UNIT II DISCRETE TIME SYSTEM ANALYSIS 6+6 Z-transform and its properties, inverse z-transforms; difference equation – Solution by ztransform, application to discrete systems - Stability analysis, frequency response –Convolution – Discrete Time Fourier transform , magnitude and phase representation

1. WEBINAR ON DISCRETE TIME SYSTEM
ANALYSIS
Day 5 – 24/7/2020
1
K.Vijay Anand - Associate Professor
Department of Electronics and Instrumentation Engineering
R.M.K Engineering College

2. Recap:
SOLUTION OF DIFFERENCE EQUATION

3. Agenda
Convolution :
Linear Convolution
 Circular Convolution

4. CONVOLUTION

5. DIFFERENCE BETWEEN LINEAR AND CIRCULAR
LINEAR CONVOLUTION
If x(n) have L number of samples
the h(n) have M number of
samples the result y(n)wil have
N=L+M-1 number of samples.
CIRCULAR CONVOLUTON
If x(n) have L number of samples
the h(n) have M number of
samples the result y(n)wil have
N=MAX[L,M] number of samples

6. CONVOLUTION
• LINEAR CONVOLUTION
• CIRCULAR CONVOLUTION

7. LINEAR CONVOLUTION
• GRAPHICAL METHOD
• TABULATION METHOD
• MATRIX MULTIPLICATION METHOD

26. Problem 1
• Determine the output response y(n)=(1,1,1) & x(n)=(1,2,3,1)

30. CIRCULAR CONVOLUTION
• CONCENTRIC CIRCLE METHOD
• MATRIX MULTIPLICATION METHOD

31. STEPS INVOLVED IN CONCENTRIC CIRCLE METHOD
STEP 1 : ARRANGE THE VALUES OF X (N) ANTICLOCKWISE DIRECTON IN OUTER CRCLE
STEP 2 : ARRANGE VALUE OF H (N) IN CLOCKWISE DIRECCTION IN INNER CIRCLE
STEP 3 : MULTIPLY IN CORRESPODING VALUES AND ALL THE ADD VALUES
STEP 4 : IF THE VALUES OF N IS POSITIVE ROTATE IN INNER CIRCLE IN ANTICOCKWISE DIRECTION AND REPEAT STEP 3
STEP 5: REPEAT STEP 4 UNTIL ALL THE ELEMENT ARE ARRANGED AS IN THE INITIAL STEP

32. STEPSINVOLVED IN MATRIX MULTIPICATION METHOD
• ARRANGE THE VALES OF X1(N)& X2(N) IN MATRIX FORMS OF
FOLLOWS

33. • IN THE ABOVE REPRENTATON X2(N) IS ROTATED CIRCULARLY AND IT IS
REPRESENTED IN NXN FORM
• X1 (N) IS REPRESENTED IN COLUMN MATRIX FORM
• WHEN THE BOTH OF THESE X1(N ) AND X2(N) ARE MULTIPLIED,X3(N)
CAN BE OBTANED COLUMN MATRIX FORM

34. PROBLEM 1
FIND THE CIRCULAR CONVOLUTION OF TWO FINITE DURATION SEQUENCES
X1(n)={1,-1,-2,3,-1} X2(n)={1,2,3}
SOLUTION
GIVEN DATA X1(n)={1,-1,-2,3,-1} AND X2(n)={1,2,3}
THE LENGTH OF X1(n)=5
THE LENGTH OF X2(n)=3
THE LENGTH OF X3(n)=MAX[5,3] =5
SO TWO ZEROS WILL BE APPENDED TO THE X2(n) SEQUENCE
• (IE) NEW X2(N)={1,2,3,0,0}

40. PROBLEM 2
• FIND THE CIRCULAR CONVOLUTON OF THE TWO SEQUENCES
X1(N)={1,2,2,1} AND X2(N)={1,2,3,1} BY USING
• CONCENTRIC CIRCLE METHOD
• MATRIX METHOD
SOLUTION
GIVEN DATA X1(N)={1,2,2,1} X2(N)={1,2,3,1}
CONCENTRIC CIRCLE METHOD
HERE VALUE OF N =4 SINCE LENGTH OF M L=4
STEP 1 ARRANGE THE ELEMENT OF X1(N)&X2(N)

44. X1(N)={1,2,2,1} X2(N)={1,2,3,1}

45. PROBLEM :
• GIVEN SEQUENCES X1(N) ={1,2,3,4};X2(N)={1,1,2,2,} FIND X3(N) SUCH
THAT X3(K)=X1(k).X2(K)
• SOLUTION
• X3(N)=IDFT[X3(K)] = IDFT[X1(k).X2(K)
= X1 (N) N X2 (N)
DO CIRCULAR CONVOLUTION TO FIND X3 (N)
X1 (N) ={1,2,3,4}
x2 (N) ={1,1,2,2}

49. PROBLEM
• CONSIDER THE SEQUENCES X1(N)={0,1,2,3,4}
X2(N)={0,1,0,0}.DETERMINE A SEQUENCES Y(N) SO THAT
Y(K)=X1(K).X2(K)
• SOLUTION
• USING CIRCULAR CONVOLUTION THE VALUE OF Y(n) CAN FOUND
OUT BY USING X1(N)&X2(N)

54. QUIZ QUESTIONS

60. THANK YOU