The document provides an overview of the overlap save and overlap add methods used in digital signal processing for filtering long data sequences. It outlines the steps for implementing each method, including examples, and discusses the differences between the two techniques. Both methods are explained in relation to their operations on blocks of data and their applicability, concluding that the choice between methods can be based on specific needs.

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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