ECE3101HW13Sol
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California State University, Fullerton *
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Course
3301
Subject
Electrical Engineering
Date
Jan 9, 2024
Type
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33
Uploaded by CorporalPower10655
1
ECE 3101 HOMEWORK #13 Solutions © Dr. James Kang, Professor, ECE Department, Cal Poly Pomona 1. Find the discrete-time Fourier transform (DTFT), X(
θ
), and plot |X(
θ
)| and ∠
X(
θ
) for 0 ≤ θ
≤ π
for the following signals. Find X(z) from the z-transform table, and set z = e
j
θ
. (a) x(k) = (0.8)
k
u(k). X(
θ
) = 1/(1-0.8e
-j
θ
) (b) x(k) = (0.8)
|k|
. X(
θ
) = (1-0.8
2
)/[(1-0.8e
-j
θ
)(1-0.8e
j
θ
)] 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
1
2
3
4
5
6
θ
/
π
|X(
θ
)|
Magnitude
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-0.4
-0.3
-0.2
-0.1
0
θ
/
π
∠
X(
θ
)/
π
Phase
2
(c) x(k) = k(0.8)
k
u(k). X(
θ
) = 0.8e
-j
θ
/(1-0.8e
-j
θ
)
2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
2
4
6
8
10
θ
/
π
|X(
θ
)|
Magnitude
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1
-0.5
0
0.5
1
θ
/
π
∠
X(
θ
)/
π
Phase
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
5
10
15
20
25
θ
/
π
|X(
θ
)|
Magnitude
3
(d) x(k) = k
2
(0.8)
k
u(k). X(
θ
) = (0.8e
-j
θ
+0.8
2
e
-j2
θ
)/(1-0.8e
-j
θ
)
3
2. Find the discrete-time Fourier transform (DTFT), X(
θ
), and plot |X(
θ
)| and ∠
X(
θ
) for 0 ≤ θ
≤ π
for the following signals. 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1
-0.8
-0.6
-0.4
-0.2
0
θ
/
π
∠
X(
θ
)/
π
Phase
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
50
100
150
200
θ
/
π
|X(
θ
)|
Magnitude
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1
-0.5
0
0.5
1
θ
/
π
∠
X(
θ
)/
π
Phase
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4
Find X(z) from the z-transform table, and set z = e
j
θ
. (a) x(k) = (0.8)
k
cos(
π
k/4) u(k). X(
θ
) = [1-
cos(π/4)0.8exp(
-j
θ
)]/[1-2cos(
π
/4)0.8exp(-j
θ
)+0.8
2
exp(-j2
θ
)] = [1-0.5657exp(-j
θ
)]/[1-1.1314exp(-j
θ
)+0.64exp(-j2
θ
)] (b) x(k) = (0.8)
k
sin(
π
k/4) u(k). X(
θ
) = [sin(π/4)0.8exp(
-j
θ
)]/[1-2cos(
π
/4)0.8exp(-j
θ
)+0.8
2
exp(-j2
θ
)] = [0.5657exp(-j
θ
)]/[1-1.1314exp(-j
θ
)+0.64exp(-j2
θ
)] 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
1
2
3
θ
/
π
|X(
θ
)|
Magnitude
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-0.4
-0.2
0
0.2
0.4
θ
/
π
∠
X(
θ
)/
π
Phase
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
1
2
3
θ
/
π
|X(
θ
)|
Magnitude
5
(c) x(k) = k (0.8)
k
cos(
π
k/4) u(k). X(
θ
) = [cos(π/4)0.8exp(
-j
θ
)-2
×
0.8
2
×
exp(-j2
θ
)+cos(
π
/4)
×
0.8
3
×
exp(-j3
θ
)]/[1-
2cos(
π
/4)0.8exp(-j
θ
)+0.8
2
exp(-j2
θ
)]
2 [0.5657exp(-j
θ
)-1.62
×
exp(-j2
θ
)+0.362
×
exp(-j3
θ
)]/[1-1.1314exp(-j
θ
)+0.64exp(-j2
θ
)]
2
(d) x(k) = k (0.8)
k
sin(
π
k/4) u(k). X(
θ
) = [sin
(π/4)0.8exp(
-j
θ
)-sin(
π
/4)
×
0.8
3
×
exp(-j3
θ
)]/[1-2cos(
π
/4)0.8exp(-j
θ
)+0.8
2
exp(-
j2
θ
)]
2 [0.5657exp(-j
θ
)-0.362
×
exp(-j3
θ
)]/[1-1.1314exp(-j
θ
)+0.64exp(-j2
θ
)]
2
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1
-0.5
0
θ
/
π
∠
X(
θ
)/
π
Phase
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
5
10
θ
/
π
|X(
θ
)|
Magnitude
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1
-0.5
0
0.5
1
θ
/
π
∠
X(
θ
)/
π
Phase
6
3. Find the 8-point (N = 8) discrete Fourier transform (DFT), X(n), and plot |X(n)| and ∠
X(n) for 0 ≤ n ≤ 7 for the following signals.
