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Finding the Inverse Laplace Transform

Module by: Carlos E. Davila. E-mail the author

Finding the Inverse Laplace Transform

Using Transform Tables

The inverse Laplace transform, given by

x ( t ) = 1 2 π j ∫ σ - j ∞ σ + j ∞ X ( s ) e s t d s x ( t ) = 1 2 π j ∫ σ - j ∞ σ + j ∞ X ( s ) e s t d s
(1)

can be found by directly evaluating the above integral. However since this requires a background in the theory of complex variables, which is beyond the scope of this book, we will not be directly evaluating the inverse Laplace transform. Instead, we will utilize the Laplace transform pairs and properties . Consider the following examples:

Example 3.1 Find the inverse Laplace transform of

X ( s ) = e - 10 s s + 5 X ( s ) = e - 10 s s + 5
(2)

By looking at the table of Laplace transform properties we find that multiplication by e-10se-10s corresponds to a time delay of 10 sec. Then from the table of Laplace transform pairs , we see that

1 s + 5 1 s + 5
(3)

corresponds to the Laplace transform of the exponential signal e-5tu(t)e-5tu(t). Therefore we must have

x ( t ) = e - 5 ( t - 10 ) u ( t - 10 ) x ( t ) = e - 5 ( t - 10 ) u ( t - 10 )
(4)

Example 3.2 Find the inverse Laplace transform of

X ( s ) = 1 ( s + 2 ) 2 X ( s ) = 1 ( s + 2 ) 2
(5)

First we note that from the table of Laplace transform pairs , the Laplace transform of tu(t)tu(t) is

1 s 2 1 s 2
(6)

Then using the ss-shift property in the table of Laplace transform properties gives

x ( t ) = t e - 2 t u ( t ) x ( t ) = t e - 2 t u ( t )
(7)

Also, the same answer may be arrived at by combining the Laplace transform of e-2tu(t)e-2tu(t) with the multiplication by tt property.

Partial Fraction Expansions

Partial fraction expansions are useful when we can express the Laplace transform in the form of a rational function,

X ( s ) = b q s q + b q - 1 s q - 1 + ⋯ + b 1 s + b 0 a p s p + a p - 1 s p - 1 + ⋯ + a 1 s + a 0 = B ( s ) A ( s ) X ( s ) = b q s q + b q - 1 s q - 1 + ⋯ + b 1 s + b 0 a p s p + a p - 1 s p - 1 + ⋯ + a 1 s + a 0 = B ( s ) A ( s )
(8)

A rational function is a ratio of two polynomials. The numerator polynomial B(s)B(s) has order qq, i.e., the largest power of ss in this polynomial is qq, while the denominator polynomial has order pp. The partial fraction expansion also requires that the Laplace transform be a proper rational function, which means that q<pq<p. Since B(s)B(s) and A(s)A(s) can be factored, we can write

X ( s ) = ( s - β 1 ) ( s - β 2 ) ⋯ ( s - β q ) ( s - α 1 ) ( s - α 2 ) ⋯ ( s - α p ) X ( s ) = ( s - β 1 ) ( s - β 2 ) ⋯ ( s - β q ) ( s - α 1 ) ( s - α 2 ) ⋯ ( s - α p )
(9)

The βi,i=1,2,...,qβi,i=1,2,...,q are the roots of B(s)B(s), and are called the zeros of X(s)X(s). The roots of A(s)A(s), are αi,i=1,...,pαi,i=1,...,p and are called the poles of X(s)X(s). If we evaluate X(s)X(s) at one of the zeros we get X(βi)=0,i=1,...,qX(βi)=0,i=1,...,q. Similarly, evaluating X(s)X(s) at a pole gives1X(αi)=±∞,i=1,...,pX(αi)=±∞,i=1,...,p. The partial fraction expansion of a Laplace transform will usually involve relatively simple terms whose inverse Laplace transforms can be easily determined from a table of Laplace transforms. We must consider several different cases which depend on whether the poles are distinct.

