4_HWSolution

# 4_HWSolution - HW EEE 407 DUE NOV 3. SOLUTION Do the...

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HW EEE 407 DUE NOV 3. SOLUTION Do the following problems from the book Chapter 3; 10, 11 Pbm 10 Ch 3 2 12 2 2 2 1 () , () , (1 0.9 ) ( 0.9) ( 1) (0 . 9 ) (1 ) zz Hz Xz z Yz == = −− = 1 1 9 . 0 1 )] ( [ Re )] ( [ Re ) ( ) ( ) ( = = + = + = z n z n ss tr z Y z s z Y z s n y n y n y 22 2 11 () R e s ( 1 ) . 9 ) ) . 9 ) ) . n ss n n z yn z z z z + ⎡⎤ = ⎢⎥ ⎣⎦ 2 1 2 0.9 0.9 21 2 0.9 0.9 100 9) e s ( 0 . 9 ) . 9 ) ) . 9 ) ) 0.9 0.9 (2 ) ) ) ) (1 ) ( 0 . 1 ) ( 0 . 1 ) z tr n n nn n d z z z d z dz z z dz z z z = ++ + = + = + ____________________________________________________________________________________ Pbm 11 Ch 3 1 1 1 ()*( ) ()* ( ) () ( ) |1 / || || | , | |1 xx xx rn x n xn a u n aun R z XzXz a z a a zaz a =− = > > < ____________________________________________________________________________________ Chapter 4 3,5,6 4.3 Show that odd symmetry result in a zero at 0. zero a have we at e H H implies which H H then z For z H z z H symmetry odd n L h n h j L 0 0 ) ( ) 1 ( ) 1 ( ) 1 ( 1 ) ( ) ( ) ( ) ( 0 1 = = = = = = = ____________________________________________________________________________________

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4,5 Develop an expression for the ideal impulse response of a bandpass filter with cutoff frequency at π /3 and π /2. [] /3 /2 11 () | | 22 sin( / 2) sin( /3) 2 0.5sinc( / 2) 0.333 sinc B P F jn h n ed e e jn eee e n n jn n nn ππ π ΩΩ −− ⎡⎤ =Ω + = + = ⎢⎥ ⎣⎦ −+ −= = ∫∫ ____________________________________________________________________________________ 4.6 0.4 0.4 1 ( ) 0.4sinc(0.4 ) 3 3 2 LP hn n n = h LP (n) =[ -0.0624 0.0935 0.3027 0.4000 0.3027 0.0935 -0.0624] -3< n< 3 a) b n = h LP (n-3) ; rectangular window b) Triangular Window samples: w(n)=[¼, ½, ¾, 1, ¾, ½, ¼] -3< n< 3 b n = h LP (n-3)*w(n-3) ; triangular window h LP (n = [0.0156 0.0468 0.2270 0.4000 0.2270 0.0468 -0.0156] c) Hamming window w(n)=[0.0800 0.3100 0.7700 1.0000 0.7700 0.3100 0.0800] -3< n< 3 b n = h LP (n-3)*w(n-3) ; Hamming window h=[-0.0050 0.0290 0.2331 0.4000 0.2331 0.0290 -0.0050]
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -400 -300 -200 -100 0 Normalized frequency (Nyquist == 1) Phase (degrees) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -100 -50 0 50 Normalized frequency (Nyquist == 1) Magnitude Response (dB) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -600 -400 -200 0 Normalized frequency (Nyquist == 1) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -30 -20 -10 0 Normalized frequency (Nyquist == 1) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -600 -400 -200 0 Normalized frequency (Nyquist == 1) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 -60 -40 -20 0 Normalized frequency (Nyquist == 1) freq responses rectangular, triangular, hamming

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____________________________________________________________________________________ Chapter 5 1, 2, 3, 4, 5 5.1
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## This note was uploaded on 11/02/2009 for the course EEE DSP taught by Professor Ap during the Fall '09 term at ASU.

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4_HWSolution - HW EEE 407 DUE NOV 3. SOLUTION Do the...

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