Inverse z transform partial fraction expansion examples

Determine IZT of the following :

X(Z) = 1- 1/2 Z-1 / 1- 1/4 Z-2

Step 1: Find we will whether a given function is the proper form or not.
Here a0 = 1 M=1 and N=2.

Since M<N the function is in the proper form.

Step 2: The given function is 

X(Z) = 1- 1/2 Z-1 / 1- 1/4 Z-2

Multiply numerator and denominator  Z2

X(Z) = Z2 - 1/2 Z1 /  Z2 - 1/4 

Step 3 :  We will obtain the equation of X(Z)/Z as follows :

X(Z) = Z(Z - 1/2)  /  Z2 - 1/4 

X(Z)/Z = (Z - 1/2)  /  Z2 - 1/4 

Step 4: We will the root of denominator  Z2- 1/4 

Here Z2 - 1/4  =  Z2 - (1/2)2


Root is ( Z - 1/2)  and ( Z + 1/2) 

Thus the poles are   ( Z - 1/2) = (Z - P1) = P1 = 1/2

 ( Z + 1/2) = (Z - P2) = P2 = - 1/2

Step 5: Thus the equation can be written as  


X(Z)/Z = (Z - 1/2)  /  ( Z - 1/2) ( Z + 1/2) 

           = 1 / ( Z + 1/2) 

Step 6: The partial fraction expansion form is 

X(Z)/Z = A /  (Z - P1) Thus A=1 we do not have to calculate this value.

Step 7:

X(Z)/Z = 1 / ( Z + 1/2) 


X(Z) = Z /  ( Z + 1/2) 

This function is causal if |Z| > 1/2. The given ROC is  |Z| > 1/2. 

So,

IZT { Z /  Z - PK  } = (PK)n u(n)

x(n) = (-1/2)n u(n)

Z transform stability

In Z transform in DSP that is a necessary and sufficient condition for the system to be BIBO stable is given as below. This article gives information about Z transform stability to better understand this topic.

n=∞
∑   |h(n)| <  ∞
n=-∞

        n=∞
H(Z) = ∑   h(n)  z-n
        n=-∞

              n=∞
|H(Z)| =| ∑   h(n)  z-n |
              n=-∞

The magnitude of the overall sum is less than the sum of the magnitude of individual terms

              n=∞
|H(Z)| < ∑   |h(n)|  | z-n |
              n=-∞

If H(z) is evaluated on the unit circle, then | z-n | = 1

              n=∞
|H(Z)| < ∑   |h(n)|  
              n=-∞

If the system is BIBO stable, then 

n=∞
 ∑   |h(n)|   < ∞
 n=-∞

|H(z)| < ∞

 This condition requires that the unit circle should be present in the ROC full form of H(Z).

Z transform examples

Find Z transform of F[K] =Cos(ak) u[k] :

F(z) = Z (cos(ak) u(k) )

= Z ( ½ (ejak+e-jak) u(K) )

= ½ (Z (ejak  u(k) ) + Z (e-jak) u(K) )

=½ ( Z/ Z-eja + Z/ Z-e-ja)

= ½ ( Z/ Z-eja * Z-e-ja/ Z-e-ja + Z/ Z-e-ja * Z-e-ja/ Z-e-ja )

=½ ( Z2- Ze-ja / Z2- Zeja – Ze-ja +1  + Z2- Zeja / Z2- Zeja – Ze-ja +1)

=½ (2Z2 – (Zeja – Ze-ja) / Z2- Zeja – Ze-ja +1)

= ½(2Z2 – 2Zcos(a)/ Z2-2Zcos(a) +1 

= Z (Z-cos(a)) / Z2-2Zcos(a) +1 

Z transform properties

In this section, we will study the properties of the Z transform. We know that x (n) and X (Z) are Z transform pair and denoted as,

               Z
If  x (n)   ↔   X(Z)  

1. Linearity :

 In this property 

                    Z                                  Z
If  x1 (n) is   ↔   X1(Z)  and  x2 (n)   ↔  X2(Z)  then
              
