We talked about splitting up an output into a transient and steady state representations. Now we are going to talk about zero-input and zero-state responses of a system. Consider the following system:

\begin{align*}
y(n) - 0.9y(n-1) = x(n) \quad \lr{\mbox{system is causal}}\\
\end{align*}
 The order of the system is the number of delay elements required for implementation. So we need one initial condition (in this case) to determine the response. What is $y(-1) \neq 0$? In other words, this means that you could pass in the zero signal $x(n) = 0$ and you won't get a zero output; the system is not ZIZO. 

\begin{problem}

What is the zero input response of the system?

\end{problem}


\begin{solution}

\begin{align*}
\mbox{let } x(n) = 0 \quad n \geq 0 \\
\end{align*}

The system's causality implies that we should write $y(n)$ in terms of its past values and solve in the forward direction ( $n=0,1,2,\dots$ ):

\begin{align*}
y(n) = 0.9y(n-1) + x(n) \\
\end{align*}

Remember that we have let $x(n) = 0$ $ \forall $ $n \geq 0$. 

\begin{align*}
y(0) &= (0.9)   y(-1) \\
y(1) &= (0.9)^2 y(-1) \\
y(2) &= (0.9)^3 y(-1) \\
&\vdots  \\
y(n) &= (0.9)^{n+1} y(-1) = y_{ZIR}(n)\\
\end{align*}

This result is called the \emph{zero-input response}, or zero-input solution (homogeneous solution). We have turned off the input but don't get a zero output.

\end{solution}


\begin{problem}

What if $y(-1) = 0$ and $x(n) = u(n)$? In other words, what is the zero-state response of the system?

\end{solution}

\begin{solution}

First we need to find the transform everything into the transform domain, and find the equation for the system. We then pass a unit step through this system.

\begin{align*}
y(n) &= 0.9y(n-1) + x(n) \quad \abs{z} \gt 0.9 \\
\hat{Y}(z) &= z^{-1}0.9\hat{Y}(z) + \hat{X}(z) \\
\hat{Y}(z)(1 - 0.9z^{-1}) &= \hat{X}(z) \\
\hat{H}(z) &= \frac{1}{1 - 0.9z^{-1}} \\
\hat{H}(z) &= \frac{z}{z - 0.9} \\
\end{align*}

We know that the Z-transform of the unit step:

\begin{align*}
\hat{X} &= \frac{1}{1 - z^{-1}} \quad \abs{z} \gt 1 \\ 
\hat{X} &= \frac{z}{z - 1} \quad \abs{z} \gt 1 \\ 
\end{align*}

Convolution in the time domain is multiplication in the Z-domain:

\begin{align*}
\hat{Y}(z) &= \hat{H}(z) \hat{X}(z) \\
\hat{Y}(z) &= \frac{z^2}{(z - 0.9)(z-1)} \quad \abs{z} \gt 1 \\
\end{align*}

Now we break $\hat{Y}(z)$ into its components: 
\begin{align*}
\hat{Y}(z) &= \frac{z^2}{(z - 0.9)(z-1)} \\
\frac{\hat{Y}(z)}{z} &= \frac{A}{(z-0.9)} + \frac{B}{(z-1)} \\
\frac{z}{(z - 0.9)(z-1)} &= \frac{A}{(z-0.9)} + \frac{B}{(z-1)} \\
A &= \left. \frac{z}{z-1} \right|_{z=0.9} = -9\\
B &= \left. \frac{z}{z-0.9} \right|_{z=1} = 10\\
\end{align*}

Now putting it all together:
\begin{align*}
\frac{\hat{Y}(z)}{z} &= \frac{A}{(z-0.9)} + \frac{B}{(z-1)} \\
\frac{\hat{Y}(z)}{z} &= \frac{-9}{(z-0.9)} + \frac{10}{(z-1)} \\
\hat{Y}(z) &= \frac{-9z}{(z-0.9)} + \frac{10z}{(z-1)} \\
y(n) &=-9(0.9)^n u(n) +10 u(n) = y_{ZSR}(n) \\
\end{align*}

This result is the \emph{zero-state response} of the system, (a.k.a. the particular solution).

