Solving LTI ODEs with Unilateral Laplace Transform

Hello! Welcome to the fourth lesson in our module on "System Analysis in the s-Domain."

In our last lesson, we explored how the properties of the bilateral Laplace transform, specifically the Region of Convergence (ROC), allow us to analyze fundamental system characteristics like causality and stability from the pole-zero plot. We saw that for a causal and stable LTI system, all poles must lie in the left-half of the s-plane. This was a powerful analytical tool.

Today, we shift from analysis to direct problem-solving. We will focus on one of the most important applications of the Laplace transform: solving linear time-invariant (LTI) differential equations. You'll find this method significantly streamlines the process compared to classical techniques, especially when dealing with initial conditions. To do this, we will introduce the unilateral Laplace transform.

By the end of this lesson, you will be able to apply the unilateral Laplace transform to find the complete response of an LTI system described by a differential equation, given a specific input and initial conditions.

The Unilateral Laplace Transform

While the bilateral transform we've used so far is defined by an integral from $- \infty$ to $\infty$ , most practical problems, especially in circuit analysis, involve signals that start at $t = 0$ and have known initial states at that time (e.g., initial voltage on a capacitor, initial current in an inductor). For these scenarios, the unilateral Laplace transform is the more natural tool.

It is defined as:

L {x (t)} = X (s) = \int_{0}^{\infty} x (t) e^{- s t} d t

The key difference is the lower integration limit of 0. This means the transform only considers the signal for $t \geq 0$ , automatically assuming $x (t) = 0$ for $t < 0$ . This seemingly small change has a profound impact on how we handle derivatives, which is the key to solving differential equations.

Historically, the British engineer Oliver Heaviside developed a similar "operational calculus" in the late 19th century to solve problems in electrical transmission lines. His methods, though initially lacking mathematical rigor, were incredibly effective. The Laplace transform provided the formal mathematical foundation for Heaviside's powerful techniques.

The Key Property: The Differentiation Theorem

The "magic" of using the Laplace transform for differential equations comes from the differentiation theorem. Let's derive it for the first derivative.

Given the definition of the unilateral transform $L {\frac{d y ( t )}{d t}} = \int_{0}^{\infty} \frac{d y ( t )}{d t} e^{- s t} d t$ , we can use integration by parts, where $\int u d v = uv - \int v d u$ .

Let:

$u = e^{- s t}$ $⟹ d u = - s e^{- s t} d t$
$d v = \frac{d y ( t )}{d t} d t$ $⟹ v = y (t)$

Substituting these gives:

\int_{0}^{\infty} \frac{d y ( t )}{d t} e^{- s t} d t = [y (t) e^{- s t}]_{0}^{\infty} - \int_{0}^{\infty} y (t) (- s e^{- s t}) d t

= t \to \infty lim y (t) e^{- s t} - y (0) e^{0} + s \int_{0}^{\infty} y (t) e^{- s t} d t

Assuming $s$ is in the ROC, the limit term goes to zero. The integral is just the definition of $Y (s)$ . This leaves us with the fundamental result:

Differentiation Theorem:

L {\frac{d y ( t )}{d t}} = s Y (s) - y (0)

Notice how the initial condition $y (0)$ is automatically incorporated! We can apply this theorem again to find the transform of the second derivative:

L {\frac{d ^{2} y ( t )}{d t ^{2}}} = L {\frac{d}{d t} (y^{'} (t))} = s L {y^{'} (t)} - y^{'} (0)

= s (s Y (s) - y (0)) - y^{'} (0)

L {\frac{d ^{2} y ( t )}{d t ^{2}}} = s^{2} Y (s) - sy (0) - y^{'} (0)

This pattern continues for higher-order derivatives. This theorem transforms the calculus operation of differentiation in the time domain into a simple algebraic multiplication by $s$ in the frequency domain, with initial conditions appearing as additive terms.

The Method for Solving LTI Differential Equations

The process for solving an $n$ -th order LTI differential equation with initial conditions is a straightforward algorithm:

Transform: Take the unilateral Laplace transform of every term in the differential equation.
Substitute: Apply the differentiation theorem to the derivative terms. This converts the differential equation into an algebraic equation in terms of $Y (s)$ .
Incorporate Initial Conditions: Plug in the given numerical values for $y (0), y^{'} (0), \dots, y^{(n - 1)} (0)$ .
Solve: Algebraically rearrange the equation to solve for $Y (s)$ . The result will be a rational function of $s$ .
Inverse Transform: Find the time-domain solution $y (t)$ by computing the inverse Laplace transform of $Y (s)$ , typically using partial fraction expansion.

Let's see this method in action.

Example 1: Homogeneous Equation

First, consider a simple case with no external input (a homogeneous equation). This isolates the core mechanics of the method.

Using Laplace Transforms to solve Differential Equations ***full example***

The following video by Dr. Trefor Bazett provides a crystal-clear walkthrough of solving a second-order homogeneous differential equation with initial conditions. It perfectly illustrates the 5-step method we just outlined.

Please watch the entire video (around 9 minutes). Pay close attention to how he transforms the equation, substitutes the initial conditions, solves for Y(s), and then uses partial fractions to find the final time-domain solution y(t).

This example shows how the initial conditions determine the specific coefficients in the final solution, which consists of the system's natural modes ( $e^{2 t}$ and $e^{- t}$ ).

Example 2: Non-Homogeneous Equation (System with an Input)

Now, let's tackle a more complete problem: finding the output of an LTI system given both initial conditions (initial state) and an input signal $x (t)$ .

