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Chap6StudentSolutions

# 01595286 vo f 120 2 0001427607 sinc0001427607 f

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Unformatted text preview: − 60) + δ ( f + 60)] ∗ 0.001427607 sinc(0.001427607 f )e − j 2πf ( 0.01595286) 2 Solutions 6-36 M. J. Roberts - 8/16/04 0.01523906 − t − j 2πft e 0.1e −10 − j 2πf 0 1 − j 2π ( f − 60)( 0.01595286) Vo1 ( f ) = 120 2 + 0.001427607 sinc(0.001427607( f − 60))e 2 1 − j 2π ( f + 60)( 0.01595286) + 2 0.001427607 sinc(0.001427607( f + 60))e − 0.01523906 − j 2πf 0.01523906 0.1 e 1 e − −10 − j 2πf −10 − j 2πf 1 − j 2π ( f − 60)( 0.01595286) Vo1 ( f ) = 120 2 + 0.001427607 sinc(0.001427607( f − 60))e 2 1 − j 2π ( f + 60)( 0.01595286) + 2 0.001427607 sinc(0.001427607( f + 60))e 0.01523906 − 0.1 e − j 2πf 0.01523906 1 − e 10 + j 2πf 1 − j 2π ( f − 60)( 0.01595286) Vo1 ( f ) = 120 2 + 0.001427607 sinc(0.001427607( f − 60))e 2 1 − j 2π ( f + 60)( 0.01595286) + 2 0.001427607 sinc(0.001427607( f + 60))e The CTFT of the actual periodic response is the product of this CTFT with the CTFT of f 60 comb(60 t) which is comb . Therefore 60 0.01523906 − 0.1 1− e e − j 2πf 0.01523906 10 + j 2πf 1 f − j 2π ( f − 60)( 0.01595286) Vo ( f ) = 120 2 + 0.001427607 sinc(0.001427607( f − 60))e comb 60 2 1 − j 2π ( f + 60)( 0.01595286) + 2 0.001427607 sinc(0.001427607( f + 60))e Solutions 6-37 M. J. Roberts - 8/16/04 |V ( f )| o dB 50 -600 600 f Phase of V ( f ) o π -600 600 f -π 43. Create a discrete-space image consisting of 96 by 96 pixels. Let the image be a “checkerboard” consisting of 8 by 8 alternating black-and-white squares. (a) Filter the image row-by-row and then column-by-column with a DT filter whose impulse response is n h[ n ] = 0.2(0.8) u[ n ] and display the image on the screen using the imagesc command in MATLAB. After defining the checkerboard we can filter it by convolving it with the impulse response using the MATLAB conv function. Notice how this lowpass spatial filter blurs the edges. A lowpass filter does not allow any fast transitions to occur. (b) Filter the image row-by-row and then column-by-column with a DT fi...
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