McSinyx

Given two discrete-time systems $A$ and $B$ connected in cascade to form a new system $C = x \mapsto B(A(x))$, we examine the following properties:

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} Tags: fun math Nguyễn Gia Phong, 2020-04-15