Distinct Poles:

When all of the poles are distinct (i.e. αi≠αj,i≠jαi≠αj,i≠j) then we can use the following partial fraction expansion:

X ( s ) = A 1 s - α 1 + A 2 s - α 2 + ⋯ + A p s - α p X ( s ) = A 1 s - α 1 + A 2 s - α 2 + ⋯ + A p s - α p
(10)

The coefficients, Ai,i=1,...,pAi,i=1,...,p can then be found using the following formula

A i = X ( s ) ( s - α i ) s = α i , i = 1 , ... , p A i = X ( s ) ( s - α i ) s = α i , i = 1 , ... , p
(11)

Equation Equation 11 is easily derived by clearing fractions in Equation 10. The inverse Fourier transform of X(s)X(s) can then be easily found since each of the terms in the right-hand side of Equation 10 is the Laplace transform of an exponential signal. This method is called the cover up method.

Example 3.3 Find the inverse Laplace transform of

X ( s ) = 2 s - 10 s 2 + 3 s + 2 = 2 s - 10 ( s + 1 ) ( s + 2 ) X ( s ) = 2 s - 10 s 2 + 3 s + 2 = 2 s - 10 ( s + 1 ) ( s + 2 )
(12)

Since the poles are α1=-1α1=-1 and α2=-2α2=-2 are distinct, we have the expansion

X ( s ) = A 1 s + 1 + A 2 s + 2 X ( s ) = A 1 s + 1 + A 2 s + 2
(13)

Using Equation 10 then gives:

A 1 = X ( s ) ( s + 1 ) s = - 1 = 2 s - 10 s + 2 s = - 1 = - 12 A 1 = X ( s ) ( s + 1 ) s = - 1 = 2 s - 10 s + 2 s = - 1 = - 12
(14)

and

A 2 = X ( s ) ( s + 2 ) s = - 2 = 2 s - 10 s + 1 s = - 2 = 14 A 2 = X ( s ) ( s + 2 ) s = - 2 = 2 s - 10 s + 1 s = - 2 = 14
(15)

Therefore, we get:

X ( s ) = - 12 s + 1 + 14 s + 2 X ( s ) = - 12 s + 1 + 14 s + 2
(16)

The inverse Laplace transform of X(s)X(s) can be found by looking up the inverse transform of each of the terms in the right-hand side of Equation 16 giving

x ( t ) = - 12 e - t u ( t ) + 14 e - 2 t u ( t ) x ( t ) = - 12 e - t u ( t ) + 14 e - 2 t u ( t )
(17)

Repeated Poles:

Let's consider the case when each pole is repeated,

X ( s ) = B ( s ) ( s - α 1 ) p 1 ( s - α 2 ) p 2 ⋯ ( s - α r ) p r X ( s ) = B ( s ) ( s - α 1 ) p 1 ( s - α 2 ) p 2 ⋯ ( s - α r ) p r
(18)

where p1+p2+⋯+pr=pp1+p2+⋯+pr=p. In this case the partial fraction expansion goes like this:

X ( s ) = A 1 , 1 s - α 1 + A 1 , 2 ( s - α 1 ) 2 + ⋯ + A 1 , p 1 ( s - α 1 ) p 1 + A 2 , 1 s - α 2 + A 2 , 2 ( s - α 2 ) 2 + ⋯ + A 2 , p 2 ( s - α 2 ) p 2 + ⋯ + A r , 1 s - α r + A r , 2 ( s - α r ) 2 + ⋯ + A r , p r ( s - α r ) p r X ( s ) = A 1 , 1 s - α 1 + A 1 , 2 ( s - α 1 ) 2 + ⋯ + A 1 , p 1 ( s - α 1 ) p 1 + A 2 , 1 s - α 2 + A 2 , 2 ( s - α 2 ) 2 + ⋯ + A 2 , p 2 ( s - α 2 ) p 2 + ⋯ + A r , 1 s - α r + A r , 2 ( s - α r ) 2 + ⋯ + A r , p r ( s - α r ) p r
(19)

We'll look at two methods. In the first method, the coefficients can be found using the following formula

A i , p i - k = 1 k ! d k d s k X i ( s ) s = α i A i , p i - k = 1 k ! d k d s k X i ( s ) s = α i
(20)

where i=1,2,...,ri=1,2,...,r, k=0,1,...,pi-1k=0,1,...,pi-1 and

X i ( s ) = X ( s ) ( s - α i ) p i X i ( s ) = X ( s ) ( s - α i ) p i
(21)

Note that the computation of Ai,piAi,pi does not require any differentiation, since k=0k=0.

