Discrete Time Fourier Transform(DTFT) exists for energy and power signals. Z-transform also exists for neither energy nor Power (NENP) type signal, up to a certain extent only. The replacement \(z=e^{jw}\) is used for Z-transform to DTFT conversion only for absolutely summable signal.
So, the Z-transform of the discrete time signal x(n) in a power series can be written as −
$$X(z) = \sum_{n-\infty}^\infty x(n)Z^{-n}$$The above equation represents a two-sided Z-transform equation.
Generally, when a signal is Z-transformed, it can be represented as −
$$X(Z) = Z[x(n)]$$Or $$x(n) \longleftrightarrow X(Z)$$
If it is a continuous time signal, then Z-transforms are not needed because Laplace transformations are used. However, Discrete time signals can be analyzed through Z-transforms only.
Mapping of S Plane onto Z Plane - Video tutorial
Region of Convergence
Region of Convergence is the range of complex variable Z in the Z-plane. The Z- transformation of the signal is finite or convergent. So, ROC represents those set of values of Z, for which X(Z) has a finite value.
Properties of ROC
- ROC does not include any pole.
- For right-sided signal, ROC will be outside the circle in Z-plane.
- For left sided signal, ROC will be inside the circle in Z-plane.
- For stability, ROC includes unit circle in Z-plane.
- For Both sided signal, ROC is a ring in Z-plane.
- For finite-duration signal, ROC is entire Z-plane.
The Z-transform is uniquely characterized by −
- Expression of X(Z)
- ROC of X(Z)
Signals and their ROC
| x(n) | X(Z) | ROC |
|---|---|---|
| $$\delta(n)$$ | $$1$$ | Entire Z plane |
| $$U(n)$$ | $$1/(1-Z^{-1})$$ | Mod(Z)>1 |
| $$a^nu(n)$$ | $$1/(1-aZ^{-1})$$ | Mod(Z)>Mod(a) |
| $$-a^nu(-n-1)$$ | $$1/(1-aZ^{-1})$$ | Mod(Z)<Mod(a) |
| $$na^nu(n)$$ | $$aZ^{-1}/(1-aZ^{-1})^2$$ | Mod(Z)>Mod(a) |
| $$-a^nu(-n-1)$$ | $$aZ^{-1}/(1-aZ^{-1})^2$$ | Mod(Z)<Mod(a) |
| $$U(n)\cos \omega n$$ | $$(Z^2-Z\cos \omega)/(Z^2-2Z \cos \omega +1)$$ | Mod(Z)>1 |
| $$U(n)\sin \omega n$$ | $$(Z\sin \omega)/(Z^2-2Z \cos \omega +1)$$ | Mod(Z)>1 |
Example
Let us find the Z-transform and the ROC of a signal given as \(x(n) = \lbrace 7,3,4,9,5\rbrace\), where origin of the series is at 3.
Solution − Applying the formula we have −
\(X(z) = \sum_{n=-\infty}^\infty x(n)Z^{-n}\)
\(= \sum_{n=-1}^3 x(n)Z^{-n}\)
\(= x(-1)Z+x(0)+x(1)Z^{-1}+x(2)Z^{-2}+x(3)Z^{-3}\)
\(= 7Z+3+4Z^{-1}+9Z^{-2}+5Z^{-3}\)
ROC is the entire Z-plane excluding Z = 0, ∞, -∞
DSP - Z-Transform Properties
In this chapter, we will understand the basic properties of Z-transforms.
Linearity
It states that when two or more individual discrete signals are multiplied by constants, their respective Z-transforms will also be multiplied by the same constants.
Mathematically,
$$a_1x_1(n)+a_2x_2(n) = a_1X_1(z)+a_2X_2(z)$$Proof − We know that,
\(X(Z) = \sum_{n=-\infty}^\infty x(n)Z^{-n}\)\(= \sum_{n=-\infty}^\infty (a_1x_1(n)+a_2x_2(n))Z^{-n}\)
\(= a_1\sum_{n = -\infty}^\infty x_1(n)Z^{-n}+a_2\sum_{n = -\infty}^\infty x_2(n)Z^{-n}\)
\(= a_1X_1(z)+a_2X_2(z)\) (Hence Proved)
Here, the ROC is \(ROC_1\bigcap ROC_2\).
