Using Laplace transforms to analyse diff circuits responses
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Laplace transforms enable interpretation and manipulation of different signals by viewing these signals as either time domain signals/pulse or else frequency domain representations. A number of examples are presented in these solutions showing how such tranforms may be maniupluted to better understand circuit driving forces. A number of questions are presented to ask to derive the Laplace transform functions or else identify the reverse Laplace transform time domain functions
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Solution Summary
These solutions show how to use Laplace transforms to identify frequency responses of cicruits to time domain waveforms and also use the Inverse Laplace transofrms to work from the frequency domain to identify equivalent time domain waveforms. A number of Questions are posed and example solutions worked through in detail.
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Question: Find the Laplace transform of exp(-3t).Cos(4t)
Solution:
Using the shifting theorem for Laplace transforms below
L{exp*(at).f(t)} = F(s - a) (1)
Where we assign Cos(4t) as the function f(t)
The general Laplace transform for Cos(wt) is s/(s^2 + w^2) s > 0
Therefore by substitution and noticing that w = 4, we get the Laplace transform for Cos(4t) as
L{Cos (4t)} = s/{s^2 + (4)62 } = s/{s2 + 16} (2)
Application of the Laplace transform shifting theorem, to solve for exp(-3t).Cos(4t), suggests we can replace (s) in (2) by {s-(-3)} or {s + 3}.
When we do this we have
L{exp(-3t). Cos(4t)} = {s + 3}/{(s + 3)2 + 16} (3)
Simplifying (3) we then get
L{exp(-3t).Cos(4t)} = {s + 3}/{s2 + 6s + 9 + 16}
L{exp(-3t).Cos(4t)} = {s + 3}/{s2 + 6s + 25}
Question: The inverse Laplace Transform of s/(s2 - 1) is given by which of the following functions?
A) ½*e-t {1 + e-2t }
B) ½*Sin(2t)
C) ½*Cos(2t)
D) ½*exp(-t).{1 + exp(2t)}
Solution:
? Lets work back we try each answer function in turn and find its Laplace transform
? Examine answer (A) ½*exp(-t).{1 + exp(-2t) } = {exp(-t) + exp(-3t) }/2
Applying the general Laplace transform L{exp(at)} = 1/{s - a} s > a
Where in our case: exp(-t) (a = -1), exp(-3t) (a = -3)
L{½*exp(-t) {1 + exp(-2t)} = L{{exp(-t) + exp(-3t) }/2} = ½*L{{exp(-t) } + {exp(-3t)}}
...
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