Circuit Application of the Fourier Transform

The Fourier transform generalizes the phasor technique to nonperiodic functions. Therefore, we apply Fourier transforms to circuits with non-sinusoidal excitations in exactly the same way we apply phasor techniques to circuits with sinusoidal excitations. Thus, Ohm's law is still valid:
$$V(\omega)=Z(\omega) I(\omega) \tag{1}$$
where $ V(\omega) $ and $ I(\omega) $ are the Fourier transforms of the voltage and current and $ Z(\omega) $ is the impedance. We get the same expressions for the impedances of resistors, inductors, and capacitors as in phasor analysis, namely,
$$\begin{split} &R \Longrightarrow R\\ &L \Longrightarrow j \omega L\\ &C \Longrightarrow \frac{1}{j \omega C} \\ \end{split} \tag{2}$$
Once we transform the functions for the circuit elements into the frequency domain and take the Fourier transforms of the excitations, we can use circuit techniques such as voltage division, source transformation, mesh analysis, node analysis, or Thevenin's theorem, to find the unknown response (current or voltage). Finally, we take the inverse Fourier transform to obtain the response in the time domain.
Although the Fourier transform method produces a response that exists for $ -\infty < t < \infty $, Fourier analysis cannot handle circuits with initial conditions.
The transfer function is again defined as the ratio of the output response $ Y(\omega) $ to the input excitation $ X(\omega) $, that is,
$$H(\omega)=\frac{Y(\omega)}{X(\omega)} \tag{3}$$
$$Y(\omega)=H(\omega) X(\omega) \tag{4}$$
The frequency-domain input-output relationship is portrayed in Fig. $ 1 $. Equation (4) shows that if we know the transfer function and the input, we can readily find the output.
Fig. 1: Input-output relationship of a circuit in the frequency-domain.
The relationship in Eq. (3) is the principal reason for using the Fourier transform in circuit analysis. Notice that $ H(\omega) $ is identical to $ H(s) $ with $ s=j \omega $. Also, if the input is an impulse function [i.e., $ x(t)=\delta(t) $ ], then $ X(\omega)=1 $, so that the response is
indicating that $ H(\omega) $ is the Fourier transform of the impulse response $ h(t) $.
Example 1: Find $ v_{o}(t) $ in the circuit of Fig. $ 2 $ for $ v_{i}(t)=2 e^{-3 t} u(t) $.
Fig. 2: For Example 1.
The Fourier transform of the input voltage is
$$V_{i}(\omega)=\frac{2}{3+j \omega}$$
and the transfer function obtained by voltage division is
$$H(\omega)=\frac{V_{o}(\omega)}{V_{i}(\omega)}=\frac{1 / j \omega}{2+1 / j \omega}=\frac{1}{1+j 2 \omega}$$
$$V_{o}(\omega)=V_{i}(\omega) H(\omega)=\frac{2}{(3+j \omega)(1+j 2 \omega)}$$
$$V_{o}(\omega)=\frac{1}{(3+j \omega)(0.5+j \omega)}$$
By partial fractions,
$$V_{o}(\omega)=\frac{-0.4}{3+j \omega}+\frac{0.4}{0.5+j \omega}$$
Taking the inverse Fourier transform yields
$$v_{o}(t)=0.4\left(e^{-0.5 t}-e^{-3 t}\right) u(t)$$
Example 2: Using the Fourier transform method, find $ i_{o}(t) $ in Fig. $ 3 $ when $ i_{s}(t)= $ $ 10 \sin 2 t \mathrm{~A} $.
Fig. 3: For Example 2.
By current division,
$$H(\omega)=\frac{I_{o}(\omega)}{I_{s}(\omega)}=\frac{2}{2+4+2 / j \omega}=\frac{j \omega}{1+j \omega 3}$$
If $ i_{s}(t)=10 \sin 2 t $, then
$$I_{s}(\omega)=j \pi 10[\delta(\omega+2)-\delta(\omega-2)]$$
$$I_{o}(\omega)=H(\omega) I_{s}(\omega)=\frac{10 \pi \omega[\delta(\omega-2)-\delta(\omega+2)]}{1+j \omega 3}$$
We resort to the inverse Fourier transform formula in Eq. (A)
$$\bbox[10px,border:1px solid grey]{f(t)=\mathcal{F}^{-1}[F(\omega)]=\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{j \omega t} d \omega} \tag{A}$$
and write
$$i_{o}(t)=\mathcal{F}^{-1}\left[I_{o}(\omega)\right]=\frac{1}{2 \pi} \int_{-\infty}^{\infty} \frac{10 \pi \omega[\delta(\omega-2)-\delta(\omega+2)]}{1+j \omega 3} e^{j \omega t} d \omega$$
We apply the sifting property of the impulse function, namely,
$$\delta\left(\omega-\omega_{0}\right) f(\omega)=f\left(\omega_{0}\right)$$
$$\int_{-\infty}^{\infty} \delta\left(\omega-\omega_{0}\right) f(\omega) d \omega=f\left(\omega_{0}\right)$$
and obtain
$$\begin{aligned}i_{o}(t) &=\frac{10 \pi}{2 \pi}\left[\frac{2}{1+j 6} e^{j 2 t}-\frac{-2}{1-j 6} e^{-j 2 t}\right] \\&=10\left[\frac{e^{j 2 t}}{6.082 e^{j 80.54^{\circ}}}+\frac{e^{-j 2 t}}{6.082 e^{-j 80.54^{\circ}}}\right] \\&=1.644\left[e^{j\left(2 t-80.54^{\circ}\right)}+e^{-j\left(2 t-80.54^{\circ}\right)}\right] \\&=3.288 \cos \left(2 t-80.54^{\circ}\right) \mathrm{A}\end{aligned}$$

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