low-pass filter: Definition from Answers.com

Wikipedia: low-pass filter

A low-pass filter is a filter that passes low frequency signals but attenuates (reduces the amplitude of) signals with frequencies higher than the cutoff frequency. The actual amount of attenuation for each frequency varies from filter to filter. It is sometimes called a high-cut filter, or treble cut filter when used in audio applications.

The concept of a low-pass filter exists in many different forms, including electronic circuits (like a hiss filter used in audio), digital algorithms for smoothing sets of data, acoustic barriers, blurring of images, and so on. Low-pass filters play the same role in signal processing that moving averages do in some other fields, such as finance; both tools provide a smoother form of a signal which removes the short-term oscillations, leaving only the long-term trend.

Examples of low pass filters

A low-pass electronic filter realized by an RC circuit

Acoustic

A stiff physical barrier tends to reflect higher sound frequencies, and so acts as a low-pass filter for transmitting sound. When music is playing in another room, the low notes are easily heard, while the high notes are attenuated.

Electronic

Electronic low-pass filters are used to drive subwoofers and other types of loudspeakers, to block high pitches that they can't efficiently broadcast.

Radio transmitters use lowpass filters to block harmonic emissions which might cause interference with other communications.

An integrator is another example of a low-pass filter.

DSL splitters use low-pass and high-pass filters to separate DSL and POTS signals sharing the same pair of wires.

Low-pass filters also play a significant role in the sculpting of sound for electronic music as created by analogue synthesisers. See subtractive synthesis.

Ideal and real filters

An ideal low-pass filter completely eliminates all frequencies above the cut-off frequency while passing those below unchanged. The transition region present in practical filters does not exist in an ideal filter. An ideal low pass filter can be realized mathematically (theoretically) by multiplying a signal by the rectangular function in the frequency domain or, equivalently, convolution with a sinc function in the time domain.

However, the ideal filter is impossible to realize without also having signals of infinite extent, and so generally needs to be approximated for real ongoing signals, because the sinc function's support region extends to all past and future times. The filter would therefore need to have infinite delay, or knowledge of the infinite future and past, in order to perform the convolution. It is effectively realizable for pre-recorded digital signals by assuming extensions of zero into the past and future, but even that is not typically practical.

Real filters for real-time applications approximate the ideal filter by truncating and windowing the infinite impulse response to make a finite impulse response; applying that filter requires delaying the signal for a moderate period of time, allowing the computation to "see" a little bit into the future. This delay is manifested as phase shift. Greater accuracy in approximation requires a longer delay.

The Whittaker–Shannon interpolation formula describes how to use a perfect low-pass filter to reconstruct a continuous signal from a sampled digital signal. Real digital-to-analog converters use real filter approximations.

Electronic low-pass filters

The frequency response of a first-order filter

There are a great many different types of filter circuits, with different responses to changing frequency. The frequency response of a filter is generally represented using a Bode plot.

A first-order filter, for example, will reduce the signal amplitude by half (about –6 dB) every time the frequency doubles (goes up one octave). The magnitude Bode plot for a first-order filter looks like a horizontal line below the cutoff frequency, and a diagonal line above the cutoff frequency. There is also a "knee curve" at the boundary between the two, which smoothly transitions between the two straight line regions. See RC circuit.

A second-order filter does a better job of attenuating higher frequencies. The Bode plot for this type of filter resembles that of a first-order filter, except that it falls off more quickly. For example, a second-order Butterworth filter will reduce the signal amplitude to one fourth its original level every time the frequency doubles (–12 dB per octave). Other second-order filters may roll off at different rates initially depending on their Q factor, but approach the same final rate of –12 dB per octave. See RLC circuit.

Third- and higher-order filters are defined similarly. In general, the final rate of rolloff for an n-order filter is 6n dB per octave.

On any Butterworth filter, if one extends the horizontal line to the right and the diagonal line to the upper-left (the asymptotes of the function), they will intersect at exactly the "cutoff frequency". The frequency response at the cutoff frequency in a first-order filter is –3 dB below the horizontal line. The various types of filters — Butterworth filter, Chebyshev filter, Bessel filter, etc. — all have different-looking "knee curves". Many second-order filters are designed to have "peaking" or resonance, causing their frequency response at the cutoff frequency to be above the horizontal line. See electronic filter for other types.

The meanings of 'low' and 'high' — that is, the cutoff frequency — depend on the characteristics of the filter. The term "low-pass filter" merely refers to the shape of the filter's response; a high-pass filter could be built that cuts off at a lower frequency than any low-pass filter – it is their responses that set them apart. Electronic circuits can be devised for any desired frequency range, right up through microwave frequencies (above 1000 MHz) and higher.

