We derived the differentiation property previously using the inverse transform formula.

\begin{nicebox}
\begin{align*}
\frac{d}{dt} x(t) \ltp s\hat{X}(s) \\
\end{align*}
$RoC \supseteq RoC(x)$
\end{nicebox}

Now we can solve this another way. Suppose we have a differentiator:

$$
e^{st} \to \fbox{$\frac{d}{dt}$} \to se^{st}
$$

We get $x(t) = e^{st}$, and we get $y(t) = se^{st}$. We know that when we pass an exponential as input, we get the system function multiplied by that exponential. We get a scaled version of the same exponential:

\begin{align*}
y(t) &= \infint h(\tau) x(t-\tau) d\tau \\
y(t) &= \infint h(\tau) e^s{t-\tau} d\tau \quad \mbox{ let } x(t) = e^{st} \\
y(t) &= \lr{ \infint h(\tau) e^{-st} d\tau } e^{st} \\
y(t) &= \hat{H}(s) e^{st} \\ 
\end{align*}

Using this information, and the convolution property ($\hat{Y}(s) = \hat{H}(s)\hat{X}(s)$), we can see that $\hat{H}(s) = s$.

\begin{align*}
y(t) &= (x*h)(t) = \frac{d}{dt} x(t) \\
\hat{Y}(s) &= \hat{H}(s)\hat{X}(s) \\
\hat{Y}(s) &= s\hat{X}(s) \\
\end{align*}

A side note on the differentiator: The time domain representation is the \emph{unit doublet}. 

\begin{align*}
h(t) &= \dot{\delta}(t) = \frac{d}{dt} \delta(t) \\
\end{align*}


More on the Differentiation Property

The dual of differentiation in the time domain is differentiation in the Laplace domain. 

\begin{align*}
\hat{X}(s) &= \infint x(t) e^{-st} dt \\
\frac{d}{ds} \hat{X}(s) &= \frac{d}{s} \infint x(t) e^{-st} dt \\
\frac{d}{ds} \hat{X}(s) &= \infint x(t) \frac{d}{ds} e^{-st} dt \\
\frac{d}{ds} \hat{X}(s) &= \infint \lr{-tx(t)} e^{-st} dt \\
\end{align*}

This gives us the Frequency Differentiation property:

\begin{nicebox}
\begin{align*}
-tx(t) \ltp \frac{d}{ds}\hat{X}(s) \\
\end{align*}
$RoC = RoC(x)$
\end{nicebox}

\begin{example}

Given a transfer function $\hat{Y}(s) = \frac{1}{(s+a)^2}$, determine the time domain representation of the signal, $y(t)$. Assume the signal is causal.

\begin{align*}
\hat{Y}(s) &= \frac{1}{(s+a)^2} \\
\hat{Y}(s) &= -\frac{d}{ds}\frac{1}{(s+a)} \\
\end{align*}

We already know the inverse transform of the causal signal whose Laplace transform is $\frac{1}{s + a}$. This signal is given by $e^{-at}u(t)$, where the $RoC = \{ s \st \Real{s} \gt -\Real{a} \}$.

\begin{align*}
\iltrans{\hat{Y}(s)} &= -\lr{ -t e^{-at}u(t) } \\
y(t) &= t e^{-at}u(t) \\
\end{align*}

We know have a double pole at $-a$. But the $RoC$ stays the same.

\end{example}

Frequency Differentiation

Lets compare continuous time and discrete time representations of shifted delta functions:

\begin{align*}
\delta(n) &\ztp 1 & \delta(t) &\ltp 1 \\
\delta(n-N) &\ztp z^{-N}  & \delta(t-T) &\ltp e^{-sT}\\
\end{align*}

We can see that the $Z$-transform will give us a rational function in $z$ when we shift, whereas the Laplace Transform will not produce a rational function in $s$. A rational function in $s$ would be polynomial in $s$ over another polynomial in $s$:

$$
\frac{P(s)}{Q(s)} = \frac{ b_m s^m + b_{m-1} s^{m-1} + \cdots + b_1 s + b_0 }{a_n s^n + a_{n-1} s^{n-1} + \cdots + a_1 s + a_0}
$$