(a) x(k) = (0.8)
k
[u(k) −
u(k
−
8)]. x(k) X(n) |X(n)| ∠
X(n) 1.0000 4.1611 4.1611 0 0.8000 0.7106 - 0.9256i 1.1669 -0.9160 0.6400 0.5075 - 0.4060i 0.6499 -0.6747 0.5120 0.4702 - 0.1699i 0.4999 -0.3467 0.4096 0.4623 - 0.0000i 0.4623 -0.0000 0.3277 0.4702 + 0.1699i 0.4999 0.3467 0.2621 0.5075 + 0.4060i 0.6499 0.6747 0.2097 0.7106 + 0.9256i 1.1669 0.9160 0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0
5
10
15
θ
/
π
|X(
θ
)|
Magnitude
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
-1
-0.5
0
0.5
1
θ
/
π
∠
X(
θ
)/
π
Phase
0
1
2
3
4
5
6
7
0
1
2
3
4
5
n
|X
n
|
Magnitude Response
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7
(b) x(k) = k(0.8)
k
[u(k) −
u(k
−
8)]. x(k) X(n) |X(n)| ∠
X(n) 0 9.9337 9.9337 0 0.8000 -2.2793 + 0.8376i 2.4284 2.7894 1.2800 -1.2145 + 0.5656i 1.3397 2.7057 1.5360 -0.9975 + 0.2519i 1.0288 2.8942 1.6384 -0.9511 - 0.0000i 0.9511 -3.1416 1.6384 -0.9975 - 0.2519i 1.0288 -2.8942 1.5729 -1.2145 - 0.5656i 1.3397 -2.7057 1.4680 -2.2793 - 0.8376i 2.4284 -2.7894 (c) x(k) = k
2
(0.8)
k
[u(k) −
u(k
−
8)]. 0
1
2
3
4
5
6
7
-1
-0.5
0
0.5
1
n
∠
X
n
Phase Response
0
1
2
3
4
5
6
7
0
2
4
6
8
10
n
|X
n
|
Magnitude Response
0
1
2
3
4
5
6
7
-4
-2
0
2
4
n
∠
X
n
Phase Response
8
x(k) X(n) |X(n)| ∠
X(n) 0 42.4268 42.4268 0 0.8000 -7.7726 +16.1120i 17.8889 2.0203 2.5600 -5.4436 + 5.8920i 8.0218 2.3167 4.6080 -5.3346 + 2.3577i 5.8324 2.7254 6.5536 -5.3253 - 0.0000i 5.3253 -3.1416 8.1920 -5.3346 - 2.3577i 5.8324 -2.7254 9.4372 -5.4436 - 5.8920i 8.0218 -2.3167 10.2760 -7.7726 -16.1120i 17.8889 -2.0203 4. Find the 8-point (N = 8) discrete Fourier transform (DFT), X(n), and plot |X(n)| and ∠
X(n) for 0 ≤ n ≤ 7 for the following signals.
(a) x(k) = [1 1 1 1 1 1 1 1]. 𝑋𝑋
(
𝑛𝑛
) =
∑
𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑁𝑁−1
𝑗𝑗=0
=
𝑒𝑒
−𝑗𝑗0𝑗𝑗
2𝜋𝜋
𝑁𝑁
+
𝑒𝑒
−𝑗𝑗1𝑗𝑗
2𝜋𝜋
𝑁𝑁
+
𝑒𝑒
−𝑗𝑗2𝑗𝑗
2𝜋𝜋
𝑁𝑁
+
𝑒𝑒
−𝑗𝑗3𝑗𝑗
2𝜋𝜋
𝑁𝑁
+
𝑒𝑒
−𝑗𝑗4𝑗𝑗
2𝜋𝜋
𝑁𝑁
+
𝑒𝑒
−𝑗𝑗5𝑗𝑗
2𝜋𝜋
𝑁𝑁
+
𝑒𝑒
−𝑗𝑗6𝑗𝑗
2𝜋𝜋
𝑁𝑁
+
𝑒𝑒
−𝑗𝑗7𝑗𝑗
2𝜋𝜋
𝑁𝑁
X(0) = 1+1+1+1+1+1+1+1 = 8 For n > 0, X(n) = 0. x(k) X(n) |X(n)| ∠
X(n) 1.0000 8.0000 8.0000 0 1.0000 -0.0000 + 0.0000i 0.0000 0 1.0000 -0.0000 - 0.0000i 0.0000 0 1.0000 -0.0000 + 0.0000i 0.0000 0 1.0000 0 - 0.0000i 0.0000 0 0
1
2
3
4
5
6
7
0
10
20
30
40
50
n
|X
n
|
Magnitude Response
0
1
2
3
4
5
6
7
-4
-2
0
2
4
n
∠
X
n
Phase Response
9
1.0000 -0.0000 - 0.0000i 0.0000 0 1.0000 -0.0000 - 0.0000i 0.0000 0 1.0000 0.0000 + 0.0000i 0.0000 0 ∠
X(n) = 0 (b) x(k) = [1 1 1 1 0 0 0 0]. 𝑋𝑋
(
𝑛𝑛
) =
� 𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑁𝑁−1
𝑗𝑗=0
=
𝑒𝑒
−𝑗𝑗0𝑗𝑗
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗1𝑗𝑗
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2𝑗𝑗
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗3𝑗𝑗
2𝜋𝜋
8
X(0) = 1 +
𝑒𝑒
−𝑗𝑗1
×
0
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
0
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗3
×
0
2𝜋𝜋
8
= 1 + 1 + 1 + 1 = 4
X(1) = 1 +
𝑒𝑒
−𝑗𝑗1
×
1
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
1
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗3
×
1
2𝜋𝜋
8
= 1 – j2.4142 = 2.6131e
-j1.1781 X(2) = 1 +
𝑒𝑒
−𝑗𝑗1
×
2
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
2
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗3
×
2
2𝜋𝜋
8
= 0 X(3) = 1 +
𝑒𝑒
−𝑗𝑗1
×
3
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
3
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗3
×
3
2𝜋𝜋
8
= 1 – j0.4142 = 1.0824e
-j0.3927
X(4) = 1 +
𝑒𝑒
−𝑗𝑗1
×
4
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
4
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗3
×
4
2𝜋𝜋
8
= 0 X(5) = 1 +
𝑒𝑒
−𝑗𝑗1
×
5
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
5
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗3
×
5
2𝜋𝜋
8
= 1 + j0.4142 = 1.0824e
j0.3927
X(6) = 1 +
𝑒𝑒
−𝑗𝑗1
×
6
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
6
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗3
×
6
2𝜋𝜋
8
= 0 X(7) = 1 +
𝑒𝑒
−𝑗𝑗1
×
7
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
7
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗3
×
7
2𝜋𝜋
8
= 1 + j2.4142 = 2.6131e
j1.1781
x(k) X(n) |X(n)| ∠
X(n) 1.0000 4.0000 4.0000 0 0
1
2
3
4
5
6
7
0
2
4
6
8
n
|X
n
|
Magnitude Response
0
1
2
3
4
5
6
7
-1
-0.5
0
0.5
1
n
∠
X
n
Phase Response
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10 1.0000 1.0000 - 2.4142i 2.6131 -1.1781 1.0000 -0.0000 - 0.0000i 0.0000 0 1.0000 1.0000 - 0.4142i 1.0824 -0.3927 0 0 - 0.0000i 0.0000 0 0 1.0000 + 0.4142i 1.0824 0.3927 0 0.0000 - 0.0000i 0.0000 0 0 1.0000 + 2.4142i 2.6131 1.1781 (c) x(k) = [1 1 -1 -1 1 1 -1 -1]. 𝑋𝑋
(
𝑛𝑛
) =
� 𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑁𝑁−1
𝑗𝑗=0
=
𝑒𝑒
−𝑗𝑗0𝑗𝑗
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗1𝑗𝑗
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗2𝑗𝑗
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3𝑗𝑗
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4𝑗𝑗
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗5𝑗𝑗
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗6𝑗𝑗
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗7𝑗𝑗
2𝜋𝜋
8
X(0) = 1 +
𝑒𝑒
−𝑗𝑗1
×
0
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗2
×
0
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
0
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
0
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗5
×
0
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗6
×
0
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
0
2𝜋𝜋
8
= 0
X(1) = 1 +
𝑒𝑒
−𝑗𝑗1
×
1
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗2
×
1
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
1
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
1