Example 3.4 Find the inverse Laplace transform of

X ( s ) = s - 1 ( s + 2 ) 2 X ( s ) = s - 1 ( s + 2 ) 2
(22)

Here we have a single repeated pole at s=-2s=-2. The expansion is therefore given by

X ( s ) = A 1 , 1 s + 2 + A 1 , 2 ( s + 2 ) 2 X ( s ) = A 1 , 1 s + 2 + A 1 , 2 ( s + 2 ) 2
(23)

Using Equation 20, we begin with k=0k=0 which corresponds to

A 1 , 2 = X ( s ) ( s + 2 ) 2 s = - 2 = s - 1 s = - 2 = - 3 A 1 , 2 = X ( s ) ( s + 2 ) 2 s = - 2 = s - 1 s = - 2 = - 3
(24)

Next, we set k=1k=1 in Equation 20

A 1 , 1 = d d s X ( s ) ( s + 2 ) 2 s = - 2 = d d s s - 1 s = - 2 = 1 A 1 , 1 = d d s X ( s ) ( s + 2 ) 2 s = - 2 = d d s s - 1 s = - 2 = 1
(25)

The partial fraction expansion is then given by

X ( s ) = 1 s + 2 - 3 ( s + 2 ) 2 X ( s ) = 1 s + 2 - 3 ( s + 2 ) 2
(26)

Therefore,

x ( t ) = e - 2 t u ( t ) - 3 t e - 2 t u ( t ) x ( t ) = e - 2 t u ( t ) - 3 t e - 2 t u ( t )
(27)

In the second method, the coefficients Ai,pi,i=1,...,rAi,pi,i=1,...,r can be found via the cover up method. The remaining coefficients, Ak,pi,i=1,...,r,k=1,...,pi-1Ak,pi,i=1,...,r,k=1,...,pi-1 can be found by substituting values of ss that are not equal to one of the poles in Equation 19. This leads to a system of linear equations which can be used to solve for the remaining coefficients. This method is generally preferable if the order of each repeated pole as well as the number of poles is sufficiently small so that the number of unknown coefficients is at most two for hand calculations.

Example 3.5 Find the inverse Laplace transform of:

X ( s ) = s ( s + 1 ) 3 = A 1 , 1 s + 1 + A 1 , 2 s + 1 2 + A 1 , 3 s + 1 3 X ( s ) = s ( s + 1 ) 3 = A 1 , 1 s + 1 + A 1 , 2 s + 1 2 + A 1 , 3 s + 1 3
(28)

Using the cover-up method we can find A1,3A1,3 as follows

A 1 , 3 = s s = - 1 = - 1 A 1 , 3 = s s = - 1 = - 1
(29)

So we are left with

X ( s ) = s ( s + 1 ) 3 = A 1 , 1 s + 1 + A 1 , 2 s + 1 2 - 1 s + 1 3 X ( s ) = s ( s + 1 ) 3 = A 1 , 1 s + 1 + A 1 , 2 s + 1 2 - 1 s + 1 3
(30)

Setting s=0s=0 in Equation 30 leads to

A 1 , 1 + A 1 , 2 = 1 A 1 , 1 + A 1 , 2 = 1
(31)

and setting s=-2s=-2 in Equation 30 gives

- A 1 , 1 + A 1 , 2 = 1 - A 1 , 1 + A 1 , 2 = 1
(32)

These choices of ss were used to simplify the linear equations to the greatest extent possible. The solution to Equation 31 and Equation 32 is easily found to be A1,1=0A1,1=0 and A1,2=1A1,2=1. The partial fraction expansion is given by

X ( s ) = 1 s + 1 2 - 1 s + 1 3 X ( s ) = 1 s + 1 2 - 1 s + 1 3
(33)

Using the corresponding Laplace transform pairs leads to

x ( t ) = t e - t u ( t ) - 1 2 t 2 e - t u ( t ) x ( t ) = t e - t u ( t ) - 1 2 t 2 e - t u ( t )
(34)

Distinct and Repeated Poles:

If a Laplace transform contains both distinct and repeated poles, then we would combine the expansions in Equation 10 and Equation 19. Perhaps the easiest way to indicate this is by way of an example:

Example 3.6 Find the inverse Laplace transform of

X ( s ) = s + 2 ( s + 1 ) ( s + 3 ) ( s + 5 ) 2 = A 1 s + 1 + A 2 s + 3 + A 3 , 1 s + 5 + A 3 , 2 ( s + 5 ) 2 X ( s ) = s + 2 ( s + 1 ) ( s + 3 ) ( s + 5 ) 2 = A 1 s + 1 + A 2 s + 3 + A 3 , 1 s + 5 + A 3 , 2 ( s + 5 ) 2
(35)

The coefficients corresponding to the distinct poles can be found using Equation 11:

A 1 = X ( s ) ( s + 1 ) s = - 1 = s + 2 ( s + 3 ) ( s + 5 ) 2 s = - 1 = 1 32 A 1 = X ( s ) ( s + 1 ) s = - 1 = s + 2 ( s + 3 ) ( s + 5 ) 2 s = - 1 = 1 32
(36)
A 2 = X ( s ) ( s + 3 ) s = - 3 = s + 2 ( s + 1 ) ( s + 5 ) 2 s = - 3 = 1 8 A 2 = X ( s ) ( s + 3 ) s = - 3 = s + 2 ( s + 1 ) ( s + 5 ) 2 s = - 3 = 1 8
(37)

The coefficient A3,2A3,2 corresponding to the double pole at s=-5s=-5 can be found using Equation 20 with k=0k=0:

A 3 , 2 = X ( s ) ( s + 5 ) 2 s = - 5 = s + 2 ( s + 1 ) ( s + 3 ) s = - 5 = - 3 8 A 3 , 2 = X ( s ) ( s + 5 ) 2 s = - 5 = s + 2 ( s + 1 ) ( s + 3 ) s = - 5 = - 3 8
(38)

The remaining coefficient, A3,1A3,1 can be found using Equation 20 with k=1k=1:

A 3 , 1 = d d s X ( s ) ( s + 5 ) 2 s = - 5 = d d s s + 2 ( s + 1 ) ( s + 3 ) s = - 5 = ( s 2 + 4 s + 3 ) - ( s + 2 ) ( 2 s + 4 ) ( s 2 + 4 s + 3 ) 2 s = - 5 = - 5 32 A 3 , 1 = d d s X ( s ) ( s + 5 ) 2 s = - 5 = d d s s + 2 ( s + 1 ) ( s + 3 ) s = - 5 = ( s 2 + 4 s + 3 ) - ( s + 2 ) ( 2 s + 4 ) ( s 2 + 4 s + 3 ) 2 s = - 5 = - 5 32
(39)

Alternately, A3,1A3,1 can be computed by substituting the values obtained for A1,A2A1,A2 and A3,2A3,2 back into Equation 35 and then substituting an arbitrary value for ss that does not equal one of the poles as indicated earlier, like s=0s=0. This leads to a simple equation whose only unknown is A3,1A3,1. The partial fraction of X(s)X(s) is then given by:

X ( s ) = s + 2 ( s + 1 ) ( s + 3 ) ( s + 5 ) 2 = 1 32 s + 1 + 1 8 s + 3 - 5 32 s + 5 - 3 8 ( s + 5 ) 2 X ( s ) = s + 2 ( s + 1 ) ( s + 3 ) ( s + 5 ) 2 = 1 32 s + 1 + 1 8 s + 3 - 5 32 s + 5 - 3 8 ( s + 5 ) 2
(40)

Applying the inverse Laplace transform to each of the individual terms in Equation 40 and using linearity gives:

x ( t ) = 1 32 e - t u ( t ) + 1 8 e - 3 t u ( t ) - 5 32 e - 5 t u ( t ) - 3 8 t e - 5 t u ( t ) x ( t ) = 1 32 e - t u ( t ) + 1 8 e - 3 t u ( t ) - 5 32 e - 5 t u ( t ) - 3 8 t e - 5 t u ( t )
(41)

The following example looks at a case where X(s)X(s) is a rational function, but is not proper.