The shifting property of the Laplace Transform,
$$x(t-T) \ltp \hat{X}(s) e^{-sT}$$

tells us that we produce a transform which is not rational. Recall that in discrete time, a rational $Z$-transform can be produce by linear combinations of $\alpha^n u(n)$, $\alpha^n u(-n)$, $n^k \alpha^n u(n)$, $n^k \alpha^n u(-n)$, where $k \in \N$, and shifted versions thereof. Note that the differentiation property for Laplace Transforms, 

$$ \frac{d}{dt} x(t) \ltp s\hat{X}(s) $$

tells us that we can produce a rational transform in $s$. In continuous time, a rational Laplace transform can be produce by linear combinations of $e^{-\alpha t} u(t)$, $e^{-\alpha t}u(-t)$, $t^k e^{-\alpha t}u(t)$, $t^k, e^{-\alpha t}u(-t)$, where $k \in \N$, and differentiated versions thereof with respect to $t$. The only difference between CT and DT when producing rational transforms is we have shifted versions in DT and differentiated versions in CT.

\begin{example}

Consider two causal signals, one in discrete time, and one in continuous time: $y(n) = \alpha y(n-1) + x(n)$ and $\dot{y}(t) = -\alpha y(t) + x(t)$. One is a first order difference equation, the other a first order differential equation. 

\begin{align*}
y(n) &= \alpha y(n-1) + x(n) & \dot{y}(t) &= -\alpha y(t) + x(t) \\
\hat{Y}(z) &= \alpha z^{-1}\hat{Y}(z) + \hat{X}(z) & s\hat{Y}(s) &= -\alpha \hat{Y}(s) + \hat{X}(s) \\
\hat{H}(z) &= \frac{1}{1 - \alpha z^{-1}} & \hat{H}(s) &= \frac{1}{s + \alpha} \\
h(n) &= \alpha^n u(n) & h(t) &= e^{-\alpha t}u(t) \\
\end{align*}

\end{example

\begin{example}

Consider a first order RC-circuit. The voltage across its capacitor is governed by the following equation: $y(t) = \frac{1}{C} \int_{-\infty}^t g(\tau) d\tau$, where $g(t)$ is the current, and $C$ is the capacitance.

% \img{images/circuits/combo/combo.ps}


Ohm's law tells us that $I = \frac{V}{R}$, where $V$ is the voltage drop across the circuit, and $R$ is the resistance. In our case, $I = g(t)$, and $V = x(t) - y(t)$. This gives us $g(t) = \frac{x(t) - y(t)}{R}$. Now we can solve for $y(t)$:

\begin{align*}
y(t) &= \frac{1}{C} \int_{-\infty}^t g(\tau) d\tau \\
y(t) &= \frac{1}{C} \int_{-\infty}^t \frac{x(\tau) - y(\tau)}{R} d\tau \\
y(t) &= \frac{1}{RC} \int_{-\infty}^t x(\tau)d\tau  -  \frac{1}{C} \int_{-\infty}^t y(\tau) d\tau \\
\frac{d}{dt} y(t) &= \frac{d}{dt} \frac{1}{RC} \int_{-\infty}^t x(\tau)d\tau  -  \frac{d}{dt} \frac{1}{RC} \int_{-\infty}^t y(\tau) d\tau \\
\end{align*}

Now we have an equation we can solve using Laplace Transforms:

\begin{align*}
\dot{y}(t) &= \frac{1}{RC}x(t) - \frac{1}{RC}y(t) \\
s\hat{Y}(s) &= \frac{1}{RC} \hat{X}(s) - \frac{1}{RC} \hat{Y}(s) \\
\hat{Y}(s) \lr{ s + \frac{1}{RC}} &= \frac{1}{RC} \hat{X}(s)  \\
\hat{H}(s) &= \frac{\frac{1}{RC}}{ \lr{ s + \frac{1}{RC}}} \\
\end{align*}

So we find that the system function is $\hat{H}(t) = \frac{\frac{1}{RC}}{ \lr{ s + \frac{1}{RC}}} $ with $RoC(h) = \{ s \st \Real{s} \gt -\frac{1}{RC} \}$.

% \img{images/roc/laplace/circuit/causal.ps}


We know the time domain representation (assuming causal):

\begin{align*}
\hat{H}(s) &= \frac{\frac{1}{RC}}{ \lr{ s + \frac{1}{RC}}} \\
h(t) &= \frac{1}{RC} e^{-\frac{1}{RC}t}u(t) \\
\end{align*}

Here the $RoC$ includes the $i\omega$-axis, thus, 

$$
H(\omega) = \left. \hat{H}(s) \right|_{s = i\omega}
$$

We can find the magnitude response to determine what type of filter this circuit is:

$$
\abs{H(\omega)} = \frac{\frac{1}{RC}}{\abs{i\omega + \frac{1}{RC}}}
$$

% \img{images/roc/laplace/circuit/mag.ps}


At $i\omega = 0$, $\abs{H(\omega)}$ is at a maximum. As $\omega$ approaches $\pm \infty$, the magnitude is 0. This first order circuit is thus a low-pass filter.

