Free Books    Introduction to Digital Filters  

Frequency Response

The frequency response of an LTI filter may be defined as the spectrum of the output signal divided by the spectrum of the input signal. In this section, we show that the frequency response of any LTI filter is given by its transfer function $ H(z)$ evaluated on the unit circle, i.e., $ H(e^{j\omega T})$. We then show that this is the same result we got using sine-wave analysis in Chapter 1.

Beginning with Eq.$ \,$(6.4), we have

$\displaystyle Y(z) = H(z)X(z) $

where X(z) is the z transform of the filter input signal $ x(n)$, $ Y(z)$ is the z transform of the output signal $ y(n)$, and $ H(z)$ is the filter transfer function.

A basic property of the z transform is that, over the unit circle $ z=e^{j\omega T}$, we find the spectrum [84].8.1To show this, we set $ z=e^{j\omega T}$ in the definition of the z transform, Eq.$ \,$(6.1), to obtain

$\displaystyle X(e^{j\omega T}) = \sum_{n=-\infty}^\infty x(n) e^{-j\omega T n} $

which may be recognized as the definition of the bilateral discrete time Fourier transform (DTFT) when $ T$ is normalized to 1 [59,84]. When $ x$ is causal, this definition reduces to the usual (unilateral) DTFT definition:

DTFT$\displaystyle _\omega(x) \isdefs \sum_{n=0}^\infty x(n) e^{-j\omega n} \protect$ (8.1)

Applying this relation to $ Y(z)=H(z)X(z)$ gives

$\displaystyle Y(e^{j\omega T}) = H(e^{j\omega T})X(e^{j\omega T}). \protect$ (8.2)

Thus, the spectrum of the filter output is just the input spectrum times the spectrum of the impulse response $ H(e^{j\omega T})$. We have therefore shown the following:
$\textstyle \parbox{0.8\textwidth}{\emph{The frequency response of a linear tim... ...uated on the unit circle in the $z$ plane, \textit{i.e.}, $H(e^{j\omega T})$.}}$
This immediately implies the following:
$\textstyle \parbox{0.8\textwidth}{\emph{The frequency response of an LTI filter equals the discrete-time Fourier transform of the impulse response.}}$
We can express this mathematically by writing

$\displaystyle \zbox {H(e^{j\omega T}) = \mbox{{\sc DTFT}}_{\omega T}(h).} $

By Eq.$ \,$(7.2), the frequency response specifies the gain and phase shift applied by the filter at each frequency. Since $ e$, $ j$, and $ T$ are constants, the frequency response $ H(e^{j\omega T})$ is only a function of radian frequency $ \omega$. Since $ \omega$ is real, the frequency response may be considered a complex-valued function of a real variable. The response at frequency $ f$ Hz, for example, is $ H(e^{j2\pi f T})$, where $ T$ is the sampling period in seconds. It might be more convenient to define new functions such as $ H^\prime(\omega) \isdeftext H(e^{j\omega T})$ and write simply $ Y^\prime(\omega) = H^\prime(\omega) X^\prime(\omega)$ instead of having to write $ e^{j\omega T}$ so often, but doing so would add a lot of new functions to an already notation-rich scenario. Furthermore, writing $ H(e^{j\omega T})$ makes explicit the connection between the transfer function and the frequency response.

Notice that defining the frequency response as a function of $ e^{j\omega T}$ places the frequency ``axis'' on the unit circle in the complex $ z$ plane, since $ \left\vert e^{j\omega T}\right\vert=1$. As a result, adding multiples of the sampling frequency to $ \omega$ corresponds to traversing whole cycles around the unit circle, since

$\displaystyle e^{j(\omega + k 2\pi f_s)T} = e^{j(\omega T + k 2\pi)} = e^{j\omega T}, $

whenever $ k$ is an integer. Since every discrete-time spectrum repeats in frequency with a ``period'' equal to the sampling rate, we may restrict $ \omega T$ to one traversal of the unit circle; a typical choice is $ -\pi \leq \omega T < \pi$ [ $ \omega T\in[-\pi,\pi)$]. For convenience, $ \omega T \in[-\pi,\pi]$ is often allowed.

We have seen that the spectrum is a particular slice through the transfer function. It is also possible to go the other way and generalize the spectrum (defined only over the unit circle) to the entire $ z$ plane by means of analytic continuation (§D.2). Since analytic continuation is unique (for all filters encountered in practice), we get the same result going either direction.

Because every complex number $ z$ can be represented as a magnitude $ r=\vert z\vert$ and angle $ \theta=\angle z$, viz., $ z=r\exp(j\theta)$, the frequency response $ H(e^{j\omega T})$ may be decomposed into two real-valued functions, the amplitude response $ \vert H(e^{j\omega T})\vert$ and the phase response $ \angle H(e^{j\omega T})$. Formally, we may define them as follows:

Next Section:
Amplitude Response
Previous Section:
Problems