(New page: == Z transform == Z transform is a general form of DTFT. If <span class="texhtml">''x''[''n'']</span> is a discrete periodic funtion, DFT of this function is <math>x[k] = \sum_{n=0}^{N-1}...)
 
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Let's say <span class="texhtml">''x''[''n'']</span> is discrete nonperiodic function. Nonperiodic function is also a function with period <math>\infty</math>. Therefore DTFT of this function can be DFT with <math>N=\infty</math>. <math>\lim_{N\to\infty}\sum_{n=0}^{N-1}x[n]e^{-j\frac{{2\pi}k n}{N}}=\sum_{n=0}^{\infty}x[n]e^{-j{\omega}n}</math>, where <math>\omega=\lim_{N\to\infty}\frac{{2\pi}k}{N}</math>  
 
Let's say <span class="texhtml">''x''[''n'']</span> is discrete nonperiodic function. Nonperiodic function is also a function with period <math>\infty</math>. Therefore DTFT of this function can be DFT with <math>N=\infty</math>. <math>\lim_{N\to\infty}\sum_{n=0}^{N-1}x[n]e^{-j\frac{{2\pi}k n}{N}}=\sum_{n=0}^{\infty}x[n]e^{-j{\omega}n}</math>, where <math>\omega=\lim_{N\to\infty}\frac{{2\pi}k}{N}</math>  
  
<math>X(w)=\sum_{n=0}^{\infty}x[n]e^{-j{\omega}n}</math> ---&gt; ==DTFT==
+
<math>X(w)=\sum_{n=0}^{\infty}x[n]e^{-j{\omega}n}</math> ---&gt; DTFT  
  
 
The difference between DFT and DTFT is that DFT has a discrete function with k and DTFT is a continuous function with <span class="texhtml">ω</span>.  
 
The difference between DFT and DTFT is that DFT has a discrete function with k and DTFT is a continuous function with <span class="texhtml">ω</span>.  
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Let's generalize the DTFT. By substitution of real value frequency <span class="texhtml">ω</span> into complex frequency value <span class="texhtml">''s'' = σ + ''i''ω</span>, DTFT is now discrete function of Laplace transform.  
 
Let's generalize the DTFT. By substitution of real value frequency <span class="texhtml">ω</span> into complex frequency value <span class="texhtml">''s'' = σ + ''i''ω</span>, DTFT is now discrete function of Laplace transform.  
  
<math> X(s)=sum_{n=0}^{\infty}x[n]e^{-sn}</math> ---&gt; ==Laplace Transform==
+
<math> X(s)=\sum_{n=0}^{\infty}x[n]e^{-sn}</math> ---&gt; Laplace Transform  
  
 
When <span class="texhtml">''e''<sup>''s''</sup> = ''z''</span>, it is a Z transform.  
 
When <span class="texhtml">''e''<sup>''s''</sup> = ''z''</span>, it is a Z transform.  
  
<math> X(z)=Z[x[n]]=sum_{n=0}^{\infty}x[n]z^{-n}</math> ---&gt; ==Z Transform==
+
<math> X(z)=Z[x[n]]=\sum_{n=0}^{\infty}x[n]z^{-n}</math> ---&gt; Z Transform  
  
 
The first reason why we are using z transform instead of laplace is becuase it is easier to calculate the geometric series by substitution of variable from s to z. Also properties of power series with differential equation is useful.  
 
The first reason why we are using z transform instead of laplace is becuase it is easier to calculate the geometric series by substitution of variable from s to z. Also properties of power series with differential equation is useful.  
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As the variable <span class="texhtml">''e''<sup>''s''</sup></span> is substituted with z, region of convergence changes from the complex plane. The region of convergence in the complex plane for Z transform is outside region of a unit circle <span class="texhtml"> | ''z'' |  = ''r'' = 1</span>, where r is radius of a unit circle.  
 
As the variable <span class="texhtml">''e''<sup>''s''</sup></span> is substituted with z, region of convergence changes from the complex plane. The region of convergence in the complex plane for Z transform is outside region of a unit circle <span class="texhtml"> | ''z'' |  = ''r'' = 1</span>, where r is radius of a unit circle.  
  
