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Overlap save method and overlap add method in dsp

The document is a lecture note on digital signal processing, specifically focusing on the overlap save and overlap add methods used for linear convolution with finite impulse response (FIR) filters. It explains how to evaluate the discrete convolution of long signals with FIR filters by dividing the input signal into blocks and provides examples for each method. The content includes detailed examples demonstrating the calculations necessary for applying both the overlap save and overlap add techniques.

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DIGITAL SIGNAL PROCESSING By Mrs.R.Chitra, Assistant Professor(SS), Department of ECE, School of Engineering, Avinashilingam Institute for Home Science and Higher Education for Women, Coimbatore
Topics to be Covered: 1. OVERLAP SAVE METHOD 2. OVERLAP ADD METHOD 3. LINEAR CONVOLUTION
OVERLAP SAVE METHOD:  Overlap–save is the traditional name for an efficient way to evaluate the discrete convolution between a very long signal x(n) and a finite impulse response FIR filter h(n)  In this method the input sequence is divided into blocks of data of size N=L+M-1.where L= length of an input sequence and M=the length of an impulse response  Each block consist of last (M-1) data points of previous block followed by L new data points to form a data sequence of length N=L+M-1.  For first block, M-1 points are set to zero. The input sequence can be divided into blocks as X1(n) = {0,0,x(0),x(1),x(2)} M-1=2zeros
X2(n) = {x(1),x(2),x(3),x(4),x(5)} Last two data points from previous block X3(n) = {x(4),x(5),x(6),x(7),x(8)} X4(n) = {x(7),x(8),x(9),x(10),x(11)} X5(n) = {x(10),x(11),x(12),x(13),x(14)} X6(n) = {x(13),x(14),0,0,0} Thus, Y1(n)=x1(n) h(n)={y1(0),y1(1),y1(2),y1(3),y1(4)} Y2(n)=x2(n) h(n)={y2(0),y2(1),y2(2),y2(3),y2(4)}N N
Y3(n)=x3(n) h(n)={y3(0),y3(1),y3(2),y3(3),y3(4)} Y4(n)=x4(n) h(n)={y4(0),y4(1),y4(2),y4(3),y4(4)} Y5(n)=x5(n) h(n)={y5(0),y5(1),y5(2),y5(3),y5(4)} Y6(n)=x6(n) h(n)={y6(0),y6(1),y6(2),y6(3),y6(4)} Therefore, the output blocks are abutted together to get Y(n)={y1(2),y1(3),y1(4),y2(2),y2(3),y2(4),y3(2),y3(3), y3(4),y4(4),y5(2),y5(3),y5(4),y6(2),y6(3)} N N N N
EXAMPLE 1: Determine the output of linear FIR filter whose impulse response is h(n)={1,2,3} and the input signal is x(n)= {1,2,3,4,5,6,7,8,9} using over lap save method. SOLUTION: Given, x(n)= {1,2,3,4,5,6,7,8,9} & h(n)={1,2,3} Therefore, L=9;M=3; Adding M-1 zeros in x1(n),we get x1(n)={0,0,1,2,3} x2(n)={2,3,4,5,6} x3(n)={5,6,7,8,9} x4(n)={8,9,0,0,0} h(n)={1,2,3,0,0}
By circular convolution, y1(n)= x1(n) h(n) y2(n)= x2(n) h(n) N N                                                  10 4 1 9 12 3 2 1 0 0 13300 01230 00123 30012 23001                                                  28 22 16 25 29 6 5 4 3 2 13300 01230 00123 30012 23001
y3(n)= x3(n) h(n) y4(n)= x4(n) h(n) N N                                                  46 40 34 43 47 9 8 7 6 5 12300 01230 00123 30012 23001                                                  0 27 42 25 8 0 0 0 9 8 12300 01230 00123 30012 23001