Example 3.7 Find the inverse Laplace transform of

X ( s ) = s 2 + 6 s + 1 s 2 + 5 s + 6 X ( s ) = s 2 + 6 s + 1 s 2 + 5 s + 6
(42)

Here since q=p=2q=p=2, we cannot perform a partial fraction expansion. First we must perform a long division, this leads to:

X ( s ) = 1 + s - 5 s 2 + 5 s + 6 = 1 + s - 5 ( s + 2 ) ( s + 3 ) X ( s ) = 1 + s - 5 s 2 + 5 s + 6 = 1 + s - 5 ( s + 2 ) ( s + 3 )
(43)

where s-5s-5 is the remainder resulting from the long division. The quotient of 1 is called a direct term. In general, the direct term corresponds to a polynomial in ss. The partial fraction expansion is performed on the quotient term, which is always proper:

s - 5 ( s + 2 ) ( s + 3 ) = A 1 s + 2 + A 2 s + 3 s - 5 ( s + 2 ) ( s + 3 ) = A 1 s + 2 + A 2 s + 3
(44)

Using Equation 10 gives

A 1 = s - 5 s + 3 s = - 2 = - 7 A 1 = s - 5 s + 3 s = - 2 = - 7
(45)
A 2 = s - 5 s + 2 s = - 3 = 8 A 2 = s - 5 s + 2 s = - 3 = 8
(46)

So we have

X ( s ) = 1 - 7 s + 2 + 8 s + 3 X ( s ) = 1 - 7 s + 2 + 8 s + 3
(47)

and

x ( t ) = δ ( t ) - 7 e - 2 t u ( t ) + 8 e - 3 t u ( t ) x ( t ) = δ ( t ) - 7 e - 2 t u ( t ) + 8 e - 3 t u ( t )
(48)

Complex Conjugate Poles:

Some poles occur in complex conjugate pairs as in the following example:

Example 3.8 Find the output of a filter whose impulse response is h(t)=e-5tu(t)h(t)=e-5tu(t) and whose input is given by x(t)=cos(2t)u(t)x(t)=cos(2t)u(t). Since the output is given by y(t)=x(t)*h(t)y(t)=x(t)*h(t), its Laplace transform is Y(s)=X(s)H(s)Y(s)=X(s)H(s). Therefore using the table of Laplace transform pairs we have

X ( s ) = s s 2 + 4 X ( s ) = s s 2 + 4
(49)

and

H ( s ) = 1 s + 5 H ( s ) = 1 s + 5
(50)

which leads to

Y ( s ) = s ( s 2 + 4 ) ( s + 5 ) = s ( s + j 2 ) ( s - j 2 ) ( s + 5 ) = A 1 s + j 2 + A 2 s - j 2 + A 3 s + 5 Y ( s ) = s ( s 2 + 4 ) ( s + 5 ) = s ( s + j 2 ) ( s - j 2 ) ( s + 5 ) = A 1 s + j 2 + A 2 s - j 2 + A 3 s + 5
(51)

The poles are at s=j2,-j2s=j2,-j2 and -5, all of which are distinct, so equation Equation 10 applies:

A 1 = Y ( s ) ( s + j 2 ) s = - j 2 = s ( s - j 2 ) ( s + 5 ) s = - j 2 = - j 2 - j 4 ( 5 - j 2 ) = 5 + j 2 58 A 1 = Y ( s ) ( s + j 2 ) s = - j 2 = s ( s - j 2 ) ( s + 5 ) s = - j 2 = - j 2 - j 4 ( 5 - j 2 ) = 5 + j 2 58
(52)

The second coefficient is

A 2 = Y ( s ) ( s - j 2 ) s = j 2 = 5 - j 2 58 A 2 = Y ( s ) ( s - j 2 ) s = j 2 = 5 - j 2 58
(53)

The calculations for A2A2 where omitted but it is easy to see that A2A2 will be the complex conjugate of A1A1 since all of the terms in A2A2 are the complex conjugates of those in A1A1. Therefore, when there are a pair of complex conjugate poles, we need only calculate one of the two coefficients and the other will be its complex conjugate. The last coefficient corresponding to the pole at s=-5s=-5 is found using

A 3 = Y ( s ) ( s + 5 ) s = - 5 = s ( s 2 + 4 ) s = - 5 = - 5 29 A 3 = Y ( s ) ( s + 5 ) s = - 5 = s ( s 2 + 4 ) s = - 5 = - 5 29
(54)

