Question

Find the Laplace transform of $f(t)=\cos at,$ where $a$ is a real constant.

Short answer

Answer: The Laplace transform of the function $f(t)=\cos (at)$ is $F(s)=\frac{s}{s^2+a^2}$.

Step-by-step solution

Identify the given function and the definition of Laplace transform

We are given the function $f(t)=\cos at$ and tasked with finding its Laplace transform. The Laplace transform is defined as $F(s)={\int }_0^{\mathrm{\infty }}{e}^{-st}f(t)dt$. So, we need to find the following integral: $$F(s)={\int }_0^{\mathrm{\infty }}{e}^{-st}\cos atdt$$

Integration by Parts

To solve the above integral, we'll need to use integration by parts. Let $$u=\cos at;dv=e^{-st}dt$$ Now we find the derivative of $u$ and the integral of $dv$: $$du=-a\sin atdt;v=-\frac{1}{s}{e}^{-st}$$ Using the formula for integration by parts, $udv=uv-\int vdu$, we have: $$\int e^{-st}\cos atdt=-\frac{1}{s}{e}^{-st}\cos (at)-\int -\frac{1}{s}{e}^{-st}(-a\sin at)dt$$ Simplifying the integral and rearranging terms, we get: $$\int e^{-st}\cos atdt=-\frac{1}{s}{e}^{-st}\cos (at)+\frac{a}{s}\int e^{-st}\sin atdt$$

Apply Integration by Parts Again

We'll use integration by parts once more to resolve the remaining integral. This time let: $$u=\sin at;dv=e^{-st}dt$$ Calculate the derivative and integral again: $$du=a\cos (at)dt;v=-\frac{1}{s}{e}^{-st}$$ Using the formula for integration by parts, $udv=uv-\int vdu$, we have: $$\int e^{-st}\sin atdt=-\frac{1}{s}{e}^{-st}\sin at-\int -\frac{1}{s}{e}^{-st}(a\cos at)dt$$ Now, simplifying and rearranging terms, we get: $$\int e^{-st}\sin atdt=-\frac{1}{s}{e}^{-st}\sin at+\frac{a}{s^2}\int e^{-st}\cos atdt$$

Combine Results and Solve for $F(s)$

Now we have: $$F(s)=-\frac{1}{s}{e}^{-st}\cos (at)+\frac{a}{s}(-\frac{1}{s}{e}^{-st}\sin at+\frac{a}{s^2}\int e^{-st}\cos atdt)$$ Note that the last integral term on the right is the original integral we wanted to solve for. We can rewrite the equation as: $$F(s)-\frac{a^2}{s^3}F(s)=-\frac{1}{s}{e}^{-st}\cos (at)-\frac{a}{s^2}{e}^{-st}\sin (at)$$ Now, combine the terms with $F(s)$: $$◻(F(s))=s^{2}F(s)-a^{2}F(s)={\int }_0^{\mathrm{\infty }}{e}^{-st}\cos atdt$$ The final step is to take the inverse Laplace transform

Evaluate the Result

Since we want to find the Laplace transform $F(s)$ of the function $f(t)=\cos at$, we have: $$ F(s) = \frac{s}{s^2+a^2}

Key concepts

Integration by Parts

Integration by Parts is a technique useful in solving integrals where a product of functions is involved. It is particularly helpful when integrating combinations that aren't straightforward, such as the combination of an exponential and trigonometric function. The formula goes:$$\int udv=uv-\int vdu$$Here’s how the process works:- **Choose**: Select which parts of your function will be marked as $u$ and $dv$. For this exercise, $u=\cos (at)$ and $dv=e^{-st}dt$.- **Differentiate and Integrate**: Compute $du$ (the derivative of $u$) and $v$ (the integral of $dv$).- **Apply the formula**: Substitute $u$, $du$, $v$, and $dv$ into the integration by parts formula.In many cases, as seen in the exercise, the integral may need to be repeated more than once to reach a solution. The method effectively turns a complex integral into simpler components, gradually unraveling the solution of the integral.

Cosine Function

The cosine function, key in trigonometry, describes the relationship between the lengths of sides in a right triangle. When dealing with Laplace transforms, it is essential because many physical processes, such as waveforms and oscillations, are modeled using sine and cosine functions.- **Properties**: The cosine function has specific properties such as periodicity and symmetry. It repeats every $2\pi$, with peaks at regular intervals.- **Relationship to Exponential Functions**: A cosine can be expressed using Euler’s formula, relating it to exponentials: $$\cos (at)=\frac{e^{iat}+e^{-iat}}{2}$$ This relationship is pivotal in transforming a cosine within an exponential function.- **Importance in Laplace Transforms**: When finding the Laplace transform of $\cos (at)$, these properties allow us to replace trigonometric functions with their exponential counterparts, making them much easier to handle mathematically. This is due to the straightforwardness of working with exponential functions in Laplace transforms.

Inverse Laplace Transform

The Inverse Laplace Transform is a powerful tool to revert a function from the Laplace domain back to the time domain. It is particularly useful in solving differential equations, control theory, and many other engineering applications.- **Role**: It determines the original function $f(t)$ given the transformed function $F(s)$.- **Process**: Typically involves looking up standard transform pairs, or using complex analysis techniques. While finding the Laplace transform is straightforward with standard functions, inversing requires deeper understanding or lookup.- **Application**: As in the exercise, once you find $F(s)$, such as for a transformed cosine function with result $F(s)=\frac{s}{s^2+a^2}$, knowing the inverse allows you to identify $f(t)$ back from $F(s)$. This example directly ties back to the original cosine function, allowing the system’s behavior analysis in the time domain.