ssrs 2d barcode (e) A parallel connection of first- and second-order systems realized in direct form I1 in Software

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(e) A parallel connection of first- and second-order systems realized in direct form I1
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(a) Writing the system function as a ratio of polynomials in z - I ,
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H (z) =
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it follows that the direct form I realization of H ( z ) is as follows:
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realization of H ( z ) ,we have
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IMPLEMENTATION O F DISCRETE-TIME SYSTEMS
[CHAP. 8
(c) Using a cascade of first- and second-order systems realized in direct form 11, we have a choice of either pairing the zero with the first-order factor in the denominator or with the second-order factor. Although it does not make a difference from a computational point of view, because the zero is closer to the pair of complex poles than to the pole at z = 0.7, we will pair the zero with the second-order factor. With this pairing, the realization of H ( z ) is as follows:
(d) If we change the direct form 11 systems in part ( r )to transposed direct form 11, we have the realization shown below.
x(n) 0
y(n)
,,z-I
(e) For a parallel structure, H ( z ) must be expanded using a partial fraction expansion:
H (z)= 1 + 0.875zr1 A + Bz-I + 0.2zr1 + 0 . 9 z r 2 ) ( l - 0 . 7 ~ - I )- 1 + 0 . 2 -~I + 0.9zr2
C 1 -0.7~-I
The constants A, B , and C may be found as follows. Recombining the two terms in the partial fraction expansion as follows.
and equating the coefficients in the numerator of this expression with the numerator of H ( z ) , we have the following three equations in the three unknowns A , B , and C :
Solving for A , B, and C we find
and, therefore, the partial fraction expansion is
CHAP. 81
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS Thus. parallel structure for H(z) is shown below.
Consider the filter structure shown in the figure below.
x(n) 0
~ ( n )
Find the system function and the unit sample response of this system.
For the three nodes labeled in the flowgraph below.
Node I
Node 3
w(n)
/I I
we have the following node equations:
Using z-transforms, the first equation becomes
Taking the z-transform of the second equation, and substituting the expression above for W ( z ) ,we have
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS Finally, taking the z-transform of the last equation, we find
[CHAP. 8
Therefore, the system function is
H ( z )=
and the unit sample response is h ( n ) = 2(0.2)"u(n)
+ 1.82-'
- 0.22-' - I)
+ 1.8(0.2)"-'u(n
Find a direct form I1 realization for the following network:
To solve this problem, we begin by writing the node equations for each of the adders in the network. If we label the three nodes that are adders as in the figure below,
x01)
"NO)
" h(2)
z- I
Node 1
Node 2 dl)
a(2)
- y(n)
Node 3
and denote the output of the first node by wl(n), and the output of the second by wz(n),we have the following three equations for the three node variables,
Substituting the first equation into the second, we have
Then, substituting this equation into the third equation, we have
CHAP. 81
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
Thus, the direct form I1 structure for this system is as shown below.
Find a transposed direct form I1 realization for the system described by the difference equation I y ( n ) = 2 Y( n - I ) - i y ( n - 2 ) + x ( n ) 4 J x ( n - 1) and write down the set of difference equations that corresponds to this realization.
The transposed direct form I1 realization for this system is as follows:
With the node variables v l ( n )and v,(n) as labeled in the network above, the difference equations that describe this network are as follows: vl ( n ) = x ( n ) v d n - I ) v z ( n )= - f x ( n ) 5 v l ( n )- :vl(n - I )
v ( n )= v ~ ( n )
Find the system function and the unit sample response for the following network, and draw an equivalent direct form 11 structure:
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
[CHAP. 8
This network is a parallel connection of two first-order systems, plus a feed-through connection with a gain of 2. Therefore, the system function is the sum of three system functions:
Thus, the unit sample response is
To find an equivalent direct form 11 realization, H ( z ) is first expressed as follows:
Therefore, the direct form 11 structure is as shown in the following figure:
Find the system function for the following network, and determine the conditions o n the coefficients a ( ] ) ,
a(2),a(3),and a(4) that will ensure the stability of the system:
The first thing to observe is that this system corresponds to a feedback network of the form shown in the following figure:
x(n)
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