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Laplace Transforms

Self-guided notes

What is a Laplace Transform?

It's obviously given in the equation book as the following F(s)=∫e−stf(t)dtF(s) = \int{e^{-st}f(t)} {dt} but what does this mean and where does it come from

Let's look at a simple function f(t)f(t) where f(t)f(t) is analytic so then the Taylor series can be represented by f(t)=∑n=0∞f(n)(0)n!tn f(t) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} t^n then we can let a(n)=f(n)(0)n!a(n) = \frac{f^{(n)}(0)}{n!} so that now f(t)=∑n=0∞a(n)tn f(t) = \sum_{n=0}^{\infty} a(n) t^n

if you have a linear system L[y]=g(t)L[y] = g(t) where g(t) is a piece wise function we would normally have to solve two diffrent intervals seperately

  • eg. see the following problem g(t)={2.40<t<101.2t>10g(t) = \begin{cases} \text{2.4} & \text{0<t<10} \\ \text{1.2} & \text{t>10} \\ \end{cases} then logically x′(t)+3x/500=g(t) let’s assume that x(0)=30x'(t) +3x/500 = g(t) \text{ let's assume that } x(0) = 30
  • To solve this we must observe that an integrating factor of the form μ=e3/500∗t\mu = e^{3/500*t} then multiplying this to both sides it becomes ddt(e3/500∗tx(t))=e3/500∗tg(t)\frac{d} {dt} (e^{3/500*t} x(t)) = e^{3/500*t}g(t) then we integrate to get x(t)=(∫e3/500∗tg(t)dt)/e3t/500x(t) = (\int{ e^{3/500*t}g(t) dt})/ e^{3t/500}now we would have to plug in two g(t)'s and solve to get a piecewise result for x

    • Excercises

    1. apply laplace transform to this f(t)={30<= t <= 26-t2<tf(t) = \begin{cases} \text{3} & \text{0<= t <= 2} \\ \text{6-t} & \text{2<t} \\ \end{cases}

    We need to use the unit step function to represent f(t)=3+u(t−2)(3−t)f(t) = 3 + u(t-2)(3-t) then we appy the laplace transform L{f(t)}=L{3}+L{u(t−2)(3−t)}=∫0∞3e−st dt+∫2∞(3−t)e−st dt=3/s+∫2∞(3−t)e−st dt \mathcal{L}\{ f(t) \} = \mathcal{L}\{ 3\} + \mathcal{L}\{ u(t-2)(3-t)\} = \int_0^\infty 3 e^{-st} \, dt + \int_2^\infty (3-t) e^{-st} \, dt = 3/s + \int_2^\infty (3-t) e^{-st} \, dt

    Now this integral is tricky, so we can do it by substitution: let w=t−2\text{let } w = t-2 then logically t=w+2 since if t=2 then w=0 logically the integral becomes ∫0∞(1−w)e−s(w+2) dw=e−2s∫0∞(1−w)e−ws) dwt = w + 2 \text{ since if } t = 2 \text{ then } w = 0 \text{ logically the integral becomes } \int_0^\infty (1-w) e^{-s(w+2)} \, dw = e^{-2s}\int_0^\infty (1-w) e^{-ws)} \, dw

    Now we can do some splitting e−2s∫0∞(1)e−ws dw−e−2s∫0∞(w)e−ws) dw=e−2s∗(1/s−1/s2)=F(s)e^{-2s}\int_0^\infty (1) e^{-ws} \, dw - e^{-2s}\int_0^\infty (w) e^{-ws)} \, dw = e^{-2s}* (1/s - 1/s^{2}) = F(s)

    F(s)=e−2s(1/s−1/s2) F(s) = e^{-2s} (1/s - 1/s^{2})

    Laplace transforms and inverses are mainly just tabling to be honest so we'll do something more intresting. Solve these IVP's

    Physical Interpretation Problem

    suppose there's an equation y′′−7y′+10y=0;y(0)=0;y′(0)=−3y'' - 7y' + 10y = 0; y(0) = 0; y'(0) = -3 solve it using laplace transforms

