Consider a circuit with a resistor and capacitor in series.

%\img{images/circuits/combo/combo.ps}

Assume the capacitor is initially uncharged. The voltage across the capacitor is given by


$$ y(t) = \frac{1}{C} \int_{-\infty}^t f(\tau) d\tau $$

where $c$ is the capacitance, and $f(\tau)$ is the current. If we look at a resistor, the voltage across the resistor is given by 

$$ g(t) = R f(t) $$

where $R$ is the resistance and $f(t)$ is the current. Thus, the current can also be written as $f(t) = \frac{g(t)}{R}$. The voltage across the resistor can also be given by $x-y$, thus we can solve for the equation that governs the current:

\begin{align*}
g(t) &= x(t) - y(t) \\
Rf(t) &= x(t) - y(t) \\
f(t) &= \frac{x(t)-y(t)}{R}
\end{align*}

Now plug $f(t)$ into the voltage equation for the capacitor:

\begin{align*}
y(t) &= \frac{1}{C} \int_{-\infty}^t f(\tau) d\tau \\
y(t) &= \frac{1}{C} \int_{-\infty}^t \frac{x(\tau)-y(\tau)}{R} d\tau \\
\frac{d}{dt}y(t) &= \frac{d}{dt}\frac{1}{C} \int_{-\infty}^t \frac{x(\tau)-y(\tau)}{R} d\tau \\
\frac{d}{dt}y(t) &= \frac{1}{RC} \lr{x(t)-y(t)} \\
RC\frac{d}{dt}y(t) + y(t) &= x(t) \\
\end{align*}

We can also the equation that governs the current as
$$ RC \dot{y} + y = x $$

We can determine the frequency response using a few methods:
\begin{enumerate}
\item Letting $x(t) = e^{i\omega t}$, and $y(t) = H(\omega) e^{i\omega t}$
\item Using the Continuous-time Fourier Transform (CTFT)
\end{enumerate}

First, lets solve for the frequency response using the first method:

\begin{align*}
RC\dot{y}(t) + y(t) &= x(t) \\
RCi\omega H(\omega)e^{i\omega t} + H(\omega)e^{i\omega t} &= e^{i\omega t} \\
RCi\omega H(\omega) + H(\omega) &= 1 \\
H(\omega) &= \frac{1}{1 + i\omega RC}
\end{align*}

Now for the CTFT method. Recall the synthesis equation for the CTFT:

\begin{nicebox}
$$x(t) = \frac{1}{2\pi} \infint X(\omega) e^{i\omega t}d\omega $$
\end{nicebox}

Lets determine the fourier transform pair for differentiation in the time domain:

\begin{align*}
x(t) = \frac{1}{2\pi} \infint X(\omega) e^{i\omega t}d\omega \\
\frac{d}{dt}x(t) = \frac{1}{2\pi} \frac{d}{dt}\infint X(\omega) e^{i\omega t}d\omega \\
\frac{d}{dt}x(t) = \frac{1}{2\pi} \infint X(\omega) \frac{d}{dt}e^{i\omega t}d\omega \\
\frac{d}{dt}x(t) = \frac{1}{2\pi} \infint X(\omega) i\omega e^{i\omega t}d\omega \\
\frac{d}{dt}x(t) = \frac{1}{2\pi} \infint \tilde{X}{\omega} e^{i\omega t}d\omega \\
\end{align*}

This tells us that differentiation in the time domain corresponds to multiplication of $i\omega$ in the frequency domain:

\begin{nicebox}
$$ \frac{d}{dt}x(t) \ftp i\omega X(\omega) $$
\end{nicebox}

Using this new relationship, lets solve for $H(\omega)$:

\begin{align*}
RC\dot{y}(t) + y(t) &= x(t) \\
RCi\omega Y(\omega) + Y(\omega) &= X(\omega) \\
RCi\omega H(\omega) + H(\omega) &= 1 \quad \mbox{since } H(\omega) =\frac{Y(\omega)}{X(\omega)} \\
H(\omega) &= \frac{1}{1 + i\omega RC}
\end{align*}

How do we plot $\abs{H(\omega)}$? What type of filter is it? Note that in continuous-time, to analyze this function, we will not use the unit circle. Lets first massage the equation for $H(\omega)$:

\begin{align*}
H(\omega) &= \frac{1}{1 + i\omega RC} \\
H(\omega) &= \frac{1/RC}{1/RC + i\omega} \\
H(\omega) &= \frac{1/RC}{i\omega - \lr{-1/RC}}
\end{align*}

Now lets look at the vector that this creates:

%\img{images/iomegaaxis/overrc/overrc.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-3,-3)(3,3)
    

% y-axis
\rput(0.3,2.5){ $i\omega$ }
\psline{->}(0,-1.5)(0,2.5)

% x-axis
\rput(2.5,0.3){ $\Re$ }
\psline{->}(-2.5,0)(2.5,0)


\rput(-2,0.4){$-1/RC$}
\psline(-2,-0.2)(-2,0.2)
\rput(0.4,-1){$i\omega$}
\psline(-0.2,-1)(0.2,-1)

 % new vector 
\rput(-1.7,-0.8){$i\omega-\left(-1/RC\right)$}
\psline[linewidth=1.5 pt]{->}(-2,0)(0,-1)
\end{pspicture}
\end{center}


We can see that at $\omega=0$, the vector $i\omega - \lr{-1/RC}$ is the shortest. As $\omega \to \infty$, the magnitude response decays. To get the curvature of the plot, consider the change in magnitude over time. 