The equation now looks like this:
$a_{n} \frac{d ^{n} y}{d t ^{n}} + \dots + a_{0} y (t) = b_{m} \frac{d ^{m} x}{d t ^{m}} + \dots + b_{0} x (t)$
When we transform this, we will get an expression for $Y (s)$ that depends on both the initial conditions and the transform of the input, $X (s)$ .

Find output of LTI system with initial conditions y(0)=1, y'(0)=1 for given x(t)=e^(-2t)u(t)

This next video solves a non-homogeneous differential equation, which represents finding the total response of a system to an input signal, starting from a non-zero initial state. The example also involves a repeated pole in the partial fraction expansion, which is a common scenario.

Please watch the entire video (around 18 minutes). Notice how the Laplace transform of the input, X(s), appears on the right side of the algebraic equation. Also, pay attention to the technique used to find the coefficients for the repeated pole in the partial fraction expansion.

The solution $y (t)$ you obtain this way is the total response of the system. It naturally includes both the zero-input response (due to the initial conditions) and the zero-state response (due to the input signal).

Test your understanding!

Take the differential equation $y^{'} (t) + 2 y (t) = x (t)$ with initial condition $y (0) = 3$ and input $x (t) = u (t)$ (the unit step function).

Transform the equation into an algebraic equation for $Y (s)$ . Remember that $L {u (t)} = 1/ s$ .
Solve for $Y (s)$ .
(Challenge) Find $y (t)$ by taking the inverse Laplace transform.

Show answer

Transforming the equation gives:
$L {y^{'} (t)} + 2 L {y (t)} = L {x (t)}$
$(s Y (s) - y (0)) + 2 Y (s) = X (s)$
Substitute $y (0) = 3$ and $X (s) = 1/ s$ :
$s Y (s) - 3 + 2 Y (s) = \frac{1}{s}$
$Y (s) (s + 2) = 3 + \frac{1}{s} = \frac{3 s + 1}{s}$
$Y (s) = \frac{3 s + 1}{s ( s + 2 )}$
Use partial fraction expansion: $Y (s) = \frac{A}{s} + \frac{B}{s + 2}$ .
- $A = s Y (s) ∣_{s = 0} = \frac{1}{2}$
- $B = (s + 2) Y (s) ∣_{s = - 2} = \frac{- 5}{- 2} = \frac{5}{2}$
  So, $Y (s) = \frac{1/2}{s} + \frac{5/2}{s + 2}$ .
  The inverse transform is $y (t) = \frac{1}{2} u (t) + \frac{5}{2} e^{- 2 t} u (t)$ for $t \geq 0$ .

Application to Circuit Analysis

For someone with your background in electrical engineering, the most direct application of this method is in circuit analysis. An LTI circuit described by resistors, inductors, and capacitors is governed by linear constant-coefficient differential equations. The Laplace transform method allows us to analyze these circuits using simple algebraic techniques, similar to DC analysis with resistors.

The key is to transform the constitutive relations for each component:

Resistor: $v (t) = R i (t) ⟹ V (s) = R I (s)$
Inductor: $v (t) = L \frac{d i ( t )}{d t} ⟹ V (s) = s L I (s) - L i (0)$
Capacitor: $i (t) = C \frac{d v ( t )}{d t} ⟹ I (s) = s C V (s) - C v (0)$

These transformed equations show that in the s-domain, components are represented by their impedance ( $R$ , $s L$ , $1/ s C$ ) along with voltage or current sources that represent the initial conditions.

The following reading explains this powerful concept in detail.

THE UNll..ATERAL LAPLACE TRANSFORM ( ...

This reading from a classic MIT text, 'THE UNILATERAL LAPLACE TRANSFORM', demonstrates how to translate a time-domain circuit into a frequency-domain equivalent. This is a cornerstone of analog system analysis.

Please read Section 2.5, 'Circuit Analysis with the Laplace Transform' (pages 54-61). Focus on Figure 2.5-1, which summarizes the frequency-domain models for R, L, and C components, including their initial conditions. Then, study Example 2.5-1 to see how a simple RC circuit is solved using this method.

By transforming the entire circuit into the s-domain, you can use familiar algebraic circuit analysis techniques (like node-voltage analysis, mesh analysis, Thevenin's theorem, etc.) to solve for any desired voltage or current transform, and then convert the result back to the time domain.

Conclusion

Today, we've added a powerful problem-solving technique to our toolkit. We moved from the analytical bilateral transform to the practical unilateral transform, which is purpose-built for solving initial value problems.

Key Takeaways:

The unilateral Laplace transform, $\int_{0}^{\infty}$ , is ideal for problems involving signals starting at $t = 0$ with known initial conditions.
The differentiation theorem ( $L {y^{'}} = s Y (s) - y (0)$ ) is the core engine that converts differential equations in time into algebraic equations in frequency.
This method provides the total response of a system, seamlessly combining the effects of initial conditions (zero-input response) and the driving input (zero-state response).
In circuit analysis, this translates to replacing components with their s-domain impedances and representing initial conditions as constant sources, reducing AC/transient analysis to a problem resembling DC circuit analysis.

Preview of the Next Lesson:
We've now seen how to find the full time-domain response $y (t)$ of a system. In the next lesson, we will focus on characterizing the LTI system itself. We will formally define the transfer function, $H (s)$ , and explore its relationship to two other fundamental system descriptions: the impulse response $h (t)$ and the frequency response $H (j ω)$ . This will unite many of the concepts we've discussed so far.

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