Discrete Time Signals and Systems

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Discrete Time Signals and Systems

Signal classification

• Periodic and non-periodic
• Odd and even signals
• energy signal and power signal

basic signal

• Impulse function
• unit step function
• Ramp function

Operation on signal

 

Periodic and Aperiodic Discrete-Time Sinusoids
$$x(n)=Acos[2\pi f_{0}n]=x(n+N)=Acos[2\pi f_0(n+N)]=Acos[2\pi f_{0}n+2\pi f_{0}N]$$
$2\pi f_{0}N=2\pi k$ just $f_0=\frac{k}{N}$

n is integer: Periodic

n is not integer: Aperiodic

Periodic judgment of composite signals

1. Find N for each signal

   if $\frac{N_1}{N_2}=rationalnumber$ it is Periodic

2. Find the lowest common multiple($ LCM(N_1,N_2)$)

   Periodic is $LCM(N_1,N_2)$

Odd and even signals

odd: $x_0(t)=\frac{1}{2}[x(t)-x(-t)]$

even: $x_e(t)=\frac{1}{2}[x(t)+x(-t)]$

$x(t)=x_0(t)+x_e(t)$

energy signal and power signal

energy: $E={\sum }_{n=-\infty }^{\infty }|x[n]{|}^2$

power:

Periodic: $P_{\infty }={\lim }_{N\to \infty }\frac{1}{2N+1}{\sum }_{n=-\infty }^{+\infty }|x[n]{|}^2$

Aperiodic: $P_x=\frac{1}{N}{\sum }_{n=0}^{N-1}|x[n]{|}^2$

energy signal:energy is finite,power is zero

power signal:energy is infinite,power is finite

 

find the energy and power for

• Impulse function
• unit step function
• Ramp function

 

1. Time Shifting(left is +;right is -)
2. Time-scale
3. Time Reversal

 

System of discrete signal

• Linear systems and nonlinear systems

• Causal and Acausal Systems

• Time-varying and time-invariant systems

• static system and dynamic system

• Stable and unstable systems

• convolution

• convolution sum

• circular convolution

• Stability of linear time-invariant systems

Linear systems and nonlinear systems

• Linear systems satisfy uniformity and superposition
• A system that satisfies uniformity and superposition is a linear system

Must have：$x(n)=y(n)=0$

Four steps to solve problems:

1. x1 to y1 = F(x1)
2. x2 to y2 = F(x2)
3. y3 = ay1 + by2
4. F(ax1 +bx2) = y4

if y4 = y3 ,the system is linear

 

Causal and non-causal Systems

casual system: The output depends only on present and past signals

Acausal Systems：The output depends on at least one future input

eg：Y(n) = x(-n) is non-casual (at n = -1)

even and odd is non-casual

ps: anti-casual system: output only upon “only future input” for all time

 

Time-varying and time-invariant systems

Time-varying system(TVS):

y(n) = F[x(n)] = x(n)cos(2$w\pi$n)

y(n,k) = F[x(n-k)] = x(n-k)cos(2$w\pi$n)

y(n-k) $\ne$ y(n,k)

y(n) to y(n,k):only change n for x(n)

time-invariant systems :

y(n,k) = y(n-k)

eg:y(n-k) = sin (x(n-k))

y change n for all n

 

static system and dynamic system

static system (memory-less system):

output only depends on now input for all time

dynamic system：

output depends on past and/or future inputs

 

Stable and unstable systems

Stable system: BIBO

bounded input to bounded output

unstable systems:

bounded input to unbounded output

eg: $y(n)=y^2(n-1)+2\delta (n)$

 

convolution

• time-invariant systems

$y(n)={\sum }_{k=-\infty }^{\infty }x(k)h(n-k)$

$y(n)=x(n)\ast h(n)$

1. time reversal
2. shifting of h(-k) to h(n-k)
3. multiply x(k)h(n-k)
4. Sum

• linear convolution
• circular convolution

matrix method:

x(n) = {1,2,3,4} h(n)={0,2,2,2}

length of x(n):4 samples;k

length of h(n):4 samples;m

length of y(n): = k + m - 1 = 4 + 4 - 1 = 7 samples

 

circular convolution

x(n) = {2,1,2} h(n) = {1,2,3}

$y(n)={\sum }_{k=-\infty }^{\infty }x(k)h(n-k)$

 

