EE-323 DIIGITAL SIGNAL PROCESSING
LAB REPORT 02
6TH SEMESTER (SPRING 2018)
Submitted to: Engr. Komal Munir
Submitted by: Syeda Ume Rubab
16-EE-03
Data presentation (0-5) Data Analysis and interpretation (0-5)
DEPARTMENT OF ELECTRICAL ENGINEERING
UNIVERSITY OF ENGINEERING AND TECHNOLOGY
TAXILA
LAB No. 02
Evaluation of Z-transform and Pole/Zero Plots
Objective:
To understand how to perform Z-Transform in MATLAB. To understand different
MATLAB command related to the evaluation of transfer functions, poles, zeros and
their plots.
Introduction:
Theory:
NAME: SYEDA UME RUBAB Reg #:16-EE-03
NAME: SYEDA UME RUBAB Reg #:16-EE-03
Lab Tasks:
1. Write a user-defined MATLAB function to compute Z-transform through following
relation:
∞
𝑋(𝑧) = ∑ 𝑥(𝑛)𝑧 −𝑛
𝑛=−∞
ℎ𝑒𝑟𝑒 𝑥(𝑛)𝑖𝑠 𝑑𝑖𝑠𝑐𝑟𝑒𝑡𝑒 − 𝑡𝑖𝑚𝑒 𝑖𝑛𝑝𝑢𝑡 𝑎𝑛𝑑 𝑧 𝑖𝑠 𝑠𝑦𝑚𝑏𝑜𝑙𝑖𝑐 𝑣𝑎𝑟𝑖𝑏𝑙𝑒
Code:
function X=z_transform(x)
syms z;
sum=0;
n=length(x)-1;
for i=0:1:n;
X=x(i+1)*z^(-1*i);
sum=sum+X;
end
X=sum
end
2. Use the function defined in Task 1 to compute Z-transform of the following signals
i. x(n) = cos(2πFn) here F = 10Hz
Code:
F=10;
x=cos(2*pi*F*n);
z_transform(x)
Output:
X =1/z + 1/z^2 + 1/z^3 + 1/z^4 + 1
ii. x(n) = [2,4,5,7,1]
Code:
x=[2 4 5 7 1];
z_transform(x)
Output:
X = 4/z + 5/z^2 + 7/z^3 + 1/z^4 + 2
NAME: SYEDA UME RUBAB Reg #:16-EE-03
iii. x(n) = [0,0,1,2,5,7,0,1]
Code:
x=[0 0 1 2 5 7 0 1];
z_transform(x)
Output:
X = 1/z^2 + 2/z^3 + 5/z^4 + 7/z^5 + 1/z^7
3. Use MATLAB command ‘ztrans’ to compute Z-transform of following (use ‘syms’
command to define inputs).
𝒌𝑻 𝒇𝒐𝒓 𝒌 ≥ 𝟎
i. 𝒂𝒌 = { 𝒔
𝟎 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
Code:
syms a k T;
a=k*T;
X=ztrans(a)
Output:
X= (T*z)/(z - 1)^2
ii. x[n]= an
Code:
syms a n;
x=a^n;
X=ztrans(x)
Output:
X = -z/(a - z)
4. We know that a transfer function is a ratio of polynomials with ascending power of 𝑧-1.
Transfer functions are specified by their numerator and denominator polynomials. In
MATLAB, NUM and DEN are row vectors listing the numerator and denominator
coefficients in descending order of z by default. For example, the polynomial 𝑧2 + 2𝑧 +
10 is specified as [1 2 10]. For discrete-time transfer functions, it is highly
recommended to make the length of the numerator and denominator equal to ensure
correct results. Define vectors in MATLAB to express following transfer function with
numerator and denominator polynomials.
4 − 5𝑧 −1
𝐻(𝑧) =
8 + 3𝑧 −1 + 𝑧 −2
NAME: SYEDA UME RUBAB Reg #:16-EE-03
Use ‘tf” command to generate transfer function from NUM and DEN vectors like
tf(NUM,DEN,1). Use ‘roots’ command to compute roots of NUM and DEN
polynomials. Show the pole/zero plot of H(z) using ‘zplane’ command.
Code:
num=[4 -5 0]
den=[8 3 1]
z=tf(num,den,1)
x=roots(num)
y=roots(den)
zplane(x,y)
Output:
z=
4 z^2 - 5 z
---------------
8 z^2 + 3 z + 1
Sample time: 1 seconds
Discrete-time transfer function.
x=
0
1.2500
y=
-0.1875 + 0.2997i
-0.1875 - 0.2997i
NAME: SYEDA UME RUBAB Reg #:16-EE-03
5. Consider the following transfer function:
4𝑧 −1 (1 − 0.5𝑧 −1 )(1 − 2𝑧 −1 )
𝐻(𝑧) =
(1 + 0.3𝑧 −1 )(1 − 0.4𝑧 −1 )(1 + 0.6𝑧 −1 )
Represent this transfer function in vector and object form and then computer zeros,
poles and gain using ‘tf2zp’ command and plot them using ‘pzmap’ command.
Code:
a=conv([0 4],[1 -0.5]);
num=conv([a],[1 -2])
b=conv([1 0.3],[1 -0.4]);
den=conv([b],[1 0.6])
H=tf(num,den,1)
[z,p,k]=tf2zp(num,den)
pzmap(H)
Output:
num = 0 4 -10 4
den = 1.0000 0.5000 -0.1800 -0.0720
H = 4 z^2 - 10 z + 4
------------------------------
z^3 + 0.5 z^2 - 0.18 z - 0.072
z = 2.0000
0.5000
p = 0.4000
-0.6000
-0.3000
k=4
NAME: SYEDA UME RUBAB Reg #:16-EE-03
After Lab Questions:
1. What are the differences between MATLAB commands ‘zplane’ and pzmap?
2. How complex variable ‘z’ is related to complex variable ‘s’? Give expression.
3. Give two practical applications of Z-Transform.
4. How the values of real part (σ) of ‘s’ be translated into stability conditions in z-domain?
Discussion/ Conclusion:
NAME: SYEDA UME RUBAB Reg #:16-EE-03
