Let u(t) denote the unit step function. The bilateral Laplace transform of the function f(t) = e${}^t$ u(−t) is ____.

The question asks us to find the bilateral Laplace transform of the function $f(t)=e^{t}u(-t)$, where $u(t)$ denotes the unit step function. Understanding the properties of the unit step function and the definition of the bilateral Laplace transform is crucial for solving this problem.

Understanding the Function f(t)

First, let's analyze the given function $f(t)=e^{t}u(-t)$.

- The unit step function $u(t)$ is defined as: $$u(t)=\{\begin{array}{ll}1& \text{for }t\ge 0\\ 0& \text{for }t<0\end{array}$$
- Therefore, $u(-t)$ means we replace $t$ with $-t$ in the definition: $$u(-t)=\{\begin{array}{ll}1& \text{for }-t\ge 0\\ 0& \text{for }-t<0\end{array}$$ This simplifies to: $$u(-t)=\{\begin{array}{ll}1& \text{for }t\le 0\\ 0& \text{for }t>0\end{array}$$
- Now, we can write $f(t)=e^{t}u(-t)$ as: $$f(t)=\{\begin{array}{ll}{e}^t\cdot 1& \text{for }t\le 0\\ e^t\cdot 0& \text{for }t>0\end{array}$$ So, $$f(t)=\{\begin{array}{ll}{e}^t& \text{for }t\le 0\\ 0& \text{for }t>0\end{array}$$

Bilateral Laplace Transform Definition

The bilateral Laplace transform $F(s)$ of a function $f(t)$ is defined by the integral:

$$F(s)={\int }_{-\infty }^{\infty }f(t)e^{-st}dt$$

Setting up the Integral

Since $f(t)=e^t$ for $t\le 0$ and $f(t)=0$ for $t>0$, the integral limits will change:

$$F(s)={\int }_{-\infty }^0{e}^t{e}^{-st}dt+{\int }_0^{\infty }0\cdot e^{-st}dt$$

The second integral is zero, so we only need to evaluate the first part:

$$F(s)={\int }_{-\infty }^0{e}^{(1-s)t}dt$$

Evaluating the Integral

Now, we integrate $e^{(1-s)t}$ with respect to $t$:

$$F(s)={\left[\frac{e^{(1-s)t}}{1-s}\right]}_{-\infty }^0$$

Applying the limits:

$$F(s)=\frac{e^{(1-s)\cdot 0}}{1-s}-\underset{A\to -\infty }{\lim }\frac{e^{(1-s)A}}{1-s}$$

$$F(s)=\frac{e^0}{1-s}-\underset{A\to -\infty }{\lim }\frac{e^{(1-s)A}}{1-s}$$

$$F(s)=\frac{1}{1-s}-\underset{A\to -\infty }{\lim }\frac{e^{(1-s)A}}{1-s}$$

Determining the Region of Convergence (ROC)

For the integral to converge, the term ${\lim }_{A\to -\infty }{e}^{(1-s)A}$ must be equal to zero. Let $s=\sigma +j\omega$, where $\sigma$ is the real part of $s$ and $\omega$ is the imaginary part.

Then, $1-s=1-(\sigma +j\omega )=(1-\sigma )-j\omega$.

So, $e^{(1-s)A}=e^{((1-\sigma )-j\omega )A}=e^{(1-\sigma )A}{e}^{-j\omega A}$.

For ${\lim }_{A\to -\infty }{e}^{(1-s)A}$ to be zero, the term $e^{(1-\sigma )A}$ must tend to zero as $A\to -\infty$. This happens if and only if the exponent $(1-\sigma )A$ tends to $-\infty$. Since $A$ is approaching $-\infty$, for $(1-\sigma )A$ to be $-\infty$, the term $(1-\sigma )$ must be positive.

Therefore, we require:

$1-\sigma >0$

$1>\sigma$

So, the Region of Convergence (ROC) is $\text{Re}\{s\}<1$.

Final Laplace Transform Result

Given that the ROC is $\text{Re}\{s\}<1$, the limit term becomes zero:

$$\underset{A\to -\infty }{\lim }\frac{e^{(1-s)A}}{1-s}=0$$

Thus, the bilateral Laplace transform $F(s)$ is:

$$F(s)=\frac{1}{1-s}$$

This can also be written as:

$$F(s)=\frac{-1}{-(1-s)}=\frac{-1}{s-1}$$

Combining this with the Region of Convergence, the bilateral Laplace transform of $f(t)=e^{t}u(-t)$ is $\frac{-1}{s-1}$ with $\text{real part of }s<1$.

This matches the third option.