Stability of linear time-invariant systems

$S={\sum }_{k=-\infty }^{\infty }|h(n)|<\infty$

it is stability

eg: $h(n)=(0.8{)}^{n}u(n+2)$

 

Fourier Series

• Fourier Series basic concepts
• Discrete time fouriew series (DTFS)

Fourier Series basic concepts

Condition for existence of Fourier Series

1. Finite Number of maxima and minima over the time period
2. Finite number of discontinuities over the time period T
3. Signal should be absolutely integrable over the range of time period T

period function:

$f(x)=a_0+{\sum }_{n=1}^{\infty }(a_{n}cos(n{w}_{0}t)+b_{n}sin(n{w}_{0}t)$

$a_0=\frac{1}{T_1}{\int }_0^{T_1}f(t)dt$

$a_n=\frac{2}{T_1}{\int }_0^{T_1}f(t)\cos n{w}_{1}tdt$

$b_n=\frac{2}{T_1}{\int }_0^{T_1}f(t)\sin n{w}_{1}tdt$

eg:

 

Discrete time fouriew series (DTFS)

$x_n={\sum }_{k=0}^{N-1}c(k)e^{Jk\frac{2\pi }{N}n}$

$c(k)=\frac{1}{N}{\sum }_{n=0}^{N-1}x(n)e^{-Jk\frac{2\pi }{N}n}$

 

CTFS

$x(w)={\int }_{-\infty }^{+\infty }x(t)e^{-Jwt}dt$

Inverse transformation:$x(t)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }x(w)e^{jwt}dw$

 

Continuous time Fourier transform (CTFT)

• basic concept
• fourier fransform
• magnitude specturm
• phase spectrum

properties of CTFT

• linearity
• time shifting
• frequency shifting
• time scaling
• time differenciation
• frequency differenciation
• convolution

Parseval’s Theorem

1. Any signal is built up addition of elementary signals which are at different frequencies.

2. CTFT is used to transform the signal from time domain to frequency domain.

3. With the help of CTFT,plot the amplitude and phase spectrum.

4. CTFT can be efxpressed as:

   $x(w)={\int }_{-\infty }^{+\infty }x(t)e^{-Jwt}dt$

   Inverse transformation:$x(t)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }x(w)e^{jwt}dw$

 

Dirichlet’s conditions

1. f(t) should be absolutely integrable.
2. f(t) should have a finite number of maxima and minima over any finite interval.
3. f(t) should have finite number of discontinues over any finite interval

These conditions are sufficient but not necessary Condition.

If the conditions are not met, there is not necessarily no Fourier transform.

 

lecture(77)

• fourier fransform
• magnitude specturm
• phase spectrum

 

lecture(78)

Inverse transformation for function

important

 

properties of CTFT

linearity

$\mathcal{F}[f(t)]=F(\omega ),\mathcal{F}[g(t)]=G(\omega )$

$\begin{array}{rl}\mathcal{F}[\alpha f(t)+\beta g(t)]& =\alpha F(\omega )+\beta G(\omega )\\ {\mathcal{F}}^{-1}[\alpha F(\omega )+\beta G(\omega )]& =\alpha f(t)+\beta g(t)\end{array}$

time shifting

$\mathcal{F}[f(t)]=F(\omega )$

$\begin{array}{rl}\mathcal{F}[f(t-t_0)]& =e^{-j\omega t_0}F(\omega )\\ {\mathcal{F}}^{-1}[F(\omega -{\omega }_0)]& =e^{j{\omega }_{0}t}f(t)\end{array}$