y(n)= Therefore, y(n)={1,4,10,16,22,28,34,40,46,42,27} 12 9 1 4 10 29 25 16 22 28 8 25 42 27 0 Add y3(n) y2(n) y1(n) 47 43 34 40 46 y4(n) Discard Discard Discard Discard
EXAMPLE 2: Determine the output of linear FIR filter whose impulse response is h(n)={1,-3,5} and the input signal is x(n)= {-1,4,7,3,-2,9,10,12,-5,8} using over lap save method. SOLUTION: Given, x(n)= {-1,4,7,3,-2,9,10,12,-5,8} & h(n)={1,-3,5} Therefore, L=9;M=3; Adding M-1 zeros in x1(n),we get x1(n)={0,0,-1,4,7} x2(n)={4,7,3,-2,9} x3(n)={-2,9,10,12,-5} x4(n)={12,-5,8,0,0} h(n)={1,-3,5,0,0}
By circular convolution, y1(n)= x1(n) h(n) y2(n)= x2(n) h(n) N N                                                           10 7 1 35 1 7 4 1 0 0 13500 01350 00135 50013 35001                                                        30 24 2 40 33 9 2 3 7 4 13500 01350 00135 50013 35001
y3(n)= x3(n) h(n) y4(n)= x4(n) h(n) N N                                                           9 27 27 10 73 5 12 10 9 2 13500 01350 00135 50013 35001                                                         40 49 83 0 1 0 0 8 5 12 13500 01350 00135 50013 35001
y(n)= Therefore, y(n)={1,7,-10,2,24,30,-27,27,9,83,-49,40} -1 35 -1 7 -10 -33 40 2 24 30 1 0 83 -49 40 Add y3(n) y2(n) y1(n) 73 -10 -27 27 9 y4(n) Discard Discard Discard Discard
OVERLAP ADD METHOD:  The length of the sequence is be Ls and the length of the impulse response is M.  The sequence is divided into blocks of data size having length L and M-1.  Zeros are appended to it make the data size of L+M-1. Let the output blocks are of the form, y1(n)={y1(0),y1(1),…….,y1(L-1),y1(L),….,y1(N-1)} y2(n)={y2(0),y2(1),……,y2(L-1),y2(L),….,y2(N-1)} y3(n)={y3(0),y3(1),…….,y3(L-1),y3(L),……,y3(N-1)} the output sequence is Y(n)={y1(0),y1(1),….,y1(L-1),y1(L)+y2(0),….,y1(N-1)+ y2(M-2),y2(M),……,y2(L)+y3(0),y2(L+1)+y3(1),….,y3(N-1)}
EXAMPLE 1: Determine the output of linear FIR filter whose impulse response is h(n)={1,2,3} and the input signal is x(n)= {1,2,3,4,5,6,7,8,9} using over lap add method. SOLUTION: Given, x(n)= {1,2,3,4,5,6,7,8,9} & h(n)={1,2,3} Therefore, L=9;M=3; Adding M-1 zero,we get x1(n)={1,2,3,0,0} x2(n)={4,5,6,0,0} x3(n)={7,8,9,0,0} h(n)={1,2,3,0,0}
By circular convolution, y1(n)= x1(n) h(n) y2(n)= x2(n) h(n) N N                                                  18 27 28 13 4 0 0 6 5 4 13300 01230 00123 30012 23001                                                  9 12 10 4 1 0 0 3 2 1 13300 01230 00123 30012 23001
y3(n)= x3(n) h(n) y(n)= Therefore, y(n)={1,4,10,16,22,28,34,40,46,42,27} N                                                  27 42 46 22 7 0 0 9 8 7 13300 01230 00123 30012 23001 1 4 10 12 9 4 13 28 27 18 7 22 46 42 27 Add Add y3(n) y2(n) y1(n)
EXAMPLE 2: Determine the output of linear FIR filter whose impulse response is h(n)={1,-3,5} and the input signal is x(n)= {-1,4,7,3,-2,9,10,12,-5,8} using over lap add method. SOLUTION: Given, x(n)= {-1,4,7,3,-2,9,10,12,-5,8} & h(n)={1,-3,5} Therefore, L=9;M=3; Adding M-1 zeros, we get x1(n)={-1,4,7,0,0} x2(n)={3,-2,9,0,0} x3(n)={10,12,-5,0,0} x4(n)={8,0,0,0,0} h(n)={1,-3,5,0,0}