\end{solution}

\begin{problem}

What if $y(-1) = 0$ and $x(n) = u(n)$? 

\end{problem}


\begin{solution}

By superposition, the general solution is the sum or the homogenous and particular solution.
\begin{align*}
y(n) &= y_{\small ZSR}(n) + y_{ZIR}(n) \\
\end{align*}

Is there a way to solve for all of these in one step? Start off by truncating both sides of the equation:
\begin{align*}
y(n) &= 0.9y(n-1) + x(n) \\
y(n)u(n) &= 0.9y(n-1)u(n) + x(n)u(n) \\
\end{align*}

Then Z-transform both sides:

\begin{align*}
\sum \limits_{n=-\infty}^{\infty} y(n)u(n) z^{-n} &= \sum \limits_{n=-\infty}^{\infty}0.9y(n-1)u(n) z^{-n} + \sum \limits_{n=-\infty}^{\infty}x(n)u(n) z^{-n} \\
\sum \limits_{n=0}^{\infty} y(n)z^{-n} &= \sum \limits_{n=0}^{\infty}0.9y(n-1) z^{-n} + \sum \limits_{n=0}^{\infty}x(n) z^{-n} \\
\end{align*}

We need to add a new definition to our toolbox, the \emph{Unilateral Z-transform}:
\begin{nicebox}
\begin{align*}
\mathcal{Y}(z) \bydef \sum \limits_{n=0}^{\infty} y(n) z^{-n}
\end{align*}
\end{nicebox}


This is used in contrast to the \emph{Bilateral Z-transform} which is what we have been using up until this point.

\begin{align*}
\sum \limits_{n=0}^{\infty} y(n)z^{-n} &= 0.9\sum \limits_{n=0}^{\infty}y(n-1) z^{-n} + \sum \limits_{n=0}^{\infty}x(n) z^{-n} \\
\mathcal{Y}(z) &= 0.9\sum \limits_{n=-1}^{\infty}y(m) z^{-(m+1)} + \mathcal{X}(z) \quad \mbox{let } m = n -1\\
\mathcal{Y}(z) &=  z^{-1}0.9\sum \limits_{m=-1}^{\infty}y(m) z^{-m} + \mathcal{X}(z) \\
\mathcal{Y}(z) &= 0.9y(-1) + z^{-1}0.9\sum \limits_{m=0}^{\infty}y(m) z^{-m} + \mathcal{X}(z) \quad \mbox{ (evaluate first term in sum)}  \\
\mathcal{Y}(z) &= 0.9y(-1) + z^{-1}0.9\mathcal{Y}(z) + \mathcal{X}(z)  \\
\end{align*}

We have successfully captured the initial conditions, $y(-1)$ in the formula. Note that for causal systems, $\hat{X}(z) = \mathcal{X}(z)$, the bilateral Z-transform is equal to the unilateral Z-transform.o

\begin{align*}
\mathcal{Y}(z) &= 0.9y(-1) + z^{-1}0.9\mathcal{Y}(z) + \mathcal{X}(z)  \\
\mathcal{Y}(z)(1 - 0.9z^{-1}) &= 0.9y(-1) + \mathcal{X}(z)  \\
\mathcal{Y}(z) &= \frac{0.9y(-1)}{(1 - 0.9z^{-1})} + \frac{1}{(1 - 0.9z^{-1})}\mathcal{X}(z)  \\
\mathcal{Y}(z) &= \frac{0.9y(-1)}{(1 - 0.9z^{-1})} + \hat{H}(z)\mathcal{X}(z)  \\
\mathcal{Y}(z) &= \frac{0.9y(-1)}{(1 - 0.9z^{-1})} + \hat{H}(z)\hat{X}(z)  \\
\mathcal{Y}(z) &= \frac{0.9y(-1)}{(1 - 0.9z^{-1})} + \hat{Y}_{ZSR}(z)  \\
\mathcal{Y}(z) &= \hat{Y}_{ZIR}(n) + \hat{Y}_{ZSR}(z)  \\
\end{align*}

