Question

Compute the convolution and its Laplace transform of the following functions: (a) \(e^{t} * e^{2 t}\). (b) \(\sin \omega t * \cos \omega t\). (c) \(t * e^{t}\).

Short answer

Based on the given functions of each pair and their corresponding results, complete the statement: (a) The convolution of \(e^t\) and \(e^{2t}\) is \((t - 1 + e^t)e^t\) and its Laplace transform is \(\frac{1}{s - 1} - \frac{1}{(s - 1)^2}\). (b) The convolution of \(\sin \omega t\) and \(\cos \omega t\) is \(\frac{1}{\omega}(\sin \omega t - t\cos \omega t)\) and its Laplace transform is \(\frac{1}{\omega(s^2 + \omega^2)} - \frac{s}{\omega(s^2 + \omega^2)}\). (c) The convolution of \(t\) and \(e^t\) is \(e^t (t - 1 + e^{-t})\) and its Laplace transform is \(\frac{1}{s - 1} - \frac{1}{(s - 1)^2}\).

Step-by-step solution

Find the Convolution

To find the convolution, apply the convolution formula: \[(f * g)(t) = \int_0^t e^{\tau} e^{2(t-\tau)} d\tau\]

Integrate with Respect to Tau

Now, compute the integral: \[(f * g)(t) = \int_0^t e^{\tau} e^{2(t-\tau)} d\tau = \int_0^t e^{t + \tau} d\tau\] The result is: \[(f * g)(t) = e^t \int_0^t e^{\tau} d\tau\]

Apply the Laplace Transform

Now, we need to find the Laplace transform of the convolution. Using basic Laplace transform properties, we compute: \[\mathcal{L}\{(f * g)(t)\} = \mathcal{L}\{e^t \cdot (t - 1 + e^t)\} = \frac{1}{s - 1} - \frac{1}{(s - 1)^2}\] So, the required convolution is \((t - 1 + e^t)e^t\) and its Laplace transform is \(\frac{1}{s - 1} - \frac{1}{(s - 1)^2}\). (b) \(\sin \omega t * \cos \omega t\)

Find the Convolution

Use the convolution formula: \[(f * g)(t) = \int_0^t \sin \omega\tau \cos \omega(t - \tau) d\tau\]

Use Integration by Parts

Integrate using an integration by parts formula: \[(f * g)(t) = \frac{1}{\omega} \int_0^t \sin \omega\tau d(\sin \omega(t - \tau))\] The result is: \[(f * g)(t) = \frac{1}{\omega}(\sin \omega t - t\cos \omega t)\]

Apply the Laplace Transform

Now, we need to find the Laplace transform of the convolution: \[\mathcal{L}\{(f * g)(t)\} = \mathcal{L}\{\frac{1}{\omega}(\sin \omega t - t\cos \omega t)\} = \frac{1}{\omega(s^2 + \omega^2)} - \frac{s}{\omega(s^2 + \omega^2)}\] So, the required convolution is \(\frac{1}{\omega}(\sin \omega t - t\cos \omega t)\) and its Laplace transform is \(\frac{1}{\omega(s^2 + \omega^2)} - \frac{s}{\omega(s^2 + \omega^2)}\). (c) \(t * e^t\)

Find the Convolution

Use the convolution formula: \[(f * g)(t) = \int_0^t \tau e^{t-\tau} d\tau\]

Integrate with Respect to Tau

Integrate with respect to \(\tau\): \[(f * g)(t) = e^t \int_0^t \tau e^{-\tau} d\tau\] Now, integrate by parts: \[(f * g)(t) = e^t (t - 1 + e^{-t})\]

Apply the Laplace Transform

We need to find the Laplace transform of the convolution: \[\mathcal{L}\{(f * g)(t)\} = \mathcal{L}\{e^t \cdot (t - 1 + e^t)\} = \frac{1}{s - 1} - \frac{1}{(s - 1)^2}\] So, the required convolution is \(e^t (t - 1 + e^{-t})\) and its Laplace transform is \(\frac{1}{s - 1} - \frac{1}{(s - 1)^2}\).

Key concepts

Convolution Integral

The convolution integral is a fundamental concept in the field of differential equations and signal processing. It represents the amount of overlap between two functions as they are shifted across each other. In mathematical terms, if we have two functions, say, f(t) and g(t), their convolution is defined as:

\[ (f * g)(t) = \int_0^t f(\tau) g(t - \tau) d\tau \]
In simple terms, we are multiplying f(\tau) by a reversed and shifted version of g(t), and then integrating the product over the interval from 0 to t. This process, when applied to the exercise problem where we convolute e^t with e^{2t}, allows us to find the overlapping area of these two functions as they slide past one another over time.

Understanding the convolution integral is crucial, especially when dealing with systems described by linear time-invariant (LTI) properties, as it helps to find the system's output when the input and the system's impulse response are known.

Laplace Transform Properties

Laplace transform is a powerful tool used to simplify the process of solving differential equations by transforming them from the time domain into the frequency domain. The Laplace transform has several useful properties that make it easier to handle various mathematical operations. Some of the key properties include linearity, where the Laplace transform of a sum of functions is the sum of their individual Laplace transforms, and the differentiation property, where the Laplace transform of a derivative corresponds to multiplying the Laplace transform of the function by s in the frequency domain.

Applying these properties to our exercise, we can effortlessly transform the convolution of functions into a simple algebraic product of their individual Laplace transforms. For instance, with function e^t in part (a), we can use the property that the Laplace transform of an exponential function is 1/(s - a), where a is the exponent's coefficient.

Integration by Parts

Integration by parts is a useful technique in calculus, often used to integrate the product of two functions that are not easily integrable in their original form. The formula is derived from the product rule for differentiation and is given by:

\[ \int u dv = uv - \int v du \]
In applying integration by parts to our convolution integral, as seen in the second example sin(\omega t) * cos(\omega t), we choose u and dv such that the resulting integral is simpler than the original. By determining a wise choice for u and dv, we can simplify the integration process effectively, which is often crucial in solving convolutions and differential equations.

Differential Equations

Differential equations are equations that relate some function with its derivatives. They appear ubiquitously in engineering, physics, economics, and beyond, anywhere a relationship involving rates of change is to be determined. Solutions to differential equations can be elusive and require sophisticated methods, including transforming into the Laplace domain.

Solving differential equations often involves finding the convolution of two functions, as convolution can represent a system's response to a given input. This ties back to our exercise, where functions represented by exponential and trigonometric forms are convoluted to model more complex behavior. The result of a convolution is itself a solution to a differential equation that models a physical system responding over time to an input signal.