Inverse Laplace transform is used when we want to convert the known Laplace equation into the timedomain equation.
Even if we have the table conversion from Laplace transform properties, we still need to so some equation simplification to match with the table.
The Inverse Laplace Transform
Given F (s), how do we transform it back to the time domain and obtain the corresponding f (t)? By matching entries in Table.(2) in the ‘Laplace Transform Properties‘ (let’s put that table in this post as Table.1 to ease our study)
we avoid using Equation.(5) in ‘Laplace Transform Definition’ to find f (t).
Suppose F (s) has the general form of
where N(s) is the numerator polynomial and D(s) is the denominator polynomial. The roots of N(s) = 0 are called the zeros of F (s), while
the roots of D(s) = 0 are the poles of F (s). Although Equation.(1) is similar in form to Equation.(3) in ‘Transfer Function’, here F (s) is the Laplace transform of a function, which is not necessarily a transfer function.
We use partial fraction expansion to break F (s) down into simple terms whose inverse transform we obtain from Table.(1).
Thus, finding the inverse Laplace transform of F (s) involves two steps.
Steps to Find the Inverse Laplace Transform :

Decompose F (s) into simple terms using partial fraction expansion.

Find the inverse of each term by matching entries in Table.(1).
Let us consider the three possible forms F (s ) may take and how to apply the two steps to each form.
Laplace Transform Simple Poles
A simple pole is the firstorder pole. If F ( s ) has only simple poles, then D (s ) becomes a product of factors, so that
where s = −p_{1}, −p_{2},…, −p_{n }are the simple poles, and p_{i} ≠ p_{j }for all i ≠ j (i.e., the poles are distinct). Assuming that the degree of N(s) is less than the degree of D(s), we use partial fraction expansion to decompose F(s) in Equation.(2) as
The expansion coefficients k_{1}, k_{2},…,k_{n }are known as the residues of F(s). There are many ways of finding the expansion coefficients.
One way is using the residue method. If we multiply both sides of the Equation.(3) by (s + p_{1}), we obtain
Since p_{i} ≠ p_{j}, setting s = −p_{1} in Equation.(4) leaves only k_{1} on the righthand side of Equation.(4). Hence,
Thus, in general,
This is known as Heaviside’s theorem. Once the values of k_{i }are known, we proceed to find the inverse of F(s) using Equation.(3).
Since the inverse transform of each term in Equation.(3) is
L^{−1} [k/(s + a)] = ke^{−}^{a}tu(t),
then, from Table 15.1 in the ‘Laplace Transform Properties’,
Laplace Transform Repeated Poles
Suppose F(s) has n repeated poles at s = −p. Then we may represent
F(s) as
where F_{1}(s) is the remaining part of F(s) that does not have a pole at s = −p. We determine the expansion coefficient k_{n }as
as we did above. To determine k_{n −1}, we multiply each term in Equation.(8) by (s + p)^{n }and differentiate to get rid of k_{n}, then evaluate the result at s = −p to get rid of the other coefficients except k_{n−1}. Thus, we obtain
Repeating this gives
The mth term becomes
where m = 1, 2,…,n − 1. One can expect the differentiation to be difficult to handle as m increases. Once we obtain the values of k_{1}, k_{2},…,k_{n }by partial fraction expansion, we apply the inverse transform
to each term in the righthand side of Equation.(8) and obtain
Laplace Transform Complex Poles
A pair of complex poles is simple if it is not repeated; it is a double or multiple poles if repeated. Simple complex poles may be handled the same as simple real poles, but because complex algebra is involved the result is always cumbersome.
An easier approach is a method known as completing the square. The idea is to express each complex pole pair (or quadratic term) in D(s) as a complete square such as
(s + α)^{2} + β^{2}
and then use Table.(1) to find the inverse of the term.
Since N(s) and D(s) always have real coefficients and we know that the complex roots of polynomials with real coefficients must occur in conjugate pairs, F(s) may have the general form
where F_{1}(s) is the remaining part of F(s) that does not have this pair of complex poles. If we complete the square by letting
and we also let
then Equation.(15) becomes
From Table.(1), the inverse transform is
The sine and cosine terms can be combined. Whether the pole is simple, repeated, or complex, a general approach that can always be used in finding the expansion coefficients is the method of algebra.
To apply the method, we first set F(s) = N(s)/D(s) equal to an expansion containing unknown constants. We multiply the result through by a common denominator. Then we determine the unknown constants by equating coefficients (i.e., by algebraically solving a set of simultaneous equations for these coefficients at like powers of s).
Another general approach is to substitute specific, convenient values of s to obtain as many simultaneous equations as the number of unknown coefficients, and then solve for the unknown coefficients. We must make sure that each selected value of s is not one of the poles of F(s). The example below illustrates this idea.
Inverse Laplace Transform Examples
Let us review the laplace transform examples below:
Inverse Laplace Transform Example 1
Find the inverse Laplace transform of
Solution:
The inverse transform is given by
where Table.(1) has been consulted for the inverse of each term.
Inverse Laplace Transform Example 2
Find f (t) given that
Solution:
Unlike in the previous example where the partial fractions have been provided, we first need to determine the partial fractions. Since there are three poles, we let
where A, B, and C are the constants to be determined. We can find the constants using two approaches.
METHOD 1 : Residue method
METHOD 2 : Algebraic method
Multiplying both sides of Equation.(2.1) by s(s + 2)(s + 3) gives
or
Equating the coefficients of like powers of s gives
Thus A = 2, B = −8, C = 7, and Equation.(2.1) becomes
By finding the inverse transform of each term, we obtain
Inverse Laplace Transform Example 3
Calculate v (t ) given that
Solution:
While the previous example is on simple roots, this example is on repeated roots. Let
METHOD 1 : Residue method
METHOD 2 : Algebraic method
Equating coefficients,
Solving these simultaneous equations gives A = 1, B = −14, C = 22, D = 13, so that
Taking the inverse transform of each term, we get
Inverse Laplace Transform Example 4
Find the inverse transform of the frequencydomain function in
Solution:
In this example, H(s) has a pair of complex poles at s2 + 8s + 25 = 0 or s = −4 ± j3. We let
We now determine the expansion coefficients in two ways.
METHOD 1 : Combination of methods.
We can obtain A using the method of residue,
Although B and C can be obtained using the method of residue, we will not do so, to avoid complex algebra. Rather, we can substitute two specific values of s [say s = 0, 1, which are not poles of F (s)] into Equation.(4.1).
This will give us two simultaneous equations from which to find B and C. If we let s = 0 in Equation.(4.1), we obtain
Since A = 2, Equation.(4.2) gives C = −10. Substituting s = 1 into Equation.(4.2) gives
But A = 2, C = −10, so that Equation.(4.3) gives B = −2.
METHOD 2 : Algebraic method.
Multiplying both sides of Equation.(4.1) by (s + 3)(s2 + 8s + 25) yields
Equating coefficients,
That is, B = −2, C = −10. Thus
Taking the inverse of each term, we obtain
It is alright to leave the result this way. However, we can combine the cosine and sine terms as
Next, we determine the coefficient A and the phase angle θ:
Thus,
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