# Fourier transform of a RC circuit

The following example of a RC circuit describes the use of the fourier transform in order to receive the output voltage across the capacitor.

My questions are:

• Have I used the convolution correct?
• How can I solve this problem numerically using the inverse fast fourier transform (iFFT) and receive the same result?

The wanted result of the convolution is the impulse response for the voltage across the capacitor which is:
$$v_C\left(t\right)=V_0\left(1-e^{-\frac{t}{RC}}\right)$$

Input signal $V_{in}\left(\omega\right)$ (rectangular function) and capacitor transfer function $H_c\left(\omega\right)$: \begin{align} V_{in}\left(\omega\right)&=V_0\,\tau\;\text{sinc}\left(\frac \tau2 \omega\right)\\ H_C\left(\omega\right)&=\frac{1}{i\,\omega\,RC+1} \end{align}

Output signal $V_C\left(\omega\right)$ is a convolution of $H_C\left(\omega\right)\cdotp V_{in}\left(\omega\right)$: \begin{align} V_C\left(\omega\right)&=H_C\left(\omega\right)\cdotp V_{in}\left(\omega\right)\\ v_C\left(t\right)&=h_C\left(t\right)* v_{in}\left(t\right) \end{align}

The convolution in time domain with the corresponding functions where $\sigma$ is the Heaviside function leads to the correct answer: \begin{align} h_C\left(t\right)&=\frac{1}{RC}\cdot e^{-\frac{t}{RC}}\cdot\sigma\left(t\right)\\ v_{in}\left(t\right)&=V_0\left(\sigma\left(t+\frac \tau2\right)-\sigma\left(t-\frac \tau2\right)\right)= \begin{cases} V_0, & |t|\le\tau/2 \\ 0, & \text{every other t} \end{cases}\\ \\ v_C\left(t\right)&=h\left(t\right)*x\left(t\right) =\int_{-\infty}^\infty h_C\left(t\right)\cdotp v_{in}\left(t-u\right)\,du\\ &=\frac{V_0}{RC}\int_{0}^\tau e^{-\frac{u}{RC}}\,du\\\\ v_C\left(t\right)&=V_0\left(1-e^{-\frac{\tau}{RC}}\right) \end{align}