Z transform inverse

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

Mathematically it can be represented as : 

x(n) = z^-1 X(Z)

x(n) = Z-1 X(Z)

Here x(n) is the signal in time domain and X(Z) is the signal in the frequency domain to be represented.

And here also define if we want to represent the above equation in integral format then we can write it as

x(n) = (1/2∏j) ∫ X(z) Z-1 dz

Here the integral is over close path C. This is within the region of conversions (ROC) of the x(z) and it does contain the origin. Now here this article also gives the information about how to find inverse Z transform.

Method to find Inverse Z - transform :

We follow the following four-way to determine the inverse Z transformation system.

• Long division method
• Partial fraction expansion method
• Contour or residue integral method

Inverse Z transform using long division method

While carrying out the long division method, it always converts the given expression in the simplest form. That means as far as possible, it should be in form of Z divided by some polynomial. |Z| > 1 it is causal sequence. For the causal sequence the denominator polynomial should have maximum power of Z transform on its left. Thus the given expression is in the total proper form.

Find inverse Z transform of  X(z) is = Z/Z-1   |Z|>1 

X(Z) = 1 + Z^-1 +  Z^-2  +  Z^-3+..........

             ∞

X(Z) = ∑ x(n) Z^-n

          n=0 

X(Z) is = x(0) + x(1) Z^-1 + x(2) Z^-2  + x(3) Z^-3

^{X(n) =  {}x(0) ,  x(1), x(2), x(3) }

x(n) = { 1, 1, 1, 1 }

x(n) = u(n)

Z^-1{Z^-1/Z-1} = u(n)

This is a standard Z transform pair.

Z transform partial fraction expansion

Partial fraction expansion method is possible only when Z transform definition which is rational in nature, That means they are expressed of two polynomials.

X(Z) = N(Z) / D(Z)

^{= b}₀ ^{+  b}₁  Z^-1  + ^b₂  Z^{-2   + …….+ b}_M Z^{-M  /  a}₀ ^{+  a}₁  Z^-1  + ^a₂  Z^{-2   + …….+ a}_N Z^-N

  ^b₀ ^{,  b}₁  ^, b₂   ^{, b}_M ⁼  Coefficients of numerator 

 ^a₀ ^, a₁  ^,  a₂  ..... ^a_N  ⁼  Coefficients of Denominator

M = Degree of numerator 

N =  Degree of denominator

N(Z) = Numerator  polynomial

D(Z) = Denominator polynomial

Step to follow the partial fraction expansion method :

Step 1: Check whether the given function is the proper form or not. The function is said to be in proper form when the following conditions are satisfied. 

• The coefficient ^a₀ in the above equation should be equal to 1. If  ^a₀  not equal to 1 then the polynomial is adjusted accordingly.
• In equation ^a₀  not equal to 1 and the degree of numerator should be less than the degree of the denominator(M<N). If this condition is satisfied then the long division is carried out to make M<N.

Step 2: Multiply the numerator and denominator by Z^N. That means to convert the function in term of the positive power of Z.

Step 3 :  Obtain  the equation X(Z) / Z

Step 4: Factorize the denominator and obtain the roots. Then the denominator will be in the form 

(Z - P₁ ) (Z - P₂ ) (Z - P₃ )............(Z -P_N )

Here P₁  P₂  P₃  ........P_N is called as poles.

Step 5 :  Write down the equation in partial expansion form as follows :

X(Z) / Z =  A₁ /  Z - P₁  + A₂ / Z - P₂  + .............+   A_N / Z - P_N

A_1, A_2,  A_{3 .....}A_N  are coefficient. the coefficient A_K is  calculated as 

 A_K =  (Z - P_K ). X(Z) / Z  | where Z=P_K

Step 6 : By calculating  transfer  of a_1, a_2,  a_{3 .....}a_n   z to R.H.S of equation. Now we standard from pair to obtain inverse Z transform

x(n) = IZT { Z /  Z - P_K  } = (P_K)ⁿ u(n) if ROC : |Z| > | P_K |
that means causal sequence 

AND 

x(n) = IZT { Z /  Z - P_K  } = - (P_K)ⁿ u(-n-1) if ROC : |Z| < | P_K |
that means anticausal sequence 

Now let us check it out the example of partial fraction method to learn more details about this article.

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) = P_{1 =} 1/2

 ( Z + 1/2) = (Z - P₂) = P_{2 =} - 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 - P_K  } = (P_K)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 (e^jak+e^-jak) u(K) )

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

=½ ( Z/ Z-e^ja + Z/ Z-e^-ja)

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

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

=½ (2Z² – (Ze^ja – Ze^-ja) / Z²- Ze^ja – 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 ɖ
• If x(n) is anticausal then ROC is interior part of a circle of radius say ɖ
• If x(n) is  two-sided sequence then ROC is the intersection of two circles of radii ɖ 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

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.

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)