0.5 Lab 5a - digital filter design (part 1)

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Questions or comments concerning this laboratory should be directedto Prof. Charles A. Bouman, School of Electrical and Computer Engineering, Purdue University, West Lafayette IN 47907;(765) 494-0340; bouman@ecn.purdue.edu

Introduction

This is the first part of a two week laboratory in digital filter design.The first week of the laboratory covers some basic examples of FIR and IIR filters, and then introduces the conceptsof FIR filter design. Then the second weekcovers systematic methods of designing both FIR and IIR filters.

Background on digital filters

In digital signal processing applications, it is often necessary to change the relative amplitudesof frequency components or remove undesired frequencies of a signal.This process is called filtering . Digital filters are used in a variety of applications.In Laboratory 4, we saw that digital filters may be used in systems that perform interpolation and decimationon discrete-time signals. Digital filters are also used in audio systemsthat allow the listener to adjust the bass (low-frequency energy) and the treble (high frequency energy) of audio signals.

Digital filter design requires the use of both frequency domain and time domain techniques.This is because filter design specifications are often given in the frequency domain, but filters are usually implementedin the time domain with a difference equation. Typically, frequency domain analysis is done using the Z-transform andthe discrete-time Fourier Transform (DTFT).

In general, a linear and time-invariant causal digital filter with input $x\left(n\right)$ and output $y\left(n\right)$ may be specified by its difference equation

$y\left(n\right)=\sum _{i=0}^{N-1}{b}_{i}x\left(n-i\right)-\sum _{k=1}^{M}{a}_{k}y\left(n-k\right)$

where ${b}_{i}$ and ${a}_{k}$ are coefficients which parameterize the filter. This filter is said to have $N$ zeros and $M$ poles. Each new value of the output signal, $y\left(n\right)$ , is determined by past values of the output, and by present and past valuesof the input. The impulse response, $h\left(n\right)$ , is the response of the filter to an input of $\delta \left(n\right)$ , and is therefore the solution to the recursive difference equation

$h\left(n\right)=\sum _{i=0}^{N-1}{b}_{i}\delta \left(n-i\right)-\sum _{k=1}^{M}{a}_{k}h\left(n-k\right)\phantom{\rule{4pt}{0ex}}.$

There are two general classes of digital filters: infinite impulse response (IIR) and finite impulse response (FIR).The FIR case occurs when ${a}_{k}=0$ , for all $k$ . Such a filter is said to have no poles, only zeros.In this case, the difference equation in [link] becomes

$h\left(n\right)=\sum _{i=0}^{N-1}{b}_{i}\delta \left(n-i\right)\phantom{\rule{4pt}{0ex}}.$

Since [link] is no longer recursive, the impulse response has finite duration $N$ .

In the case where ${a}_{k}\ne 0$ , the difference equation usually represents an IIR filter.In this case, [link] will usually generate an impulse response which has non-zero values as $n\to \infty$ . However, later we will see that for certain valuesof ${a}_{k}\ne 0$ and ${b}_{i}$ , it is possible to generate an FIR filter response.

The Z-transform is the major tool used for analyzing the frequency response of filters and their differenceequations. The Z-transform of a discrete-time signal, $x\left(n\right)$ , is given by

$X\left(z\right)=\sum _{n=-\infty }^{\infty }x\left(n\right){z}^{-n}\phantom{\rule{4pt}{0ex}}.$

where $z$ is a complex variable. The DTFT may be thought of as a special case of the Z-transformwhere $z$ is evaluated on the unit circle in the complex plane.

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Ali
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learn
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da
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Giriraj
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Damian
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Professor
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Professor
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LITNING
scanning tunneling microscope
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Rafiq
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brayan
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Damian
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Kyle
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biomolecules are e building blocks of every organics and inorganic materials.
Joe
how did you get the value of 2000N.What calculations are needed to arrive at it
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