By circular convolution, y1(n)= x1(n) h(n) y2(n)= x2(n) h(n) N N                                                          45 37 30 11 3 0 0 9 2 3 13500 01350 00135 50013 35001                                                        35 1 10 7 1 0 0 7 4 1 13500 01350 00135 50013 35001
y3(n)= x3(n) h(n) y4(n)= x4(n) h(n) N N                                                          25 75 9 18 10 0 0 5 12 10 13500 01350 00135 50013 35001                                                        0 0 40 24 8 0 0 0 0 8 13500 01350 00135 50013 35001
y(n)= Therefore, y(n)={1,7,-10,2,24,30,-27,27,9,83,-49,40} 1 7 -10 -1 35 3 -11 30 -37 45 8 -24 40 0 0 Add Add y3(n) y2(n) y1(n) 10 -18 9 75 -25 Add Add y4(n)
LINEAR CONVOLUTION FROM CIRCULAR CONVOLUTION: • Let us consider two finite duration sequences x(n) and h(n). The duration of x(n) is L and y(n) is M samples. The linear convolution of x(n) and h(n) is given by the formula where y(n) is a finite duration of sequence of L+M-1 samples. The circular convolution of x(n) and h(n) give N samples where N=Max(L,M). Consider two sequences x(n) and h(n) having sequence length L=5 and M=3 respectively as shown in the figure. x(n) L=5 0 1 2 3 4     k knhkxny )()()(
h(n) M=3 0 1 2 The linear convolution of x(n) and h(n) consist of 5+3-1=7 points. 3 3 3 y(n)=x(n)*h(n) 2 2 1 1 0 1 2 3 4 5 6 The circular convolution of x(n) and h(n) consist of 5 points short of M-1=2 points. Therefore the circular convolution will contain corrupted points due to time domain aliasing. These points are first (M-1) points as shown in the figure. Aliasing y(0)&y(5) y(1)&y(6) 0 1 2 3 4
EXAMPLE:
Thus, linear convolution is done.
THANK YOU!!!

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Overlap save method and overlap add method in dsp

  • 1.
    DIGITAL SIGNAL PROCESSING By Mrs.R.Chitra, AssistantProfessor(SS), Department of ECE, School of Engineering, Avinashilingam Institute for Home Science and Higher Education for Women, Coimbatore
  • 2.
    Topics to beCovered: 1. OVERLAP SAVE METHOD 2. OVERLAP ADD METHOD 3. LINEAR CONVOLUTION
  • 3.
    OVERLAP SAVE METHOD: Overlap–save is the traditional name for an efficient way to evaluate the discrete convolution between a very long signal x(n) and a finite impulse response FIR filter h(n)  In this method the input sequence is divided into blocks of data of size N=L+M-1.where L= length of an input sequence and M=the length of an impulse response  Each block consist of last (M-1) data points of previous block followed by L new data points to form a data sequence of length N=L+M-1.  For first block, M-1 points are set to zero. The input sequence can be divided into blocks as X1(n) = {0,0,x(0),x(1),x(2)} M-1=2zeros
  • 4.
    X2(n) = {x(1),x(2),x(3),x(4),x(5)} Lasttwo data points from previous block X3(n) = {x(4),x(5),x(6),x(7),x(8)} X4(n) = {x(7),x(8),x(9),x(10),x(11)} X5(n) = {x(10),x(11),x(12),x(13),x(14)} X6(n) = {x(13),x(14),0,0,0} Thus, Y1(n)=x1(n) h(n)={y1(0),y1(1),y1(2),y1(3),y1(4)} Y2(n)=x2(n) h(n)={y2(0),y2(1),y2(2),y2(3),y2(4)}N N
  • 5.