% \img{images/graphs/ctlowpass/graph.ps}

\end{example}

Rational Laplace Transforms

Consider the causal system represented by the following $n^{th}$ order differential equation:

\begin{align*}
a_n \frac{d^n}{dt^n}y(t) + a_{n-1} \frac{d^{n-1}}{dt^{n-1}}y(t) + \cdots + a_1 \frac{d}{dt} y(t) + a_0 y(t) = b_n \frac{d^n}{dt^n}x(t) + b_{n-1} \frac{d^{n-1}}{dt^{n-1}}x(t) + \cdots + b_1 \frac{d}{dt} x(t) +  b_0 x(t) \\
\end{align*}

We can use the generalized differentiation property of the Laplace Transform:

\begin{nicebox}
$$
\frac{d^k}{dt^k} x(t) \ltp s^k \hat{X}(s) 
$$
\end{nicebox}

This allows us to solve for the system function $\hat{H}(s) = \frac{\hat{Y}(s)}{\hat{X}(s)}$ of an $n^{th}$ order differential equation:

\begin{align*}
\hat{H}(s) &= \frac{ b_n s^n + b_{n-1} s^{n-1} + \cdots + b_1 s + b_0}{a_n s^n + a_{n-1} s^{n-1} + \cdots + a_1 s + a_0}
\end{align*}

Generalized Differentiation

Given the following system function, determine all of the possible inverse Laplace Transforms:

$$
\hat{H}(s) = \frac{ (s-1) }{ (s+1)(s-2) } 
$$

We can see that the system has two poles, one at -1 and the other at 2. The system has one zero at 1, and the other at $\infty$. We know that the system can be broken up into components:

\begin{align*}
\hat{H}(s) = \frac{A}{s+1} + \frac{B}{s-2} \\
\end{align*}

We can find the coefficients using the Heaviside method:

\begin{align*}
A &= \left. (s+1)\hat{H}(s) \right|_{s = -1} = \frac{2}{3} \\
B &= \left. (s-2)\hat{H}(s) \right|_{s = 2} = \frac{1}{3} \\
\hat{H}(s) &= \frac{2/3}{s+1} + \frac{1/3}{s-2} \\
\end{align*}


We now have three cases with non-trivial regions of convergence. The first is where we have two right-sided signals. In this case $RoC = \{ s \st \Real{s} \gt 2 \}$:

% \imgsize{0.75}{images/roc/laplace/allpossible/causal.ps}

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This system is causal but not BIBO stable. Here the possible inverse transform is 

\begin{align*}
h(t) &= \frac{2}{3} e^{-t}u(t) + \frac{1}{3} e^{2t}u(t) \\
\end{align*}

Here we have $ \frac{1}{3} e^{2t}u(t)$ is the unstable term. The next possibility is where we have two left-sided signals. In this case $RoC =  \{ s \st \Real{s} \lt -1 \}$:

% \imgsize{0.75}{images/roc/laplace/allpossible/anticausal.ps}

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This system is anti-causal and not BIBO stable. Here the inverse transform is

\begin{align*}
h(t) &= -\frac{2}{3} e^{-t}u(-t) - \frac{1}{3} e^{2t}u(-t) \\
\end{align*}

In this case we have $- \frac{1}{3} e^{2t}u(-t)$ is the stable term. The last possibility is where we have right and left sided signals whose regions of convergence have a non-trivial overlap. In this case $RoC = \{ s \st -1 \lt \Real{s} \lt 2 \}$:

% \imgsize{0.75}{images/roc/laplace/allpossible/leftright.ps}

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This system is two-sided and is BIBO stable. Here the inverse transform is 

\begin{align*}
h(t) &= \frac{2}{3} e^{-t}u(t) - \frac{1}{3} e^{2t}u(-t) \\
\end{align*}

Now both terms are stable, which can be verified by the fact that the $RoC$ includes the $i\omega$-axis implying BIBO stability.