This gives

Y ( s ) = 5 + j 2 58 s + j 2 + 5 - j 2 58 s - j 2 - 5 29 s + 5 Y ( s ) = 5 + j 2 58 s + j 2 + 5 - j 2 58 s - j 2 - 5 29 s + 5
(55)

We can now easily find the inverse Laplace transform of each individual term in the right-hand side of Equation 55:

y ( t ) = 5 + j 2 58 e - j 2 t u ( t ) + 5 - j 2 58 e j 2 t u ( t ) - 5 29 e - 5 t u ( t ) y ( t ) = 5 + j 2 58 e - j 2 t u ( t ) + 5 - j 2 58 e j 2 t u ( t ) - 5 29 e - 5 t u ( t )
(56)

At this point, we are technically done, however the first two terms in y(t)y(t) are complex and also happen to be complex conjugates of each other. So we can simplify further by noting that

5 + j 2 58 e - j 2 t u ( t ) + 5 - j 2 58 e j 2 t u ( t ) = 2 Re 5 - j 2 58 e j 2 t u ( t ) = 2 Re 0 . 0928 e - j 0 . 3805 e j 2 t u ( t ) = 0 . 1857 cos ( 2 t - 0 . 3805 ) u ( t ) 5 + j 2 58 e - j 2 t u ( t ) + 5 - j 2 58 e j 2 t u ( t ) = 2 Re 5 - j 2 58 e j 2 t u ( t ) = 2 Re 0 . 0928 e - j 0 . 3805 e j 2 t u ( t ) = 0 . 1857 cos ( 2 t - 0 . 3805 ) u ( t )
(57)

The simplified answer is given by

y ( t ) = 0 . 1857 cos ( 2 t - 0 . 3805 ) u ( t ) - 0 . 1724 e - 5 t u ( t ) y ( t ) = 0 . 1857 cos ( 2 t - 0 . 3805 ) u ( t ) - 0 . 1724 e - 5 t u ( t )
(58)

We note that the answer contains a transient term, -0.1724e-10tu(t)-0.1724e-10tu(t), and a steady-state term 0.1857cos(2t-0.3805)0.1857cos(2t-0.3805). The steady-state term corresponds to the sinusoidal steady-state response of the filter (see Chapter 3). It can be readily seen that the frequency response of the filter is

H ( j Ω ) = 1 5 + j Ω H ( j Ω ) = 1 5 + j Ω
(59)

and therefore H(j2)=0.1857H(j2)=0.1857 and ∠H(j2)=-0.3805∠H(j2)=-0.3805.

While the above example provides some insight into the sinusoidal steady-state response, the number of complex arithmetic calculations can be tedious. We repeat the example using an alternative expansion involving complex conjugate poles:

1 s 2 + b s + c = A 1 s + A 2 s 2 + b s + c 1 s 2 + b s + c = A 1 s + A 2 s 2 + b s + c
(60)

where it has been assumed that b2-4c<0b2-4c<0 (otherwise, we have distinct or repeated real poles). As mentioned above, the expansion in Equation 60 can be combined with expansions for distinct or repeated poles.

Example 3.9

Y ( s ) = s ( s 2 + 4 ) ( s + 5 ) = A 1 s + A 2 ( s 2 + 4 ) + A 3 s + 5 Y ( s ) = s ( s 2 + 4 ) ( s + 5 ) = A 1 s + A 2 ( s 2 + 4 ) + A 3 s + 5
(61)

Using the cover up method gives

A 3 = s s 2 + 4 s = - 5 = - 5 29 A 3 = s s 2 + 4 s = - 5 = - 5 29
(62)

Clearing fractions in Equation 61 gives:

s = ( A 1 s + A 2 ) ( s + 5 ) - 5 29 ( s 2 + 4 ) s = ( A 1 s + A 2 ) ( s + 5 ) - 5 29 ( s 2 + 4 )
(63)

Setting s=0s=0 in Equation 63 gives A2=429A2=429. Substituting this value back into Equation 63 and setting s=1s=1 leads to A1=529A1=529. The resulting Laplace transform is:

Y ( s ) = 5 29 s + 4 29 ( s 2 + 4 ) - 5 29 s + 5 Y ( s ) = 5 29 s + 4 29 ( s 2 + 4 ) - 5 29 s + 5
(64)