Time Shifting
Time shifting property depicts how the change in the time domain in the discrete signal will affect the Z-domain, which can be written as;
$$x(n-n_0)\longleftrightarrow X(Z)Z^{-n}$$Or $$x(n-1)\longleftrightarrow Z^{-1}X(Z)$$
Proof −
Let \(y(P) = X(P-K)\)
\(Y(z) = \sum_{p = -\infty}^\infty y(p)Z^{-p}\)
\(= \sum_{p = -\infty}^\infty (x(p-k))Z^{-p}\)
Let s = p-k
\(= \sum_{s = -\infty}^\infty x(s)Z^{-(s+k)}\)
\(= \sum_{s = -\infty}^\infty x(s)Z^{-s}Z^{-k}\)
\(= Z^{-k}[\sum_{s=-\infty}^\infty x(m)Z^{-s}]\)
\(= Z^{-k}X(Z)\) (Hence Proved)
Here, ROC can be written as Z = 0 (p>0) or Z = ∞(p<0)
Example
U(n) and U(n-1) can be plotted as follows
Z-transformation of U(n) cab be written as;
$$\sum_{n = -\infty}^\infty [U(n)]Z^{-n} = 1$$
Z-transformation of U(n-1) can be written as;
$$\sum_{n = -\infty}^\infty [U(n-1)]Z^{-n} = Z^{-1}$$
So here \(x(n-n_0) = Z^{-n_0}X(Z)\) (Hence Proved)
Time Scaling
Time Scaling property tells us, what will be the Z-domain of the signal when the time is scaled in its discrete form, which can be written as;
$$a^nx(n) \longleftrightarrow X(a^{-1}Z)$$Proof −
Let \(y(p) = a^{p}x(p)\)
\(Y(P) = \sum_{p=-\infty}^\infty y(p)Z^{-p}\)
\(= \sum_{p=-\infty}^\infty a^px(p)Z^{-p}\)
\(= \sum_{p=-\infty}^\infty x(p)[a^{-1}Z]^{-p}\)
\(= X(a^{-1}Z)\)(Hence proved)
ROC: = Mod(ar1) < Mod(Z) < Mod(ar2) where Mod = Modulus
Example
Let us determine the Z-transformation of \(x(n) = a^n \cos \omega n\) using Time scaling property.
Solution −
We already know that the Z-transformation of the signal \(\cos (\omega n)\) is given by −
$$\sum_{n=-\infty}^\infty(\cos \omega n)Z^{-n} = (Z^2-Z \cos \omega)/(Z^2-2Z\cos \omega +1)$$
Now, applying Time scaling property, the Z-transformation of \(a^n \cos \omega n\) can be written as;
\(\sum_{n=-\infty}^\infty(a^n\cos \omega n)Z^{-n} = X(a^{-1}Z)\)
\(= [(a^{-1}Z)^2-(a^{-1}Z \cos \omega n)]/((a^{-1}Z)^2-2(a^{-1}Z \cos \omega n)+1)\)
\(= Z(Z-a \cos \omega)/(Z^2-2az \cos \omega+a^2)\)
Successive Differentiation
Successive Differentiation property shows that Z-transform will take place when we differentiate the discrete signal in time domain, with respect to time. This is shown as below.