% \img{images/graphs/respmag/respmag.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-3,-3)(3,3)
    
\fileplot[linewidth=1.5pt]{magresp.dat}
% function name
\rput(3.14159265,1){$\left|H(\omega)\right|$}

% x-axis
\psline(-3.54159265, 0)(3.54159265, 0)
\rput(3.54159265,-0.3){$\omega$}


\rput(0.85840735, 1.2){$(1/\sqrt{2})$}
\rput(-1.14159265, 1.2){$(1/\sqrt{2})$}
\rput(0.85840735, -0.4){$1/RC$}
\rput(-1.14159265, -0.4){$-1/RC$}


\psline[linestyle=dashed](0.85840735, 0)(0.85840735, 0.85)
\psline[linestyle=dashed](-1.14159265, 0)(-1.14159265, 0.85)
\psline[linestyle=dashed](-1.14159265, 0.85)(0.85840735, 0.85)


% begin x-axis labels:
\psline(-3.14159265, -0.1)(-3.14159265, 0.1)
\psline(-1.14159265, -0.1)(-1.14159265, 0.1)
\psline(0.85840735, -0.1)(0.85840735, 0.1)
\psline(2.85840735, -0.1)(2.85840735, 0.1)
% end x-axis labels:

\end{pspicture}
\end{center}

When is the magnitude response have a value of $\frac{1}{\sqrt{2}}$? Using pythagorean theorem, we can see that this occurs at $\omega = \pm \frac{1}{RC}$. 

\begin{align*}
\abs{H(\omega)} &= \abs{\frac{1/RC}{i\omega - \lr{-1/RC}}} \\
\abs{H(1/RC)} &= \frac{\abs{1/RC}}{\abs{i \lr{1/RC} - \lr{-1/RC}}} \\
\abs{H(1/RC)} &= \frac{1}{\abs{i + 1}} = \frac{1}{\sqrt{2}}
\end{align*}

To increase the bandwidth of this signal, what do we do? Note that we can decrease either $R$ or $C$ and $\frac{1}{RC}$ will increase. Now, to plot the phase response of the system:

\begin{align*}
\angle H(\omega) &= \angle \frac{1/RC}{i\omega - \lr{-1/RC}} \\
\angle H(\omega) &= \angle \lr{1/RC} - \angle \lr{i\omega - \lr{-1/RC}} \\
\angle H(\omega) &= 0 - \tan^{-1}(\omega RC) \\
\angle H(\omega) &=  \tan^{-1}(\omega RC) \\
\end{align*}

% \img{images/graphs/respmag/freq/phaseresp.ps}

\psset{unit=1cm}
\begin{center}
\begin{pspicture}(-3,-3)(3,3)
    
\fileplot[linewidth=1.5pt]{phaseresp.dat}
% function name
\rput(3.14159265,0.99783018390619){$\angle H(\omega)$}

% x-axis
\psline{->}(-3.54159265, 0)(3.54159265, 0)
\rput(3.54159265,-0.3){$\omega$}

% y-axis
\psline{->}(0,-2)(0,2)
% begin x-axis labels:
\psline(-0.1, -1.570796325)(0.1, -1.570796325)
\rput(3.2, -1.570796325){$-\frac{\pi}{2}$}
\psline(-0.1, 1.570796325)(0.1, 1.570796325)
\rput(3.2, 1.570796325){$\frac{\pi}{2}$}
% end y-axis labels:

\psline[linestyle=dashed](-3.14, -1.570796325)(3.14, -1.570796325)
\psline[linestyle=dashed](-3.14, 1.570796325)(3.14, 1.570796325)


\end{pspicture}
\end{center}

What if we take the voltage across the resistor instead of the capacitor?

% \img{images/circuits/combo/variation/combo.ps}

If $\omega = 0$, the capacitor opens, no current around the loop, and no voltage drop across the resistor. Therefore, there is zero gain. In other words, $G(0)=0$. If $\omega \to \infty$, the capacitor shorts, becoming a short circuit, so the voltage drop across the resistor, $y$, becomes $x$. In other words, as $\omega \to \infty$, $G(\omega) \to 1$. So this is a high pass filter. 

% \img{images/circuits/combo/both/combo.ps}

The voltage drop across the circuit is the sum of the voltage drops across the resistor and the capacitor, $y_r + y_c$.

\begin{align*}
x(t) &= y_r(t) + y_c(t) \\
X(\omega) &= Y_r(\omega) + Y_c(\omega) \\
X(\omega) &= G(\omega)X(\omega) + G(\omega)X(\omega) \\
1 &= G(\omega) + H(\omega) \\
G(\omega) &= 1 - H(\omega) = 1 - \frac{1}{1+i\omega RC} = \frac{i\omega RC}{1 + i\omega RC} \\
G(\omega) &= \frac{i\omega RC}{1 + i\omega RC}
\end{align*}

Note that as $\omega \to \infty$, $i\omega RC$ dominates the denominator and the 1 is negligible. Thus, as $\omega \to \infty$, $G(\omega) \to 1$.


Basic Circuits