circular convolution
(1) The length of the linear convolution of the finite length sequence is equal to the sum of the lengths of the two sequences minus one ( N 1 + N 2 − 1 N_1+N_2-1N1+N2−1)
x 1 ( n ) and x 2 ( n ) x_1(n) \ and \ x_2(n)x1( n ) and x 2( n ) LL_Circular convolution of point L :
- Put x 1 ( n ) and x 2 ( n ) x_1(n) \ and \ x_2(n)x1( n ) and x 2( n ) are both extended toLLSequence of L points, i.e. zero padding
- Then L-point circular convolution is: y ( n ) = [ ∑ m = 0 L − 1 x 1 ( m ) x 2 ( ( n − m ) ) L ] RL ( n ) = [ ∑ m = 0 L − 1 x 1 ( m ) x 2 ( ( n + r L − m ) ) L ] RL ( n ) y(n) = [\sum_{m=0}^{L-1}x_1(m)x_2((nm ))_L]R_L(n)=[\sum_{m=0}^{L-1}x_1(m)x_2((n+rL-m))_L]R_L(n)and ( n )=[m=0∑L−1x1(m)x2((n−m))L]RL(n)=[m=0∑L−1x1(m)x2((n+rL−m))L]RL( n ) y ( n ) = [ ∑ r = − ∞ ∞ y 1 ( n + r L ) ] RL ( n ) , y 1 ( n ) is linear convolution y(n) = [\sum_{r=- \infty}^{\infty}y_1(n+rL)]R_L(n), \ \ \ y_1(n) is linear convolutionand ( n )=[r=−∞∑∞y1(n+rL)]RL(n), y1( n ) is a linear convolution soLLL point circular convolutiony ( n ) y(n)y ( n ) is linear convolutiony 1 ( n ) y_1(n)y1( n ) inLLL is the principal value sequenceof the periodic extension sequence of the period
From (1) we know that y 1 ( n ) has N 1 + N 2 − 1 y_1(n) has N_1+N_2-1y1( n ) have N1+N2−1 nonzero value, then the period of the continuationLLL must satisfyL ≥ N 1 + N 2 − 1 L \geq N_1 +N_2 -1L≥N1+N2−1
At this time, the continuation of each period will not overlap,y ( n ) y(n)The firstN 1 + N 2 − 1 of y ( n ) N_1+N_2-1N1+N2−1 value is exactlyy 1 ( n ) y_1(n)y1All non-zero values of ( n )
Conclusion : If L ≥ N 1 + N 2 − 1 L \geq N_1 +N_2 -1L≥N1+N2−1 , L-point circular convolution can represent linear convolution, and the mathematical expression is as follows
example:
