{\pi}\left(\frac{1}{7}\right)(20 \mathrm{~V}) \sin 7 \omega t \\& +\frac{4}{\pi}\left(\frac{1}{9}\right)(20 \mathrm{~V}) \sin 9 \omega t+\frac{4}{\pi}\left(\frac{1}{11}\right)(20 \mathrm{~V}) \sin 11 \omega t\end{aligned}
$$V=25.465 \sin \omega t+8.488 \sin 3 \omega t+5.093 \sin 5 \omega t+3.638 \sin 7 \omega t+2.829 \sin 9 \omega t+2.315 \sin 11 \omega t$$
Eq. (1):
\begin{aligned}V_{\mathrm{rms}} & =\sqrt{V_{0}^{2}+\frac{V_{m_{1}}^{2}+V_{m_{2}}^{2}+V_{m_{3}}^{2}+V_{m_{4}}^{2}+V_{m_{5}}^{2}+V_{m_{6}}^{2}}{2}} \\& =\sqrt{(0 \mathrm{~V})^{2}+\frac{(25.465 \mathrm{~V})^{2}+(8.488 \mathrm{~V})^{2}+(5.093 \mathrm{~V})^{2}+(3.638 \mathrm{~V})^{2}+(2.829 \mathrm{~V})^{2}+(2.315 \mathrm{~V})^{2}}{2}} \\& =\mathbf{1 9 . 6 6 \mathrm { V }}\end{aligned}
The solution differs less than $0.4 \mathrm{~V}$ from the correct answer of $20 \mathrm{~V}$. However, each additional term in the Fourier series will bring the result closer to the 20-V level. An infinite number would result in an exact solution of $20 \mathrm{~V}$.
Example 3: The input to the circuit of Fig. 5 is the following:
$$e=12+10 \sin 2 t$$
a. Find the current $i$ and the voltages $v_{R}$ and $v_{C}$.
b. Find the rms values of $i, v_{R}$, and $v_{C}$.
c. Find the power delivered to the circuit.
Fig. 5: For Example 3.
Solution:
a. Redraw the original circuit as shown in Fig. 6. Then apply superposition:
Fig. 6: Circuit of Fig. 5 with the components of the Fourier series input.
1. For the 12-V dc supply portion of the input, $I=0$ since the capacitor is an open circuit to dc when $v_{C}$ has reached its final (steady-state) value. Therefore,
$$V_{R}=I R=0 \mathrm{~V} \text { and } \quad V_{C}=12 \mathrm{~V}$$
2. For the ac supply,
$$\begin{array}{c}\mathbf{Z}=3 \Omega-j 4 \Omega=5 \Omega \angle-53.13^{\circ} \\\text { and } \mathbf{I}=\frac{\mathbf{E}}{\mathbf{Z}}=\frac{\frac{10}{\sqrt{2}} \mathrm{~V} \angle 0^{\circ}}{5 \Omega \angle-53.13^{\circ}}=\frac{2}{\sqrt{2}} \mathrm{~A} \angle+53.13^{\circ} \\\mathbf{V}_{R}=(I \angle \theta)\left(R \angle 0^{\circ}\right)=\left(\frac{2}{\sqrt{2}} \mathrm{~A} \angle+53.13^{\circ}\right)\left(3 \Omega \angle 0^{\circ}\right) \\=\frac{6}{\sqrt{2}} \mathrm{~V} \angle+53.13^{\circ}\end{array}$$
and
\begin{aligned}\mathbf{V}_{C}=(I \angle \theta)\left(X_{C} \angle-90^{\circ}\right) & =\left(\frac{2}{\sqrt{2}} \mathrm{~A} \angle+53.13^{\circ}\right)\left(4 \Omega \angle-90^{\circ}\right) \\& =\frac{8}{\sqrt{2}} \mathrm{~V} \angle-36.87^{\circ}\end{aligned}
In the time domain,
$$i=0+2 \sin \left(2 t+53.13^{\circ}\right)$$
Note that even though the dc term was present in the expression for the input voltage, the dc term for the current in this circuit is zero:
$$v_{R}=0+6 \sin \left(2 t+53.13^{\circ}\right)$$
and
$$v_{C}=12+8 \sin \left(2 t-36.87^{\circ}\right)$$
b.
$$I_{rms} = \sqrt{(0)^2 + { (2 A)^2 \over 2}} = \sqrt{2} V = 1.414$$
$$V_{R_{rms}} = \sqrt{(0)^2 + { (6 V)^2 \over 2}} = \sqrt{18} V = 4.243$$
$$V_{C_{rms}} = \sqrt{(12)^2 + { (8 A)^2 \over 2}} = \sqrt{176} V = 13.26$$
c.
$$P = I_{rms}^2R = ({ 2 \over \sqrt{2}} A )^2 (3 \Omega) = 6 W$$