Question

Use the convolution theorem to show that if \(f(t) \leftrightarrow F(s)\) then $$ \int_{0}^{t} f(\tau) d \tau \leftrightarrow \frac{F(s)}{s} $$

Short answer

Question: Prove that the Laplace transform of the integral of a function, f(t), from 0 to t is equal to the division of the Laplace transform of the given function by 's', i.e., $$ \int_{0}^{t} f(\tau) d \tau \leftrightarrow \frac{F(s)}{s} $$

Step-by-step solution

Define the function for convolution

We will first define a function, g(t), which we will convolute with the given function f(t) such that it results in the integral of f(t) from 0 to t. We define g(t) as: $$ g(t) = \begin{cases} 1, & \text{if}\ 0 \leq t \leq 1 \\ 0, & \text{elsewhere} \end{cases} $$

Compute the convolution integral of f(t) and g(t)

Now let's compute the convolution of f(t) and g(t): $$ h(t) = (f*g)(t) = \int_{-\infty}^{\infty} f(\tau) g(t-\tau) d\tau $$ Since g(t) is defined to be 1 between 0 and t, and 0 elsewhere, we can focus our convolution integral to only go from 0 to t: $$ h(t) = \int_{0}^{t} f(\tau) g(t-\tau) d\tau = \int_{0}^{t} f(\tau) d\tau $$

Applying the Convolution Theorem

The convolution theorem states that the Laplace transform of the convolution of two functions is equal to the product of their Laplace transforms. Thus, we have: $$ \mathcal{L}\{h(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\} $$ Substitute the expression for h(t) from step 2: $$ \mathcal{L}\left\{\int_{0}^{t} f(\tau) d\tau\right\} = F(s) \cdot G(s) $$

Compute the Laplace transform of g(t)

We need to find the Laplace transform of g(t) to proceed: $$ G(s) = \mathcal{L}\{g(t)\} = \int_{0}^{1} e^{-st} dt $$ Integrating with respect to t gives: $$ G(s) = \left[\frac{-1}{s}e^{-st}\right]_0^1 = \frac{1}{s}(1-e^{-s}) $$

Use G(s) to determine the Laplace transform of the integral

From step 3, we have: $$ \mathcal{L}\left\{\int_{0}^{t} f(\tau) d\tau\right\} = F(s) \cdot G(s) $$ Substitute the expression for G(s) found in step 4: $$ \mathcal{L}\left\{\int_{0}^{t} f(\tau) d\tau\right\} = F(s) \cdot \frac{1}{s}(1-e^{-s}) $$ However, as s approaches infinity, e^{-s} approaches 0: $$ \lim_{s \to \infty} e^{-s} = 0 $$ Thus, we can simplify our expression to: $$ \mathcal{L}\left\{\int_{0}^{t} f(\tau) d\tau\right\} = \frac{F(s)}{s} $$ By using the convolution theorem, we have proven the desired Laplace transform relation: $$ \int_{0}^{t} f(\tau) d \tau \leftrightarrow \frac{F(s)}{s} $$

Key concepts

Laplace Transform

The Laplace Transform is a powerful tool used in various fields such as mathematics, engineering, and physics. It transforms a time-domain function into a frequency-domain function, making it easier to solve differential equations and analyze systems. The transform is defined as follows:
\[\mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt\]
where \( f(t) \) is the original function, \( s \) is a complex number, and \( \mathcal{L}\{f(t)\} \) is the Laplace transform of \( f(t) \).

Some benefits of using the Laplace Transform include:

• It converts complex differential equations into simpler algebraic equations.
• It helps to solve initial value problems much more easily.
• The Laplace Transform can be used to analyze linear time-invariant systems.
• It simplifies the process of finding system responses to various inputs.

In the context of the convolution theorem, the Laplace Transform plays a crucial role in converting the convolution of two functions in the time domain into multiplication in the frequency domain.

Integral of a Function

An integral of a function represents the accumulation of quantities, such as areas under curves. There are two main types: definite and indefinite integrals. For a given function \( f(t) \), the definite integral from \( a \) to \( b \) is:
\[\int_{a}^{b} f(t) \, dt\]
This integral calculates the net area between the curve and the x-axis, from \( t = a \) to \( t = b \). If no endpoints are specified, it's an indefinite integral. This represents a family of functions including a constant \( C \):
\[\int f(t) \, dt = F(t) + C\]
where \( F(t) \) is an antiderivative of \( f(t) \).

Within the context of the convolution theorem problem, the objective was to calculate \( \int_{0}^{t} f(\tau) \, d\tau \), which is the area under the curve of the function \( f(\tau) \) from \( 0 \) to \( t \). This integral is directly linked with the convolution integral, such that using the Laplace Transform simplifies its computation when paired with another function.

Convolution Integral

The convolution integral is a mathematical operation on two functions, \( f(t) \) and \( g(t) \), producing a third function, \( h(t) \). It combines two signals to produce a new signal, representing transformations such as filtering, mixing, or spreading.
The convolution of \( f(t) \) and \( g(t) \) is defined as:
\[(f*g)(t) = \int_{0}^{t} f(\tau)g(t-\tau) \, d\tau\]
Here, the limits of the integral focus on the effective range where both functions overlap.

In terms of properties, the convolution operation is:

• Commutative: \( f*g = g*f \)
• Associative: \( f*(g*h) = (f*g)*h \)
• Distributive: \( f*(g + h) = f*g + f*h \)

The convolution theorem is crucial as it enables us to switch between time-domain convolution and frequency-domain multiplication by utilizing the Laplace Transform. This method allows for efficient problem-solving, particularly when determining how functions affect systems over time.