    Let's just transform both sides L{y′′}+L{−7y′}+L{10y}=L{0}=0\mathcal{L}\{ y''\} + \mathcal{L}\{ -7y' \} + \mathcal{L}\{ 10y\} = \mathcal{L}\{0\} = 0

    s2Y(s)−sy(0)−y′(0)−7sY(s)+7y(0)+Y(0)=0 using our initial conditions and rearranging Y(s)(s2−7s+10)+3=0s^{2}Y(s)−sy(0)−y′(0) -7sY(s) +7y(0) + Y(0) = 0 \text{ using our initial conditions and rearranging } Y(s)(s^{2}−7s+10)+3=0

    Y(s)=−3(s−5)(s−2)Y(s) = \frac{-3}{(s-5)(s-2)} Now we want to decompose fraction −3=A(s−2)+B(s−5)A=−1;B=1 Now you just table to get y(t)=e2t−e5t−3=A(s−2)+B(s−5) A = -1; B =1 \text{ Now you just table to get } y(t) = e^{2t}−e^{5t}

    Physical Interpretation Problem

    Suppose there's a water tank where the inflow increases gradually after t = 1. The height of water \(y(t)\) is modeled by y′′+2(t−1)y′−2y=0;y(0)=0;y′(0)=0y'' + 2(t-1)y' - 2y = 0; y(0) = 0; y'(0) = 0 Solve it using Laplace transforms

    Let's just transform both sides L{y′′}+2L{(t−1)y′}−2L{y}=L{0}=0\mathcal{L}\{y''\} + 2 \mathcal{L}\{(t-1)y'\} - 2 \mathcal{L}\{y\} = \mathcal{L}\{0\} = 0

    s2Y(s)−sy(0)−y′(0)+2(−dds[sY(s)−y(0)]−Y(s))−2Y(s)=0s^2 Y(s) - s y(0) - y'(0) + 2 \left(-\frac{d}{ds} [s Y(s) - y(0)] - Y(s)\right) - 2 Y(s) = 0

    s2Y(s)−0−0−2dds[sY(s)]−2Y(s)−2Y(s)=0s^2 Y(s) - 0 - 0 - 2\frac{d}{ds}[s Y(s)] - 2 Y(s) - 2 Y(s) = 0

    Combine terms: −2dds[sY(s)]+s2Y(s)−4Y(s)=0 \text{Combine terms: } -2\frac{d}{ds}[s Y(s)] + s^2 Y(s) - 4 Y(s) = 0

    Solve this first-order ODE for Y(s) (separation of variables): Y(s)=Cs2(s+2) \text{Solve this first-order ODE for } Y(s) \text{ (separation of variables): } Y(s) = \frac{C}{s^2 (s+2)}

    Now decompose fraction: Cs2(s+2)=As+Bs2+Ds+2, solve for constants \text{Now decompose fraction: } \frac{C}{s^2 (s+2)} = \frac{A}{s} + \frac{B}{s^2} + \frac{D}{s+2} , \text{ solve for constants}

    Using Laplace table, invert to get y(t)=Bt+A+De−2t=(linear term + exponential) \text{Using Laplace table, invert to get } y(t) = B t + A + D e^{-2t} = \text{(linear term + exponential)}

    use laplace to solve this system x′+y=0;  x(0)=0,  x+y′=1−u(t−2);  y(0)=0x' + y = 0; \; x(0) = 0, \; x + y' = 1 - u(t-2); \; y(0) = 0

    So..., we do laplace both sides
    L{x′+y}=0\mathcal{L}\{x' + y\} = 0
    ∫0∞x′(t) e−st dt+∫0∞y(t) e−st dt=0 \int_{0}^{\infty} x'(t)\, e^{-st}\, dt + \int_{0}^{\infty} y(t)\, e^{-st}\, dt = 0 ∫0∞x′(t)e−st dt \int_{0}^{\infty} x'(t)e^{-st}\,dt
    Use integration by parts:
    u=e−st,dv=x′(t) dt u = e^{-st}, \quad dv = x'(t)\,dt du=−se−st dt,v=x(t) du = -s e^{-st}\,dt, \quad v = x(t)
    Apply \(\int u\,dv = uv - \int v\,du\):
    ∫0∞x′(t)e−st dt=[x(t)e−st]0∞+s∫0∞x(t)e−st dt \int_{0}^{\infty} x'(t)e^{-st}\,dt = \left[ x(t)e^{-st} \right]_{0}^{\infty} + s \int_{0}^{\infty} x(t)e^{-st}\,dt
    Evaluate the boundary term:
    [x(t)e−st]0∞=0−x(0)=−x(0) \left[ x(t)e^{-st} \right]_{0}^{\infty} = 0 - x(0) = -x(0)
    Final result:
    ∫0∞x′(t)e−st dt=s∫0∞x(t)e−st dt−x(0) \int_{0}^{\infty} x'(t)e^{-st}\,dt = s \int_{0}^{\infty} x(t)e^{-st}\,dt - x(0)

