how to generate barcode in ssrs report THE LAPLACE TRANSFORM METHOD in Software

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THE LAPLACE TRANSFORM METHOD
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[CHAP. 16
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(a) I s 1=s and the unit-step response is V s 1 RLs 1 1 2 Ls 1 s RLCs C s  2 !2 d 1 t v t e sin !d t u t C!d s   1 2 1 !d 2RC LC
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where 1 ;  RC (b) I s 1 and the unit-impulse response is V s 1 1 C s  2 !2 d 1 t v t e !d cos !d t  sin !d t u t C!d and
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16.1 Find the Laplace transform of e at cos !t, where a is a constant.
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1 Applying the de ning equation l f t 0 f t e st dt to the given function, we obtain 1 l e at cos !t cos !te s a t dt 0 " #1 s a cos !te s a t e s a t ! sin !t s a 2 !2 0 s a s a 2 !2
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If l f t F s , show that l e at f t F s a .
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By de nition, l f t 1
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Apply this result to Problem 16.1.
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f t e st dt F s . Then, 1 1 l e at f t e at f t e st dt f t e s a t dt F s a
Applying (5) to line 6 of Table 16-1 gives l e at cos !t as determined in Problem 16.1. s a s a 2 !2
Find the Laplace transform of f t 1 e at , where a is a constant.
l 1 e at 1 e st dt e s a t dt 0 0 0  1 1 1 1 1 a e s a t e st s s a s s a s s a 0 1 e at e st dt 1 1
CHAP. 16]
THE LAPLACE TRANSFORM METHOD
Another Method  t  1= s a a a l a e d a s s s a 0
Find l 1
Using the method of partial fractions,
1 2 s s a2
1 A B C s s2 a2 s s a s a and the coe cients are A Hence, 1 1 1 1 1 1 B 2 C s s a s a 2a s s a s a 2a2 s2 a2 s 0 a2 " # " # " #   1 1=a2 1=2a2 1=2a2 l 1 l 1 l 1 l 1 s s a s a s s2 a2
The corresponding time functions are found in Table 16-1:   1 1 1 1 1 2 2 e at 2 eat l s s2 a2 a 2s 2a   1 1 eat e at 1 2 cosh at 1 2 2 2 a a a Another Method By lines 11 and 14 of Table 16-1, " #   2 2 t sinh a cosh a t 1 1 1= s a l 2 cosh at 1 d s a a2 a 0 0
Find l
s 1 s s2 4s 4
Using the method of partial fractions, we have s 1 A B B2 1 s s 2 s 2 2 s s 2 2 s 1 s 1 1 1 A B2 s s 2 2 s 2 2 s 0 4 s 2 1 B1 s 2 4 2s s 2 2 s 2 1  1     1  s 1 1 1 4 1 4 1 2 l l l l s s 2 s s2 4s 4 s 2 2
Then and Hence,
The corresponding time functions are found in Table 16-1:   s 1 1 1 1 l 1 e 2t te 2t 4 4 2 s s2 4s 4
THE LAPLACE TRANSFORM METHOD
[CHAP. 16
In the series RC circuit of Fig. 16-5, the capacitor has an initial charge 2.5 mC. At t 0, the switch is closed and a constant-voltage source V 100 V is applied. Use the Laplace transform method to nd the current.
The time-domain equation for the given circuit after the switch is closed is   t 1 Ri t Q0 i  d V C 0 or 10i t   t 1 2:5 10 3 i  d V 50 10 6 0 (6)
Q0 is opposite in polarity to the charge which the source will deposit on the capacitor. Taking the Laplace transform of the terms in (6), we obtain the s-domain equation 10I s 2:5 10 3 I s 100 6 s 6 s s 50 10 50 10 I s 15 s 2 103 (7)
The time function is now obtained by taking the inverse Laplace transform of (7):   3 15 15e 2 10 t A i t l 1 s 2 103
Fig. 16-5
Fig. 16-6
In the RL circuit shown in Fig. 16-6, the switch is in position 1 long enough to establish steadystate conditions, and at t 0 it is switched to position 2. Find the resulting current.
Assume the direction of the current as shown in the diagram. i0 50=25 2 A. The time-domain equation is 25i 0:01 Taking the Laplace transform of (9), 25I s 0:01sI s 0:01i 0 100=s Substituting for i 0 , 25I s 0:01sI s 0:01 2 100=s 100 0:02 104 2 s 0:01s 25 0:01s 25 s s 2500 s 2500 11 10 di 100 dt The initial current is then
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