# 3.10 Frequency response

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Recall from [link] that the convolution integral

$y\left(t\right)={\int }_{-\infty }^{\infty }x\left(\tau \right)h\left(t-\tau \right)d\tau$

has the Fourier Transform:

$Y\left(j\Omega \right)=H\left(j\Omega \right)X\left(j\Omega \right)$

where $H\left(j\Omega \right)$ and $X\left(j\Omega \right)$ are the Fourier Transforms of $h\left(t\right)$ and $x\left(t\right)$ , respectively. Solving for $H\left(j\Omega \right)$ gives the frequency response :

$H\left(j\Omega \right)=\frac{Y\left(j\Omega \right)}{X\left(j\Omega \right)}$

The frequency response, the Fourier Transform of the impulse response of a filter, is useful since it gives a highly descriptive representation of the properties of the filter. The frequency response can be considered to be the gain of the filter, expressed as a function of frequency. The magnitude of the frequency response evaluated at $\Omega ={\Omega }_{0}$ , $|H\left(j{\Omega }_{0}\right)|$ gives the factor the frequency component of $x\left(t\right)$ at $\Omega ={\Omega }_{0}$ would be scaled by. The phase of the frequency response at $\Omega ={\Omega }_{0}$ , $\angle H\left(j{\Omega }_{0}\right)$ gives the phase shift the component of $x\left(t\right)$ at $\Omega ={\Omega }_{0}$ would undergo. This idea will be discussed in greater detail in [link] . A lowpass filter is a filter which only passes low frequencies, while attenuating or filtering out higher frequencies. A highpass filter would do just the opposite, it would filter out low frequencies and allow high frequencies to pass. [link] shows examples of these various filter types.

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