Now we can transform these responses back to the time domain:
\begin{align*}
\hat{Y}_{ZIR}(z) &= \frac{0.9y(-1)}{(1 - 0.9z^{-1})}  \\
y_{ZIR}(n) &= 0.9y(-1)(0.9)^n u(n) \\
y_{ZIR}(n) &= y(-1)(0.9)^{n+1} u(n) \\
\end{align*}

\begin{align*}
\hat{Y}_{ZSR}(z) &= \hat{H}(z)\hat{X}(z) \\
y_{ZSR}(n) &= -9(0.9)^n u(n) + 10 u(n) \\
\end{align*}

Now for the final result:

$$
y(n) = y(-1)(0.9)^{n+1} u(n) - 9(0.9)^n u(n) + 10 u(n) 
$$

Here we find all of the various responses:

\begin{align*}
y_{ZIR}(n) &= y(-1)(0.9)^{n+1} u(n) \\
y_{ZSR}(n) &= -9(0.9)^n u(n) + 10 u(n) \\
y_{tr}(n) &= y(-1)(0.9)^{n+1} u(n) -9(0.9)^n u(n)\\
y_{ss}(n) &= 10 u(n) \\
\end{align*}

Note that the steady-state response is equal to the transfer function evaluated at $z = 1$:

\begin{align*}
y_{ss}(n) &= \left. \hat{H}(z) \right|_{z=1} = 10 u(n)
\end{align*}

The natural response of the system, the contribution from the pole (which is inside of the unit circle), dies out. What remains is the response from the input $x(n) = 1$ $\forall n$. In other words, $H$ is BIBO, so the natural response or response due to the poles of the system decays, so for $n$ sufficiently large, the system can't distinguish between $x(n) = 1$ $\forall n$ and $x(n) = u(n)$.

\end{solution}


\begin{example}


Given the following system, find zero-input response.
\begin{align*}
y(n) - \alpha y(n-1) + \beta y(n-2) &= x(n) \\
\end{align*}

Take the unilateral Z-transform:

\begin{align*}
\mathcal{Y}(z) - \alpha \sum \limits_{n=0}^{\infty} y(n-1)z^{-n} + \beta \sum \limits_{n=0}^{\infty} y(n-2)z^{-n} = \mathcal{X}(z) \\
\mathcal{Y}(z) - \alpha z^{-1} \sum \limits_{m=-1}^{\infty} y(m)z^{-m} + \beta z^{-2} \sum \limits_{\ell=-2}^{\infty} y(\ell)z^{-\ell} = \mathcal{X}(z) \\
\mathcal{Y}(z) - \alpha y(-1) - \alpha z^{-1} \sum \limits_{m=0}^{\infty} y(m)z^{-m} + \beta y(-2) + \beta y(-1) z^{-1} + \beta z^{-2} \sum \limits_{\ell=0}^{\infty} y(\ell)z^{-\ell} = \mathcal{X}(z) \\
\mathcal{Y}(z) - \alpha y(-1) - \alpha z^{-1} \mathcal{Y}(z) + \beta y(-2) + \beta y(-1) z^{-1} + \beta z^{-2} \mathcal{Y}(z) = \mathcal{X}(z) \\
\end{align*}

Since we are finding the zero-input response (homogeneous solution), we set $x(n) = 0 \ztp 0$:
\begin{align*}
\mathcal{Y}(z) - \alpha y(-1) - \alpha z^{-1} \mathcal{Y}(z) + \beta y(-2) + \beta y(-1) z^{-1} + \beta z^{-2} \mathcal{Y}(z) &= 0 \\
\mathcal{Y}(z)\lr{ 1 - \alpha z^{-1} + \beta z^{-2}  }  &= \alpha y(-1) - \beta y(-2) - \beta y(-1) z^{-1} \\
\mathcal{Y}(z)  &= \frac{\alpha y(-1) - \beta y(-2) - \beta y(-1) z^{-1}}{\lr{ 1 - \alpha z^{-1} + \beta z^{-2}  }} \\
y_{ZIR}(n)  &= \iztrans{\frac{\alpha y(-1) - \beta y(-2) - \beta y(-1) z^{-1}}{\lr{ 1 - \alpha z^{-1} + \beta z^{-2}  }}} \\
\end{align*}

