Therefore, 1 3y1 + 9yo - 1, 11-5yo - 5yı 6. F(2) 1 1 (4.325) z +1 10 z+3 15 z – 2 and taking the inverse transform gives Yk = 16(3y1 + 9yo – 1)(-1)* +/10(1 – 5yo – 5y1)(-3)* + 1/152*. (4.326)
Protection System
A system that protects electrical systems from faults by isolating the problematic part from the remainder of the system, preventing power from being cut from healthy elements, improving system dependability and efficiency is the protection system. Protection devices are the equipment that are utilized to implement the protection system.
Predictive Maintenance System
Predictive maintenance technologies are designed to assist in determining the state of in-service equipment so that maintenance can be scheduled. Predictive maintenance is the application of information; proactive maintenance approaches examine the condition of equipment and anticipate when it should maintain. The purpose of predictive maintenance is to forecast when equipment will fail (depending on a variety of parameters), then prevent the failure through routine and corrective maintenance.Condition monitoring is the continual monitoring of machines during process conditions to maintain optimal machine use, which is necessary for predictive maintenance. There are three types of condition monitoring: online, periodic, and remote. Finally, remote condition monitoring allows the equipment observed from a small place and data supplied for analysis.
Preventive Maintenance System
To maintain the equipment and materials on a regular basis in order to maintain those running conditions and reduce unnecessary shutdowns due to unexpected equipment failure is called Preventive Maintenance (PM).
Show me the steps of determine red
![Difference Equations
Example F
Taking the z-transform of each term of the second-order equation
Yk+2 + a1Yk+1+a2Yk = Rµ
(4.313)
gives
[z?F(2) – 2?yo – zYyı] + a1[zF(z) – zyo] + azF(z) = G(2),
(4.314)
or
G(2) + yoz2
+ (a140 + Y1)z
z2 + a1z + a2
F(2) =
(4.315)
where G(z) = Z(Rk).
Let a1 = 0, a2 = 4, and R.
equation
= 0. This corresponds to the difference
Yk+2 – 4yk = 0.
(4.316)
The z-transform is, from equation (4.315), given by the expression
F(2)
zYo + Y1
(z + 2)(z – 2)
2y0 – Y1
1
2y0 + Y1
1
(4.317)
4
z + 2
4
z – 2
From Table 4.1, we can obtain the inverse transform of F(z); doing this gives
Yk = /¼(2yo – Y1)(-2)* +¼(2y0 + Y1)2*.
(4.318)
Since yo and Yı are arbitrary, this gives the general solution of equation
(4.316).
The equation
Yk+2 +4yk+1+3yk
2k
(4.319)
%3D
corresponds to a1 = 4, a2 = 3, and R = 2k. Using the fact that
G(2) = Z(2*)
(4.320)
z – 2
we can rewrite equation (4.315), for this case, as
F(2)
Yo(z + 4)
(z + 1)(z + 3)
Y1
1
(4.321)
(z + 1)(z + 3)
(z – 2)(z + 1)(z + 3)*
The three terms on the right-hand side of this equation have the following
partial-fraction expansions:
1
1
1
1
1
(4.322)
(z + 1)(z + 3)
2 z + 1
2 z +3
z + 4
3
1
1
1
(4.323)
(z + 1)(z + 3)
2 z + 1
2 z + 3
1
1
1
1
1
1
(4.324)
(z – 2)(z + 1)(z +3)
15 z – 2
6 z +1
10 z + 3
LINEAR DIFFERENCE EQUATIONS
157
Therefore,
1 1
+
15 z – 2
1 Зу1 + 9yo — 1
1 1- 5y0 – 5y1
(4.325)
6
z + 1
10
z + 3
and taking the inverse transform gives
Yk = /6(3y1 + 9y0 – 1)(-1)* + !/10(1 – 5y0 – 5y1)(-3)* + 1/152*. (4.326)
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