Electronics and Communication Engineering - Networks Analysis and Synthesis - Discussion
Discussion Forum : Networks Analysis and Synthesis - Section 1 (Q.No. 22)
22.
, i(t) =Answer: Option
Explanation:
i(t) = 3e-4t - 2e-3t
Discussion:
4 comments Page 1 of 1.
Ruhi said:
10 years ago
I didn't understood that can anyone explain this?
(2)
Tejashri said:
1 decade ago
How you take the value of numerator as 3&2?
Balmukund said:
1 decade ago
Put s=-3 for (s+4) & s=-4 for (s+3).
Rajesh said:
9 years ago
Given I(s) = s+1/(s+4)(s+3).
Split it into partial fractions s+1/(s+4)(s+3) = A/(s+4) +B/(s+3).
Now (s+1)/(s+4)(s+3) = (A(s+3) + B(s+4))/(s+4)(s+3).
(s+1) = A(s+3) + B(s+4).
Now to get A value put s = -4 because B(-4+4) = 0.
-4+1 = -A,
A = 3,
So, A = 3.
Now to get B value put s=-3 because A(-3+3) = 0.
-3+1= B,
B = -2.
So we got A is 3 and B is -2.
Substitute in the equation A/(s+4) + B/(s+3).
i.e = 3/(s+4) -2/(s+3).
Split it into partial fractions s+1/(s+4)(s+3) = A/(s+4) +B/(s+3).
Now (s+1)/(s+4)(s+3) = (A(s+3) + B(s+4))/(s+4)(s+3).
(s+1) = A(s+3) + B(s+4).
Now to get A value put s = -4 because B(-4+4) = 0.
-4+1 = -A,
A = 3,
So, A = 3.
Now to get B value put s=-3 because A(-3+3) = 0.
-3+1= B,
B = -2.
So we got A is 3 and B is -2.
Substitute in the equation A/(s+4) + B/(s+3).
i.e = 3/(s+4) -2/(s+3).
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