E4810 HW3 Solution
2.68
The causal LTI system is described by y[n] = p0 x[n] + p1 x[n 1] d1 y[n 1]
where x[n], y[n] are the input and output respectively.
The differential equation of its inverse syst
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E4810 HW8 Solution
7.57
(a)
The magnitude response of G1(z)= H(zM)F1(z) is as follows.
| F1 (e jw ) |
1
0
w
(2 ws )
M
wp
M
| H (e jwM ) |
1
0
wp
M
ws
M
(2 ws )
M
| G1 (e jw ) |
1
wp
M
ws
M
The pass-ba
E4810 HW7 Solution
3.75
(a)
Q hA[n] = 0.3 [n] [n 1] + 0.3 [n 2] H A (e jw ) = 0.3 e jw + 0.3e 2 jw
=> | H A (e jw ) |=| e jw (0.6 cos w 1) |=| 1 0.6 cos w | , A ( w) = w
Q hB [n] = 0.3 [n] + [n 1] + 0
E4810 HW10 Solution
10.36
wp = 0.45 , ws = 0.6 , p = 0.2043 and s = 0.0454
The weighting function used in Parks-McClellan algorithm is
1, in the passband
1, 0 w 0.45
=> W ( w) =
W ( w) =
4.5 , 0.
E4810 HW11 Solution
11.17
Let the input be normal order and the output be in the bit-reversed order in the
radix-2 DIT FFT algorithm
11.21
To let M DFT samples by using N-points DFT with an N-point ra
E4810 HW4 Solution
3.19
(a) y1[n] = 1, N n N ; = 0 , otherwise
Y1 (e jw ) =
y [n]e
n =
jwn
=
1
N
e
jwn
= e jwN
e
jw ( 2 N +1)
e
n= N
jw
1
=
1
(2 N + 1) w
)
2
sin( w / 2)
sin(
(b) y2 [n] = 1, 0
E4810 HW5 Solution
6.1
Q x[n] = 0 , n < 0
X ( z) =
x[n]z
n =
n
= x[n]z n = x[0] + x[n]z n
n =1
n =0
=> lim X ( z ) = x[0] + lim x[n]z n = x[0]
z >
z >
n =1
6.5
(a)
(i)Q x1[n] = (0.3) n u[n + 1]
z
Q.1
each
Determine the number of vertices and edges and find the in-degree and out-degree of
vertex for the given directed multi-graph.
(5)
Q.2
Use an adjacency list to represent the given graph.
(5)