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗5
×
1
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗6
×
1
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
1
2𝜋𝜋
8
= 0
X(2) = 1 +
𝑒𝑒
−𝑗𝑗1
×
2
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗2
×
2
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
2
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
2
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗5
×
2
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗6
×
2
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
2
2𝜋𝜋
8
= 4-j4 0
1
2
3
4
5
6
7
0
1
2
3
4
n
|X
n
|
Magnitude Response
0
1
2
3
4
5
6
7
-3
-2
-1
0
1
2
n
∠
X
n
Phase Response
11 X(3) = 1 +
𝑒𝑒
−𝑗𝑗1
×
3
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗2
×
3
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
3
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
3
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗5
×
3
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗6
×
3
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
3
2𝜋𝜋
8
= 0 X(4) = 1 +
𝑒𝑒
−𝑗𝑗1
×
4
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗2
×
4
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
4
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
4
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗5
×
4
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗6
×
4
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
4
2𝜋𝜋
8
= 0 X(5) = 1 +
𝑒𝑒
−𝑗𝑗1
×
5
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗2
×
5
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
5
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
5
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗5
×
5
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗6
×
5
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
5
2𝜋𝜋
8
= 0 X(6) = 1 +
𝑒𝑒
−𝑗𝑗1
×
6
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗2
×
6
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
6
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
6
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗5
×
6
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗6
×
6
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
6
2𝜋𝜋
8
= 4+j4 X(7) = 1 +
𝑒𝑒
−𝑗𝑗1
×
7
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗2
×
7
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
7
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
7
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗5
×
7
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗6
×
7
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
7
2𝜋𝜋
8
= 0 x(k) X(n) |X(n)| ∠
X(n) 1.0000 0 0 0 1.0000 0.0000 - 0.0000i 0.0000 0 -1.0000 4.0000 - 4.0000i 5.6569 -0.7854 -1.0000 -0.0000 + 0.0000i 0.0000 0 1.0000 0 0 0 1.0000 0.0000 - 0.0000i 0.0000 0 -1.0000 4.0000 + 4.0000i 5.6569 0.7854 -1.0000 -0.0000 - 0.0000i 0.0000 0 0
1
2
3
4
5
6
7
0
2
4
6
n
|X
n
|
Magnitude Response
12 (d) x(k) = [1 -1 1 -1 1 -1 1 -1]. 𝑋𝑋
(
𝑛𝑛
) =
� 𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑁𝑁−1
𝑗𝑗=0
=
𝑒𝑒
−𝑗𝑗0𝑗𝑗
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗1𝑗𝑗
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2𝑗𝑗
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3𝑗𝑗
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4𝑗𝑗
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗5𝑗𝑗
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗6𝑗𝑗
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗7𝑗𝑗
2𝜋𝜋
8
X(0) = 1
− 𝑒𝑒
−𝑗𝑗1
×
0
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
0
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
0
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
0
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗5
×
0
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗6
×
0
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
0
2𝜋𝜋
8
= 0
X(1) = 1
− 𝑒𝑒
−𝑗𝑗1
×
1
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
