Migrate *the rest* of the math blogs

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+rss = "Properties of cascade connected systems analyzed via anonymous functions"
+date = Date(2020, 4, 15)
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+@def tags = ["system", "fun", "anonymous"]
+
+# System Cascade Connection
+
+Given two discrete-time systems $A$ and $B$ connected in cascade
+to form a new system $C = x \mapsto B(A(x))$.
+
+## Linearity
+
+If $A$ and $B$ are linear, i.e. for all signals $x_i$ and scalars $a_i$,
+
+\[\begin{aligned}
+  A\left(n \mapsto \sum_i a_i x_i[n]\right) = n \mapsto \sum_i a_i A(x_i)[n]\\
+  B\left(n \mapsto \sum_i a_i x_i[n]\right) = n \mapsto \sum_i a_i B(x_i)[n]
+\end{aligned}\]
+
+then $C$ is also linear
+
+\[\begin{aligned}
+  C\left(n \mapsto \sum_i a_i x_i[n]\right)
+  &= B\left(A\left(n \mapsto \sum_i a_i x_i[n]\right)\right)\\
+  &= B\left(n \mapsto \sum_i a_i A(x_i)[n]\right)\\
+  &= n \mapsto \sum_i a_i B(A(x_i))[n]\\
+  &= n \mapsto \sum_i a_i C(x_i)[n]
+\end{aligned}\]
+
+## Time Invariance
+
+If $A$ and $B$ are time invariant,
+i.e. for all signals $x$ and integers $k$,
+
+\[\begin{aligned}
+  A(n \mapsto x[n - k]) &= n \mapsto A(x)[n - k]\\
+  B(n \mapsto x[n - k]) &= n \mapsto B(x)[n - k]
+\end{aligned}\]
+
+then $C$ is also time invariant
+
+\[\begin{aligned}
+  C(n \mapsto x[n - k])
+  &= B(A(n \mapsto x[n - k]))\\
+  &= B(n \mapsto A(x)[n - k])\\
+  &= n \mapsto B(A(x))[n - k]\\
+  &= n \mapsto C(x)[n - k]
+\end{aligned}\]
+
+## LTI Ordering
+
+If $A$ and $B$ are linear and time-invariant, there exists
+signals $g$ and $h$ such that for all signals $x$,
+$A = x \mapsto x * g$ and $B = x \mapsto x * h$, thus 
+
+\[B(A(x)) = B(x * g) = x * g * h = x * h * g = A(x * h) = A(B(x))\]
+
+or interchanging $A$ and $B$ order does not change $C$.
+
+## Causality
+
+If $A$ and $B$ are causal,
+i.e. for all signals $x$, $y$ and any choise of integer $k$,
+
+\[\begin{aligned}
+  \forall n < k, x[n] = y[n]\quad
+  \Longrightarrow &\;\begin{cases}
+  \forall n < k, A(x)[n] = A(y)[n]\\
+  \forall n < k, B(x)[n] = B(y)[n]
+  \end{cases}\\
+  \Longrightarrow &\;\forall n < k, B(A(x))[n] = B(A(y))[n]\\
+  \Longleftrightarrow &\;\forall n < k, C(x)[n] = C(y)[n]
+\end{aligned}\]
+
+then $C$ is also causal.
+
+## BIBO Stability
+
+If $A$ and $B$ are stable, i.e. there exists a signal $x$
+and scalars $a$ and $b$ that for all integers $n$,
+
+\[\begin{aligned}
+  |x[n]| < a &\Longrightarrow |A(x)[n]| < b\\
+  |x[n]| < a &\Longrightarrow |B(x)[n]| < b
+\end{aligned}\]
+
+then $C$ is also stable, i.e. there exists a signal $x$
+and scalars $a$, $b$ and $c$ that for all integers $n$,
+
+\[\begin{aligned}
+  |x[n]| < a\quad
+  \Longrightarrow &\;|A(x)[n]| < b\\
+  \Longrightarrow &\;|B(A(x))[n]| < c\\
+  \Longleftrightarrow &\;|C(x)[n]| < c
+\end{aligned}\]