    sL{x}−x(0)+L{y}=0s\mathcal{L}\{x\} - x(0) + \mathcal{L}\{y\} = 0

    Since x(0)=0,x(0)= 0,

    L{x′+y}=0=sL{x}+L{y}\mathcal{L}\{x' + y\} = 0 = s\mathcal{L}\{x \} +\mathcal{L}\{y \}

    Now take Laplace of the second equation:
    L{x+y′}=L{1−u(t−2)} \mathcal{L}\{x + y'\} = \mathcal{L}\{1 - u(t-2)\} L{x}+L{y′}=1s−e−2ss \mathcal{L}\{x\} + \mathcal{L}\{y'\} = \frac{1}{s} - \frac{e^{-2s}}{s}
    Use the derivative rule:
    L{y′}=sL{y}−y(0) \mathcal{L}\{y'\} = s\mathcal{L}\{y\} - y(0)
    Since \(y(0)=0\):
    L{x}+sL{y}=1−e−2ss \mathcal{L}\{x\} + s\mathcal{L}\{y\} = \frac{1 - e^{-2s}}{s}
    So the system becomes:
    sX+Y=0X+sY=1−e−2ss \begin{aligned} sX + Y &= 0 \\ X + sY &= \frac{1 - e^{-2s}}{s} \end{aligned}
    Solve the system:
    Y=−sX Y = -sX X+s(−sX)=1−e−2ss X + s(-sX) = \frac{1 - e^{-2s}}{s} X(1−s2)=1−e−2ss X(1 - s^2) = \frac{1 - e^{-2s}}{s} X(s)=1−e−2ss(1−s2) X(s) = \frac{1 - e^{-2s}}{s(1 - s^2)} Y(s)=−sX(s)=−(1−e−2s)1−s2 Y(s) = -sX(s) = \frac{-(1 - e^{-2s})}{1 - s^2}
    Prepare for inverse Laplace:
    X(s)=1s(1−s2)−e−2ss(1−s2) X(s) = \frac{1}{s(1 - s^2)} - \frac{e^{-2s}}{s(1 - s^2)} Y(s)=−11−s2+e−2s1−s2 Y(s) = \frac{-1}{1 - s^2} + \frac{e^{-2s}}{1 - s^2}

    Find the taylor series for f(t)=e−t2 about t= 0 then assuming that the laplace transform of f(t) can be computed term by term find an expansion for L{f(x)} in powers of 1/sf(t) = e^{-t^{2}}\text{ about t= 0 then assuming that the laplace transform of f(t) can be computed term by term find an expansion for } \mathcal{L}\{f(x)\} \text{ in powers of } 1/s
    e−t2=1−t2+t4/2!−t6/3!+t8/4!.... And so on ....e^{-t^{2}} = 1- t^2 + t^4/2! -t^6/3! + t^8/4! .... \text{ And so on ....}

    L{e−t2}=L{1}+L{−t2}...I’m not writing the rest...\mathcal{L}\{e^{-t^{2}}\} = \mathcal{L}\{1\} + \mathcal{L}\{-t^2\} \text{...I'm not writing the rest...}

    L{f(t)}=L{e−t2}=1s−2s3+12s5−120s7 No need for the rest you get the gist .... \mathcal{L}\{f(t)\} = \mathcal{L}\{e^{-t^{2}}\} = \frac{1}{s} - \frac{2}{s^3} + \frac{12}{s^5} - \frac{120}{s^7} \text{ No need for the rest you get the gist ....}