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z-Transform Analysis of Discrete-Time Filters

Module by: Mark A. Davenport. E-mail the author

zz-transform analysis of discrete-time filters

The zz-transform might seem slightly ugly. We have to worry about the region of convergence, and we haven't even talked about how to invert it yet (it isn't pretty). However, in the end it is worth it because it is extremely useful in analyzing digital filters with feedback. For example, consider the system illustrated below

Figure 1
A flowchart of a system.

We can analyze this system via the equations

v [ n ] = b 0 x [ n ] + b 1 x [ n - 1 ] + b 2 x [ n - 2 ] v [ n ] = b 0 x [ n ] + b 1 x [ n - 1 ] + b 2 x [ n - 2 ]
(1)

and

y [ n ] = v [ n ] + a 1 y [ n - 1 ] + a 2 y [ n - 2 ] . y [ n ] = v [ n ] + a 1 y [ n - 1 ] + a 2 y [ n - 2 ] .
(2)

More generally,

v [ n ] = N k = 0 b k x [ n - k ] v [ n ] = N k = 0 b k x [ n - k ]
(3)

and

y [ n ] = M k = 1 a k y [ n - k ] + v [ n ] y [ n ] = M k = 1 a k y [ n - k ] + v [ n ]
(4)

or equivalently

N k = 0 b k x [ n - k ] = y [ n ] - M k = 1 a k y [ n - k ] . N k = 0 b k x [ n - k ] = y [ n ] - M k = 1 a k y [ n - k ] .
(5)

In general, many LSI systems satisfy linear difference equations of the form:

M k = 0 a k y [ n - k ] = N k = 0 b k x [ n - k ] . M k = 0 a k y [ n - k ] = N k = 0 b k x [ n - k ] .
(6)

What does the zz-transform of this relationship look like?

Z k = 0 M a k y [ n - k ] = Z k = 0 M b k x [ n - k ] k = 0 M a k Z { y [ n - k ] } = k = 0 N b k Z { x [ n - k ] } . Z k = 0 M a k y [ n - k ] = Z k = 0 M b k x [ n - k ] k = 0 M a k Z { y [ n - k ] } = k = 0 N b k Z { x [ n - k ] } .
(7)

Note that

Z y [ n - k ] = n = - y [ n - k ] z - n = m = - y [ m ] z - m · z - k = Y ( z ) z - k . Z y [ n - k ] = n = - y [ n - k ] z - n = m = - y [ m ] z - m · z - k = Y ( z ) z - k .
(8)

Thus the relationship above reduces to

k = 0 M a k Y ( z ) z - k = k = 0 N b k X ( z ) z - k Y ( z ) k = 0 M a k z - k = X ( z ) k = 0 N b k z - k Y ( z ) X ( z ) = k = 0 N b k z - k k = 0 M a k z - k k = 0 M a k Y ( z ) z - k = k = 0 N b k X ( z ) z - k Y ( z ) k = 0 M a k z - k = X ( z ) k = 0 N b k z - k Y ( z ) X ( z ) = k = 0 N b k z - k k = 0 M a k z - k
(9)

Hence, given a system like the one above, we can pretty much immediately write down the system's transfer function, and we end up with a rational function, i.e., a ratio of two polynomials in zz. Similarly, given a rational function, it is easy to realize this function in a simple hardware architecture. We will focus exclusively on such rational functions in this course.

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