\end{example}


Zero-State and Zero-Input Responses of LTI Systems

For the same reasons we discussed Z-transforms as a way to manipulate functions where the Fourier Transform could not (in discrete time), we now discuss the Laplace Transform for continuous time.

\begin{nicebox}
The \emph{Laplace Transform} of a function $h(t)$ is defined as 
\begin{align*}
\hat{H}(s) &= \int_{-\infty}^{\infty} h(t) e^{-st} dt \quad s \in \C \\
\end{align*}
where $ s = \sigma + i\omega$. The associated Region of Convergence is
\begin{align*}
RoC(h) &\bydef \{ s = \sigma + i\omega \in \C \st \int_{-\infty}^{\infty} \abs{h(t)e^{-st}} dt \lt \infty \} \\
\end{align*}
\end{nicebox}


We can now handle systems that don't have Fourier Transforms. Because $s = \sigma + i\omega$, we can ``veer off'' the $i\omega$ axis:

\begin{align*}
\hat{H}(s) &= \int_{-\infty}^{\infty} h(t) e^{-st} dt \\
\hat{H}(s) &= \int_{-\infty}^{\infty} h(t) e^{-(\sigma + i\omega)t} dt \\
\hat{H}(s) &= \int_{-\infty}^{\infty} h(t) e^{-\sigma t} e^{-i\omega t} dt \\
\hat{H}(s) &= \ftrans{h(t) e^{-\sigma t}} \\
\end{align*}

If $h(t)$ is growing exponentially, we can tame that growth with $e^{-\sigma t}$. In the definition of the Laplace transform, we see that $h(t)e^{-st}$ must be absolutely integrable. Does the $\omega$ is $s = \sigma + i\omega$ matter?

\begin{align*}
&\int_{-\infty}^{\infty} \abs{h(t)e^{-st}} dt \\
&\int_{-\infty}^{\infty} \abs{h(t)e^{-\sigma t}}\abs{e^{-i\omega t}} dt \\
&\int_{-\infty}^{\infty} \abs{h(t)e^{-\sigma t}} dt \\
\end{align*}

Clearly only $\sigma$ matters for absolute integrability since $e^{-i\omega t} = 1$. So what determines the $RoC$ is $\sigma = \Real{s}$.

\begin{claim}
Right-sided signals have right-sided $RoC$s.
\end{claim}

\begin{proof}

Let $x$ be right-sided such that $x(t) = 0 $ $ \forall t \lt T$. Then
\begin{align*}
\hat{X}(s) &= \int_{T}^{\infty}e^{-st} \\
\hat{X}(s) &= \int_{T}^{0}e^{-st} + \int_{0}^{\infty}e^{-st} \\
\end{align*}

We only care about the integral with limits from 0 to $\infty$, since the other integral is finite and thus converges.

\begin{align*}
&\mbox{Suppose } \sigma_0 \in RoC(x) \\
&\Rightarrow \int_0^\infty \abs{x(t) e^{-\sigma_0 t} } dt \lt \infty \\
\end{align*}

Let $\sigma_1 \gt \sigma_0$ be arbitrary. Then we have the following relations:

\begin{align*}
\sigma_1 &\gt \sigma_0 \\
-\sigma_1 &\lt -\sigma_0 \\
-\sigma_1t &\lt -\sigma_0t \quad (\mbox{for } t \in (0,\infty))\\
e^{-\sigma_1t} &\lt e^{-\sigma_0t} \\
\end{align*}

