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Find inverse z-transform of We get, Using table, Example 7 Example 8. Ghulam Muhammad King Saud University 12 Inverse z- Transform: Examples Find inverse. Z Transform Pairs Time Domain. Z Domain z z-1 k (unit impulse) 1 1 γk † (unit step) z (z) z1 1 1 (z) 1z ak z za 1 1 1 z a e-bTk bT z ze 1 bT 1 1 z e k 2 z z1 1 1 2 z 1z sin(bk) 2 zsin(b) z 2zcos(b) 1 1 12 z sin(b) 1 2z cos(b) z cos(bk) 2 z z cos(b) z 2zcos(b) 1 1 12 1 z cos(b) 1 2z cos(b) z aksin(bk) 22 azsin(b) z 2azcos(b) a 1 1 2 2 az. DSP - Z-Transform Inverse - If we want to analyze a system, which is already represented in frequency domain, as discrete time signal then we go for Inverse Z-transformation.
Table of Laplace and Z-transforms X(s) x(t) x(kT) or x(k) X(z) 1. – – Kronecker delta δ0(k) 1 k = 0 0 k ≠ 0 1 2. – – δ0(n-k) 1 n = k 0 n ≠ k z-k 3. S 1 1(t) 1(k) 1 1 1 −z− 4. S +a 1 e-at e-akT 1 1 1 −e−aT z− 5. 2 1 s t kT 2 1 1 1 − − −z Tz 6. 3 2 s t2 (kT)2 1 3 2 1 1 1 1 − − − − + z T z z 7. Added Oct 13, 2017 by tygermeow in Engineering. Simplest form of Z-Transform. This super basic widget just gives you an open window to use as your hand calculator and save you the trouble of tedious programming repetition for simple results.
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If we want to analyze a system, which is already represented in frequency domain, as discrete time signal then we go for Inverse Z-transformation.
Mathematically, it can be represented as;
$$x(n) = Z^{-1}X(Z)$$where x(n) is the signal in time domain and X(Z) is the signal in frequency domain.
If we want to represent the above equation in integral format then we can write it as
$$x(n) = (frac{1}{2Pi j})oint X(Z)Z^{-1}dz$$![Table Table](https://image.slidesharecdn.com/tablatrasformadaz-160918145150/95/tabla-trasformada-z-1-638.jpg?cb=1474210329)
Here, the integral is over a closed path C. This path is within the ROC of the x(z) and it does contain the origin.
Methods to Find Inverse Z-Transform
When the analysis is needed in discrete format, we convert the frequency domain signal back into discrete format through inverse Z-transformation. We follow the following four ways to determine the inverse Z-transformation.
- Long Division Method
- Partial Fraction expansion method
- Residue or Contour integral method
Long Division Method
In this method, the Z-transform of the signal x (z) can be represented as the ratio of polynomial as shown below;
$$x(z)=N(Z)/D(Z)$$Now, if we go on dividing the numerator by denominator, then we will get a series as shown below
$$X(z) = x(0)+x(1)Z^{-1}+x(2)Z^{-2}+..quad..quad..$$The above sequence represents the series of inverse Z-transform of the given signal (for n≥0) and the above system is causal.
However for n<0 the series can be written as;
$$x(z) = x(-1)Z^1+x(-2)Z^2+x(-3)Z^3+..quad..quad..$$Partial Fraction Expansion Method
Here also the signal is expressed first in N (z)/D (z) form.
If it is a rational fraction it will be represented as follows;
$x(z) = b_0+b_1Z^{-1}+b_2Z^{-2}+..quad..quad..+b_mZ^{-m})/(a_0+a_1Z^{-1}+a_2Z^{-2}+..quad..quad..+a_nZ^{-N})$
The above one is improper when m<n and an≠0
If the ratio is not proper (i.e. Improper), then we have to convert it to the proper form to solve it.