Example

Assume that we have a causal, stable system that is rational in $s$. If we pass a complex exponential as input, we know that we get the frequency response of the system multiplied by the same exponential:

$$
x(t) = e^{i\omega t} \to \fbox{$H$} \to y(t) = H(\omega) e^{i\omega t}
$$

What is we apply the toneburst, $e^{i\omega t} u(t)$, a suddenly-applied sinusoid? The system is describable by a LCCDE, 

$$
\hat{H}(s) = \frac{ B(s) }{ A(s) } = \frac{ B(s) }{ (s-p_1) (s-p_2) \cdots (s-p_n) } 
$$

The stability of the system implies that all of the poles are to the left of the $i\omega$-axis. In other words, the stability implies $\Real{p_\ell} \lt 0 $ $ \forall \ell$. We can now apply the toneburst to this system. We know that the transform of the toneburst is given by

$$
e^{i\omega_0 t} u(t) \ltp \frac{1}{s - i\omega_0} \quad \Real{s} \gt 0 
$$ 

We also know that convolution in the time domain corresponds to multiplication in the frequency domain, thus

\begin{align*}
\hat{Y}(s) &= \hat{H}(s) \hat{X}(s) \\
\hat{Y}(s) &= \frac{ B(s) }{ A(s) } = \frac{ B(s) }{ (s-p_1) (s-p_2) \cdots (s-p_n)(s - i\omega_0) } \quad RoC \supseteq RoC(h) \cap RoC(x) \\
\end{align*}

The new region of convergence is $\Real{s} \gt 0$. Lets apply partial fraction decomposition to determine the time domain expression. For simplicity, assume that there are no duplicate poles.

\begin{align*}
\hat{Y}(s) &= \frac{R_1}{s - p_1} + \frac{R_2}{s - p_2} + \cdots + \frac{R_n}{s - p_n} + \frac{Q}{s - i\omega_0} \\
y(t) &= R_1 e^{p_1 t}u(t) + R_2 e^{p_2 t} u(t) + \cdots + R_n e^{R_n t} u(t) + Q e^{i\omega_0 t} u(t)  \\
\end{align*}

Since all poles $P_\ell$ were to the left of the $i\omega$-axis, and the pole from $Q e^{i\omega_0 t} u(t)$ is the $i\omega$-axis, we can see why the intersection of the regions of convergence is $\{ s \st \Real{s} \gt 0 \}$; the rightmost pole is zero. In this case, all of the poles that were to the left of the $i\omega$-axis are apart of the \emph{transient response}. This is because all $p_\ell \lt 0 \quad \forall \ell$, so the terms associated with these poles decay.

\begin{align*}
y_{tr}(t) &= R_1 e^{p_1 t}u(t) + R_2 e^{p_2 t} u(t) + \cdots + R_n e^{R_n t} u(t) \\
y_{ss}(t) &=  Q e^{i\omega_0 t} u(t) \\
\end{align*}

When we look at the transient response, we see that the terms emanate from the stable poles of the system. We can see that the \emph{steady state response} is due to the pole of the input. The steady state response's pole is to the right of the right-most pole of the transient state's. When you introduce an input with a pole to the right of the systems right-most pole it then becomes steady state response. \\
 
How can we find $Q$? 

\begin{align*}
\hat{Y}(s) &= \frac{R_1}{s - p_1} + \frac{R_2}{s - p_2} + \cdots + \frac{R_n}{s - p_n} + \frac{Q}{s - i\omega_0} \\
\hat{Y}(s) &= \sum \limits_{\ell = 1}^{n} \frac{R_\ell}{s - p_\ell} + \frac{Q}{s - i\omega_0} \\
(s - i\omega_0)\hat{Y}(s) &= \sum \limits_{\ell = 1}^{n} \frac{R_\ell(s-i\omega_0)}{s - p_\ell} + Q \\
\end{align*}

If we evaluate $\hat{Y}(s)(s - i\omega_0)$, we can get the coefficient $Q$:

\begin{align*}
Q &= \left. (s - i\omega_0)\hat{Y}(s) \right|_{s = i\omega_0} \\
Q &= \left. \frac{ B(s) }{ (s-p_1) (s-p_2) \cdots (s-p_n) } \right|_{s = i\omega_0} \\
Q &= \left.\hat{H}(s) \right|_{s = i\omega_0} \\
Q &= H(\omega_0) \\
\end{align*}