$$\frac{dx(n)}{dn} = (1-Z^{-1})X(Z)$$Proof −
Consider the LHS of the equation − \(\frac{dx(n)}{dn}\)
\(= \frac{[x(n)-x(n-1)]}{[n-(n-1)]}\)
\(= x(n)-X(n-1)\)
\(= x(Z)-Z^{-1}x(Z)\)
\(= (1-Z^{-1})x(Z)\) (Hence Proved)
ROC: R1< Mod (Z) <R2
Example
Let us find the Z-transform of a signal given by \(x(n) = n^2u(n)\)
By property we can write
\(Zz[nU(n)] = -Z\frac{dZ[U(n)]}{dz}\)
\(= -Z\frac{d[\frac{Z}{Z-1}]}{dZ}\)
\(= Z/((Z-1)^2\)
\(= y(let)\)
Now, Z[n.y] can be found out by again applying the property,
\(Z(n,y) = -Z\frac{dy}{dz}\)
\(= -Z\frac{d[Z/(Z-1)^3]}{dz}\)
\(= Z(Z+1)/(Z-1)^2\)
Convolution
This depicts the change in Z-domain of the system when a convolution takes place in the discrete signal form, which can be written as −
$$x_1(n)*x_2(n) \longleftrightarrow X_1(Z).X_2(Z)$$
Proof −
\(X(Z) = \sum_{n = -\infty}^\infty x(n)Z^{-n}\)
\(= \sum_{n=-\infty}^\infty[\sum_{k = -\infty}^\infty x_1(k)x_2(n-k)]Z^{-n}\)
\(= \sum_{k = -\infty}^\infty x_1(k)[\sum_n^\infty x_2(n-k)Z^{-n}]\)
\(= \sum_{k = -\infty}^\infty x_1(k)[\sum_{n = -\infty}^\infty x_2(n-k)Z^{-(n-k)}Z^{-k}]\)
Let n-k = l, then the above equation cab be written as −
\(X(Z) = \sum_{k = -\infty}^\infty x_1(k)[Z^{-k}\sum_{l=-\infty}^\infty x_2(l)Z^{-l}]\)
\(= \sum_{k = -\infty}^\infty x_1(k)X_2(Z)Z^{-k}\)
\(= X_2(Z)\sum_{k = -\infty}^\infty x_1(Z)Z^{-k}\)
\(= X_1(Z).X_2(Z)\) (Hence Proved)
ROC:\(ROC\bigcap ROC2\)
Example
Let us find the convolution given by two signals
\(x_1(n) = \lbrace 3,-2,2\rbrace\) ...(eq. 1)
\(x_2(n) = \lbrace 2,0\leq 4\quad and\quad 0\quad elsewhere\rbrace\) ...(eq. 2)
Z-transformation of the first equation can be written as;
\(\sum_{n = -\infty}^\infty x_1(n)Z^{-n}\)
\(= 3-2Z^{-1}+2Z^{-2}\)
Z-transformation of the second signal can be written as;
\(\sum_{n = -\infty}^\infty x_2(n)Z^{-n}\)
\(= 2+2Z^{-1}+2Z^{-2}+2Z^{-3}+2Z^{-4}\)
So, the convolution of the above two signals is given by −
\(X(Z) = [x_1(Z)^*x_2(Z)]\)
\(= [3-2Z^{-1}+2Z^{-2}]\times [2+2Z^{-1}+2Z^{-2}+2Z^{-3}+2Z^{-4}]\)
\(= 6+2Z^{-1}+6Z^{-2}+6Z^{-3}+...\quad...\quad...\)
Taking the inverse Z-transformation we get,
\(x(n) = \lbrace 6,2,6,6,6,0,4\rbrace\)
Initial Value Theorem
If x(n) is a causal sequence, which has its Z-transformation as X(z), then the initial value theorem can be written as;
$$X(n)(at\quad n = 0) = \lim_{z \to \infty} X(z)$$
Proof − We know that,
$$X(Z) = \sum_{n = 0} ^\infty x(n)Z^{-n}$$
Expanding the above series, we get;
\(= X(0)Z^0+X(1)Z^{-1}+X(2)Z^{-2}+...\quad...\)
\(= X(0)\times 1+X(1)Z^{-1}+X(2)Z^{-2}+...\quad...\)
In the above case if Z → ∞ then \(Z^{-n}\rightarrow 0\) (Because n>0)
Therefore, we can say;
\(\lim_{z \to \infty}X(z) = X(0)\) (Hence Proved)
Final Value Theorem
Final Value Theorem states that if the Z-transform of a signal is represented as X(Z) and the poles are all inside the circle, then its final value is denoted as x(n) or X(∞) and can be written as −
$$X(\infty) = \lim_{n \to \infty}X(n) = \lim_{z \to 1}[X(Z)(1-Z^{-1})]$$
Conditions −
- It is applicable only for causal systems.