Therefore $e^{-\sigma_1t}$ decays faster than $e^{-\sigma_0t}$. Now we use this to show how this affects absolute integrability:

\begin{align*}
e^{-\sigma_1t} &\lt e^{-\sigma_0t} \\
\abs{x(t)e^{-\sigma_1t}} &\lt \abs{x(t)e^{-\sigma_0t}} \\
\int_0^\infty \abs{x(t)e^{-\sigma_1t}} dt &\lt \int_0^\infty \abs{x(t)e^{-\sigma_0t}} dt \lt \infty \\
\int_0^\infty \abs{x(t)e^{-\sigma_1t}} dt &\lt \infty \\
\end{align*}

Therefore if $\sigma_1 \gt \sigma_0$ and $\sigma_0 \in RoC(x)$, then $\sigma_1 \in RoC(x)$.

\end{proof}

Below is the $RoC$ of a right-sided signal. In this case the $RoC$ includes the $i\omega$-axis, or $\Imag{s}$-axis:

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Here is an $RoC$ for an left-sided system. In this case the $RoC$ includes the $i\omega$-axis:

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\psline{->}(0,-2.5)(0,2.5)

\end{pspicture}
\end{center}


Here is an example $RoC$ for a two-sided system:

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\rput(-1,-2.4){$\sigma_0$}

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\rput(2.5,0.3){ $\Real{s}$ }
\psline{->}(-2.5,0)(2.5,0)

% y-axis
\rput(0.6,2.5){ $\Imag{s}$ }
\psline{->}(0,-2.5)(0,2.5)

\end{pspicture}
\end{center}


\begin{example}

Find the Laplace Transform for the function $h(t) = e^{-2t} u(t)$.
% \img{images/graphs/expun/graph.ps}

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\begin{align*}
\hat{H}(s) &= \int_{-\infty}^{\infty} e^{-2t} u(t) e^{-st} dt \\
\hat{H}(s) &= \int_{0}^{\infty} e^{-2t}e^{-st} dt \\
\hat{H}(s) &= \int_{0}^{\infty} e^{-(2+s)t} dt \\
\hat{H}(s) &= \left. \frac{e^{-(2+s)t}}{-(2+s)} \right|_{0}^{\infty} \\
\hat{H}(s) &= \lim_{t \to \infty} \frac{e^{-(2+s)t}}{-(2+s)} - \frac{1}{-(2+s)} \\
\hat{H}(s) &= \frac{1}{(2+s)} \\
\end{align*}

In order for absolute integrability, $\Real{ -(2+s) } \lt 0$. Otherwise the limit in the above formulae would not have converged to zero as $t$ approaches $\infty$. In other words,
\begin{align*}
\Real{-(2+s)} \lt 0 \\
-2 - \sigma \lt 0 \\
-2 \lt \sigma \\
\end{align*}

So $RoC(h) : \{ s \st \Real{s} \gt -2 \}$:

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\psline{->}(0,-2.5)(0,2.5)

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Notice that the Fourier Transform for this function is related to the Laplace Transform because the $RoC$ includes the $i\omega$-axis:

\begin{align*}
\ftrans{h(t)} &= \frac{1}{2 + i\omega} \\
\ltrans{h(t)} &= \frac{1}{2 + s} \\
\ftrans{h(t)} &= \left. \ltrans{h(t)} \right|_{s = i\omega} \\
\end{align*}

\end{example}

The Laplace Transform

What if we shift a signal in time? What happens to its Laplace Transform? 

\begin{align*}
y(t) &= x(t + T) \quad T \gt 0 \\
\hat{Y}(s) &= \int_{-\infty}^{\infty} x(t + T)e^{-st}dt  \\
\hat{Y}(s) &= \int_{-\infty}^{\infty} x(t)e^{-s(t-T)}dt \quad \mbox{let } t = t + T \\
\hat{Y}(s) &= e^{sT}\int_{-\infty}^{\infty} x(t)e^{-st}dt \\
\hat{Y}(s) &= e^{sT}\hat{X}(s) \\
\end{align*}

This result is the Shifting Property of the Laplace Transform:

\begin{nicebox}
\begin{align*}
x(t) &\ltp \hat{X}(s) \\
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{example}