                                 DTFT
a1 x1 (n)   + a2 x2 (n)    ↔    a1 X1 (Z)   + a2 X2 (Z)  

2. Time shifting : 

 If 
               Z
x (n) is    ↔    X(Z)  then 

              Z
x(n-k)    ↔    Z –k  X(Z)  

3. Scaling in the Z domain :

  If 

              Z                                          Z
x (n) is   ↔    X(Z)  then  a n  x(n)      ↔    X ( Z / a)

4. Time reversal :

 If 
              Z
x (n) is   ↔    X(Z)  then 
  
             Z
x (-n)    ↔    X(Z-1)   

5. Differentiation :

If 
                Z
x (n)   is  ↔    X(Z)  then 

               Z
n x (n)    ↔   -Z  dX(Z) / dZ  


6. Convolution theorem :

                   Z                                      Z
If  x1 (n) is  ↔   X1(Z)  and  x2 (n)     ↔     X2(Z)  then

                                Z
  x1 (n) * x2 (n)  is    ↔     X1(Z) . X2(Z) 

Z transform roc

Definition of ROC : 

The ROC stands for the region of convergence of X(Z) is set for all the values of Z for which X(Z)  attains a finite value. Every time when we find Z transform we must indicate its ROC.

The significance of ROC :
  • ROC will decide whether a system is stable or unstable
  • ROC also determine the type of sequence that means: Causal or noncausal, Finite or infinite
Properties of ROC for Z transform :
  • The ROC is a ring, whose center is at an origin
  • ROC cannot contain any pole
  • If x(n) is causal then ROC is exterior part of a circle of radius say É–
  • If x(n) is anticausal then ROC is interior part of a circle of radius say É–
  • If x(n) is  two-sided sequence then ROC is the intersection of two circles of radii É– and β
  • The ROC must be connected region
  • If ROC of X(Z) include unit circle then and then only the Fourier transform of DT sequence x(n) converges
  • For a finite duration sequence, x(n) the ROC is the entire Z plane except Z = 0 and Z = ∞
Summary of ROC :
  • For Causal system, ROC is entire Z plane except |Z| = 0 
  • For the anti-causal system, ROC is the entire Z plane except |Z| = ∞
  • For two-sided sequence ROC is entire Z plane except  |Z| = 0  and |Z| = ∞

z transform convolution

Statement :

If  x1(n) * x2(n) ↔ X1(Z)  * X2(Z)         


Proof :
  
              ∞
y (n)   = ∑      x(k) h(n-k)
            k=-∞

                 ∞           ∞
y(z)  =       ∑       [   ∑   x(k) h(n-k) ] Z-n
                n=-∞    k=-∞

                  ∞                         ∞
y(z)  =       ∑      x(k) Z-k      ∑  h(n-k)  Z-(n-k)
                k=-∞                n=-∞

n=l and k=0 in second terms;

                  ∞                       ∞
y(z)  =       ∑      x(k) Z-k    ∑  h(l)  Z-l
                k=-∞                l=-∞


Y(z) = X(Z) H(Z)

Application of z transform

The field of signal processing is essentially a field of signal analysis in which they are reduced to their mathematical component and evaluated. One important concept of signal processing is that of the Z transform method which converts unwieldy sequence into a form that can be easily dealt with. Z transform is used for much signal processing application. Here this article gives information about the application of Z transform to know more details about it.
  • Pole-zero description of the discrete-time system
  • Analysis of linear discrete signal
  • Use to analysis digital filter
  • Used to find the frequency response
  • Obtain impulse response estimation
  • Determine the difference equation
  • Analysis of discrete signal
  • Calculation of  a signal to control system
  • Direct computer evolution of frequency response
  • Voice transmission
  • Enhance the electrical as well as mechanical energy to provide a dynamic nature of the system
  • Geometric evolution of frequency response
  • Frequency response estimation via the FFT
  • Used to simulate the continuous systems
  • It also helps in system design, analysis and also checks the system stability
  • For automatic control in telecommunication
  • Used in designing digital filters

Project topic for electronics engineering students

Full forms related to electronics engineering terms

Important definition for electronics / electrical engineering students

Important books to refer for electronics engineering students

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Advantages of Z transform

The Z transform provides a valuable technique for analysis design of discrete time signals as well as discrete time LTI systems. This article gives information about the advantages and disadvantages of Z transform to know more details about it.

