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OVERLAP SAVE
OVERLAP ADD
METHOD
Gourab Chandra Ghosh
Surendra Institute of Engineering &
Management
CONTENTS:
1. Why overlap save & overlap add method ?
2. Steps to perform overlap save method.
3. Example of overlap save method.
4. Steps to perform overlap add method.
5. Example of overlap add method.
6. Difference between overlap save & add method.
7. Reference
8. Q & A
2
FILTERING USING DFT:
1. In practical application we often come across linear filtering of long
data sequences.
2. DFT involves operation on block of data.
3. The block size of data should minimum because digital processors
have limited memory.
4. Two methods of filtering:
3
FILTERING USING DFT:
1. In practical application we often come across linear filtering of long
data sequences.
2. DFT involves operation on block of data.
3. The block size of data should minimum because digital processors
have limited memory.
4. Two methods of filtering:
 OVERLAP SAVE METHOD
 OVERLAP ADD METHOD
4
5
OVERLAP SAVE METHOD
STEP-1:
Determine length ‘M’, which is the length of the impulse
response data sequences i.e. h[n] & determine ‘M-1’.
STEP-2:
Given input sequence x[n] size of DFT is ‘N’.
let assume, N=5
6
OVERLAP SAVE METHOD
STEP-3:
Determine the length of the new data, ‘L’
=> N=(L+M-1)
STEP-4:
Pad ‘L-1’ zeros to h[n]
7
OVERLAP SAVE METHOD
STEP-3:
Determine the length of the blocks, ‘L’
STEP-4:
Pad ‘L-1’ zeros to h[n]
8
h[n] ‘L-1’ zeros (padded)
OVERLAP SAVE METHOD
Input Data Sequence x[n]
9
x1[n]
x2[n]
x4[n]
x3[n]
M-1 zeros
M-1 data
M-1 data
M-1 data
L L LL
OVERLAP SAVE METHOD
STEP-4:
Perform Circular Convolution of h[n] & blocks of x[n]
i.e. y1[n]= x1[n] h[n]
y2[n]= x2[n] h[n]
y3[n]= x3[n] h[n]
y4[n]= x4[n] h[n]
10
N
N
N
N
OVERLAP SAVE METHOD
11
y1[n]
y2[n]
y4[n]
y3[n]
M-1 data
M-1 data
M-1 data
M-1 data
X
X
X
X
X=> discard
OVERLAP SAVE METHOD
12
y1[n]
y2[n]
y4[n]
y3[n]
M-1 data
M-1 data
M-1 data
Final Output Data Sequence y[n]
M-1 data
X
X
X
X
X=> discard
OVERLAP SAVE EXAMPLE
Ques. Given x[n]={3,-1,0,1,3,2,0,1,2,1} & h[n]={1,1,1}
 Let, N=5
Length of h[n], M= 3
Therefore, M-1= 2
We know,
 N=(L+M-1)
 5=L+3-1
 L=3
∴Pad L-1=2 zeros with h[n] i.e. h[n]={1,1,1,0,0}
13
OVERLAP SAVE EXAMPLE
3 -1 0 1 3 2 0 1 2 1
14
0 0 3 -1 0
-1 0 1 3 2
3 2 0 1 2
1 2 1 0 0
x1[n]
x2[n]
x4[n]
x3[n]
M-1(=2) no. of previous data
L(=3) no. of new data
OVERLAP SAVE EXAMPLE
Performing yk[n]= xk[n] h[n], where k=1,2,3,4
1. y1[n]= {-1,0,3,2,2}
2. y2[n]= {4,1,0,4,6}
3. y3[n]= {6,7,5,3,3}
4. y4[n]= {1,3,4,3,1}
15
N
OVERLAP SAVE EXAMPLE
-2 -1 0 1 2 3 4 5 6 7 8 9 10 11
16
3 2 2
0 4 6
