barcode generator for ssrs the closed-loop system function is in Software

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the closed-loop system function is
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TRANSFORM ANALYSIS O F SYSTEMS
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[CHAP. 5
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Fig. 5-6.
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A feedback network.
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If H ( z ) and G ( z ) are rational functions of z,
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the closed-loop system function may be written as
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Therefore, the poles of the closed-loop system Q ( z ) are the roots of the equation
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With the appropriate order and coefficients for G ( z ) , the poles may be placed anywhere in the z-plane.
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Suppose that we have an unstable system with system function
H (z) =
Placed in a feedback network with
I 1 - 1.22-'
G(z) = K the system function of the closed-loop system is
which has a pole at z = 1.2/(1
+ K). Therefore, this system will be stable for all K
s 0.2.
Solved Problems
System Function
If the input to a linear shift-invariant system is
the output is
~ ( n= 6 ( ; ) " u ( n ) - 6 ( : ) " u ( n ) )
Find the system function, H ( z ) , and determine whether or not the system is stable and/or causal.
CHAP. 51
TRANSFORM ANALYSIS OF SYSTEMS
In order to find the system function, recall that H ( z ) = Y ( z ) / X ( z ) . Because we are given both x ( n ) and y ( n ) , all that is necessary to find H ( z ) is to evaluate the z-transform of x ( n ) and y ( n ) and divide. With
Then, For the region of convergence of H ( z ) , we have two possibilities. Either lzl z $ or Izl < Because the region of convergence of Y ( z ) is Izl > and includes the intersection of the regions of convergence of X ( z ) and H ( z ) , the region of convergence of H ( z ) must be lzl > Because the region of convergence of H ( z ) includes the unit circle, h ( n ) is stable, and because the region of convergence is the exterior of a circle and includes z = m, h ( n ) is causal.
When the input to a linear shift-invariant system is
the output is
A n ) = [4(;)"
Find the unit sample response of the system.
- 3(-:)"]u(n)
One approach that we may use to solve this problem is to evaluate H ( z ) = Y ( z ) / X ( z ) and then compute the inverse z-transform. Note, however, that we are given the response of the system to a step with an amplitude of 2, and we are asked to find the unit sample response. Because
if we let s ( n ) be the step response, it follows from linearity that
h ( n ) = s ( n ) - s(n - 1)
Therefore, from the response given above, we have
A causal linear shift-invariant system is characterized by the difference equation
y(n) = by(n - 1 )
+ i y ( n - 2) + x(n) - x(n - 1 )
Find the system function, H ( z ) , and the unit sample response, h ( n ) .
To find the system function, we take the z-transform of the difference equation,
Y(z)= f z - ' ~ ( z ) $ Z - ~ Y ( Z ) X(z) - z-Ix(z)
TRANSFORM ANALYSIS OF SYSTEMS
[CHAP. 5
Therefore, the system function is
Because the system is causal, the region of convergence is lzl z f . To find the unit sample response, we perform a partial fraction expansion of H ( z ) ,
where
Therefore. and the unit sample response is
A causal linear shift-invariant system has a system function 1 z-' I - iz-l
H (z) =
Find the z-transform of the input, x(n), that will produce the output
I y(n) = - 7 ( z )I "
~ ( n - +(2)"u(-n )
To find the input to a linear shift-invariant filter that will produce a given output y ( n ) , we use the relationship Y(z) = H(z)X(z) to solve for X(z):
Computing the z-transform of y(n), we have Y (z) = - -----+ A 1-I 1 - 22--'
I -4 -
I (1
+ fz-'
- iz-l)(l - 2z-1)
Therefore,
X(z) =
(1 (1
+ iz-I)(1
- ;z-1)
- $z-')(l - 2z-')(I + z-I)
1- 2
B C + ------ + --I - 2z-1 I + z-1
CHAP. 51
TRANSFORM ANALYSIS OF SYSTEMS
where
Because h ( n ) is causal, the region of convergence for H ( z ) is Izl > With the region of convergence of Y ( z ) the annulus < 15 1 i2, the region of convergence of X ( z ) is < IzJ i1. Therefore,
Show that if h ( n ) is real, and H ( ) is rational, z
- pkz-9
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