4.3 Finding the inverse laplace transform  (Page 3/3)

Alternately, $A_{3,1}$ can be computed by substituting the values obtained for $A_1,A_2$ and $A_{3,2}$ back into [link] and then substituting an arbitrary value for $s$ that does not equal one of the poles as indicated earlier, like $s=0$ . This leads to a simple equation whose only unknown is $A_{3,1}$ . The partial fraction of $X\left(s\right)$ is then given by:

Applying the inverse Laplace transform to each of the individual terms in [link] and using linearity gives:

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

Example 3.7 Find the inverse Laplace transform of

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

where $s-5$ 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 $s$ . The partial fraction expansion is performed on the quotient term, which is always proper:

Using [link] gives

So we have

and

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 $h\left(t\right)=e^{-5t}u\left(t\right)$ and whose input is given by $x\left(t\right)=cos\left(2t\right)u\left(t\right)$ . Since the output is given by $y\left(t\right)=x\left(t\right)*h\left(t\right)$ , its Laplace transform is $Y\left(s\right)=X\left(s\right)H\left(s\right)$ . Therefore using the table of Laplace transform pairs we have

and

which leads to

The poles are at $s=j2,-j2$ and -5, all of which are distinct, so equation [link] applies:

The second coefficient is

The calculations for $A_2$ where omitted but it is easy to see that $A_2$ will be the complex conjugate of $A_1$ since all of the terms in $A_2$ are the complex conjugates of those in $A_1$ . 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 $s=-5$ is found using

This gives

We can now easily find the inverse Laplace transform of each individual term in the right-hand side of [link] :

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

The simplified answer is given by

We note that the answer contains a transient term, $-0.1724e^{-10t}u\left(t\right)$ , and a steady-state term $0.1857cos(2t-0.3805)$ . 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

and therefore $\left|H,\left(,j,2,\right)\right|=0.1857$ and $\angle H\left(j2\right)=-0.3805$ .

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:

where it has been assumed that $b^2-4c<0$ (otherwise, we have distinct or repeated real poles). As mentioned above, the expansion in [link] can be combined with expansions for distinct or repeated poles.

Example 3.9

Using the cover up method gives

Clearing fractions in [link] gives:

Setting $s=0$ in [link] gives $A_2=\frac{4}{29}$ . Substituting this value back into [link] and setting $s=1$ leads to $A_1=\frac{5}{29}$ . The resulting Laplace transform is:

Using the table of Laplace transforms then leads to

Comparing this answer with [link] , 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

with $r=\sqrt{a^2+b^2}$ and $\phi =arctan\frac{b}{a}$ . 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 $X\left(s\right)=\frac{1}{s}$ and whose system function is given by

Multiplying $X\left(s\right)$ and $H\left(s\right)$ gives

Clearing fractions gives:

Setting $s=0$ leads to a quick solution for $A_1$ , however two subsequent substitutions are needed to find $A_2$ and $A_3$ . A slightly faster way of solving for the coefficients in [link] is to rearrange the right hand side in terms of different powers of $s$ (see second line). Then equate the coefficients of like powers of $s$ on both sides of the equation to solve for the coefficients. For example equating the constant terms leads to $1=3A_1$ which gives $A_1=\frac{1}{3}$ . The coefficients of $s$ on either side of the equation are related by $0=2A_1+A_3$ which leads to $A_3=-\frac{2}{3}$ . Similarly, equating the coefficients of $s^2$ gives $0=A_1+A_2$ which leads to $A_2=-\frac{1}{3}$ . So we have:

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

After a bit more massaging we get

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