Ch6 Fourier

Discrete-time filters in continuous frequency space

Back to your textbook, in Chapter 6:

Here we will work with discrete signals (using time as the index) and discrete filters, but be interested in their continuous frequency representations.

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Imagine we have a signal \(x(t)\) which is defined for \(t\) is the interval \([0, T]\)

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We will imagine that \(x(t)\) which is defined over all \(t\) but periodic of period \(T\). So the function has frequency \(f=1 / T\) and angular frequency \(\omega=2 \pi / T\)

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A discrete filter is a sequence of complex numbers \(\mathbf{h}=\left\{h_{n}\right\}_{n=-\infty}^{\infty}\).

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This will be a finite impulse response (FIR) filter if \(h_n\) is 0 for \(n<0\) or \(n>L\) for some L.

In this case we say \(\mathbf{h}\) has length \(L\) and express it in the form \(\mathbf{h}=\left\{h_{n}\right\}_{n=0}^{L}\).

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Then we can take its discrete time Fourier transform (DTFT) as the Fourier series \(X(\mathbf{h})(\zeta)=\sum_{n=-\infty}^{\infty} h_{n} e^{i \zeta}, \zeta \in \mathbb{R}\).

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Suppose we have a discrete signal \(x(t)\) in sampling periods \(T_{s}\), i.e., at \(t_{n}=n T_{s}\). Define \(x_{n}=x\left(t_{n}\right)\)

If we use our filter on our signal, we get:

\[ y_{n}=h_{0} x_{n}+h_{1} x_{n-1}+\cdots+h_{L} x_{n-L}, \quad n \in \mathbb{Z} \]

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To understand the effect of the filter, examine its action on a single Fourier mode of the signal, with a possible phase shift \(\phi\):

\(x_{m}(t)=c_{m} e^{i m(\omega t+\phi)}\)

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At individual points in time, this gives

\(x_{m, n}=c_{m} e^{i m \omega\left(n T_{s}+\phi\right)}\)

Thus for fixed \(m\) and arbitrary \(n\), we have

\[ \begin{aligned} y_{m, n} & =h_{0} c_{m} e^{i m \omega\left(n T_{s}+\phi\right)}+h_{1} c_{m} e^{i m \omega\left((n-1) T_{s}+\phi\right)}+\cdots+h_{L} c_{m} e^{i m \omega\left((n-L) T_{s}+\phi\right)} \\ & =\left(h_{0} \cdot 1+h_{1} e^{-i m \omega T_{s}}+\cdots+h_{L} e^{-i L m \omega T_{s}}\right) c_{m} e^{i m \omega\left(n T_{s}+\phi\right)} \\ & =H\left(-m \omega T_{s}\right) x_{m, n} \end{aligned} \]

where

\(H(\zeta)=h_{0}+h_{1} e^{i \zeta}+\cdots+h_{L} e^{i L \zeta}, \quad \zeta \in \mathbb{R}\) is the DTFT of \(\mathbf{h}\).

The units of \(\zeta\) are radians per sampling period. So when \(\zeta=1\), our filter goes through half a cycle in one sampling period.

Gain

\(\left|y_{m, n}\right| \leq\left|H\left(-m \omega T_{s}\right)\right|\left|x_{m, n}\right|\)

So we define gain or attenuation of this transformation as \(G(\zeta)=|H(\zeta)|\)

We define phase rotation as \(\Theta(\zeta)=\theta\), where \(H(\zeta)=|H(\zeta)| e^{i \theta}\)

The filter \(h\) will attenuate the signal at frequencies where \(|H(\zeta)|<1\) and amplify the signal at frequencies where \(|H(\zeta)|>1\). It will phase shift the signal by \(\Theta(\zeta)\).

High-pass and low-pass filters

The FIR filter \(\mathbf{h}=\left\{h_{k}\right\}_{k=0}^{L}\) with discrete time Fourier transform \(H(\zeta)\) is a lowpass filter if \(|H(0)|=1\) and \(|H(\pi)|=0\); \(\mathbf{h}\) is a highpass filter if \(|H(0)|=0\) and \(|H(\pi)|=1\).