Circular Convolution using Graphical method

1. Circular Convolution using
Graphical method
- SARANG JOSHI

2. )()()]()([ 2121 kXkXnxnx N
Circular Convolution






−= 
−
=
1
0
21 ))(()()(
N
n
Nnmxnxmy
For m=0,1,2………N-1

3. 





−=
=
=
3
0
421 ))(()()0(
0;mfor
n
nxnxy

4. Circular representation
x1(1)=1
x1(0)=0
x1(3)=3
x1(2)=2 x1(n)
x1(n)={0, 1 , 2 , 3 }

5. x2(0)=4
x2(3)=7
x2(2)=6
x2(1)=5
x2(n)
x2(n)={4 , 5 , 6 , 7 }

6. x2(0)=4
x2(-1)=5
x2(-2)=6
x2(-3)=7
x2((-n))
x2(-n)=x2(N-n)
x2(-1)=x2(3)
x2(-2)=x2(2)
x2(-3)=x2(1)

7. 





−=
=
=
3
0
421 ))(()()0(
0;mfor
n
nxnxy
))3(()3())2(()2(
))1(()1())0(()0()0(
2121
2121
NN
NN
xxxx
xxxxy
−+−
+−+−=

8. x2(0)=4
x2(-1)=5
x2(-2)=6
x2(-3)=7
x1(0)=0
x1(3)=3
x1(2)=2
x1(1)=1
1x7=7
2x6=12 0x4=0
3x5=15






−=
=
=
3
0
421 ))(()()0(
0;mfor
n
nxnxy
y(0)=0+7+12+15
=34
x2((-n))
x1(n)

9. 





−=
=
=
3
0
421 ))1(()()1(
1;mfor
n
nxnxy
))2(()3())1(()2(
))0(()1())1(()0()1(
421421
421421
−+−
++=
xxxx
xxxxy

10. x2(1)=5
x2(2)=6
x2(3)=7
x2(0)=4
x2((-n+1))
x2(1-n)=x2(-n+1)
i.e. shift x2(-n) by 1 sample anticlockwise

11. 





−=
=
=
3
0
421 ))1(()()1(
1;mfor
n
nxnxy
x2(1)=5
x2(2)=6
x2(3)=7
x2(0)=4
x1(0)=0
x1(3)=3
x1(2)=2
x1(1)=1
1x4=4
2x7=14 0x5=0
3x6=18
y(1)=0+4+14+18
=36
x2((-n+1))
x1(n)

12. 





−=
=
=
3
0
421 ))2(()()2(
2;mfor
n
nxnxy
))1(()3())0(()2(
))1(()1())2(()0()2(
421421
421421
−+
++=
xxxx
xxxxy

13. x2(2)=6
x2(3)=7
x2(0)=4
x2(1)=5
x2((-n+2))
x2(2-n)=x2(-n+2)
i.e. shift x2(-n) by 2 samples anticlockwise

14. 





−=
=
=
3
0
421 ))2(()()2(
2;mfor
n
nxnxy
x2(2)=6
x2(3)=7
x2(0)=4
x2(1)=5
x1(0)=0
x1(3)=3
x1(2)=2
x1(1)=1
1x5=5
2x4=8 0x6=0
3x7=21
y(2)=0+5+8+21
=34
x2((-n+2))
x1(n)

15. 





−=
=
=
3
0
421 ))3(()()3(
3;mfor
n
nxnxy
))0(()3())1(()2(
))2(()1())3(()0()3(
421421
421421
xxxx
xxxxy
+
++=

16. x2(3)=7
x2(0)=4
x2(1)=5
x2(2)=6
x2((-n+3))
x2(3-n)=x2(-n+3)
i.e. shift x2(-n) by 3 samples anticlockwise

17. 





−=
=
=
3
0
421 ))3(()()3(
3;mfor
n
nxnxy
x2(3)=7
x2(0)=4
x2(1)=5
x2(2)=6
x1(0)=0
x1(3)=3
x1(2)=2
x1(1)=1
1x6=6
2x5=10 0x7=0
3x4=12
y(3)=0+6+10+12
=28
x2((-n+3))
x1(n)

18. y(m)={34,36,34,28}

19. RATE, FOLLOW & SHARE
https://unacademy.com/user/jsarang70-7008

20. THANK YOU !