barcode lib ssrs Consider the fourth-order combfilter that has a system function in Software

Encoder Code128 in Software Consider the fourth-order combfilter that has a system function

Consider the fourth-order combfilter that has a system function
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(a) Draw a pole-zero diagram for H (z).
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(b) Find the value for A so that the peak gain of the filter is equal to 2.
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(c) Find a structure for this filter that requires only one multiplier.
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(a) The comb filter has four zeros on the unit circle at
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pk = e~(2k+l)n/4
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and four poles at
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ark =
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k = O . 1.2.3
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k = 0, 1,2,3
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A pole-zero diagram for H(z) is shown in the following figure:
(b) Due to the zeros on the unit circle, the magnitude of the frequency response is zero at wt=(2k l)n/4 for k = 0, 1,2, 3, and it increases monotonically until it reaches a maximum value at the frequencies that are midway between the zeros at wk = k1r/2. Therefore, in order for the peak gain to be equal to 2, we want
which implies that
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
(c) With a system function
[CHAP. 8
we may implement this system using two multiplies as shown in the figure below.
Note, however, that the multiplier may be shared as follows:
a(n)
The difference equations for this network are
The system function of an allpass filter has the form
The symmetry that exists between the numerator and denominator polynomials allows for special structures that are more efficient than direct form realizations in terms of the number of multiplies required to compute each output value y(n ).
(a) Consider the first-order allpass filter,
H(z) =
+az-I
where a! is real. Find an implementation for this system that requires two delays but only one multiplication.
(h) For a second-order allpass filter with h ( n ) real, the system function has the form
where a and B are real. Derive a structure that implements this system using four delays but only two multiplies.
(a) The direct form realization of a first-order allpass filter requires two multiplies and one delay as shown in the figure below.
CHAP. 81
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS To see how the two multiplies may be combined, consider the difference equation for this system:
Therefore, only one multiplication is necessary if we form the difference x ( n ) - y ( n - I ) prior to multiplying by a. Thus, we have the structure illustrated in the figure below that has two delays but only one multiplication.
Because this structure requires an extra delay compared to direct form, this structure is not canonic.
( b ) As with the first-order allpass filter. we may find a two-multiplier realization ol'a second-order allpass filter by
combining together terms in the difference equation for the allpass filter as follows:
Thus, only two multiplications are required if we can form the differencess(n - 1 ) - y ( n - 1 ) and . r ( n ) - y ( n -2) prior to performing any multiplications. A structure that accomplishes this is given in the figure below.
Note that with the additional delays, two multiplications are saved compared to a direct form implementation.
Lattice Filters
8.24 Sketch a lattice filter implementation of the FIR filter
To implement this system using a lattice filter structure, we must find the reflection coefficients that generate the polynomial H(z). First, however, it is necessary to normalize H ( z ) so that the first coefficient is unity:
Now, with we see that
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS Next. we generate the second-order system H r ( z )using the step-down recursion:
[CHAP. 8
Therefore,
r2 = 0.1905. Finally, we have
and, therefore, I-1 = 0.4. Thus, the lattice filter structure is as shown below.
Shown in the figure below is an FIR lattice filter.
((1)
Find the system function A ( = ) = F ( z ) / X ( z ) relating the input x ( n ) to the output f ( n ) . Does this system have minimum phase
(0) for
(b) Repeat part
the system function relating x ( n ) to ~ ( n ) .
( a ) To find the system function relating .u(tz) to f ' ( t l ) , we use the step-up recursion. Using the vector form of the recursion, we have for the coefficients u l ( k )
Then, with Tz = 0.4, for a (/,), have we
CHAP. 81 Finally, with
IMPLEMENTATION O F DISCRETE-TIME SYSTEMS
r3= 0.2. we have
Thus.
A(:) = 1
+ 0.78;-' + 0.54z-' + 0.2;-'
This system will have minimum phase if the zeros of A(z) are inside this unit circle. Although this could be determined by factoring A(z), because the reflection coefficients used to generate A ( : ) are bounded by I In magnitude. it follows that A ( z ) has minimum phase.
( b ) The system function A'(z) = G(-.)/X ( z ) is related to X ( z ) as follows:
all Because the zeros of A1(z)are formed by flipping the zeros of A ( )about the unit c~rcle. of the zeros of A'(:) will be o~ctside unit circle and thus will not have minimum phase. the
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