time scaling

$\mathcal{F}[f(t)]=F(\omega )$

$\mathcal{F}[f(at)]=\frac{1}{|a|}F\left(\frac{\omega }{a}\right)$

frequency shifting

$\mathcal{F}[f(t)]=F(\omega )$

$\begin{array}{rl}\mathcal{F}[f(t)e^{j{w}_{0}t}]& =F(\omega -{\omega }_0)\\ {\mathcal{F}}^{-1}[f(t)e^{-j{w}_{0}t}]& =F(\omega +{\omega }_0)\end{array}$

time differenciation

$\mathcal{F}[f(t)]=F(\omega )$

$\begin{array}{c}\mathcal{F}\left[\frac{d^{n}f(t)}{d{t}^n}\right]=(j\omega {)}^{n}F(\omega )\\ {\mathcal{F}}^{-1}\left[\frac{d^{n}F(\omega )}{d{\omega }^n}\right]=(-jt{)}^{n}f(t)\end{array}$

frequency differenciation

$\mathcal{F}[f(t)]=F(\omega )$

$\mathcal{F}[(-jt{)}^{n}f(t)]=\frac{d^{n}F(w)}{d{w}^n}$

$\mathcal{F}[tf(t)]=j\frac{dF(w)}{dw}$

convolution

$\mathcal{F}[f(t)]=F(\omega ),\mathcal{F}[g(t)]=G(\omega )$

$\mathcal{F}[f_1(t)\ast f_2(t)]=F_1(\omega )\cdot F_2(\omega )$

$\mathcal{F}[f_1(t)\cdot f_2(t)]=\frac{1}{2\pi }{F}_1(\omega )\ast F_2(\omega )$

 

Parseval’s Theorem

if x(t) CTFT to $x(w)$ or $x(f)$

$E={\int }_{-\infty }^{\infty }|x(t){|}^{2}dt={\int }_{-\infty }^{\infty }|x(f){|}^{2}df$

 

Fourier transform of Periodic signal

• spectrum of $e^{j{\omega }_{0}t}$ and $e^{-j{\omega }_{0}t}$
• spectrum of $sinwt$ and $coswt$
• Spectrum of a general periodic signal

spectrum of $e^{j{\omega }_{0}t}$ and $e^{-j{\omega }_{0}t}$

$\begin{array}{rl}\mathcal{F}[e^{j{\omega }_{0}t}]& =2\pi \delta (\omega -{\omega }_0)\\ \mathcal{F}[e^{-j{\omega }_{0}t}]& =2\pi \delta (\omega +{\omega }_0)\end{array}$

 

spectrum of $sin{w}_{0}t$ and $cos{w}_{0}t$

$\begin{array}{c}\mathcal{F}[\cos ({\omega }_{0}t)]=\mathcal{F}[\frac{e^{j{\omega }_{0}t}+e^{-j{\omega }_{0}t}}{2}]=\pi [\delta (\omega -{\omega }_0)+\delta (\omega +{\omega }_0)]\\ \mathcal{F}[\sin ({\omega }_{0}t)]=\mathcal{F}[\frac{e^{j{\omega }_{0}t}-e^{-j{\omega }_{0}t}}{2j}]=j\pi [\delta (\omega +{\omega }_0)-\delta (\omega -{\omega }_0)]\end{array}$

 

Spectrum of a general periodic signal

$f(t)={\sum }_{n=-\infty }^{\infty }F(n{w}_0)e^{jn{w}_{0}t}$

$F(w)=\mathcal{F}[f(t)]=2\pi {\sum }_{n=-\infty }^{\infty }F(n{w}_0)\delta (\omega -n{\omega }_0)$

$F(n{w}_1)=\frac{1}{T}{\int }_0^{T}f(t)e^{-jn{w}_{0}t}dt$

 

Discrete time fouriew series (DTFT)

• basic properties
• DTFT for common signals

basic properties

DTFT:$X\left(e^{j\omega }\right)={\sum }_{n=-\infty }^{+\infty }x[n]e^{-j\omega n}$

