【正文】 )N x xxxxx= = == ? \ = \ = = ? \ = \ =%%%%Q%%QEngineering college, Linyi Normal University ? The same as doing in the timedomain ? On the other hand : Therefore: ( ) , 0 1()0,( ) [ ( m od ) ] ( ( ) )( ) ( ) ( )NNX k k NXkothe rw iseX k X k N X kor X k X k R n236。239。 ＃ 239。=237。239。239。238。===%%%1010( ) ( )1( ) ( )NnkNnNnkNnX k x n Wx n X k WN=== 229。= 229。% %%%Engineering college, Linyi Normal University 1010( ) [ ( ) ] ( )1( ) [ ( ) ] ( )NnkNnNnkNnX k D FT x n x n Wx n I D FT X k X k WN==== 229。== 229。? Note: x(n) and X(k) are not defined outside the interval: 0=n=N1 and 0=k=N1 Engineering college, Linyi Normal University Properties of the DFT ? Linearity If Then Note: If the length of each sequence is : Then 3 1 2( ) ( ) ( )x n ax n bx n=+3 1 2( ) ( ) ( )X k aX k bX k=+1 1 2 2 3 3( ) , ( ) , ( )x n N x n N x n N 3 1 2m a x( , )N N N=Engineering college, Linyi Normal University While: Therefore, we should augment zero to the shorter sequence until the two sequences have the same length. 11223311 1 1012 2 2013 3 30( ) ( ) , 0 1( ) ( ) , 0 1( ) ( ) , 0 1NknNnNknNnNknNnX k x n W k NX k x n W k NX k x n W k N======= ＃ 229。= ＃ 229。= ＃ 229。Engineering college, Linyi Normal University ? Circular shift To a sequence x(n) with length N, its circular shifting is defined as: ( ) (( )) ( )NNy n x n m R n=+Engineering college, Linyi Normal University If Then ( ) ( )D F Tx n X k171。(( )) ( )( ) (( ))DFTnkNNDFTnlNNx n m W X kW x n X k l+??Engineering college, Linyi Normal University ? Circular Convolution ? Definition of circular convolution Suppose: two finiteduration sequences:x1(n) and x2(n) Note: The result of circular convolution is also a finiteduration sequence with the duration [0,N1] ? The operate steps can be divided into 3 main steps: ① To period the two sequences with period N。 ② To pute the periodic convolution of the two periodic sequences ③ To get out the duration sequence between [0,N1] 11 2 1 20( ) ( ) ( ) ( ( ) ) , 0 1NNmx n x n x m x n m n N=? ＃ 229。)()()]()([ 2121 kXkXnxnxD F T ??12( ) ( )x n x n*%%Engineering college, Linyi Normal University Engineering college, Linyi Normal University ? Circular convolution as linear convolution with aliasing ? Linear convolution of two finitelength sequences X1(n)N1 points。 x2(n)N2 points Then the linear convolution of the two sequences are Because: So: x3(n)=0 when n0 or nN1+N22 i. e. the total length of x3(n) is N1+N21 3 1 2 1 2( ) ( ) ( ) ( ) ( )mx n x n x n x m x n m165。= ?= * = 229。112212( ) 0 1( ) 0 102x m m Nx n m n m Nn N N＃ ? ?\ ＃ +Engineering college, Linyi Normal University ?Suppose When Prove: 1122( ) : 0 1( ) : 0 1x n n Nx n n N＃ ＃ 1 2 1 2 1 21 , ( ) ( ) ( ) ( )L N N x n x n x n x n? ? *{ }11 2 1 201120112012( ) ( ) ( ) ( ( ) )( ) [ ( ) ]( ) [ ( ) ][ ( ) ( ) ] 0 1LLmLmrLrmrx n x n x m x n mx m x n m r Lx m x n m r Lx n x n n L=?= = ??= ?165。= ?? 229。= +邋= +邋=* ＃ 229。Engineering college, Linyi Normal University Conclusion: If , then the circular convolution of x1(n) and x2(n) is equal to the linear convolution of the two sequences, and time aliasing in the circular convolution of two finitelength sequences can be avoided . Example: suppose x1(n)=x2(n)=u(n)u(n6) (1)pute (2)pute 12( ) ( ) 。 6x n x n N?12( ) ( ) 。 12x n x n N?12 1L N N?Engineering college, Linyi Normal University Engineering college, Linyi Normal University ? ConjugateSymmetry properties ? Definition ( ) ( )( ) ( ( ) ) ( ), [ ( ) ] ( )D F TD F TNI f x n X kthe n x n X k X N k