5 3 3
4 3 1
-1 0
4 1
6 7
1 3
X=> discard
X
X
X
X
n-
y1[n]
y1[n]
y1[n]
y1[n]
OVERLAP SAVE EXAMPLE
-2 -1 0 1 2 3 4 5 6 7 8 9 10 11
17
3 2 2
0 4 6
5 3 3
4 3 1
-1 0
4 1
6 7
1 3
X=> discard
X
X
X
X
n-
y1[n]
y1[n]
y1[n]
y1[n]
y[n] 3 2 2
OVERLAP SAVE EXAMPLE
-2 -1 0 1 2 3 4 5 6 7 8 9 10 11
18
3 2 2
0 4 6
5 3 3
4 3 1
-1 0
4 1
6 7
1 3
X=> discard
X
X
X
X
n-
y1[n]
y1[n]
y1[n]
y1[n]
y[n] 3 2 2 0 4 6
OVERLAP SAVE EXAMPLE
-2 -1 0 1 2 3 4 5 6 7 8 9 10 11
19
3 2 2
0 4 6
5 3 3
4 3 1
-1 0
4 1
6 7
1 3
X=> discard
X
X
X
X
n-
y1[n]
y1[n]
y1[n]
y1[n]
y[n] 3 2 2 0 4 6 5 3 3
OVERLAP SAVE EXAMPLE
-2 -1 0 1 2 3 4 5 6 7 8 9 10 11
20
3 2 2
0 4 6
5 3 3
4 3 1
-1 0
4 1
6 7
1 3
X=> discard
X
X
X
X
n-
y1[n]
y1[n]
y1[n]
y1[n]
y[n] 3 2 2 0 4 6 5 3 3 4 3 1
21
OVERLAP ADD METHOD
STEP-1:
Determine length ‘M’, which is the length of the impulse
response data sequences i.e. h[n] & determine ‘M-1’.
STEP-2:
Given input sequence x[n] size of DFT is ‘N’.
let assume, N=5
22
OVERLAP ADD METHOD
STEP-3:
Determine the length of the new data, ‘L’
STEP-4:
Pad ‘M-1’ zeros to xk[n]
Pad ‘L-1’ zeros to h[n]
23
xk[n] ‘M-1’ zeros (padded)
h[n] ‘L-1’ zeros (padded)
OVERLAP ADD METHOD
Input Data Sequence x[n]
24
x1[n]
x2[n]
x4[n]
x3[n]
M-1 zeros
M-1 zeros
M-1 zeros
M-1 zeros
L L LL
OVERLAP ADD METHOD
STEP-4:
Perform Circular Convolution of h[n] & blocks of x[n]
i.e. y1[n]= x1[n] h[n]
y2[n]= x2[n] h[n]
y3[n]= x3[n] h[n]
y4[n]= x4[n] h[n]
25
N
N
N
N
OVERLAP ADD METHOD
26
y1[n]
y2[n]
y4[n]
y3[n]
M-1 data
M-1 data
M-1 data
Final Output Data Sequence y[n]
M-1 data
add
add
add
OVERLAP ADD EXAMPLE
Ques. Given x[n]={3,-1,0,1,3,2,0,1,2,1} & h[n]={1,1,1}
 Let, N=5
Length of h[n], M= 3
Therefore, M-1= 2
We know,
 N=(L+M-1)
 5=L+3-1
 L=3
∴Pad L-1=2 zeros with h[n] i.e. h[n]={1,1,1,0,0}
27
OVERLAP ADD EXAMPLE
3 -1 0 1 3 2 0 1 2 1
28
3 -1 0
1 3 2
0 1 2
1 0 0
x1[n]
x2[n]
x4[n]
x3[n]
M-1(=2) no. of zeros
L(=3) no. of new data
Input sequence x[n]
0 0
0 0
0 0
0 0
OVERLAP ADD EXAMPLE
Performing yk[n]= xk[n] h[n], where k=1,2,3,4
1. y1[n]= {3,2,2,-1,0}
2. y2[n]= {1,4,6,5,2}
3. y3[n]= {0,1,3,3,2}
4. y4[n]= {1,1,1,0,0}
29
N
OVERLAP ADD EXAMPLE
0 1 2 3 4 5 6 7 8 9 10 11 12 13
30
3 2 2
6 5 2
3 3 2
1 0 0
-1 0
1 4
0 1
1 1
+=> add
+
n-
y1[n]
y1[n]
y1[n]
y1[n]
+
++
++
3 2 2y[n]
OVERLAP ADD EXAMPLE
0 1 2 3 4 5 6 7 8 9 10 11 12 13
31
3 2 2
6 5 2
3 3 2
1 0 0
-1 0
1 4
0 1
1 1
+=> add
+
n-
y1[n]
y1[n]
y1[n]
y1[n]
+
++
++
3 2 2 0 4 6y[n]
OVERLAP ADD EXAMPLE
0 1 2 3 4 5 6 7 8 9 10 11 12 13
32
3 2 2
6 5 2
3 3 2
1 0 0
-1 0
1 4
0 1
1 1
+=> add
+
n-
y1[n]
y1[n]
y1[n]
y1[n]
+
++
++
3 2 2 0 4 6 5 3 3y[n]
OVERLAP ADD EXAMPLE
0 1 2 3 4 5 6 7 8 9 10 11 12 13