Using the table of Laplace transforms then leads to

y ( t ) = 5 29 cos ( 2 t ) u ( t ) + 2 29 sin ( 2 t ) u ( t ) - 5 29 e - 5 t u ( t ) y ( t ) = 5 29 cos ( 2 t ) u ( t ) + 2 29 sin ( 2 t ) u ( t ) - 5 29 e - 5 t u ( t )
(65)

Comparing this answer with Equation 58, we see that the sum of a cosine and a sine having the same frequency is equal to a cosine at the same frequency having a certain phase shift and amplitude. In fact, it can be shown that

a cos ( Ω o t ) + b sin ( Ω o t ) = r cos ( Ω o t - φ ) a cos ( Ω o t ) + b sin ( Ω o t ) = r cos ( Ω o t - φ )
(66)

with r=a2+b2r=a2+b2 and φ=arctanbaφ=arctanba. The following example also involves complex conjugate poles and illustrates some additional tricks to solving the partial fraction expansion.

Example 3.10 Find the output of a filter whose input has Laplace transform X(s)=1sX(s)=1s and whose system function is given by

H ( s ) = 1 s 2 + 2 s + 3 H ( s ) = 1 s 2 + 2 s + 3
(67)

Multiplying X(s)X(s) and H(s)H(s) gives

Y ( s ) = 1 s ( s 2 + 2 s + 3 ) = A 1 s + A 2 s + A 3 s 2 + 2 s + 3 Y ( s ) = 1 s ( s 2 + 2 s + 3 ) = A 1 s + A 2 s + A 3 s 2 + 2 s + 3
(68)

Clearing fractions gives:

1 = A 1 ( s 2 + 2 s + 3 ) + s ( A 2 s + A 3 ) = ( A 1 + A 2 ) s 2 + ( 2 A 1 + A 3 ) s + 3 A 1 1 = A 1 ( s 2 + 2 s + 3 ) + s ( A 2 s + A 3 ) = ( A 1 + A 2 ) s 2 + ( 2 A 1 + A 3 ) s + 3 A 1
(69)

Setting s=0s=0 leads to a quick solution for A1A1, however two subsequent substitutions are needed to find A2A2 and A3A3. A slightly faster way of solving for the coefficients in Equation 69 is to rearrange the right hand side in terms of different powers of ss (see second line). Then equate the coefficients of like powers of ss on both sides of the equation to solve for the coefficients. For example equating the constant terms leads to 1=3A11=3A1 which gives A1=13A1=13. The coefficients of ss on either side of the equation are related by 0=2A1+A30=2A1+A3 which leads to A3=-23A3=-23. Similarly, equating the coefficients of s2s2 gives 0=A1+A20=A1+A2 which leads to A2=-13A2=-13. So we have:

Y ( s ) = 1 3 s - 1 3 s + 2 3 s 2 + 2 s + 3 Y ( s ) = 1 3 s - 1 3 s + 2 3 s 2 + 2 s + 3
(70)

The second term in Y(s)Y(s) does not appear in most Laplace transform tables, however, we can complete the square of s2+2s+3s2+2s+3 by taking one-half the coefficient of ss, squaring it, then adding and subtracting it to give:

s 2 + 2 s + 3 + 1 - 1 = ( s + 1 ) 2 + 2 s 2 + 2 s + 3 + 1 - 1 = ( s + 1 ) 2 + 2
(71)

After a bit more massaging we get

Y ( s ) = 1 3 s - 1 3 ( s + 1 ) ( s + 1 ) 2 + 2 - 1 3 ( s + 1 ) 2 + 2 Y ( s ) = 1 3 s - 1 3 ( s + 1 ) ( s + 1 ) 2 + 2 - 1 3 ( s + 1 ) 2 + 2
(72)

whose inverse Laplace transform is readily found from the table of Laplace transforms as

y ( t ) = 1 3 u ( t ) - 1 3 e - t cos ( 2 t ) u ( t ) - 1 3 2 e - t sin ( 2 t ) u ( t ) y ( t ) = 1 3 u ( t ) - 1 3 e - t cos ( 2 t ) u ( t ) - 1 3 2 e - t sin ( 2 t ) u ( t )
(73)

Footnotes

  1. The actual sign would need to be evaluated at some value of ss that is sufficiently close to the pole.

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EPUB is an electronic book format that can be read on a variety of mobile devices.

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My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

Module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks