ODE: First and Second Shifting Theorems

Mastering the shifting theorems of the Laplace transform is crucial for any engineer, as they provide powerful tools for solving ordinary differential equations (ODEs) that would be cumbersome or impossible to handle with standard methods. These theorems allow you to elegantly manage differential equations with exponential forcing functions and those involving time delays—common scenarios in control systems, circuit analysis, and mechanical vibrations. Understanding when and how to apply these shifts transforms a challenging problem into a straightforward algebraic one in the s-domain.

The First Shifting Theorem: Exponential Multiplication in Time

The First Shifting Theorem, also known as the frequency-shifting property, establishes a direct link between multiplication by an exponential in the time domain and a shift in the frequency domain. Formally, if the Laplace transform of a function $f(t)$ is $F(s)=\mathcal{L}\{f(t)\}$, then multiplying $f(t)$ by an exponential $e^{at}$ shifts the transform argument by $a$.

The theorem states: $$\mathcal{L}\{e^{at}f(t)\}=F(s-a)$$ where $a$ can be any real or complex constant. This is a transformative tool because it allows you to build a library of transforms. If you know the transform of $t^n$, $\sin (bt)$, or $\cos (bt)$, you can immediately find the transform of functions like $e^{at}{t}^n$ or $e^{at}\sin (bt)$ simply by replacing every $s$ in the known transform $F(s)$ with $(s-a)$.

Example: Find $\mathcal{L}\{e^{-3t}\cos (4t)\}$.

1. Recall the known transform: $\mathcal{L}\{\cos (4t)\}=\frac{s}{s^2+16}=F(s)$.
2. Here, the exponential multiplier is $e^{at}$ with $a=-3$.
3. Apply the First Shifting Theorem: $\mathcal{L}\{e^{-3t}\cos (4t)\}=F(s-(-3))=F(s+3)$.
4. Substitute $(s+3)$ for $s$ in $F(s)$:

$$\mathcal{L}\{e^{-3t}\cos (4t)\}=\frac{(s+3)}{(s+3{)}^2+16}$$

The primary engineering application is solving linear ODEs with constant coefficients and exponential forcing terms. The theorem converts the differential equation into an algebraic equation in $s$ where the forcing term's transform is easily written down by shifting a standard result.

The Second Shifting Theorem: Time Delay and Step Functions

The Second Shifting Theorem, or the time-shifting property, deals with delays in the time domain. It is intrinsically linked to the Heaviside step function, $u(t-a)$, which is 0 for $t<a$ and 1 for $t\ge a$. This function "turns on" a signal at time $t=a$.

The theorem has two essential forms. First, for a function $f(t)$ defined for $t\ge 0$, a delay by $a>0$ units is represented by $f(t-a)u(t-a)$. Its Laplace transform is: $$\mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(s)$$ Second, if you encounter $e^{-as}F(s)$ in the s-domain, its inverse transform is the delayed function: ${\mathcal{L}}^{-1}\{e^{-as}F(s)\}=f(t-a)u(t-a)$.

Example: Find the inverse Laplace transform of $\frac{e^{-2s}}{s-3}$.

1. Identify $e^{-as}$ with $a=2$ and $F(s)=\frac{1}{s-3}$.
2. Find the inverse of $F(s)$: ${\mathcal{L}}^{-1}\{\frac{1}{s-3}\}=e^{3t}=f(t)$.
3. Apply the Second Shifting Theorem: ${\mathcal{L}}^{-1}\{e^{-2s}F(s)\}=f(t-2)u(t-2)$.
4. Therefore, the inverse is $e^{3(t-2)}u(t-2)$.

This theorem is indispensable for modeling real-world engineering systems where actions are initiated after a delay, such as a switched voltage source in a circuit turning on at $t=5$ seconds, or a mechanical load being suddenly applied to a structure.

Applying the Shifts to Solve ODEs

The true power of these theorems is realized when solving non-homogeneous ODEs with discontinuous or delayed forcing functions. The process follows the standard Laplace transform method but requires careful handling of the shifts.

Solving an ODE with a Delayed Forcing Function: Consider the initial value problem: $y^{\prime \prime }+y=f(t)$, with $y(0)=0,y^{\prime }(0)=0$, where $f(t)=1$ for $t\ge \pi$ and $0$ otherwise. This is a step function delay: $f(t)=u(t-\pi )$.

1. Transform the ODE: Apply the Laplace transform to both sides, using linearity and derivative properties:

$\mathcal{L}\{y^{\prime \prime }\}+\mathcal{L}\{y\}=\mathcal{L}\{u(t-\pi )\}$. This gives $[s^{2}Y(s)-sy(0)-y^{\prime }(0)]+Y(s)=\mathcal{L}\{u(t-\pi )\}$.

1. Apply the Second Shifting Theorem: We know $\mathcal{L}\{u(t)\}=\frac{1}{s}$. Therefore, a delay by $\pi$ gives $\mathcal{L}\{u(t-\pi )\}=e^{-\pi s}\cdot \frac{1}{s}$.

1. Solve for Y(s): Substitute initial conditions and the shifted transform:

$(s^2+1)Y(s)=\frac{e^{-\pi s}}{s}$. So, $Y(s)=\frac{e^{-\pi s}}{s(s^2+1)}$.

1. Partial Fractions and Inverse Transform: First, find the inverse of the part without the delay: Let $G(s)=\frac{1}{s(s^2+1)}=\frac{1}{s}-\frac{s}{s^2+1}$. Thus, $g(t)={\mathcal{L}}^{-1}\{G(s)\}=1-\cos (t)$.

Now, since $Y(s)=e^{-\pi s}G(s)$, we apply the Second Shifting Theorem in reverse: $$y(t)={\mathcal{L}}^{-1}\{e^{-\pi s}G(s)\}=g(t-\pi )u(t-\pi )=[1-\cos (t-\pi )]u(t-\pi )$$

This solution is physically intuitive: the system is at rest until $t=\pi$, when the constant force is applied, and it responds according to its natural dynamics starting from that moment.

Recognizing When to Apply Each Theorem

Choosing the correct theorem is a critical skill. Misapplication leads to incorrect solutions.

• Apply the First Shifting Theorem when: The time-domain function is explicitly multiplied by an exponential, $e^{at}$. You will see this directly in the ODE's forcing term (e.g., $e^{-2t}\sin t$) or in the function you are trying to transform. The clue in the s-domain is that the denominator is a quadratic (or other form) that appears "shifted" from a standard pattern, like $\frac{1}{(s+5{)}^2+9}$ instead of $\frac{1}{s^2+9}$.

• Apply the Second Shifting Theorem when: The system has a clear time delay or a piecewise-defined forcing function that can be written using step functions $u(t-a)$. The tell-tale sign in the s-domain is the presence of an exponential multiplier $e^{-as}$ outside a standard transform $F(s)$. For example, $\frac{e^{-4s}}{s^2}$ or $e^{-s}\cdot \frac{3}{s+2}$.

In complex problems, you may need to use both theorems sequentially. For instance, finding the inverse transform of $\frac{e^{-5s}}{(s+1{)}^2}$ requires the Second Theorem (for $e^{-5s}$) and the First Theorem (for the fact that $\frac{1}{(s+1{)}^2}$ is a shifted version of $\frac{1}{s^2}$, corresponding to $t{e}^{-t}$).

Common Pitfalls

1. Misplacing the Shift in the First Theorem: A common error is writing $\mathcal{L}\{e^{at}f(t)\}=F(s+a)$ instead of $F(s-a)$. Remember the direction: multiplication by $e^{at}$ in time subtracts $a$ in the frequency domain. To check, consider the simple case $f(t)=1$, where $F(s)=1/s$. We know $\mathcal{L}\{e^{at}\cdot 1\}=1/(s-a)$, which is $F(s-a)$, confirming the correct rule.

1. Forgetting the Step Function in the Second Theorem's Inverse: When you compute ${\mathcal{L}}^{-1}\{e^{-as}F(s)\}$, the answer is always $f(t-a)u(t-a)$, not just $f(t-a)$. Omitting $u(t-a)$ implies the system responded before the input was applied, which is non-causal and incorrect for physical systems. The step function ensures the solution is zero for $t<a$.

1. Incorrectly Handling Partial Fractions with Delays: When performing inverse transforms on an expression like $\frac{e^{-as}}{(s-b)(s-c)}$, you must first find the partial fraction decomposition of $\frac{1}{(s-b)(s-c)}$, find its inverse $g(t)$, and then apply the delay. Do not attempt to include the $e^{-as}$ term inside the partial fraction decomposition algebra.

1. Confusing the Theorems for Similar-Looking Transforms: Recognize that $\frac{1}{(s-3{)}^2}$ and $\frac{e^{-3s}}{s^2}$ are fundamentally different. The first uses the First Shifting Theorem (inverse: $t{e}^{3t}$). The second uses the Second Shifting Theorem (inverse: $(t-3)u(t-3)$). The presence or absence of the exponential factor $e^{-as}$ is the key differentiator.

Summary

• The First Shifting Theorem ($\mathcal{L}\{e^{at}f(t)\}=F(s-a)$) handles multiplication by exponentials in the time domain by shifting the frequency variable $s$. It is essential for solving ODEs with exponentially modulated forcing functions.
• The Second Shifting Theorem ($\mathcal{L}\{f(t-a)u(t-a)\}=e^{-as}F(s)$) manages time delays via the Heaviside step function. Its inverse form is crucial for finding time-domain solutions from transforms containing $e^{-as}$.
• To solve ODEs with delays, transform the equation, express the delayed forcing function using the Second Theorem, solve algebraically for $Y(s)$, and then meticulously apply the inverse transform, remembering to attach the step function to the delayed part of the answer.
• Correct application depends on recognition: an exponential multiplier in time ($e^{at}$) points to the First Theorem, while an exponential multiplier in the s-domain ($e^{-as}$) points to the Second Theorem. Avoid the common traps of misdirected shifts and omitted step functions.