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RC Filter Analysis

Referring again to Fig.E.1, let's perform an impedance analysis of the simple RC lowpass filter.

Driving Point Impedance

Taking the Laplace transform of both sides of Eq.$ \,$(E.1) gives

$\displaystyle V_e(s) = V_R(s) + V_C(s) = R\, I(s) + \frac{1}{Cs} I(s)
$

where we made use of the fact that the impedance of a capacitor is $ 1/(Cs)$, as derived above. The driving point impedance of the whole RC filter is thus

$\displaystyle R_d(s) \isdef \frac{V_e(s)}{I(s)} = R + \frac{1}{Cs}.
$

Alternatively, we could simply note that impedances always sum in series and write down this result directly.


Transfer Function

Since the input and output signals are defined as $ v_e(t)$ and $ v_C(t)$, respectively, the transfer function of this analog filter is given by, using voltage divider rule,

$\displaystyle H(s) = \frac{V_C(s)}{V_e(s)}
= \frac{\frac{1}{Cs}}{R+\frac{1}{Cs...
...{RC}\frac{1}{s+\frac{1}{RC}}
\isdef \frac{1}{\tau} \frac{1}{s+\frac{1}{\tau}}.
$

The parameter $ \tau\isdef RC$ is called the RC time constant, for reasons we will soon see.


Impulse Response

In the same way that the impulse response of a digital filter is given by the inverse z transform of its transfer function, the impulse response of an analog filter is given by the inverse Laplace transform of its transfer function, viz.,

$\displaystyle h(t) = {\cal L}_t^{-1}\{H(s)\} = \tau e^{-t/\tau} u(t)
$

where $ u(t)$ denotes the Heaviside unit step function

$\displaystyle u(t) \isdef \left\{\begin{array}{ll}
1, & t\geq 0 \\ [5pt]
0, & t<0. \\
\end{array}\right.
$

This result is most easily checked by taking the Laplace transform of an exponential decay with time-constant $ \tau>0$:

\begin{eqnarray*}
{\cal L}_s\{e^{-t/\tau}\}
&\isdef & \int_0^{\infty}e^{-t/\tau...
...ght\vert _0^\infty\\
&=& \frac{1}{s+1/\tau} = \frac{RC}{RCs+1}.
\end{eqnarray*}

In more complicated situations, any rational $ H(s)$ (ratio of polynomials in $ s$) may be expanded into first-order terms by means of a partial fraction expansion (see §6.8) and each term in the expansion inverted by inspection as above.


The Continuous-Time Impulse

The continuous-time impulse response was derived above as the inverse-Laplace transform of the transfer function. In this section, we look at how the impulse itself must be defined in the continuous-time case.

An impulse in continuous time may be loosely defined as any ``generalized function'' having ``zero width'' and unit area under it. A simple valid definition is

$\displaystyle \delta(t) \isdef \lim_{\Delta \to 0} \left\{\begin{array}{ll} \fr...
...eq t\leq \Delta \\ [5pt] 0, & \hbox{otherwise}. \\ \end{array} \right. \protect$ (E.5)

More generally, an impulse can be defined as the limit of any pulse shape which maintains unit area and approaches zero width at time 0. As a result, the impulse under every definition has the so-called sifting property under integration,

$\displaystyle \int_{-\infty}^\infty f(t) \delta(t) dt = f(0), \protect$ (E.6)

provided $ f(t)$ is continuous at $ t=0$. This is often taken as the defining property of an impulse, allowing it to be defined in terms of non-vanishing function limits such as

$\displaystyle \delta(t) \isdef \lim_{\Omega\to\infty}\frac{\sin(\Omega t)}{\pi t}.
$

An impulse is not a function in the usual sense, so it is called instead a distribution or generalized function [13,44]. (It is still commonly called a ``delta function'', however, despite the misnomer.)


Poles and Zeros

In the simple RC-filter example of §E.4.3, the transfer function is

$\displaystyle H(s) = \frac{1}{s+1/\tau} = \frac{RC}{RCs+1}.
$

Thus, there is a single pole at $ s=-1/\tau=-RC$, and we can say there is one zero at infinity as well. Since resistors and capacitors always have positive values, the time constant $ \tau = RC$ is always non-negative. This means the impulse response is always an exponential decay--never a growth. Since the pole is at $ s=-1/\tau$, we find that it is always in the left-half $ s$ plane. This turns out to be the case also for any complex analog one-pole filter. By consideration of the partial fraction expansion of any $ H(s)$, it is clear that, for stability of an analog filter, all poles must lie in the left half of the complex $ s$ plane. This is the analog counterpart of the requirement for digital filters that all poles lie inside the unit circle.


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RLC Filter Analysis
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