==Inverse Z transform== can be stated as  
+
Inverse Z transform can be stated as  
  
 
<math> x[n]=\frac{1}{j{2\pi}}\oint{X(z)z^{n-1}dz} </math>  
 
<math> x[n]=\frac{1}{j{2\pi}}\oint{X(z)z^{n-1}dz} </math>  
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  When <math>x[n]=1 (n{\ge}0)</math>
 
  When <math>x[n]=1 (n{\ge}0)</math>
     <span class="texhtml">''x''[''n''] = 0(''n''''l''''e''''s''''s'')</span>
+
     <span class="texhtml">''x''[''n''] = 0(''n'' &lt; 0)</span>
  
 
<math>X(z)=\sum_{n=0}^{\infty}x[n]z^{n}=\sum_{n=0}^{\infty}1\cdot z^{-n}=\frac{1}{1-z^{-1}}</math>  
 
<math>X(z)=\sum_{n=0}^{\infty}x[n]z^{n}=\sum_{n=0}^{\infty}1\cdot z^{-n}=\frac{1}{1-z^{-1}}</math>  

Revision as of 09:21, 12 October 2010

Z transform

Z transform is a general form of DTFT. If x[n] is a discrete periodic funtion, DFT of this function is $ x[k] = \sum_{n=0}^{N-1} x[n]e^{-j\frac{{2\pi}k n}{N}} $

Let's say x[n] is discrete nonperiodic function. Nonperiodic function is also a function with period $ \infty $. Therefore DTFT of this function can be DFT with $ N=\infty $. $ \lim_{N\to\infty}\sum_{n=0}^{N-1}x[n]e^{-j\frac{{2\pi}k n}{N}}=\sum_{n=0}^{\infty}x[n]e^{-j{\omega}n} $, where $ \omega=\lim_{N\to\infty}\frac{{2\pi}k}{N} $

$ X(w)=\sum_{n=0}^{\infty}x[n]e^{-j{\omega}n} $ ---> DTFT

The difference between DFT and DTFT is that DFT has a discrete function with k and DTFT is a continuous function with ω.

Let's generalize the DTFT. By substitution of real value frequency ω into complex frequency value s = σ + iω, DTFT is now discrete function of Laplace transform.

$ X(s)=\sum_{n=0}^{\infty}x[n]e^{-sn} $ ---> Laplace Transform

When es = z, it is a Z transform.

$ X(z)=Z[x[n]]=\sum_{n=0}^{\infty}x[n]z^{-n} $ ---> Z Transform

The first reason why we are using z transform instead of laplace is becuase it is easier to calculate the geometric series by substitution of variable from s to z. Also properties of power series with differential equation is useful.

As the variable es is substituted with z, region of convergence changes from the complex plane. The region of convergence in the complex plane for Z transform is outside region of a unit circle | z | = r = 1, where r is radius of a unit circle.

Inverse Z transform can be stated as

$ x[n]=\frac{1}{j{2\pi}}\oint{X(z)z^{n-1}dz} $

These are the examples of Z transform.

1. Unit step

When $ x[n]=1 (n{\ge}0) $
    x[n] = 0(n < 0)

$ X(z)=\sum_{n=0}^{\infty}x[n]z^{n}=\sum_{n=0}^{\infty}1\cdot z^{-n}=\frac{1}{1-z^{-1}} $

2. Power series

$ X(z)=\sum_{n=0}^{\infty}x[n]z^{n}=\sum_{n=0}^{\infty}a^{n} z^{-n}=\frac{1}{1-az^{-1}} $

3. Exponential funtion

$ X(z)=\sum_{n=0}^{\infty}x[n]z^{n}=\sum_{n=0}^{\infty}e^{-an} z^{-n}=\sum_{n=0}^{\infty}[e^{-a} z^{-1}]^{n}=\frac{1}{1-e^{-a}z^{-n}} $

4. Sinusoidal function

$ X(z)=\sum_{n=0}^{\infty}x[n]z^{n}=\sum_{n=0}^{\infty}\frac{e^{jn{\omega}} -e^{-jn{\omega}}} {2j} z^{-n} $ $ =\frac{1}{2j} (\frac{1}{1-e^{j\omega}z^{-1}}-\frac{1}{1-e^{-j\omega}z^{-1}}) $ $ =\frac{1}{2j} (\frac{-e^{-j\omega}z^{-1}+e^{j\omega}z^{-1}}{1-e^{-j\omega}z^{-1}-e^{j\omega}z^{-1}+z^{-2}}) $ $ =\frac{z^{-1}sin(\omega)}{1-2z^{-1}cos(\omega)+z^{-2}} $

Properties of Z transform

1. Linearity

Z(a'x[n] + b'y[n]) = a'X(z) + b'Y(z)

2. Time Delay

$ Z(x[n-k])=z^{-k}[X(z)+\sum_{n=1}^{k}x[-n]z^{n}] $

3. Time Advance

$ Z(x[n+k])=z^{k}[X(z)-\sum_{n=0}^{k-1}x[n]z^{-n}] $

4. Time Convolution Theorem

Z(x[n] * y[n]) = X(z)Y(z)

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010