    Y3(n)=x3(n) h(n)={y3(0),y3(1),y3(2),y3(3),y3(4)} Y4(n)=x4(n) h(n)={y4(0),y4(1),y4(2),y4(3),y4(4)} Y5(n)=x5(n)h(n)={y5(0),y5(1),y5(2),y5(3),y5(4)} Y6(n)=x6(n) h(n)={y6(0),y6(1),y6(2),y6(3),y6(4)} Therefore, the output blocks are abutted together to get Y(n)={y1(2),y1(3),y1(4),y2(2),y2(3),y2(4),y3(2),y3(3), y3(4),y4(4),y5(2),y5(3),y5(4),y6(2),y6(3)} N N N N
  • 6.
    EXAMPLE 1: Determinethe output of linear FIR filter whose impulse response is h(n)={1,2,3} and the input signal is x(n)= {1,2,3,4,5,6,7,8,9} using over lap save method. SOLUTION: Given, x(n)= {1,2,3,4,5,6,7,8,9} & h(n)={1,2,3} Therefore, L=9;M=3; Adding M-1 zeros in x1(n),we get x1(n)={0,0,1,2,3} x2(n)={2,3,4,5,6} x3(n)={5,6,7,8,9} x4(n)={8,9,0,0,0} h(n)={1,2,3,0,0}
  • 7.
    By circular convolution, y1(n)=x1(n) h(n) y2(n)= x2(n) h(n) N N                                                  10 4 1 9 12 3 2 1 0 0 13300 01230 00123 30012 23001                                                  28 22 16 25 29 6 5 4 3 2 13300 01230 00123 30012 23001
  • 8.
    y3(n)= x3(n) h(n) y4(n)=x4(n) h(n) N N                                                  46 40 34 43 47 9 8 7 6 5 12300 01230 00123 30012 23001                                                  0 27 42 25 8 0 0 0 9 8 12300 01230 00123 30012 23001
  • 9.
    y(n)= Therefore, y(n)={1,4,10,16,22,28,34,40,46,42,27} 12 9 14 10 29 25 16 22 28 8 25 42 27 0 Add y3(n) y2(n) y1(n) 47 43 34 40 46 y4(n) Discard Discard Discard Discard
  • 10.
    EXAMPLE 2: Determinethe output of linear FIR filter whose impulse response is h(n)={1,-3,5} and the input signal is x(n)= {-1,4,7,3,-2,9,10,12,-5,8} using over lap save method. SOLUTION: Given, x(n)= {-1,4,7,3,-2,9,10,12,-5,8} & h(n)={1,-3,5} Therefore, L=9;M=3; Adding M-1 zeros in x1(n),we get x1(n)={0,0,-1,4,7} x2(n)={4,7,3,-2,9} x3(n)={-2,9,10,12,-5} x4(n)={12,-5,8,0,0} h(n)={1,-3,5,0,0}
  • 11.
    By circular convolution, y1(n)=x1(n) h(n) y2(n)= x2(n) h(n) N N                                                           10 7 1 35 1 7 4 1 0 0 13500 01350 00135 50013 35001                                                        30 24 2 40 33 9 2 3 7 4 13500 01350 00135 50013 35001
  • 12.
    y3(n)= x3(n) h(n) y4(n)=x4(n) h(n) N N                                                           9 27 27 10 73 5 12 10 9 2 13500 01350 00135 50013 35001                                                         40 49 83 0 1 0 0 8 5 12 13500 01350 00135 50013 35001
  • 13.
    y(n)= Therefore, y(n)={1,7,-10,2,24,30,-27,27,9,83,-49,40} -1 35 -17 -10 -33 40 2 24 30 1 0 83 -49 40 Add y3(n) y2(n) y1(n) 73 -10 -27 27 9 y4(n) Discard Discard Discard Discard
  • 14.
