Comb filter

In signal processing, a comb filter adds a slightly delayed version of a signal to itself, causing phase cancellations. It is similar to ring modulation. The frequency response of a comb filter consists of a series of regularly-spaced spikes, so that it looks like a comb.

In discrete-time systems, the filter implements the following formula:

<math>

y[n] = ax[n] + bx[n – \tau] + cy[n – \tau] <math>

where τ is a constant delay. The comb filter can also be implemented in the continuous-time domain. The frequency response is given by the following:

<math>H(\omega) = \frac{a + be^{-i \omega \tau}} {1 – ce^{-i \omega \tau}}<math>

The peaks of the characteristic "comb" behavior occur because the frequency response experiences periodic discontinuities. This occurs when the following condition holds:

<math>\cos(\omega \tau) = \frac{1+c^2}{2c}<math>

Applications

There are 2D and 3D comb filters implemented in hardware (and occasionally software) for NTSC television decoders. The filters work to reduce artifacts such as dot crawl.

Other versions of comb filters can be found in networked terrestrial telecommunications systems. However, comb filters are not found in deep-space telecommunications systems because of separate system design considerations.

Comb filters can be used to produce echo effects. For instance, if the delay is set to a few milliseconds, and the filter is used on sound signals, it models the effect of a cylindrical cavity. This is because such a cavity amplifies frequencies that correspond to standing waves across its width.

Derivation of the frequency response

The comb filter is a linear time-invariant system. Thus, if the input signal x(n) is an exponential of the form:

<math>x(n) = e^{i \omega n}<math>

the output signal y(n) has the form:

<math>y(n) = H(\omega) e^{i \omega n}<math>   .

If these expressions are substituted into the comb filter formula above, it becomes:

<math>H(\omega)e^{i \omega n} = ae^{i \omega n} + be^{i \omega (n-\tau)} + cH(\omega)e^{i \omega (n-\tau)}<math>

<math>H(\omega)e^{i \omega n} = ae^{i \omega n} + be^{-i \omega \tau}e^{i \omega n} + cH(\omega)e^{-i \omega \tau}e^{i \omega n}<math>

Since the exponential function is never zero, it may be divided out:

<math>H(\omega) = a + be^{-i \omega \tau} + cH(\omega)e^{-i \omega \tau}<math>

Solving for <math>H(\omega)<math> yields:

<math>H(\omega) = \frac{a + be^{-i \omega \tau}} {1 – ce^{-i \omega \tau}}<math>