Inverse transform of DTFT:$x\left[n\right]=\frac{1}{2\pi }{\int }_{2\pi }X(e^{j\omega })e^{j\omega n}d\omega$

DTFT for common signals

$x(n)=a^{n}u(n);|a|<1$

$\begin{array}{rl}\text{DTFT [X(n)]}& =\sum _{n=-\infty }^{\infty }x(n)e^{-jwn}\\ & =\sum _{n=0}^{+\infty }{a}^n{e}^{-jwn}\\ & =\sum _{n=0}^{+\infty }(a\cdot e^{-jw}{)}^n\\ & =\frac{1}{1-a{e}^{-jw}}\\ & =\frac{1}{1-a(cosw-jsinw)}\\ & =\frac{1}{(1-acosw)+jasinw}\end{array}$

magnitude:

$|x(e^{jw})|=\frac{1}{\sqrt{(1-acosw{)}^2}+(asinw{)}^2}=\frac{1}{\sqrt{1+a^2-2acosw}}$

phase:

$\angle x(e^{jw})=-arctan\frac{asinw}{1-acosw}$

eg:x(n) = {-4,3,-2,3,-4}

find:(1).$x(e^{j0})$ (2).$x(e^{jw})$ (3).${\int }_{\pi }^{-\pi }x(e^{jw})dw$

 

properties

linear

$a{x}_1\left[n\right]+b{x}_2\left[n\right]\leftrightarrow a{X}_1(e^{j\omega })+b{X}_2(e^{j\omega })$

time shifting

$x(n)\leftrightarrow x(e^{jw})$

$x(n-n_0)\leftrightarrow x(e^{jw})\cdot e^{-jw{n}_0}$

frequency shifting

$x(n)\leftrightarrow x(e^{jw})$

$x(n)\cdot e^{-jw{n}_0}\leftrightarrow x(e^{j(w-w_0)})$

Frequency domain differential

$x(n)\leftrightarrow x(e^{jw})$

$nx(n)\leftrightarrow j\frac{dx(e^{jw})}{dw}$

 

eg:

1.$y(n)-Ay(n-1)=x(n);|A|<1$

2.$y(n)-\frac{3}{4}y(n-1)+\frac{1}{8}y(n-2)=2x(n)$

3.$x(e^{jw})=e^{-jw}$ for $-\pi \le w\le \pi$

4.find convolution $x(n)={\frac{1}{2}}^{n}u(n)$ and $x(n)={\frac{1}{3}}^{n}u(n)$

 

Discrete Fourier Transform（DFT）

DFT is equally spaced sampling of DTFT

DFT:$x(n)——>x(k)$

$x(k)={\sum }_{n=0}^{N-1}x(n)e^{-j\frac{2\pi k}{N}n};k=0,1,2,3,...,N-1$

making $W_N=e^{-j\frac{2\pi }{N}}$

$x(k)={\sum }_{n=0}^{N-1}x(n)W_N^{kn}$

 

Calculation of DFT

eg:$x(k)={\sum }_{n=0}^{4-1}x(n)e^{-j\frac{2\pi k}{4}n}$

1.$x(k)={\sum }_{n=0}^{N-1}x(n)e^{-j\frac{2\pi k}{N}n};k=0,1,2,3,...,N-1$

2.matrix method

Rotation matrix of inverse DFT transform（IDFT）

 

Properties of DFT

1.linear

$X_1(k)=DFT[x_1(n)]$

$X_2(k)=DFT[x_2(n)]$

$DFT[a{x}_1(n)+b{x}_2(n)]=a{X}_1(k)+b{X}_2(k)$

2.circular convolution

DFT:$x[n]⊛h[n]\iff X[k]\cdot H[k]$

DFT:$x[n]\cdot h[n]\iff X[k]⊛H[k]$

3.periodicity

$DFT[x(n)]=x(k)$

$x(n+N)=x(n)$;for all n

$x(k+N)=x(k)$;for all k

4.cyclic frequency shift

$DFT[x(n)e^{j\frac{2\pi }{N}ln}]=x((k-l){)}_N$

(( )) is cyclic frequency

 