Advantages of Z transform :
  • Z transform is used for the digital signal
  • Both Discrete-time signals and linear time-invariant (LTI) systems can be completely characterized using Z transform
  • The stability of the linear time-invariant (LTI)  system can be determined using the Z transform
  • By calculating Z transform of the given signal, DFT and FT can be determined
  • The entire family of digital filters can be obtained from one proto-type design using Z transform
  • The solution of differential equations can be simplified or obtained by using Z transform theorem
  • Mathematical calculations are reduced using the Z transform. For example, consider the convolution operation is transformed into simple multiplication operation 
Disadvantages of Z transform :
  • Z transform cannot apply in the continuous signal

What is z transform

The Z transform has a real and imaginary part like Fourier transform. A plot of imaginary part versus the real part is called a Z plane or complex Z plane.

The pole-zero plot is the main characteristic of discrete time LTI systems full form. We can also check the stability of the system using a pole-zero plot. Z transform used for many signal processing

There are two types of Z transform :

1. Single sided Z transform
2. Double-sided Z transform 

1. Single sided Z transform :

As we know that a single-sided Z transform of discrete time signal x (n) is defined as,

                ∞
X  (Z) =  ∑     x (n)  Z -n

              n=0

Here "Z" is a complex variable. In this equation limits of summation are from the 0 too ∞. So while expanding the summation we will put only positive values of n (from the range n = 0 ton =  ∞). So this is single-sided or one-sided Z- transform.

2. Double-sided Z transform 

As we know that a double-sided Z transform of discrete time signal x (n) is defined as,

                ∞
X  (Z) =  ∑     x (n)  Z -n

              n= - ∞

While expanding the summation we will put both positive as well as negative values of "n". Thus this is called a double-sided Z transform.

How to denote Z transform?

The relationship between x(n) and X(Z) is indicated as given below :

         Z
x (n) ↔ X (Z)

Note that X (Z) is Z transform of x (n), always when Z transform of the sequence is obtained then it is denoted by capital letter, X (Z) Here arrow is bidirectional. This is because we can also obtain x (n) from X (Z) using the inverse Z transform.

The Z transform of x (n) is also denoted as in this form,

X (Z) = Z  { x(n) }

So here x (n) and  X(Z) are called as Z transform pairs.

In same way if we want to analyze a system, which is already represented in frequency domain, a discrete time signal then we go for inverse Z transformation.

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Time shifting property of z transform

Before we learn about time shifting property first let we learn about what is z transform system in digital signal processing subject.

 Statement : 

              Z
If  x(n) ↔ X (Z)

                      Z
Then x (n-k) ↔ Z -k  X (Z)

Proof :
                                 -∞
So here Z{x(n)} = X  (Z) = ∑     x (n)  Z-n
                                 +∞
Z { x(n-k) }  can be written as,


                    -∞
Z{x(n-k)} = ∑     x (n-k)  Z-n
                     +∞

Now  Z-n can be  written as  Z-n  =   Z-(n-k) Z-k  thus the equation 

                    -∞
Z{x(n-k)} = ∑     x (n-k)  Z-(n-k) Z-k
                     +∞

Since the limits of summation are in terms of  'n'  we can take Z-k outside the summation sign.

                             -∞
so Z{x(n-k)} = Z-k  ∑     x (n-k)  Z-(n-k) 
                             +∞

Now put n-k =m on R.H.S 

At  n = -∞, -∞-k = m → m =  -∞
At n = ∞, ∞-k = m  → m =  ∞

                              ∞
Z{x(n-k)} = Z-k  ∑     x (m)  Z-(m) 
                          m = -∞ 
                  
Comparing Equation : 
                           
Z{x(n-k)} = Z-k   X (Z)


= x (n-k) ↔ Z-k   X (Z)

Similarly, we can write 

= x (n+k) ↔ Z+k   X (Z)
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