- \(X(Z)(1-Z^{-1})\) should have poles inside the unit circle in Z-plane.
Proof − We know that
\(Z^+[x(n+1)-x(n)] = \lim_{k \to \infty}\sum_{n=0}^kZ^{-n}[x(n+1)-x(n)]\)
\(\Rightarrow Z^+[x(n+1)]-Z^+[x(n)] = \lim_{k \to \infty}\sum_{n=0}^kZ^{-n}[x(n+1)-x(n)]\)
\(\Rightarrow Z[X(Z)^+-x(0)]-X(Z)^+ = \lim_{k \to \infty}\sum_{n = 0}^kZ^{-n}[x(n+1)-x(n)]\)
Here, we can apply advanced property of one-sided Z-Transformation. So, the above equation can be re-written as;
$$Z^+[x(n+1)] = Z[X(2)^+-x(0)Z^0] = Z[X(Z)^+-x(0)]$$
Now putting z = 1 in the above equation, we can expand the above equation −
$$\lim_{k \to \infty}{[x(1)-x(0)+x(6)-x(1)+x(3)-x(2)+...\quad...\quad...+x(x+1)-x(k)]}$$
This can be formulated as;
\(X(\infty) = \lim_{n \to \infty}X(n) = \lim_{z \to 1}[X(Z)(1-Z^{-1})]\)(Hence Proved)
Example
Let us find the Initial and Final value of x(n) whose signal is given by
$$X(Z) = 2+3Z^{-1}+4Z^{-2}$$
Solution − Let us first, find the initial value of the signal by applying the theorem
\(x(0) = \lim_{z \to \infty}X(Z)\)
\(= \lim_{z \to \infty}[2+3Z^{-1}+4Z^{-2}]\)
\(= 2+(\frac{3}{\infty})+(\frac{4}{\infty}) = 2\)
Now let us find the Final value of signal applying the theorem
\(x(\infty) = \lim_{z \to \infty}[(1-Z^{-1})X(Z)]\)
\(= \lim_{z \to \infty}[(1-Z^{-1})(2+3Z^{-1}+4Z^{-2})]\)
\(= \lim_{z \to \infty}[2+Z^{-1}+Z^{-2}-4Z^{-3}]\)
\(= 2+1+1-4 = 0\)
Some other properties of Z-transform are listed below −
Differentiation in Frequency
It gives the change in Z-domain of the signal, when its discrete signal is differentiated with respect to time.
$$nx(n)\longleftrightarrow -Z\frac{dX(z)}{dz}$$
Its ROC can be written as;
$$r_2< Mod(Z)< r_1$$
Example
Let us find the value of x(n) through Differentiation in frequency, whose discrete signal in Z-domain is given by $$x(n)\longleftrightarrow X(Z) = log(1+aZ^{-1})$$
By property, we can write that
\(nx(n)\longleftrightarrow -Z\frac{dx(Z)}{dz}\)
\(= -Z[\frac{-aZ^{-2}}{1+aZ^{-1}}]\)
\(= (aZ^{-1})/(1+aZ^{-1})\)
\(= 1-1/(1+aZ^{-1})\)
\(nx(n) = \delta(n)-(-a)^nu(n)\)
\(\Rightarrow x(n) = 1/n[\delta(n)-(-a)^nu(n)]\)
Multiplication in Time
It gives the change in Z-domain of the signal when multiplication takes place at discrete signal level.
$$x_1(n).x_2(n)\longleftrightarrow(\frac{1}{2\Pi j})[X_1(Z)*X_2(Z)]$$
Home work
- Determine the Z-transform of the following sequences and find ROC
i. \(x(n) = (n + 2) (1/2)^n u(n)\)
ii. \(x(n) = (2/3)^n u(n) + (3/4) u(n - 1)\) - The z-transform of a signal is given by \(X(z) = \frac{1z^{-1} (1 - z^{-4})}{4(1 - z^{-1})^2}\) . Find the final value.