DTF calculates linear convolution and circular convolution

linear convilution

$[y(n)=x(n)\ast h(n)]$

then the length of x(n) is L,and h(n) is M

the length of linear convolution: N = L + M -1

1. filling the length of x(n) and h(n) to N (add 0)
2. calculates DTF[x(n)] and DTF[h(n)]
3. then$ y(k) = x(k)·h(k)$
4. calculate $IDFT[Y(k)]=y(n)$

 

cyclical convolution

$[y(n)=x(n)⊛h(n)]$

then the length of x(n) is L,and h(n) is M(add 0)

the length of cyclical convolution is :N = max(L,M)

1. filling the length of x(n) and h(n) to N
2. calculates DTF[x(n)] and DTF[h(n)]
3. then $y(k)=x(k)\cdot h(k)$
4. calculate $IDFT[Y(k)]=y(n)$

 

another way to calculate linear convolution

• Overlap-Add and Overlap-Save

Overlap-Add

then the length of $x_k(n)$ ($x_k(n)$ is not $x(n)$) is L,and h(n) is M

the length of convolution is :N = L + M - 1 (the size of N is depend on you)

eg: x ( n ) = { 3 , − 1 , 0 , 1 , 3 , 2 , 0 , 1 , 2 , 1 } x(n) = \{ 3,-1,0,1,3,2,0,1,2,1\} x(n)={3,−1,0,1,3,2,0,1,2,1} h ( n ) = { 1 , 1 , 1 } h(n) = \{ 1,1,1\} h(n)={1,1,1}

1.we take N = 6,then 6= L + 3 - 1;so L = 4.

x 1 ( n ) = { 0 , 0 , 3 , − 1 , 0 , 1 } x_1(n) = \{ 0,0,3,-1,0,1 \} x1​(n)={0,0,3,−1,0,1}

x 2 ( n ) = { 0 , 1 , 3 , 2 , 0 , 1 } x_2(n) = \{ 0,1,3,2 ,0,1\} x2​(n)={0,1,3,2,0,1}

x 3 ( n ) = { 0 , 1 , 2 , 1 , 0 , 0 } x_3(n) = \{ 0,1,2,1,0,0 \} x3​(n)={0,1,2,1,0,0}

h ( n ) = { 1 , 1 , 1 , 0 , 0 , 0 } h(n) = \{1,1,1,0,0,0\} h(n)={1,1,1,0,0,0}

2.calculate

x 1 ( n ) ⊛ h ( n ) = { 1 , 1 , 3 , 2 , 2 , 0 } x_1(n)\circledast h(n) =\{1,1,3,2,2,0\} x1​(n)⊛h(n)={1,1,3,2,2,0}

x 2 ( n ) ⊛ h ( n ) = { 1 , 2 , 4 , 6 , 5 , 3 } x_2(n)\circledast h(n)=\{1,2,4,6,5,3\} x2​(n)⊛h(n)={1,2,4,6,5,3}

x 3 ( n ) ⊛ h ( n ) = { 0 , 1 , 3 , 4 , 3 , 1 } x_3(n)\circledast h(n) =\{0,1,3,4,3,1\} x3​(n)⊛h(n)={0,1,3,4,3,1}

3.Remove first (M-1) points ,Concatenate all results

the result is { 3 , 2 , 2 , 0 , 4 , 6 , 5 , 3 , 3 , 4 , 3 , 1 } \{3,2,2,0,4,6,5,3,3,4,3,1\} {3,2,2,0,4,6,5,3,3,4,3,1}

 

Overlap-Save

then the length of $x_k(n)$ ($x_k(n)$ is not $x(n)$) is L,and h(n) is M

the length of convolution is :N = L + M - 1 (the size of N is depend on you)

eg: x ( n ) = { 3 , − 1 , 0 , 1 , 3 , 2 , 0 , 1 , 2 , 1 } x(n) = \{ 3,-1,0,1,3,2,0,1,2,1\} x(n)={3,−1,0,1,3,2,0,1,2,1} h ( n ) = { 1 , 1 , 1 } h(n) = \{ 1,1,1\} h(n)={1,1,1}

