CE 40763 DIGITAL SIGNAL PROCESSING Homework 5 – Solutions (FIR Filters & Optimal FIR Filters) ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­ Problem 1:
(a) First of all we should get 7 samples in equal distances (0-2π) , which are : 0, 2π/7, 4 π/7, 6π/7, 8π/7,
10π/7, 12π/7 , So we will have
0, 1, 0, 2, 2, 0, 1
2
2
= 1/7[2cos 2πn/7 + 4cos6πn/7]
1/7
3 0
6
= { -0.3846 , 0.2927, -0.3367 , 0.6571 , -0.3367 , 0.2927 , -0.3846 }
1
2
/
1
2
2
2
… , 0.3312 , 0.1592 , 0.3569 , 0.75 , 0.3569 , 0.1592 , 0.3312 , …
3| E1 = 2π ∑ |
∑
|
= 2π (
∑
3
3|
∑
= 1.3591
(b) Fourier series approach
3 0
6
= { -0.3312 , 0.1592, -0.3569 , 0.75 , -0.3569 , 0.1592 , -0.33 }
E2 = 2π ∑ |
3 | = 2π ( ∑
∑
3
= 1.0223
E1 – E2 = 0.3366
Problem 2:
The same as problem 1,
0, 0, 1, 0, 0, 2, 2, 0, 0, 1, 0
1/11
2
2
1/11 [2cos 4πn/11 + 4cos 10 /11
3 0
6
0.4126 , 0.1868 , 0.2734 , 0.5455 , 0.2734 , 0.1868 , 0.4126 ∑
E3 = 2π |
3|
= 2π ( ∑ |
∑
3
3|
∑
= 1.4652
E1 – E3 = - 0.1061
Problem 3:
Length = 2L + 1 = 3 then L = 1
We need at least L+2 = 3 alternations, that We know 2 of them (F = 0.125 , F = 0.25) , guess the third one
(F = 0.5),
1
1
1
cos 2 . 0.125
cos 2 . 0.25
cos 2 . 0.5
1
1
1
1
= 0
0
0.2929
0.5858
0.2929
0.2929 .3 !
0.2929
h[0] =
h[1] = /2 = 0.2929
Problem 4:
(a)
(b)
(c)
1 1
2 2
1 2N-1 = 2L + 1 => L = N-1 and we must have at least L+2 = N+1 Alternations!
Problem 5:
Problem 6:
Problem 7:
Problem 8: