Circular Convolution

Circular Convolution

• 1.
  Circular Convolution using Graphicalmethod - 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 !