1.we take N = 5,then 5 = L + 3 - 1;so L = 3.

x 1 ( n ) = { 0 , 0 , 3 , − 1 , 0 } x_1(n) = \{ 0,0,3,-1,0 \} x1​(n)={0,0,3,−1,0}

x 2 ( n ) = { − 1 , 0 , 1 , 3 , 2 } x_2(n) = \{ -1,0,1,3,2 \} x2​(n)={−1,0,1,3,2}

x 3 ( n ) = { 3 , 2 , 0 , 1 , 2 } x_3(n) = \{ 3,2,0,1,2 \} x3​(n)={3,2,0,1,2}

x 4 ( n ) = { 1 , 2 , 1 , 0 , 0 } x_4(n) = \{ 1,2,1,0,0 \} x4​(n)={1,2,1,0,0}

h ( n ) = { 1 , 1 , 1 , 0 , 0 } h(n) = \{1,1,1,0,0\} h(n)={1,1,1,0,0}

2.calculate

x 1 ( n ) ⊛ h ( n ) = { − 1 , 0 , 3 , 2 , 2 } x_1(n)\circledast h(n) =\{-1,0,3,2,2\} x1​(n)⊛h(n)={−1,0,3,2,2}

x 2 ( n ) ⊛ h ( n ) = { 4 , 1 , 0 , 4 , 6 } x_2(n)\circledast h(n)=\{4,1,0,4,6\} x2​(n)⊛h(n)={4,1,0,4,6}

x 3 ( n ) ⊛ h ( n ) = { 6 , 7 , 5 , 3 , 3 } x_3(n)\circledast h(n) =\{6,7,5,3,3\} x3​(n)⊛h(n)={6,7,5,3,3}

x 4 ( n ) ⊛ h ( n ) = { 1 , 3 , 4 , 3 , 1 } x_4(n)\circledast h(n) =\{1,3,4,3,1\} x4​(n)⊛h(n)={1,3,4,3,1}

3.Remove first (M-1) points ,Concatenate all results

the result is { 3 , 2 , 2 , 0 , 4 , 6 , 5 , 3 , 3 , 4 , 3 , 1 } \{3,2,2,0,4,6,5,3,3,4,3,1\} {3,2,2,0,4,6,5,3,3,4,3,1}

 

 

z-transformation

• Basic concepts of z-transformation
• ROC
• Properties of z-transformation
• Inverse transformation of z-transform
• stable

Basic concepts of z-transformation

• Bilateral Z Transform

• Unilateral Z Transform

Bilateral Z Transform:

X ( Z ) = Z { x [ n ] } = ∑ n = − ∞ + ∞ x [ n ] Z − n , Z ∈ R x X(Z)=Z\left\{x[n]\right\}=\sum_{n=-\infty}^{+\infty}x\left[n\right]Z^{-n}\text{,}Z\in Rx X(Z)=Z{x[n]}=∑n=−∞+∞​x[n]Z−n,Z∈Rx

Unilateral Z Transform:

X ( Z ) = Z { x [ n ] } = ∑ n = 0 + ∞ x [ n ] Z − n , Z ∈ R x X(Z)=Z\left\{x[n]\right\}=\sum_{n=0}^{+\infty}x\left[n\right]Z^{-n},Z\in Rx X(Z)=Z{x[n]}=∑n=0+∞​x[n]Z−n,Z∈Rx