Now we have our steady state response 
\begin{align*}
y_{ss}(t) &=  Q e^{i\omega_0 t} u(t) \\
y_{ss}(t) &=  H(\omega_0) e^{i\omega_0 t} u(t) \\
\end{align*}

If we apply a one-sided exponential to the system, $e^{i\omega_0 t} u(t)$, as $t$ becomes large enough, the system cannot distinguish between the one-sided exponential and the infinite duration exponential, $e^{i\omega_0 t}$. The coefficient of the steady state response should then be the frequency response of the system evaluated at the frequency of the input; this is what we knew from before when applying the infinite duration exponential as input to an LTI system:

\begin{align*}
e^{i\omega_0 t} \to \fbox{$H$} \to H(\omega_0) e^{i\omega_0 t}
\end{align*}


In other words, if $y(t) = y_{tr}(t) + y_{ss}(t)$, then as time goes on, we have

\begin{align*}
\lim_{t \to \infty} y(t) = y_{ss}(t) = e^{i\omega_0 t}u(t) = e^{i\omega_0 t}
\end{align*}

If $\omega_0 = 0$, then $x(t) = u(t)$, and then we have $y_{ss}(t) = H(0)u(t) = H(0)$. We find that we get the DC gain of the system. 


Steady State and Transient Responses of CT-LTI Systems

<h4>Example</h4>

Consider an example model of a DC motor. The system is based on simple harmonic motion; there is inertia $M$, and damping $D$. The system function is given by 

\begin{align*}
\hat{G}(s) &= \frac{1}{Ms^2 + Ds} = \frac{1/M}{s(s + D/M)} \\
\end{align*}

For this example, consider $M = 1$, and $D=16$. 


(a) is this system BIBO?


The $i\omega$-axis is not in the region of convergence. The right-most pole of the system happens to be on the $i\omega$-axis. 


(b) what is the LCCDE of the system?


\begin{align*}
\hat{G}(s) &= \frac{ \hat{Y}(s) }{ \hat{Q}(s) } \\
\frac{ \hat{Y}(s) }{ \hat{Q}(s) } &= \frac{1}{s^2 + 16} \\
\lr{s^2 + 16} \hat{Y}(s)  &= \hat{Q}(s) \\
\end{align*}

Recall the differentiation property of the laplace transform:

\begin{nicebox}
\begin{align*}
\frac{d}{dt} x(t) \ltp s\hat{X}(s) \\
\end{align*}
$RoC \supseteq RoC(x)$
\end{nicebox}

We can use this to solve for the time domain expression:

\begin{align*}
\lr{s^2 + 16} \hat{Y}(s)  &= \hat{Q}(s) \\
s^2\hat{Y}(s) + 16\hat{Y}(s)  &= \hat{Q}(s) \\
\ddot{y}(t) + 16\dot{y}(t) &= q(t) \\
\end{align*}

(c) Stabilizing the system via the feedback loop in the image below, determine the overall system function $\hat{H}(s)$


% image of feedback
% \img{images/feedback/motor/motor.ps}

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\rput(1.5,-1.75){$H$}

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$x$ is the desired position of the system, $y$ is the actual position of the system, and $\epsilon$ is the error signal, given by $x - y$. This system is called a proportional feedback controller.

\begin{align*}
\ltrans{\epsilon(t)} \hat{K}(s) \hat{G}(s) &= \hat{Y}(s) \\
\ltrans{ x(t) - y(t) } \hat{K}(s) \hat{G}(s) &= \hat{Y}(s) \\
\hat{X}(s) \hat{K}(s) \hat{G}(s) - \hat{Y}(s) \hat{K}(s) \hat{G}(s) &= \hat{Y}(s) \\
\hat{X}(s) \hat{K}(s) \hat{G}(s) &= \hat{Y}(s) ( 1 +  \hat{K}(s) \hat{G}(s) ) \\
\hat{H}(s) &= \frac{ \hat{K}(s) \hat{G}(s)  }{  1 +  \hat{K}(s) \hat{G}(s)  } \\ 
\end{align*}

In this case, we have $\hat{K}(s) = K \gt 0$. 

\begin{align*}
\hat{H}(s) &= \frac{ K \hat{G}(s)  }{  1 +  K \hat{G}(s)  } \\ 
\hat{H}(s) &= \frac{ K \frac{1}{s^2 + 16} }{ ( 1 +  K \frac{1}{s^2 + 16} ) } \\ 
\hat{H}(s) &= \frac{K}{s^2 + 16s + K} \\
\end{align*}

Another way to solve for this is using Blacks Formula, $\frac{\mbox{Forward Gain}}{1 - \mbox{Loop Gain}}$:

\begin{align*}
\frac{\mbox{Forward Gain}}{1 - \mbox{Loop Gain}} \\
\frac{K \hat{G}(s) }{1 + K\hat{G}(s) } \\
\end{align*}



(c) Is $H$ guaranteed to be BIBO for $K \gt 0$?



