# [SOLVED]Region of Convergence

#### dwsmith

##### Well-known member
Consider the signal
$x(t) = e^{-5t}\mathcal{U}(t) + e^{-\beta t}\mathcal{U}(t)$
and denote the its Laplace transform by $$X(s)$$.
What are the constraints placed on the real and imagainary parts of $$\beta$$
if the region of convergence of $$X(s)$$ is $$\text{Re} \ \{s\} > -3$$?

Let's start by taking the Laplace transform.
\begin{align*}
X(s) &= \int_0^{\infty}e^{-5t}\mathcal{U}(t)e^{-st}dt +
\int_0^{\infty}e^{-\beta t}\mathcal{U}(t)e^{-st}dt\\
&= \int_0^{\infty}e^{-5t}e^{-st}dt +
\int_0^{\infty}e^{-\beta t}e^{-st}dt\\
&= \frac{1}{s + 5} + \frac{1}{s + \text{Re} \ \{\beta\}}
\end{align*}
The first integral gives us a region of convergence of $$\sigma > -5$$. For $$\beta$$, we need the real part to be $$3$$. Then the second integral would have a region of convergence of $$\sigma_1 > -3$$. With the addition of the integrals, how does that affect the overall region of convergence? Do we just take the lesser of the two or is there something else?

#### ThePerfectHacker

##### Well-known member
This is not exactly what you are asking but it is a good starting point as you had a similar issue in your other post involving convergence of the Laplace transform. This is to clear this issue up.

Let us start with a simple example. Consider the function $f(t) = e^{at}$.
Now compute the Laplace transform,
$$L[f(t)](s) = \int_0^{\infty} e^{at} e^{-st} ~ dt$$
Write $s = \sigma + ib$ then $e^{-st} = e^{-\sigma t}e^{-ibt}$, since $|e^{-ibt}|=1$ it follows that $|e^{-st}| = e^{-\sigma t}$.

If $\sigma > a$ then $|e^{at}e^{-st}| = e^{at}e^{-\sigma t} = e^{-(\sigma - a)t}$. Note that $\sigma - a > 0$. Then we have,
$$\int_0^{\infty} |e^{at}e^{-st}| ~ dt = \int_0^{\infty} e^{-(\sigma - a)t} ~ dt = -\frac{1}{\sigma - a} e^{-(\sigma - a)t} \bigg|_0^{\infty}$$
At limit $t=\infty$ the exponential goes to zero because its exponent is of the form $e^{-\infty}$, this has to do with the crucial fact that $\sigma - a > 0$. In particular, the integral $\int_0^{\infty} e^{at}e^{-st} ~ dt$ converges absolutely* for $\sigma > a$. Recall that $\sigma = \text{Re}(s)$ and so we have shown that the Laplace transform of $e^{at}$ converges absolutely for $\text{Re}(s) > a$.

*We say $\int_0^{\infty} g(t) ~ dt$ converges absolutely when $\int_0^{\infty} |g(t)| ~ dt$ converges. In particular it means the original integral without absolute values converges also.

#### dwsmith

##### Well-known member
This is not exactly what you are asking but it is a good starting point as you had a similar issue in your other post involving convergence of the Laplace transform. This is to clear this issue up.

Let us start with a simple example. Consider the function $f(t) = e^{at}$.
Now compute the Laplace transform,
$$L[f(t)](s) = \int_0^{\infty} e^{at} e^{-st} ~ dt$$
Write $s = \sigma + ib$ then $e^{-st} = e^{-\sigma t}e^{-ibt}$, since $|e^{-ibt}|=1$ it follows that $|e^{-st}| = e^{-\sigma t}$.

If $\sigma > a$ then $|e^{at}e^{-st}| = e^{at}e^{-\sigma t} = e^{-(\sigma - a)t}$. Note that $\sigma - a > 0$. Then we have,
$$\int_0^{\infty} |e^{at}e^{-st}| ~ dt = \int_0^{\infty} e^{-(\sigma - a)t} ~ dt = -\frac{1}{\sigma - a} e^{-(\sigma - a)t} \bigg|_0^{\infty}$$
At limit $t=\infty$ the exponential goes to zero because its exponent is of the form $e^{-\infty}$, this has to do with the crucial fact that $\sigma - a > 0$. In particular, the integral $\int_0^{\infty} e^{at}e^{-st} ~ dt$ converges absolutely* for $\sigma > a$. Recall that $\sigma = \text{Re}(s)$ and so we have shown that the Laplace transform of $e^{at}$ converges absolutely for $\text{Re}(s) > a$.

*We say $\int_0^{\infty} g(t) ~ dt$ converges absolutely when $\int_0^{\infty} |g(t)| ~ dt$ converges. In particular it means the original integral without absolute values converges also.
This doesn't help with this problem though. We can either look at the integrals separate or use the triangle inequality, but I don't see an advantage of using the triangle inequality since we will still have two integrals with different regions of convergence. How does the adding of two integrals with separate regions of convergence affect the overal region of convergence?

#### ThePerfectHacker

##### Well-known member
This doesn't help with this problem though. We can either look at the integrals separate or use the triangle inequality, but I don't see an advantage of using the triangle inequality since we will still have two integrals with different regions of convergence. How does the adding of two integrals with separate regions of convergence affect the overal region of convergence?
Laplace transform of $e^{t}$ converges for $\text{Re}(s) > 1$ and Laplace transform of $e^{2t}$ converges for $\text{Re}(s) > 2$. So if $\text{Re}(s) > 2$ then it is certainly bigger than $1$ and so it converges for both. More generally, you take the larger of the two numbers.