Passive electronic realization

A passive low-pass filter showing impedances of the components

One simple electrical circuit that will serve as a low-pass filter consists of a resistor in series with a load, and a capacitor in parallel with the load. The capacitor exhibits reactance, and blocks low-frequency signals, causing them to go through the load instead. At higher frequencies the reactance drops, and the capacitor effectively functions as a short circuit. The combination of resistance and capacitance gives you the time constant of the filter $τ = R C$ (represented by the Greek letter tau). The break frequency, also called the turnover frequency or cutoff frequency (in hertz), is determined by the time constant:

$f_\mathrm{c} = {1 \over 2 \pi \tau } = {1 \over 2 \pi R C}$

or equivalently (in radians per second):

$\omega_\mathrm{c} = {1 \over \tau} = { 1 \over R C}.$

One way to understand this circuit is to focus on the time the capacitor takes to charge. It takes time to charge or discharge the capacitor through that resistor:

At low frequencies, there is plenty of time for the capacitor to charge up to practically the same voltage as the input voltage.
At high frequencies, the capacitor only has time to charge up a small amount before the input switches direction. The output goes up and down only a small fraction of the amount the input goes up and down. At double the frequency, there's only time for it to charge up half the amount.

Another way to understand this circuit is with the idea of reactance at a particular frequency:

Since DC cannot flow through the capacitor, DC input must "flow out" the path marked $V out$ (analogous to removing the capacitor).
Since AC flows very well through the capacitor — almost as well as it flows through solid wire — AC input "flows out" through the capacitor, effectively short circuiting to ground (analogous to replacing the capacitor with just a wire).

It should be noted that the capacitor is not an "on/off" object (like the block or pass fluidic explanation above). The capacitor will variably act between these two extremes. It is the Bode plot and frequency response that show this variability.

Active electronic realization

An active low-pass filter

Another type of electrical circuit is an active low-pass filter.

In this example, the cutoff frequency (in hertz) is defined as:

$f_\mathrm{c} = {1 \over 2 \pi R_2 C }$

or equivalently (in radians per second):

$\omega_\mathrm{c} = \frac{1}{R_2 C}$

The gain in the passband is $\frac{-R_2}{R_1}$ , and the stopband drops off at −6 dB per octave, as it is a first-order filter.

Many times, a simple gain or attenuation amplifier (See operational amplifier) is turned into a lowpass filter by adding the capacitor C. This decreases the frequency response at high frequencies and helps to avoid oscillation in the amplifier. For example, an audio amplifier can be made into a lowpass filter with cutoff frequency 100 kHz to reduce gain at frequencies which would otherwise oscillate. Since the audio band (what we can hear) only goes up to 20 kHz or so, the frequencies of interest fall entirely in the passband, and the amplifier behaves the same way as far as audio is concerned.

Laplace notation

Continuous-time filters can also be described in terms of the Laplace transform of their impulse response in a way that allows all of the characteristics of the filter to be easily analyzed by considering the pattern of poles and zeros of the Laplace transform in the complex plane (in discrete time, one can similarly consider the Z-transform of the impulse response).

A first-order low-pass filter can be described in Laplace notation as

$\frac{\mathrm{Output}}{\mathrm{Input}} = \frac{1}{1 + s \tau}$

where s is the Laplace transform variable and τ is the filter time constant.

Digital simulation

For a more general algorithm to convert an analog filter into a digital filter, see Bilinear transform.

The effect of a low-pass filter can be simulated on a computer by analyzing its behavior in the time domain, and then discretizing the model.

A simple low-pass RC filter

From the circuit diagram to the right, according to Kirchoff's Laws and the definition of capacitance:

V i n (t) - V o u t (t) = I (t) R

Q c (t) = C V o u t (t)

Taking the time derivative of the second equation, $I(t) = C \frac{dV_{out}}{dt}$ . Combining this with the first equation:

$V_{in}(t) - V_{out}(t) = C \left[\frac{dV_{out}}{dt}\right] R$

Now we may discretize the equation. Let us represent $V i n$ by a series of samples $x 1... n$ . We will likewise represent $V o u t$ by a series of sample $y 1... n$ at the same points in time. For simplicity we assume that the samples are taken at evenly-spaced points in time separated by $Δ t$ . Making these substitutions:

$x_i - y_i = C \left[ \frac{y_{i}-y_{i-1}}{\Delta t} \right] R$

And rearranging terms:

$y_i = x_i \left( \frac{\Delta t}{RC + \Delta t} \right) + y_{i-1} \left( \frac{RC}{RC + \Delta t} \right)$

or more succinctly,

$y_n = \alpha x + (1 - \alpha) y_{n-1}\,$

where $\alpha = \frac{\Delta t}{RC + \Delta t}$

This gives us a way to determine the output samples in terms of the input samples and the preceding output. The following algorithm will simulate the effect of a low-pass filter on a series of digital samples:

 // Return RC low-pass filter output samples, given input samples,
 // time interval dt, and time constant RC
 function lowpass(real[0..n] x, real dt, real RC)
   var real[0..n] y
   var real alpha := dt / (RC + dt)
   
   for i from 0 to n
       y[i] := alpha * x[i] + (1-alpha) * y[i-1]
   
   return y

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