 

region of convergence (ROC)

making $z=r\cdot e^{jw}$

$$\begin{gathered} x(z)={\sum }_{n=-\infty }^{\infty }x(n)z^{-n} \\ ={\sum }_{n=-\infty }^{\infty }x(n)r^{-n}{e}^{-jwn} \end{gathered}$$

if $r=1$,then $x(z)=DTFT$

eg: x ( n ) = { 1 , 2 , 3 , 4 , 0 , 1 } x(n) = \{1,2,3,4,0,1 \} x(n)={1,2,3,4,0,1}

$$\begin{gathered} x(z)={\sum }_{n=-\infty }^{\infty }x(n)z^{-n}={\sum }_{n=0}^{5}x(n)z^{-n}=x(0)z^{-0}+x(1)z^{-1}+x(2)z^{-2}+x(3)z^{-3}+x(4)z^{-4}+x(5)z^{-5} \\ =1+2z^{-1}+3z^{-2}+4z^{-3}+z^{-5} \end{gathered}$$

ROC:exist entire z-plane except $z=0$

using plot:

eg: $x(n)=\{\begin{array}{c}{a}^n;n\ge 0\\ 0;n<0\end{array}$

$$\begin{gathered} x(z)={\sum }_{n=0}^{+\infty }(a{z}^{-1}{)}^n \\ =\frac{1}{1-a{z}^{-1}}=\frac{z}{z-a} \end{gathered}$$

needing that : $|a{z}^{-1}|<1$ just ROC:$|z|>|a|$

eg : $x(n)=-a^{n}u(-n-1)$

ROC:$|a^{-1}|<1$ is just $|z|<|a|$

 

Properties of z-transformation

• linear
• time shifting
• scaling
• differential
• convolution
• initial value theorem
• terminal value theorem

linear

$x_1(n)⟶x_1(z);RO{C}_1$

$x_2(n)⟶x_2(z);RO{C}_2$

$$\begin{gathered} a{x}_1(n)+b{x}_2(n)⟶a{x}_1(z)+b{x}_2(z); \\ ROC:[RO{C}_1\cap RO{C}_2] \end{gathered}$$

time shifting

$x(n)⟶x(z)$

$x(n-n_0)⟶x(z)z^{-n_0}$

scaling

$x(n)⟶x(z);ROC:|z|>1$

$a^{n}x(n)⟶x(\frac{z}{a})$

differential

$x(n)⟶x(z)$

$nx(n)⟶-z\frac{dx(z)}{dz}$

convolution

$x_1(n)\ast x_2(n)⟶x_1(z)\cdot x_2(z)$

initial value theorem

$x(0)={\lim }_{n\to 0}x(n)={\lim }_{z\to +\infty }x(z)$

terminal value theorem

$x(+\infty )={\lim }_{n\to \infty }x(n)={\lim }_{z\to 1}(1-z^{-1})x(z)$

 

Inverse transformation of z-transform

• long division method
• partial fraction expansion method
• Residue method

long division method

partial fraction expansion method

$x(z)=\frac{z(z-\frac{1}{2})}{(z+\frac{1}{2})(z+\frac{1}{4})}$

$\frac{x(z)}{z}=\frac{z-\frac{1}{2}}{(z+\frac{1}{2})(z+\frac{1}{4})}=\frac{4}{z+\frac{1}{2}}-\frac{3}{z+\frac{1}{4}}$

$x(z)=\frac{4z}{z+\frac{1}{2}}-\frac{3z}{z+\frac{1}{4}}$

$x(n)=4(-\frac{1}{2}{)}^{n}u(n)-3(-\frac{1}{4}{)}^{n}u(n)$

Residue method

$x(z)=\frac{1}{(z-2)(z-3)}$

1.$z^{n-1}x(z)=\frac{z^{n-1}}{(z-2)(z-3)}$

2.$R_1=(z-2)\frac{z^{n-1}}{(z-2)(z-3)}{|}_{z=2}=-2^{n-1}$

3.$R_2=(z-3)\frac{z^{n-1}}{(z-2)(z-3)}{|}_{z=3}=3^{n-1}$

4.$x(n)=-2^{n-1}+3^{n-1}$

 

stability

$S={\sum }_{k=-\infty }^{\infty }|h(n)|<+\infty$

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