33
3 2 2
6 5 2
3 3 2
1 0 0
-1 0
1 4
0 1
1 1
+=> add
+
n-
y1[n]
y1[n]
y1[n]
y1[n]
+
++
++
3 2 2 0 4 6 5 3 3 4 3y[n]
OVERLAP ADD EXAMPLE
0 1 2 3 4 5 6 7 8 9 10 11 12 13
34
3 2 2
6 5 2
3 3 2
1 0 0
-1 0
1 4
0 1
1 1
+=> add
+
n-
y1[n]
y1[n]
y1[n]
y1[n]
+
++
++
3 2 2 0 4 6 5 3 3 4 3 1 0 0y[n]
OVERLAP ADD EXAMPLE
0 1 2 3 4 5 6 7 8 9 10 11 12 13
35
3 2 2
6 5 2
3 3 2
1 0 0
-1 0
1 4
0 1
1 1
+=> add
+
n-
y1[n]
y1[n]
y1[n]
y1[n]
+
++
++
3 2 2 0 4 6 5 3 3 4 3 1 0 0y[n]
OVERLAP SAVE vs ADD
METHOD
Overlap Save
 Overlapped values has to be
discarded.
 It does not require any addition.
 It can be computed using linear
convolution
Overlap Add
 Overlapped values has to be
added.
 It will involve adding a number
of values in the output.
 Linear convolution is not
applicable here.
36
CONCLUSION:
37
 Overlap Add and Save methods are almost similar.
 From the differences one can choose any of the
methods, which is suitable at that time.
REFERENCE:
 Digital Signal Processing - P. Rameshbabu
 Discrete Time Signal Processing - Oppenheim & Schafer
38
39

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Overlap Add, Overlap Save(digital signal processing)

  • 1. OVERLAP SAVE OVERLAP ADD METHOD Gourab Chandra Ghosh Surendra Institute of Engineering & Management
  • 2. CONTENTS: 1. Why overlap save & overlap add method ? 2. Steps to perform overlap save method. 3. Example of overlap save method. 4. Steps to perform overlap add method. 5. Example of overlap add method. 6. Difference between overlap save & add method. 7. Reference 8. Q & A 2
  • 3. FILTERING USING DFT: 1. In practical application we often come across linear filtering of long data sequences. 2. DFT involves operation on block of data. 3. The block size of data should minimum because digital processors have limited memory. 4. Two methods of filtering: 3
  • 4. FILTERING USING DFT: 1. In practical application we often come across linear filtering of long data sequences. 2. DFT involves operation on block of data. 3. The block size of data should minimum because digital processors have limited memory. 4. Two methods of filtering:  OVERLAP SAVE METHOD  OVERLAP ADD METHOD 4
  • 5. 5
  • 6. OVERLAP SAVE METHOD STEP-1: Determine length ‘M’, which is the length of the impulse response data sequences i.e. h[n] & determine ‘M-1’. STEP-2: Given input sequence x[n] size of DFT is ‘N’. let assume, N=5 6
  • 7. OVERLAP SAVE METHOD STEP-3: Determine the length of the new data, ‘L’ => N=(L+M-1) STEP-4: Pad ‘L-1’ zeros to h[n] 7
  • 8. OVERLAP SAVE METHOD STEP-3: Determine the length of the blocks, ‘L’ STEP-4: Pad ‘L-1’ zeros to h[n] 8 h[n] ‘L-1’ zeros (padded)
