ssrs 2d barcode CHAP. 81 in Software

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CHAP. 81
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IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
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where G(z) is the second-order system shown below.
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Therefore,
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Y(z) = X(z)
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+ G(z)Y(z)
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1 - G(z)
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H(z)=
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To find G(z), we begin by writing the node equations for this network:
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Taking the z-transform of the first two equations, we have
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Substituting the second equation into the first gives
Finally, from the last difference equation, we have
Therefore. and for H(z) we have
For stability, it is necessary and sufficient that the coefficients [a(2) +a(4)] and [a(l) triangle (see Chap. 5 ) , which requires that
+ a(3)] lie within the stability
326 8.16
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
[CHAP. 8
Find the system function of the following network:
x(n)
This system is a feedback network that has the following form:
x(n)
G(z)
-If2
Therefore, the system function is
With
we have
Find the system function of the following network:
x(n)
The system function of this network may be found by writing down the difference equations corresponding to each adder and solving these equations using z-transforms. A simpler approach, however, is to redraw the network as follows,
CHAP. 81
IMPLEMENTATION O F DISCRETE-TIME SYSTEMS
which we recognize as a cascade of three second-order networks. Therefore, the system function is the product of the system functions of each network in the cascade, and we have
Consider the network in the figure below. Redraw the flowgraph as a cascade of second-order sections in transposed direct form 11.
To implement lhis system as a cascade of second-order transformed direct form 1 networks, we must first find the 1 system function corresponding to this network. Note that this network is of the form shown in the following figure:
HI(:)
Hs(z)
XOI)
Hz(-)
~ ( n )
where
Therefore, or
- 0.2:r1
+ 0.8:r2)( 1 + 0
. 2 ~ 0.82-I) ~ ~
Therefore, the desired network is as shown in the following figure:
.~(II)
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
[CHAP. 8
A digital oscillator has a unit sample response
The system function of this oscillator is
(a) Draw a direct form I1 network for this oscillator, and show how the sinusoid may be generated with no input and the appropriate set of initial conditions. (b) In applications involving the modulation of sinusoidal carrier signals in phase quadrature, it is necessary to generate the sinusoids
Beginning with a system that has a unit sample response h ( n ) = e J n q u ( n ) , separate the difference equation for this system into its real and imaginary parts, and draw a network that will generate these signals when initialized with the appropriate set of initial conditions.
(a) A direct form I1 network for the oscillator is as follows:
sin wl
With the input x ( n ) = &n). the response is y(n) = sin[(n registers corresponding to the delays are initialized so that
+ I)w]for n
2 0. Equivalently, if the storage
the zero-input response will be a sinusoid of frequency wo.
( h ) A complex exponential sequence y(n) = e J n q u(n) is generated by the difference equation
with the initial condition y(- I ) = e - j q Writing this difference equation in terms of its real and imaginary parts, we have
This equation is equivalent to the following pair of coupled difference equations, which are formed from the real part and the imaginary part of the equation:
CHAP. 81
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS A network that implements this pair of equations is shown below.
.yi(n) = sinnwo
COS Wg
The initial condition required to generate the desired output is yc - I ) = e - J q , or
Implement the system
as a parallel network of first-order direct form structures.
Factoring the denominator of the system function, we find
To implement H ( z ) as a parallel network of first-order filters, we must express H ( z ) as a sum of first-order factors using a partial fraction expansion. Because the order of the numerator is equal to the order of the denominator, this expansion will contain a constant term,
To find the value of C. we divide the numerator polynomial by the denominator as follows:
Therefore, C = 2, and we may write H ( z ) as follows:
Finally, with
we have, for the coefficients A and B,
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
[CHAP. 8
Thus, and the parallel network for this system is as shown below.
82 .1
The system function of a discrete-time system is
Draw a signal flowgraph of this system using a cascade of second-order systems in direct form 11, and write down the set of difference equations that corresponds to this implementation.
Expressing H ( z ) as a product of two second-order systems, we have H ( z )=
+ 22-I + z-=
- z-1
+ 2z-' + z
+ 12-2 8
1 + 22-1 + 3
which leads to the following cascade implementation for H(z):
CHAP. 81
IMPLEMENTATION OF DISCRETE-TIME SYSTEMS
With w(n), vl(n). and v2(n) as labeled in the figure above, the set of difference equations for this network is:
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