Consider the causal signal $x(t) = e^{-3t}u(t)$. What is the Laplace Transform for $x(t+T)$? 
\begin{align*}
y(t) &= x(t+T) \\
y(t) &= e^{-3(t+T)}u(t+T) \\
\end{align*}

Note that the signal is no longer causal, although it is still right-sided. We know the transform for $x(t)$ and can utilize the shifting property:
\begin{align*}
x(t)&\ltp \hat{X}(s) = \frac{1}{3+s} \\
x(t+T) &\ltp e^{sT}\hat{X}(s) = \frac{e^{sT}}{3+s} \\
\end{align*}

The $RoC$ is the same in this case, except that it excludes $\infty$:
$$
RoC(y) : \{ s \st -3 \lt \Real{s} \lt \infty \} 
$$ 

\end{example}

The Shifting Property

Lets consider the signal whose Laplace Transform is $\hat{X}(s) = \frac{1}{s + 3}$. If the signal is causal, then we have

\begin{align*}
x(t) &= e^{-3t}u(t) \\
RoC(x) &= \{ s \st \Real{s} \gt -3 \} \\
\end{align*}

This signal decays as $t$ approaches $\infty$. This is a stable signal, which can be confirmed by the fact that the signal's region of convergence includes the $i\omega$-axis. If this signal signal was anti-causal, then we have

\begin{align*}
x(t) &= -e^{-3t}u(-t) \\
RoC(x) &= \{ s \st \Real{s} \lt -3 \} \\
\end{align*}

Notice that the region of convergence is flipped, and doesn't contain the $i\omega$-axis. This signal is unstable. This brings us to the statement:

\begin{claim}
$x$ is a stable signal if and only if the region of convergence contains the $i\omega$-axis.
\end{claim}

\begin{proof}

Assume that $x$ is stable. This means that $x$ is absolutely integrable.
\begin{align*}
&\infint \abs{x(t)} dt \lt \infty \\
&\infint \abs{x(t)}\abs{e^{-i\omega t}} dt \lt \infty \quad \mbox{multiply by 1}\\
&\infint \abs{x(t) e^{-(\sigma + i\omega) t}} dt \lt \infty \quad \mbox{for } \sigma = 0\\
&\infint \abs{x(t) e^{-st}} dt \lt \infty \\
&\Rightarrow \mbox{for the chosen values of $s = i\omega$, the Laplace Transform converges}
\end{align*}

Therefore the stable signal $x$'s absolute integrability implies that $i\omega \in RoC(x)$ since $\infint \abs{x(t) e^{-st}} dt \lt \infty$ for $\sigma = 0$. The reverse direction is practically the same. Assume that $i\omega \in RoC(x)$. This implies that $\infint \abs{x(t) e^{-st}} dt \lt \infty$ for $s = 0 + i\omega$.

\begin{align*}
&\infint \abs{x(t) e^{-st}} dt \lt \infty \\
&\infint \abs{x(t) e^{-i\omega t}} dt \lt \infty \quad s = i\omega \\
&\infint \abs{x(t)}\abs{e^{-i\omega t}} dt \lt \infty  \\
&\infint \abs{x(t)}dt \lt \infty \\
&\Rightarrow \mbox{ $x$ is absolutely integrable } \\
\end{align*}

Lets compare the CTFT and the Laplace Transform:

\begin{align*}
X(\omega) &= \infint x(t) e^{-i\omega t} dt \\
\hat{X}(s) &= \infint x(t) e^{-s t} dt \\
\end{align*}

If $i\omega \in RoC(x)$, then 

\begin{align*}
X(\omega) &= \left. \hat{X}(s) \right|_{s = i\omega} \\ 
\end{align*}

\end{proof}

\begin{example}

Given the following signal, 
\begin{align*}
x(t) &= e^{-3t}u(t) \ltp \frac{1}{s + 3} \quad \Real{s} \gt -3 \\
\end{align*}

we can evaluate the Laplace Transform at $s = i\omega$ to find the Fourier Transform:

\begin{align*}
\ftrans{x(t)} &= \left. \hat{X}(s) \right|_{s=i\omega} = \frac{1}{3 + i\omega} \\
\end{align*}

