1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
|
\documentclass[a4paper,12pt]{article}
\usepackage[english,vietnamese]{babel}
\usepackage{amsmath}
\usepackage{enumerate}
\usepackage{lmodern}
\title{System Cascade Connection}
\author{{\selectlanguage{vietnamese}Nguyễn Gia Phong}}
\begin{document}
\selectlanguage{english}\maketitle
Given two discrete-time systems $A$ and $B$ connected in cascade to form
a new system $C = x \mapsto B(A(x))$.
\section{Linearity}
If $A$ and $B$ are linear, i.e. for all signals $x_i$ and scalars $a_i$,
\begin{align*}
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{align*}
then $C$ is also linear
\begin{align*}
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{align*}
\section{Time Invariance}
If $A$ and $B$ are time invariant, i.e. for all signals $x$ and integers $k$,
\begin{align*}
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{align*}
then $C$ is also time invariant
\begin{align*}
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{align*}
\section{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 $C$.
\section{Causality}
If $A$ and $B$ are causal, i.e. for all signals $x$, $y$ and integers $k$,
\begin{multline*}
x[n] = y[n]\quad\forall n < k
\Longrightarrow\begin{cases}
A(x)[n] = A(y)[n]\quad\forall n < k\\
B(x)[n] = B(y)[n]\quad\forall n < k
\end{cases}\\
\Longrightarrow B(A(x))[n] = B(A(y))[n]\quad\forall n < k
\iff C(x)[n] = C(y)[n]\quad\forall n < k
\end{multline*}
then $C$ is also causal.
\section{BIBO Stability}
If $A$ and $B$ are stable, i.e. there exists a signal $x$
and scalars $a$, $b$ that
\begin{align*}
|x[n]| < a\quad\forall n \in \mathbb Z
&\Longrightarrow |A(x)[n]| < b\quad\forall n \in \mathbb Z\\
|x[n]| < a\quad\forall n \in \mathbb Z
&\Longrightarrow |B(x)[n]| < b\quad\forall n \in \mathbb Z
\end{align*}
then $C$ is also stable, i.e. there exists a signal $x$
and scalars $a$, $b$, $c$ that
\begin{align*}
|x[n]| < a\;\forall n \in \mathbb Z
&\Longrightarrow |A(x)[n]| < b\;\forall n \in \mathbb Z\\
&\Longrightarrow |B(A(x))[n]| < c\;\forall n \in \mathbb Z
\iff |C(x)[n]| < c\;\forall n \in \mathbb Z
\end{align*}
\end{document}
|
