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X ( e j ω ) = X ( z ) z = e j ω = n = - x ( n ) e - j ω n

From the definition of the Z-transform, a change of variable m = n - K shows that a delay of K samples in the time domain is equivalent to multiplication by z - K in the Z-transform domain.

x ( n - K ) Z n = - x ( n - K ) z - n = m = - x ( m ) z - ( m + K ) = z - K m = - x ( m ) z - m = z - K X ( z )

We may use this fact to re-write [link] in the Z-transform domain, by taking Z-transforms of both sides ofthe equation:

Y ( z ) = i = 0 N - 1 b i z - i X ( z ) - k = 1 M a k z - k Y ( z ) Y ( z ) 1 + k = 1 M a k z - k = X ( z ) i = 0 N - 1 b i z - i H ( z ) = Y ( z ) X ( z ) = i = 0 N - 1 b i z - i 1 + k = 1 M a k z - k

From this formula, we see that any filter which can be represented by a linear difference equation with constant coefficientshas a rational transfer function (i.e. a transfer function which is a ratio of polynomials).From this result, we may compute the frequency response of the filter by evaluating H ( z ) on the unit circle:

H ( e j ω ) = i = 0 N - 1 b i e - j ω i 1 + k = 1 M a k e - j ω k .

There are many different methods for implementing a general recursive difference equation of the form [link] . Depending on the application, some methods may be more robustto quantization error, require fewer multiplies or adds, or require less memory. [link] shows a system diagram known as the direct form implementation; it works for any discrete-time filterdescribed by the difference equation in [link] . Note that the boxes containing the symbol z - 1 represent unit delays, while a parameter written next to a signal path represents multiplication by that parameter.

Direct form implementation for a discrete-time filter described by a linear recursive difference equation.

Design of a simple fir filter

Download the files, nspeech1.mat and DTFT.m for the following section.

Location of two zeros for simple a FIR filter.

To illustrate the use of zeros in filter design, you will design a simple second order FIR filterwith the two zeros on the unit circle as shown in [link] . In order for the filter's impulse response to be real-valued,the two zeros must be complex conjugates of one another:

z 1 = e j θ
z 2 = e - j θ

where θ is the angle of z 1 relative to the positive real axis. We will see later that θ [ 0 , π ] may be interpreted as the location of the zeros in the frequency response.

The transfer function for this filter is given by

H f ( z ) = ( 1 - z 1 z - 1 ) ( 1 - z 2 z - 1 ) = ( 1 - e j θ z - 1 ) ( 1 - e - j θ z - 1 ) = 1 - 2 cos θ z - 1 + z - 2 .

Use this transfer function to determine the difference equation for this filter.Then draw the corresponding system diagram and compute the filter's impulse response h ( n ) .

This filter is an FIR filter because it has impulse response h ( n ) of finite duration. Any filter with only zeros and no poles other than thoseat 0 and ± is an FIR filter. Zeros in the transfer function represent frequencies that arenot passed through the filter. This can be useful for removing unwanted frequencies in a signal.The fact that H f ( z ) has zeros at e ± j θ implies that H f ( e ± j θ ) = 0 . This means that the filter will not pass pure sine waves at a frequencyof ω = θ .

Use Matlab to compute and plot the magnitude of the filter's frequency response | H f ( e j ω ) | as a function of ω on the interval - π < ω < π , for the following three values of θ :

  • θ = π / 6
  • θ = π / 3
  • θ = π / 2

Questions & Answers

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The nanotechnology is as new science, to scale nanometric
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
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Source:  OpenStax, Purdue digital signal processing labs (ece 438). OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10593/1.4
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