\end{example}

Stability

Lets look at the Laplace Transform of a \emph{Toneburst}, $x(t) = \cos(\omega_0 t) u(t)$:


\begin{align*}
x(t) &= \cos(\omega_0 t) u(t) \\
\hat{X}(s) &= \int_{-\infty}^{\infty} \cos(\omega_0 t) u(t) e^{-st} dt \\
\hat{X}(s) &= \int_{0}^{\infty} \cos(\omega_0 t) e^{-st} dt \\
\hat{X}(s) &= \int_{0}^{\infty} \lr{ \frac{1}{2} e^{i\omega_0 t} + \frac{1}{2}e^{-i\omega_0 t}  } e^{-st} dt \\
\hat{X}(s) &= \frac{1}{2} \int_{0}^{\infty}  e^{i\omega_0 t}  e^{-st} dt  + \frac{1}{2} \int_{0}^{\infty} e^{-i\omega_0 t}   e^{-st} dt \\
\hat{X}(s) &= \frac{1}{2} \int_{0}^{\infty}  e^{-(s - i\omega_0) t} dt  + \frac{1}{2} \int_{0}^{\infty} e^{-(s + i\omega_0) t} dt \\
\hat{X}(s) &= \left. \frac{1}{2} \frac{e^{-(s - i\omega_0) t}}{-(s - i\omega_0)} \right|_0^\infty + \left. \frac{1}{2} \frac{e^{-(s + i\omega_0) t}}{-(s + i\omega_0)} \right|_0^\infty \\ 
\hat{X}(s) &= \lim_{t \to \infty} \frac{1}{2} \frac{e^{-(s - i\omega_0) t}}{-(s - i\omega_0)} - \frac{1}{2} \frac{1}{-(s - i\omega_0)} + \lim_{t \to \infty} \frac{1}{2} \frac{e^{-(s + i\omega_0) t}}{-(s + i\omega_0)} - \frac{1}{2} \frac{1}{-(s + i\omega_0)} \\
\end{align*}

Before evaluating, we must note conditions for the limits to converge for the evaluation at infinity. The limit $\lim_{t \to \infty} \frac{1}{2} \frac{e^{-(s - i\omega_0) t}}{-(s - i\omega_0)} $ converges to 0 if and only if $\Real{s - i\omega_0} \gt 0$. The limit $\lim_{t \to \infty} \frac{1}{2} \frac{e^{-(s + i\omega_0) t}}{-(s + i\omega_0)}$ converges to 0 if and only if $\Real{s +i\omega_0} \gt 0$. Both of these simplify to $\Real{s} \gt 0$. Thus, for $\Real{s} \gt 0$ we have

\begin{align*}
\hat{X}(s) &= 0 - \frac{1}{2} \frac{1}{-(s - i\omega_0)} + 0 - \frac{1}{2} \frac{1}{-(s + i\omega_0)} \\
\hat{X}(s) &=\frac{1}{2} \frac{1}{(s - i\omega_0)} +  \frac{1}{2} \frac{1}{(s + i\omega_0)} \\
\hat{X}(s) &=\frac{1}{2} \lr{ \frac{1}{(s - i\omega_0)} +  \frac{1}{(s + i\omega_0)}} \\
\hat{X}(s) &=\frac{1}{2} \lr{ \frac{2s}{(s^2 + \omega_0^2)} } \\
\hat{X}(s) &=\frac{s}{(s^2 + \omega_0^2)}  \\
\end{align*}

This gives us the transform pair:

\begin{nicebox}
$$
\cos(\omega_0 t) u(t) \ltp \frac{s}{(s^2 + \omega_0^2)} \quad \Real{s} \gt 0
$$
\end{nicebox}

Note that we have two finite poles as $\pm i\omega$. The zeros are at 0 and $\infty$. The $RoC$ is to the right of the right-most pole, which is on the $i\omega$-axis. Here is a question: What is the Laplace Transform of $\cos(\omega_0 t)$? This doesn't have a Laplace transform, although it does have a Fourier Transform. 


