Inverse Laplace Transform Examples & Solutions

by Kenji Nakamura 47 views

Hey guys! Today, we're diving deep into the fascinating world of inverse Laplace transforms. If you're scratching your head trying to figure out how to solve these, you've come to the right place. We're going to break down three examples step-by-step, making sure you grasp the concepts along the way. So, buckle up and let's get started!

(3.1) Unraveling {L^{-1}\left{\frac{2s-4}{(s^2+s)(s^2+1)}\right}}

Okay, let's tackle our first inverse Laplace transform: {L^{-1}\left{\frac{2s-4}{(s^2+s)(s^2+1)}\right}}. This might look intimidating at first, but don't worry, we'll break it down into manageable pieces. The key here is partial fraction decomposition. This technique allows us to express a complex fraction as a sum of simpler fractions, which are easier to handle with inverse Laplace transforms. First things first, let's factor the denominator completely. We have s2+s=s(s+1){s^2 + s = s(s+1)}, so our expression becomes:

2s−4s(s+1)(s2+1){\frac{2s-4}{s(s+1)(s^2+1)}}

Now, we can decompose this into partial fractions. We'll set it up like this:

2s−4s(s+1)(s2+1)=As+Bs+1+Cs+Ds2+1{\frac{2s-4}{s(s+1)(s^2+1)} = \frac{A}{s} + \frac{B}{s+1} + \frac{Cs+D}{s^2+1}}

Our mission is to find the values of A, B, C, and D. To do this, we'll multiply both sides by the original denominator, s(s+1)(s2+1){s(s+1)(s^2+1)}. This gives us:

2s−4=A(s+1)(s2+1)+Bs(s2+1)+(Cs+D)s(s+1){2s-4 = A(s+1)(s^2+1) + Bs(s^2+1) + (Cs+D)s(s+1)}

Now, we can use some clever tricks to solve for our constants. Let's start by plugging in some convenient values for s. If we let s=0{s = 0}, we get:

−4=A(1)(1)⇒A=−4{-4 = A(1)(1) \Rightarrow A = -4}

Next, let's try s=−1{s = -1}:

−6=B(−1)(2)⇒B=3{-6 = B(-1)(2) \Rightarrow B = 3}

Great! We've found A and B. Now, let's expand the equation and collect like terms to find C and D:

2s−4=A(s3+s2+s+1)+B(s3+s)+(Cs+D)(s2+s){2s-4 = A(s^3 + s^2 + s + 1) + B(s^3 + s) + (Cs+D)(s^2+s)}

2s−4=−4(s3+s2+s+1)+3(s3+s)+(Cs3+Cs2+Ds2+Ds){2s-4 = -4(s^3 + s^2 + s + 1) + 3(s^3 + s) + (Cs^3 + Cs^2 + Ds^2 + Ds)}

2s−4=−4s3−4s2−4s−4+3s3+3s+Cs3+Cs2+Ds2+Ds{2s-4 = -4s^3 - 4s^2 - 4s - 4 + 3s^3 + 3s + Cs^3 + Cs^2 + Ds^2 + Ds}

Combining terms, we get:

2s−4=(−4+3+C)s3+(−4+C+D)s2+(−4+3+D)s−4{2s-4 = (-4 + 3 + C)s^3 + (-4 + C + D)s^2 + (-4 + 3 + D)s - 4}

Now, we can equate the coefficients of the corresponding powers of s. For the s3{s^3} term, we have:

0=−1+C⇒C=1{0 = -1 + C \Rightarrow C = 1}

For the s2{s^2} term, we have:

0=−4+C+D⇒0=−4+1+D⇒D=3{0 = -4 + C + D \Rightarrow 0 = -4 + 1 + D \Rightarrow D = 3}

Fantastic! We've found all our constants: A=−4{A = -4}, B=3{B = 3}, C=1{C = 1}, and D=3{D = 3}. Now we can rewrite our expression with the partial fractions:

2s−4s(s+1)(s2+1)=−4s+3s+1+s+3s2+1{\frac{2s-4}{s(s+1)(s^2+1)} = \frac{-4}{s} + \frac{3}{s+1} + \frac{s+3}{s^2+1}}

Time for the inverse Laplace transform! We'll take the inverse Laplace transform of each term separately:

{L^{-1}\left{\frac{2s-4}{s(s+1)(s^2+1)}\right} = L^{-1}\left{\frac{-4}{s}\right} + L^{-1}\left{\frac{3}{s+1}\right} + L^{-1}\left{\frac{s}{s^2+1}\right} + L^{-1}\left{\frac{3}{s^2+1}\right}}

Using our trusty Laplace transform table, we know:

  • {L^{-1}\left{\frac{1}{s}\right} = 1}
  • {L^{-1}\left{\frac{1}{s+a}\right} = e^{-at}}
  • {L^{-1}\left{\frac{s}{s^2+a^2}\right} = \cos(at)}
  • {L^{-1}\left{\frac{a}{s^2+a^2}\right} = \sin(at)}

Applying these, we get:

{L^{-1}\left{\frac{2s-4}{s(s+1)(s^2+1)}\right} = -4L^{-1}\left{\frac{1}{s}\right} + 3L^{-1}\left{\frac{1}{s+1}\right} + L^{-1}\left{\frac{s}{s^2+1}\right} + 3L^{-1}\left{\frac{1}{s^2+1}\right}}

=−4(1)+3e−t+cos⁡(t)+3sin⁡(t){= -4(1) + 3e^{-t} + \cos(t) + 3\sin(t)}

So, our final answer is:

{L^{-1}\left{\frac{2s-4}{(s^2+s)(s^2+1)}\right} = -4 + 3e^{-t} + \cos(t) + 3\sin(t)}

Boom! We cracked it! The partial fraction decomposition was the real MVP here. Remember, guys, practice makes perfect, so keep at it!

(3.2) Tackling {L^{-1}\left{\frac{e^{-\pi s}}{s^2+1}\right}}

Alright, let's move on to our next challenge: {L^{-1}\left{\frac{e^{-\pi s}}{s^2+1}\right}}. This one introduces a new element: the exponential term e−πs{e^{-\pi s}}. This is a classic sign that we'll be using the time-shifting property of Laplace transforms. The time-shifting property states that if {L^{-1}\left{F(s)\right} = f(t)}, then {L^{-1}\left{e^{-as}F(s)\right} = f(t-a)u(t-a)}, where u(t−a){u(t-a)} is the Heaviside step function. This function is 0 for t<a{t<a} and 1 for t≥a{t\geq a}. So, in essence, it shifts the function f(t){f(t)} to the right by a{a} units and