We need to look at the poles of the system. We can determine the poles by finding the roots of $s^2 + 16s + K$, using the quadratic formula:

\begin{align*}
\frac{-16 \pm \sqrt{ 16^2 - 4K} } { 2} \\
-8 \pm \sqrt{ 64 - K }
\end{align*}

We have two cases that we need to analyze:


I $64 - K \geq 0$



Here we have $64 - K \in \R$ and $\sqrt{ 64 - K} \lt 8$. This means that $-8 \pm \sqrt{ 64 - K} $ is always less than 0, or to the left of the $i\omega$-axis.


II  $64 - K \lt 0$.



In this case we have $\sqrt{ 64 - K} \in \C$, and we have $-8 \pm i\sqrt{K - 64}$. This is to the left of the $i\omega$-axis. 





In all cases, we have a causal, stable signal whose poles are always to the left of the $i\omega$-axis.



(e) if $x(t) = u(t)$, what is $\hat{Y}(s)$?



\begin{align*}
\hat{Y}{s} &= \hat{H}(s)\hat{X}(s) \\
\hat{Y}(s) &= \frac{ K } { (s^2 + 16s + K)s } \\
\end{align*}




(f) Let $K = 63$, what is the coefficient of the steady state response?




\begin{align*}
\hat{Y}(s) &= \frac{63}{s(s+7)(s+9)} \\
\hat{Y}(s) &= \frac{A}{s} + \frac{B}{(s+7)} + \frac{C}{(s+9)} \\
A &= \left. s \hat{Y}(s) \right|_{s = 0} = \hat{H}(0) = 1 \\
\end{align*}



Example DC Motor

\begin{nicebox}
{\bf Time-Shifting (Translation) Property }
\begin{align*}
x(t+T) &\ltp e^{sT}\hat{X}(s) \\
\end{align*}
The $RoC$ excludes $\infty$ if the signal is not causal, or excludes $-\infty$ if the signal is not anti-causal. 
\end{nicebox}

\begin{nicebox}
{\bf Convolution in the Time Domain }
\begin{align*}
h(t) &= f(t) * g(t) \\
\hat{H}(s) &= \hat{F}(s)\hat{G}(s) \\
\end{align*}
The $RoC(h) \supseteq RoC(f) \cap RoC(g)$.
\end{nicebox}

\begin{nicebox}
{\bf Modulation with a Complex Exponential }
\begin{align*}
e^{s_0 t} x(t) &\ltp \hat{X}(s-s_0) \\
\end{align*}
$RoC = RoC(x) + \Real{s_0} $
\end{nicebox}

\begin{nicebox}
{\bf Integration in the Time Domain }
\begin{align*}
 \int_{-\infty}^t x(\tau) d\tau &\ltp \frac{\hat{X}(s)}{s}
\end{align*}
$RoC(y) \supseteq RoC(x) \cap \{ s \st \Real{s} \gt 0 \} $
\end{nicebox}


\begin{nicebox}
{\bf Differentiation in the Time Domain }
\begin{align*}
\frac{d}{dt} x(t) \ltp s\hat{X}(s) \\
\end{align*}
$RoC \supseteq RoC(x)$
\end{nicebox}

\begin{nicebox}
{\bf Generalized Differentiation in the Time Domain }
\begin{align*}
\frac{d^k}{dt^k} x(t) \ltp s^k \hat{X}(s) \quad k \in \N \\
\end{align*}
$RoC \supseteq RoC(x)$
\end{nicebox}

\begin{nicebox}
{\bf Frequency Differentiation }
\begin{align*}
-tx(t) \ltp \frac{d}{ds}\hat{X}(s) \\
\end{align*}
$RoC = RoC(x)$
\end{nicebox}

\begin{nicebox}
{\bf Conjugation Property }
\begin{align*}
x^*(t) &\ltp \hat{X}^*(s^*)
\end{align*}
$RoC = RoC(x)$
\end{nicebox}

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