  • 9. OVERLAP SAVE METHOD Input Data Sequence x[n] 9 x1[n] x2[n] x4[n] x3[n] M-1 zeros M-1 data M-1 data M-1 data L L LL
  • 10. OVERLAP SAVE METHOD STEP-4: Perform Circular Convolution of h[n] & blocks of x[n] i.e. y1[n]= x1[n] h[n] y2[n]= x2[n] h[n] y3[n]= x3[n] h[n] y4[n]= x4[n] h[n] 10 N N N N
  • 11. OVERLAP SAVE METHOD 11 y1[n] y2[n] y4[n] y3[n] M-1 data M-1 data M-1 data M-1 data X X X X X=> discard
  • 12. OVERLAP SAVE METHOD 12 y1[n] y2[n] y4[n] y3[n] M-1 data M-1 data M-1 data Final Output Data Sequence y[n] M-1 data X X X X X=> discard
  • 13. OVERLAP SAVE EXAMPLE Ques. Given x[n]={3,-1,0,1,3,2,0,1,2,1} & h[n]={1,1,1}  Let, N=5 Length of h[n], M= 3 Therefore, M-1= 2 We know,  N=(L+M-1)  5=L+3-1  L=3 ∴Pad L-1=2 zeros with h[n] i.e. h[n]={1,1,1,0,0} 13
  • 14. OVERLAP SAVE EXAMPLE 3 -1 0 1 3 2 0 1 2 1 14 0 0 3 -1 0 -1 0 1 3 2 3 2 0 1 2 1 2 1 0 0 x1[n] x2[n] x4[n] x3[n] M-1(=2) no. of previous data L(=3) no. of new data
  • 15. OVERLAP SAVE EXAMPLE Performing yk[n]= xk[n] h[n], where k=1,2,3,4 1. y1[n]= {-1,0,3,2,2} 2. y2[n]= {4,1,0,4,6} 3. y3[n]= {6,7,5,3,3} 4. y4[n]= {1,3,4,3,1} 15 N
  • 16. OVERLAP SAVE EXAMPLE -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 16 3 2 2 0 4 6 5 3 3 4 3 1 -1 0 4 1 6 7 1 3 X=> discard X X X X n- y1[n] y1[n] y1[n] y1[n]
  • 17. OVERLAP SAVE EXAMPLE -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 17 3 2 2 0 4 6 5 3 3 4 3 1 -1 0 4 1 6 7 1 3 X=> discard X X X X n- y1[n] y1[n] y1[n] y1[n] y[n] 3 2 2
  • 18. OVERLAP SAVE EXAMPLE -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 18 3 2 2 0 4 6 5 3 3 4 3 1 -1 0 4 1 6 7 1 3 X=> discard X X X X n- y1[n] y1[n] y1[n] y1[n] y[n] 3 2 2 0 4 6
  • 19. OVERLAP SAVE EXAMPLE -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 19 3 2 2 0 4 6 5 3 3 4 3 1 -1 0 4 1 6 7 1 3 X=> discard X X X X n- y1[n] y1[n] y1[n] y1[n] y[n] 3 2 2 0 4 6 5 3 3
  • 20. OVERLAP SAVE EXAMPLE -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 20 3 2 2 0 4 6 5 3 3 4 3 1 -1 0 4 1 6 7 1 3 X=> discard X X X X n- y1[n] y1[n] y1[n] y1[n] y[n] 3 2 2 0 4 6 5 3 3 4 3 1
  • 21. 21
  • 22. OVERLAP ADD METHOD STEP-1: Determine length ‘M’, which is the length of the impulse response data sequences i.e. h[n] & determine ‘M-1’. STEP-2: Given input sequence x[n] size of DFT is ‘N’. let assume, N=5 22
  • 23. OVERLAP ADD METHOD STEP-3: Determine the length of the new data, ‘L’ STEP-4: Pad ‘M-1’ zeros to xk[n] Pad ‘L-1’ zeros to h[n] 23 xk[n] ‘M-1’ zeros (padded) h[n] ‘L-1’ zeros (padded)
  • 24. OVERLAP ADD METHOD Input Data Sequence x[n] 24 x1[n] x2[n] x4[n] x3[n] M-1 zeros M-1 zeros M-1 zeros M-1 zeros L L LL