Residue or Contour Integral Method
Z Transform Table Laplace
In this method, we obtain inverse Z-transform x(n) by summing residues of $[x(z)Z^{n-1}]$ at all poles. Mathematically, this may be expressed as
$$x(n) = displaystylesumlimits_{allquad polesquad X(z)}residuesquad of[x(z)Z^{n-1}]$$Here, the residue for any pole of order m at $z = beta$ is
$$Residues = frac{1}{(m-1)!}lim_{Z rightarrow beta}lbrace frac{d^{m-1}}{dZ^{m-1}}lbrace (z-beta)^mX(z)Z^{n-1}rbrace$$- Signals and Systems Tutorial
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Z-Transform has following properties:
Linearity Property
If $,x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
and $,y(n) stackrel{mathrm{Z.T}}{longleftrightarrow} Y(Z)$
Then linearity property states that
$a, x (n) + b, y (n) stackrel{mathrm{Z.T}}{longleftrightarrow} a, X(Z) + b, Y(Z)$
![Z Transform Table Z Transform Table](https://i.pinimg.com/originals/04/ac/4b/04ac4bc74eef5ec17821ad287fbe27af.png)
Time Shifting Property
If $,x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
Then Time shifting property states that
$x (n-m) stackrel{mathrm{Z.T}}{longleftrightarrow} z^{-m} X(Z)$
Multiplication by Exponential Sequence Property
If $,x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
Then multiplication by an exponential sequence property states that
$a^n, . x(n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z/a)$
Time Reversal Property
If $, x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
Then time reversal property states that
$x (-n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(1/Z)$
Differentiation in Z-Domain OR Multiplication by n Property
If $, x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
Then multiplication by n or differentiation in z-domain property states that
$ n^k x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} [-1]^k z^k{d^k X(Z) over dZ^K} $
Convolution Property
If $,x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
and $,y(n) stackrel{mathrm{Z.T}}{longleftrightarrow} Y(Z)$
Then convolution property states that
$x(n) * y(n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z).Y(Z)$
Correlation Property
If $,x (n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z)$
and $,y(n) stackrel{mathrm{Z.T}}{longleftrightarrow} Y(Z)$
Then correlation property states that
$x(n) otimes y(n) stackrel{mathrm{Z.T}}{longleftrightarrow} X(Z).Y(Z^{-1})$
Initial Value and Final Value Theorems
Initial value and final value theorems of z-transform are defined for causal signal.
Initial Value Theorem
For a causal signal x(n), the initial value theorem states that
$ x (0) = lim_{z to infty }X(z) $
This is used to find the initial value of the signal without taking inverse z-transform
Final Value Theorem
For a causal signal x(n), the final value theorem states that
$ x ( infty ) = lim_{z to 1} [z-1] X(z) $
This is used to find the final value of the signal without taking inverse z-transform.
Region of Convergence (ROC) of Z-Transform
The range of variation of z for which z-transform converges is called region of convergence of z-transform.
Properties of ROC of Z-Transforms
- ROC of z-transform is indicated with circle in z-plane.
- ROC does not contain any poles.
- If x(n) is a finite duration causal sequence or right sided sequence, then the ROC is entire z-plane except at z = 0.
- If x(n) is a finite duration anti-causal sequence or left sided sequence, then the ROC is entire z-plane except at z = ∞.
- If x(n) is a infinite duration causal sequence, ROC is exterior of the circle with radius a. i.e. |z| > a.
- If x(n) is a infinite duration anti-causal sequence, ROC is interior of the circle with radius a. i.e. |z| < a.
- If x(n) is a finite duration two sided sequence, then the ROC is entire z-plane except at z = 0 & z = ∞. Dolby atmos demo clip download.
The concept of ROC can be explained by the following example:
Example 1: Find z-transform and ROC of $a^n u[n] + a^{-}nu[-n-1]$
$Z.T[a^n u[n]] + Z.T[a^{-n}u[-n-1]] = {Z over Z-a} + {Z over Z {-1 over a}}$
$$ ROC: |z| gt a quadquad ROC: |z| lt {1 over a} $$
The plot of ROC has two conditions as a > 1 and a < 1, as you do not know a. Euromax 360i hd new software.
In this case, there is no combination ROC.
Z Transform Table Pdf
Here, the combination of ROC is from $a lt |z| lt {1 over a}$
Hence for this problem, z-transform is possible when a < 1.
Causality and Stability
Causality condition for discrete time LTI systems is as follows:
A discrete time LTI system is causal when
- ROC is outside the outermost pole.
- In The transfer function H[Z], the order of numerator cannot be grater than the order of denominator.
Stability Condition for Discrete Time LTI Systems
A discrete time LTI system is stable when
- its system function H[Z] include unit circle |z|=1.
- all poles of the transfer function lay inside the unit circle |z|=1.
Z-Transform of Basic Signals
x(t) | X[Z] |
---|---|
$delta$ | 1 |
$u(n)$ | ${Zover Z-1}$ |
$u(-n-1)$ | $ -{Zover Z-1}$ |
$delta(n-m)$ | $z^{-m}$ |
$a^n u[n]$ | ${Z over Z-a}$ |
$a^n u[-n-1]$ | $- {Z over Z-a}$ |
$n,a^n u[n]$ | ${aZ over |Z-a|^2}$ |
$n,a^n u[-n-1] $ | $- {aZ over |Z-a|^2}$ |
$a^n cos omega n u[n] $ | ${Z^2-aZ cos omega over Z^2-2aZ cos omega +a^2}$ |
$a^n sin omega n u[n] $ | $ {aZ sin omega over Z^2 -2aZ cos omega +a^2 } $ |