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<page xmlns="https://rub.parody">
<title>KaTeX demo</title>
<description>Math formulae to demo LaTeX to MathML rendering</description>
<date>2020-04-15</date>
<category>demo</category>
<category>math</category>
<markdown>
Given two discrete-time systems <m>A</m> and <m>B</m> connected in cascade
to form a new system <m>C = x \mapsto B(A(x))</m>, we examine
the following properties:
## Linearity
If <m>A</m> and <m>B</m> are linear,
i.e. for all signals <m>x_i</m> and scalars <m>a_i</m>,
<math>
\begin{aligned}
A\left(n \mapsto \sum_i a_i x_i[n]\right) = n \mapsto \sum_i a_i A(x_i)[n]\\
B\left(n \mapsto \sum_i a_i x_i[n]\right) = n \mapsto \sum_i a_i B(x_i)[n]
\end{aligned}
</math>
then <m>C</m> is also linear
<math><![CDATA[
\begin{aligned}
C\left(n \mapsto \sum_i a_i x_i[n]\right)
&= B\left(A\left(n \mapsto \sum_i a_i x_i[n]\right)\right)\\
&= B\left(n \mapsto \sum_i a_i A(x_i)[n]\right)\\
&= n \mapsto \sum_i a_i B(A(x_i))[n]\\
&= n \mapsto \sum_i a_i C(x_i)[n]
\end{aligned}
]]></math>
## Time Invariance
If <m>A</m> and <m>B</m> are time invariant,
i.e. for all signals <m>x</m> and integers <m>k</m>,
<math><![CDATA[
\begin{aligned}
A(n \mapsto x[n - k]) &= n \mapsto A(x)[n - k]\\
B(n \mapsto x[n - k]) &= n \mapsto B(x)[n - k]
\end{aligned}
]]></math>
then <m>C</m> is also time invariant
<math><![CDATA[
\begin{aligned}
C(n \mapsto x[n - k])
&= B(A(n \mapsto x[n - k]))\\
&= B(n \mapsto A(x)[n - k])\\
&= n \mapsto B(A(x))[n - k]\\
&= n \mapsto C(x)[n - k]
\end{aligned}
]]></math>
## LTI Ordering
If <m>A</m> and <m>B</m> are linear and time-invariant, there exists
signals <m>g</m> and <m>h</m> such that for all signals <m>x</m>,
<m>A = x \mapsto x * g</m> and <m>B = x \mapsto x * h</m>, thus
<math>
B(A(x)) = B(x * g) = x * g * h = x * h * g = A(x * h) = A(B(x))
</math>
or interchanging <m>A</m> and <m>B</m> order does not change <m>C</m>.
## Causality
If <m>A</m> and <m>B</m> are causal,
i.e. for all signals <m>x</m>, <m>y</m> and any choise of integer <m>k</m>,
<math><![CDATA[
\begin{aligned}
\forall n < k, x[n] = y[n]\quad
\Longrightarrow &\;\begin{cases}
\forall n < k, A(x)[n] = A(y)[n]\\
\forall n < k, B(x)[n] = B(y)[n]
\end{cases}\\
\Longrightarrow &\;\forall n < k, B(A(x))[n] = B(A(y))[n]\\
\Longleftrightarrow &\;\forall n < k, C(x)[n] = C(y)[n]
\end{aligned}
]]></math>
then <m>C</m> is also causal.
## BIBO Stability
If <m>A</m> and <m>B</m> are stable, i.e. there exists a signal <m>x</m>
and scalars <m>a</m> and <m>b</m> that for all integers <m>n</m>,
<math><![CDATA[
\begin{aligned}
|x[n]| < a &\Longrightarrow |A(x)[n]| < b\\
|x[n]| < a &\Longrightarrow |B(x)[n]| < b
\end{aligned}
]]></math>
then <m>C</m> is also stable, i.e. there exists a signal <m>x</m>
and scalars <m>a</m>, <m>b</m> and <m>c</m> that for all integers <m>n</m>,
<math><![CDATA[
\begin{aligned}
|x[n]| < a\quad
\Longrightarrow &\;|A(x)[n]| < b\\
\Longrightarrow &\;|B(A(x))[n]| < c\\
\Longleftrightarrow &\;|C(x)[n]| < c
\end{aligned}
]]></math>
</markdown>
</page>
|
