MA1035: Engineering Modeling using Dynamic Systems

Session 12: 2nd Translation Theorem (t-shift)

Prof. Alejandro Ucan-Puc

Tecnológico de Monterrey • Departamento de Ciencias

Goals

  • Introduce the 2nd Traslation Theorem.

  • Apply the 2nd Traslation Theorem in the calculation of Laplace Transforms.

  • Apply the 2nd Traslation Theorem in the solution of ODEs.

Requirements:



The unit step function in $a$ is denoted by $u_a(t)$ and it is defined as $$u_a(t)=\left\{\begin{array}{cc} 0, & 0\leq t < a \\ 1, t\geq a \end{array} \right.$$



How does the unit step function affect our functions?
Picture the graph of the functions $f(t)=e^{t-2} u_2(t),\, g(t)=e^{t-3} u_3(t),\, h(t)=e^{t-4} u_4(t).$

2nd Traslation Theorem

Theorem: Let $f(t)$ be a function such that $\mathcal{L}\{f(t)\}=F(s)$ exists. Then $$\mathcal{L}\{f(t-a)u_a(t)\}=e^{-as}F(s).$$



Theorem Implications: Now every time we calculate the Laplace Transform of scalonated function, we can use the 2nd Traslation Theorem to obtain the result.

Applying the Theorem

Example (Direct Computation):

Compute the Laplace Transform of $f(t)=e^{t-2}u_2(t)$, $\cos(t-\pi)u_{\pi}(t)$ y $(-3(t-3)^4+(t-3)^3)u_{3}(t).$



Solution

$$\mathcal{L}\{e^{t-2}u_2(t)\}=\mathcal{L}\{e^{t-2}u_2(t-2)\}=e^{-2s}\mathcal{L}\{e^t\}=e^{-2s}\frac{1}{s-1}.$$

Example (Direct Computation):

Compute the Inverse Laplace transform of $F(s)=\frac{1}{s-4}e^{-2s},$ $F(s)=\frac{s}{s^2+9}e^{-\pi s/2}.$



Solution

The Inverse laplace transform of the function without the exponential factor is $$\mathcal{L}^{-1}\left\{\frac{1}{s-4}\right\}=e^{4t}$$ since it has an exponential factor, we need to modify it with the unit step function: $$\mathcal{L}^{-1}\left\{\frac{1}{s-4}e^{-2s}\right\}=e^{4(t-2)}u_2(t-2).$$

Example (Computation with Partial Fractions)

Compute the inverse Laplace transform of $F(s)=\frac{e^{-s}}{s(s^2-4)}.$


Solution

  1. We need to do partial fraction in the fraction without the exponential factor. $$\frac{1}{s(s-2)(s+2)}=\frac{A}{s}+\frac{B}{s-2}+\frac{C}{s+2}$$ $$=\frac{-1}{4 s} + \frac{1}{8 (s + 2)} + \frac{1}{8 (s - 2)}$$
    Now, we use the 2nd traslation theorem to use express the inverse $$f(t)=\frac{-1}{4}u_1(t)+\frac{1}{8}e^{2(t+1)}u_{1}(t)+\frac{1}{8}e^{-2(t+1)}u_1(t).$$

Example (Solving ODEs):

Find the solution for $y'+y=f(t),\quad y(0)=0,\quad f(t)=\left\{\begin{array}{cc} 0, & 0\leq t < 1 \\ 5, & t\geq 1 \end{array}\right.$


Solution:

  1. First we need to express the function $f$ as a unit step function: $f(t)=5u_1(t).$

  2. We need to find the Laplace Transform of the ODE. $$\mathcal{L}\{y'\}+\mathcal{L}\{y\}=\mathcal{L}\{f(t)\}$$ $$sY(s)-y(0)+Y(s)=\frac{5}{s}$$ $$Y(s)=\frac{5e^{-s}}{s(s+1)}.$$
  3. Reduce, use partial fractions and take Inverse Laplace transform.

Example (Solving ODEs):

Find the solution for $y'+y=f(t),\quad y(0)=5,\quad f(t)=\left\{\begin{array}{cc} 0, & 0\leq t <\pi \\ 3\cos(t-\pi), & t\geq \pi \end{array}\right.$

+--

Solution:

  1. First we need to express the function $f$ as a unit step function: $f(t)=3\cos(t-\pi)u_\pi(t).$

  2. We need to find the Laplace Transform of the ODE. $$\mathcal{L}\{y'\}+\mathcal{L}\{y\}=\mathcal{L}\{f(t)\}$$ $$sY(s)-y(0)+Y(s)=\frac{3se^{-\pi s}}{s^2+1}$$ $$Y(s)=\frac{3e^{-s}}{s(s+1)}+\frac{5}{s^2+1}.$$
  3. Reduce, use partial fractions and take Inverse Laplace transform.

Summary & Next Step

Heaviside unit step function, time-delay translation L{f(t-a) U(t-a)} = e^(-as) F(s), piecewise switching inputs, and impulsive responses in engineering.

Next Session: Introduction to ODEs