Ch6 Fourier

On board, did demonstration to show what convolution looks like for two sin curves of the same vs different frequencies, and of different phases.

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.