Trigonometric Toneburst

Lets see what happens with convolution in the time domain.

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

So we have the convolution property:

\begin{nicebox}
\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}

This property is what we can use to solve for the system function $\hat{H}(s)$ when given an input $x(t)$ and an output $y(t)$: we have the relation $\hat{H}(s) = \frac{\hat{Y}(s)}{\hat{X}(s)}$.


Convolution in Time

When we integrate a function in time, what happens to its Laplace Transform?

\begin{align*}
x(t) \ltp \hat{X}(s) \\
y(t) = \int_{-\infty}^t x(\tau) d\tau 
\end{align*}

We can solve this by using the fact that $y(t) = x(t) * u(t) = \infint x(\tau) u(t - \tau) d\tau$

\begin{align*}
x(t) &\ltp \hat{X}(s) \\
u(t) &\ltp \frac{1}{s} \\
y(t) &\ltp \frac{\hat{X}(s)}{s} \quad \mbox{ (using the convolution property)} \\
\end{align*}

The $RoC(u) = \{ s \st \Real{s} \gt 0 \}$, and the new region of convergence is at least the intersection of $RoC(x)$ and $RoC(u)$. Hence, we have our Integration in Time property:

\begin{nicebox}
\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}

Integration in the Time Domain

Now, lets find out what happens when we differentiate in the time domain. Recall the formula for the inverse Laplace Transform:

\begin{align*}
x(t) &= \frac{1}{2\pi i} \int_{\sigma_0 -i\infty}^{\sigma_0 + i\infty} \hat{X}(s) e^{st} ds \\
\end{align*}

We shall use this esoteric formula to solve for the differentiation property, although we most likely will not encounter it elsewhere in this course.

\begin{align*}
x(t) &= \frac{1}{2\pi i} \int_{\sigma_0 -i\infty}^{\sigma_0 + i\infty} \hat{X}(s) e^{st} ds \\
\frac{d}{dt}x(t) &= \frac{d}{dt} \frac{1}{2\pi i} \int_{\sigma_0 -i\infty}^{\sigma_0 + i\infty} \hat{X}(s) e^{st} ds \\
\frac{d}{dt}x(t) &= \frac{1}{2\pi i} \int_{\sigma_0 -i\infty}^{\sigma_0 + i\infty} \hat{X}(s)  \frac{d}{dt}e^{st} ds \\
\frac{d}{dt}x(t) &= \frac{1}{2\pi i} \int_{\sigma_0 -i\infty}^{\sigma_0 + i\infty} \lr{s \hat{X}(s)} e^{st} ds = \iltrans{s\hat{X}(s)} \\
\end{align*}

This gives us the time 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}

Differentiation in the Time Domain

What happens if we multiply a signal in the time-domain by a complex exponential? 

\begin{align*}
x(t) &\ltp \infint x(t) e^{-st} dt \hat{X}(s) \\
e^{s_0 t} x(t) &\ltp  \infint e^{s_0 t} x(t) e^{-st} dt \\
e^{s_0 t} x(t) &\ltp  \infint x(t) e^{-(s - s_0) t} dt = \hat{X}(s - s_0)\\
\end{align*}

What happens to the region of convergence?

\begin{align*}
\alpha \lt &\Real{s} \lt \beta  \\
\alpha \lt &\Real{s - s_0} \lt \beta \\ 
\alpha + \Real{s_0} \lt &\Real{s} \lt \beta + \Real{s_0} \\
\end{align*}

Therefore the new $RoC$ is equal to $RoC(x) + \Real{s_0}$. This gives us the following property:

\begin{nicebox}
\begin{align*}
e^{s_0 t} x(t) &\ltp \hat{X}(s-s_0) \quad RoC = RoC(x) + \Real{s_0} \\
\end{align*}
\end{nicebox}

14 Week 14