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Find DFT of x(n)=2n using the 8-point DIT-FFT algorithm.
written 2.4 years ago by gravatar for prajapatijaimin prajapatijaimin • 3.3k •   modified 2.4 years ago
digital signal processing
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written 2.4 years ago by gravatar for prajapatijaimin prajapatijaimin • 3.3k

Solution:

x[n]=2n={1,2,4,8,16,32,64,128}.

enter image description here

x[k]=N/1∑n=0x[n]WknNW08=e−j2π/8⋅0=0,1.⋅N−1W18=e−j2π/8⋅1=0.707−j0.707W28=e−j2π/g2⋅2=−jW38=e−j2π/8⋅3=−0.707−j0.707

Stage I O/P.

V11[0]=x[0]+x[4]=17V11[1]=x[0]−x[4]=−15V12[0]=x[2]+x[6]=68V12[1]=x[2]−x[6]=−60V21[0]=x[1]+x[5]=34V21[1]=x[1]−x[5]=−30V22[0]=x[3]+x[7]=186V22[1]=x[3]−x[7]=−120

 Stage II o/P:- F1[k]=V11[k]+W2kNV12[k]  F1[k+N/4]=V11[k]−W2kNV12[k]F1[0]=V11[0]+W08V12[0]=85FR[k]=V11[1]+W28V12[1]=−15+j60  F2[k]=V27[k]+W2kNV22[k]  F2[k+N/4]=V21[k]−W2kNV22[k]

 Stage III opP:- x[k]=F1[k]+WkNF2[k]x[k+k/2]=F1[k]−wkNF2[k]x[0]=F1[0]+W08F2[0]=85+170=255

x[1]=F1[1]+W8F2[1]=(−15+j60)+(0.707−j0.707)=−(−30+j120)=45+j60−21.21+j84.84+j21.21=84.84

×[2]=F1[2]+w28F2[2]=−51−j(−102)=−51+j102×[3]=F1[3]+w38F2[3]=−15−j60+(−0.707−j0.707)(−30−j120)=−15−j60+21.21+j84.84+j21.21−84.84=−78.63+j46.05

x[4]=F1[0]−W08F2[0]=85−170=−85x[5]=F1[1]−W′8F2[1]=−15+j60−(0.707−j0⋅707)(−30+j120)=−15+j60+21.21−j84.84−j21.21−84.84=−78.63−j46.05x[6]=F1[2]−W22F2[2]=−51+j(−102)=−51−j102

x[7]=F1[3]−W38F2[3]=−15−j60−(−0.707−j0.707)(−30−j120)=−15−j60−21.21−j84.84−j21.21+84.84=48.63−j166.05x[k]={255,48.63+j166.05,−51+j102,−78.63+j4.05−85,−78.63−j46.05,−51−j102,+48.63−j166.05}
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