# Frequency response

In signal processing, the frequency response of an analog or digital equipment or software is some representation, usually numeric, graphical, or mathematical, of how that equipment or software impacts frequencies.

Since a frequency is fully defined by its amplitude and phase, the frequency response typically describes how the equipment, software (or any system) impacts the amplitude and phase of a set of frequencies. When describing only amplitudes, the impact is called the magnitude response of the system. When describing only phase, the impact is called the phase response of the system. Additional detail and examples of magnitude responses and phase responses is provided in their respective topics.

## Frequency response and transfer functions

In the general case, when the system is defined with a transfer function, the frequency response (both magnitude response and phase response) can be computed from the transfer function. For example, given the Z transform transfer function H(z) of a linear time-invariant systems, the magnitude of H(z), |H(z)|, z = A e-j ω, when evaluated at A = 1 (on the unit circle), produces the magnitude response of the system

$$|H(e^{-j\,\omega})|=\sqrt{Re(H(e^{-j\,\omega}))^2+Im(H(e^{-j\,\omega}))^2}$$

where Re and Im are the real and imaginary parts of H. The phase response of the system is

$$\Phi(e^{-j\,\omega})=atan2(Im(H(e^{-j\,\omega})),Re(H(e^{-j\,\omega})))$$

Similarly, the magnitude response of a system defined with its Laplace transform transfer function H(s), s = σ + j ω, evaluated at σ = 0, is |H(j ω)| and its phase response is arg(H(j ω)).

## Example: Frequency response of a finite impulse response filter

A finite impulse response filter a(k) of length N computes the output signal y(k) from the input signal x(k) with the formula

$$y(k)=\sum_{n=0}^{N-1} a(n) \, x(k-n)$$

The Z transform of both sides of the equation above produces

$$Z(y(k))=Z(\sum_{n=0}^{N-1} a(n) \, x(k-n))=Z(x(k)) \sum_{n=0}^{N-1} a(n) \, z^{-n}$$

and the transfer function of the filter is

$$H(z)=\frac{Z(y(k))}{Z(x(k))}=\sum_{k=0}^{N-1} a(k) \, z^{-k}$$

Thus, the magnitude response of the filter at the frequency f is

$$|H(f)|=\sqrt{(\sum_{k=0}^{N-1} a(k) \, cos(\frac{2\pi\,f\,k}{N}))^2+(\sum_{k=0}^{N-1} a(k) \, sin(\frac{2\pi\,f\,k}{N}))^2}$$

If the filter is one with coefficients that are symmetric around the middle, it can be shown that the phase response is as follows (see Phase response).

$$\Phi(f)=\frac{2\pi\,f}{N} \frac{N-1}{2}$$

## Example: frequency response of headphones

Occasionally, you may see the frequency response of headphones (or, say, speakers) listed as (20 Hz, 20 KHz) for example. This is a particularly simplistic representation of the frequency response and it means that the headphones are able to reproduce frequencies between 20 Hz and 20 KHz. Thus, if an input signal contains frequencies in this interval it will be reproduced by the headphones. You would not know, however, how well these headphones will do for specific frequencies. It may be that a bass at 300 Hz is reproduced, but at lower amplitude, which means that the bass will be heard through the headphones, but not very well.

### Filtered HTML

• Freelinking helps you easily create HTML links. Links take the form of [[indicator:target|Title]]. By default (no indicator): Click to view a local node.