    OVERLAP ADD METHOD: The length of the sequence is be Ls and the length of the impulse response is M.  The sequence is divided into blocks of data size having length L and M-1.  Zeros are appended to it make the data size of L+M-1. Let the output blocks are of the form, y1(n)={y1(0),y1(1),…….,y1(L-1),y1(L),….,y1(N-1)} y2(n)={y2(0),y2(1),……,y2(L-1),y2(L),….,y2(N-1)} y3(n)={y3(0),y3(1),…….,y3(L-1),y3(L),……,y3(N-1)} the output sequence is Y(n)={y1(0),y1(1),….,y1(L-1),y1(L)+y2(0),….,y1(N-1)+ y2(M-2),y2(M),……,y2(L)+y3(0),y2(L+1)+y3(1),….,y3(N-1)}
  • 15.
    EXAMPLE 1: Determinethe output of linear FIR filter whose impulse response is h(n)={1,2,3} and the input signal is x(n)= {1,2,3,4,5,6,7,8,9} using over lap add method. SOLUTION: Given, x(n)= {1,2,3,4,5,6,7,8,9} & h(n)={1,2,3} Therefore, L=9;M=3; Adding M-1 zero,we get x1(n)={1,2,3,0,0} x2(n)={4,5,6,0,0} x3(n)={7,8,9,0,0} h(n)={1,2,3,0,0}
  • 16.
    By circular convolution, y1(n)=x1(n) h(n) y2(n)= x2(n) h(n) N N                                                  18 27 28 13 4 0 0 6 5 4 13300 01230 00123 30012 23001                                                  9 12 10 4 1 0 0 3 2 1 13300 01230 00123 30012 23001
  • 17.
  • 18.
    EXAMPLE 2: Determinethe output of linear FIR filter whose impulse response is h(n)={1,-3,5} and the input signal is x(n)= {-1,4,7,3,-2,9,10,12,-5,8} using over lap add method. SOLUTION: Given, x(n)= {-1,4,7,3,-2,9,10,12,-5,8} & h(n)={1,-3,5} Therefore, L=9;M=3; Adding M-1 zeros, we get x1(n)={-1,4,7,0,0} x2(n)={3,-2,9,0,0} x3(n)={10,12,-5,0,0} x4(n)={8,0,0,0,0} h(n)={1,-3,5,0,0}
  • 19.
    By circular convolution, y1(n)=x1(n) h(n) y2(n)= x2(n) h(n) N N                                                          45 37 30 11 3 0 0 9 2 3 13500 01350 00135 50013 35001                                                        35 1 10 7 1 0 0 7 4 1 13500 01350 00135 50013 35001
  • 20.
    y3(n)= x3(n) h(n) y4(n)=x4(n) h(n) N N                                                          25 75 9 18 10 0 0 5 12 10 13500 01350 00135 50013 35001                                                        0 0 40 24 8 0 0 0 0 8 13500 01350 00135 50013 35001
  • 21.
    y(n)= Therefore, y(n)={1,7,-10,2,24,30,-27,27,9,83,-49,40} 1 7 -10-1 35 3 -11 30 -37 45 8 -24 40 0 0 Add Add y3(n) y2(n) y1(n) 10 -18 9 75 -25 Add Add y4(n)
  • 22.
    LINEAR CONVOLUTION FROMCIRCULAR CONVOLUTION: • Let us consider two finite duration sequences x(n) and h(n). The duration of x(n) is L and y(n) is M samples. The linear convolution of x(n) and h(n) is given by the formula where y(n) is a finite duration of sequence of L+M-1 samples. The circular convolution of x(n) and h(n) give N samples where N=Max(L,M). Consider two sequences x(n) and h(n) having sequence length L=5 and M=3 respectively as shown in the figure. x(n) L=5 0 1 2 3 4     k knhkxny )()()(
  • 23.
    h(n) M=3 0 12 The linear convolution of x(n) and h(n) consist of 5+3-1=7 points. 3 3 3 y(n)=x(n)*h(n) 2 2 1 1 0 1 2 3 4 5 6 The circular convolution of x(n) and h(n) consist of 5 points short of M-1=2 points. Therefore the circular convolution will contain corrupted points due to time domain aliasing. These points are first (M-1) points as shown in the figure. Aliasing y(0)&y(5) y(1)&y(6) 0 1 2 3 4
  • 24.
  • 27.
  • 28.