  • 25. OVERLAP ADD METHOD STEP-4: Perform Circular Convolution of h[n] & blocks of x[n] i.e. y1[n]= x1[n] h[n] y2[n]= x2[n] h[n] y3[n]= x3[n] h[n] y4[n]= x4[n] h[n] 25 N N N N
  • 26. OVERLAP ADD METHOD 26 y1[n] y2[n] y4[n] y3[n] M-1 data M-1 data M-1 data Final Output Data Sequence y[n] M-1 data add add add
  • 27. OVERLAP ADD EXAMPLE Ques. Given x[n]={3,-1,0,1,3,2,0,1,2,1} & h[n]={1,1,1}  Let, N=5 Length of h[n], M= 3 Therefore, M-1= 2 We know,  N=(L+M-1)  5=L+3-1  L=3 ∴Pad L-1=2 zeros with h[n] i.e. h[n]={1,1,1,0,0} 27
  • 28. OVERLAP ADD EXAMPLE 3 -1 0 1 3 2 0 1 2 1 28 3 -1 0 1 3 2 0 1 2 1 0 0 x1[n] x2[n] x4[n] x3[n] M-1(=2) no. of zeros L(=3) no. of new data Input sequence x[n] 0 0 0 0 0 0 0 0
  • 29. OVERLAP ADD EXAMPLE Performing yk[n]= xk[n] h[n], where k=1,2,3,4 1. y1[n]= {3,2,2,-1,0} 2. y2[n]= {1,4,6,5,2} 3. y3[n]= {0,1,3,3,2} 4. y4[n]= {1,1,1,0,0} 29 N
  • 30. OVERLAP ADD EXAMPLE 0 1 2 3 4 5 6 7 8 9 10 11 12 13 30 3 2 2 6 5 2 3 3 2 1 0 0 -1 0 1 4 0 1 1 1 +=> add + n- y1[n] y1[n] y1[n] y1[n] + ++ ++ 3 2 2y[n]
  • 31. OVERLAP ADD EXAMPLE 0 1 2 3 4 5 6 7 8 9 10 11 12 13 31 3 2 2 6 5 2 3 3 2 1 0 0 -1 0 1 4 0 1 1 1 +=> add + n- y1[n] y1[n] y1[n] y1[n] + ++ ++ 3 2 2 0 4 6y[n]
  • 32. OVERLAP ADD EXAMPLE 0 1 2 3 4 5 6 7 8 9 10 11 12 13 32 3 2 2 6 5 2 3 3 2 1 0 0 -1 0 1 4 0 1 1 1 +=> add + n- y1[n] y1[n] y1[n] y1[n] + ++ ++ 3 2 2 0 4 6 5 3 3y[n]
  • 33. OVERLAP ADD EXAMPLE 0 1 2 3 4 5 6 7 8 9 10 11 12 13 33 3 2 2 6 5 2 3 3 2 1 0 0 -1 0 1 4 0 1 1 1 +=> add + n- y1[n] y1[n] y1[n] y1[n] + ++ ++ 3 2 2 0 4 6 5 3 3 4 3y[n]
  • 34. OVERLAP ADD EXAMPLE 0 1 2 3 4 5 6 7 8 9 10 11 12 13 34 3 2 2 6 5 2 3 3 2 1 0 0 -1 0 1 4 0 1 1 1 +=> add + n- y1[n] y1[n] y1[n] y1[n] + ++ ++ 3 2 2 0 4 6 5 3 3 4 3 1 0 0y[n]
  • 35. OVERLAP ADD EXAMPLE 0 1 2 3 4 5 6 7 8 9 10 11 12 13 35 3 2 2 6 5 2 3 3 2 1 0 0 -1 0 1 4 0 1 1 1 +=> add + n- y1[n] y1[n] y1[n] y1[n] + ++ ++ 3 2 2 0 4 6 5 3 3 4 3 1 0 0y[n]
  • 36. OVERLAP SAVE vs ADD METHOD Overlap Save  Overlapped values has to be discarded.  It does not require any addition.  It can be computed using linear convolution Overlap Add  Overlapped values has to be added.  It will involve adding a number of values in the output.  Linear convolution is not applicable here. 36
  • 37. CONCLUSION: 37  Overlap Add and Save methods are almost similar.  From the differences one can choose any of the methods, which is suitable at that time.
  • 38. REFERENCE:  Digital Signal Processing - P. Rameshbabu  Discrete Time Signal Processing - Oppenheim & Schafer 38
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