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If f(t) is a function of t, defined for all t > 0, the Laplace transform of f(t), denote by the symbol
L{ f (t) } is defined by: ![]()
where s may be real or complex. In circuit applications we assume s = σ + jω.
The operation L{f(t)} transforms a function f(t) in the time domain into a
function F(s) in the complex frequency domain or simply the s domain.
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Solved examples: | At |
e−at |
te−at |
sin ωt |
Acosωt |
e−atcosωt |
sin(ωt + θ) |
cosh ωt |
df(t) / dt |
e−t(t2 + 12t + 10) |
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Example 1a: Obtain the Laplace transform of f(t) = At where A is a constant.
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Example 2a: Obtain the Laplace transform of f(t) = e−at where a is a constant.
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Example 3a: Obtain the Laplace transform of f(t) = te−at where a is a constant.
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Example 4a: Find the Laplace transform of f(t) = sin(ωt)
Apply integration by parts second time on the cos term on the right side of the integral:
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Example 5a: Find the Laplace transform of f(t) = A cos ωt
Apply integration by parts second time on the cos term on the right side of the integral:
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Example 6a: Find the Laplace transform of f(t) = e−at cos ωt
The Laplace trensform is defined by the integral:
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The integrals on both sides of the equation are the same and can be added.
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Example 7a: Find the Laplace transform of f(t) = sin (ωt + θ)
Apply integration by parts second time on the integral at the right side to get:
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Example 8a: Find the Laplace transform of f(t) = cosh ωt
The second term reach zero at infinity, the first term is zero only when s>|ω|
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Example 9a: Find the Laplace transform of the derivative of f'(t) = df(t) / dt.
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Example 10a: Find the Laplace transform of: f(t) = e−t (t2 + 12t + 10)
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Laplace transform by partial fraction technique.
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After aranging terms we get the following form:
![]() By comparing the coefficients we get the following set of linear equations:
Solving the set of linear equations we get: A = 1, B = 5, C =− 8, D = 0.75
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Find the denominator roots (also called poles) and rewrite the polinomials as followes:
s3 + 2s2 + s + 1 = A(s3 + 10s2 + 29s + 20) + B(s + 4)(s + 5) + C(s + 1)(s + 5) + D(s + 1)(s + 4)
After arranging the powers of s we get the values:
s3 + 2s2 + s + 1 = (A)s3 + (10A + B + C + D)s2 + (29A + 9B + 6C + 5D)s + (20A + 20B + 5C + 4D)
The coefficients of s in the right and left side of the equation must be the same, now we have to solve the system of linear equations of 4 unknowns:
And the solutions of the coefficients are: A = 1 B = 0.0833 C = 11.667 D = − 19.75
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