1
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
1
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
1
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗5
×
1
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗6
×
1
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
1
2𝜋𝜋
8
= 0
X(2) = 1
− 𝑒𝑒
−𝑗𝑗1
×
2
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
2
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
2
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
2
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗5
×
2
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗6
×
2
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
2
2𝜋𝜋
8
= 0 X(3) = 1
− 𝑒𝑒
−𝑗𝑗1
×
3
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
3
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
3
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
3
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗5
×
3
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗6
×
3
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
3
2𝜋𝜋
8
= 0 X(4) = 1
− 𝑒𝑒
−𝑗𝑗1
×
4
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
4
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
4
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
4
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗5
×
4
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗6
×
4
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
4
2𝜋𝜋
8
= 8 X(5) = 1
− 𝑒𝑒
−𝑗𝑗1
×
5
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
5
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
5
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
5
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗5
×
5
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗6
×
5
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
5
2𝜋𝜋
8
= 0 X(6) = 1
− 𝑒𝑒
−𝑗𝑗1
×
6
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
6
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
6
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
6
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗5
×
6
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗6
×
6
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
6
2𝜋𝜋
8
= 0 X(7) = 1
− 𝑒𝑒
−𝑗𝑗1
×
7
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗2
×
7
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗3
×
7
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗4
×
7
2𝜋𝜋
8
− 𝑒𝑒
−𝑗𝑗5
×
7
2𝜋𝜋
8
+
𝑒𝑒
−𝑗𝑗6
×
7
2𝜋𝜋
8
−
𝑒𝑒
−𝑗𝑗7
×
7
2𝜋𝜋
8
= 0 x(k) X(n) |X(n)| ∠
X(n) 1.0000 0 0 0 -1.0000 0.0000 - 0.0000i 0.0000 0 1.0000 0.0000 - 0.0000i 0.0000 0 -1.0000 0.0000 - 0.0000i 0.0000 0 0
1
2
3
4
5
6
7
-1
-0.5
0
0.5
1
n
∠
X
n
Phase Response
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13 1.0000 8.0000 + 0.0000i 8.0000 0 -1.0000 -0.0000 + 0.0000i 0.0000 0 1.0000 0.0000 - 0.0000i 0.0000 0 -1.0000 -0.0000 - 0.0000i 0.0000 0 5. For the following periodic signals, find the coefficients d
n
of the discrete-time Fourier series representation, and plot |d
n
| and ∠
d
n
. 𝑑𝑑
𝑗𝑗
=
1
𝑁𝑁
�
𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑗𝑗
𝑗𝑗
1
+𝑁𝑁−1
𝑗𝑗=𝑗𝑗
1
, 0
≤ 𝑛𝑛 ≤ 𝑁𝑁 −
1
𝑥𝑥
(
𝑘𝑘
) =
� 𝑑𝑑
𝑗𝑗
𝑒𝑒
𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑗𝑗
𝑁𝑁−1
𝑗𝑗=0
, 0
≤ 𝑘𝑘 ≤ 𝑁𝑁 −
1
(a) N = 10 0
1
2
3
4
5
6
7
0
2
4
6
8
n
|X
n
|
Magnitude Response
0
1
2
3
4
5
6
7
-1
-0.5
0
0.5
1
n
∠
X
n
Phase Response
14 𝑑𝑑
𝑗𝑗
=
1
𝑁𝑁
�
𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑗𝑗
𝑗𝑗
1
+𝑁𝑁−1
𝑗𝑗=𝑗𝑗
1
=
1
10
�
1
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
𝑗𝑗
2
𝑗𝑗=−2
=
1
10
�𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
−2
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
−1
)
+ 1 +
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
1
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
2
�
= 1
10
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �𝑛𝑛
2𝜋𝜋
10
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �𝑛𝑛
4𝜋𝜋
10
��
d
0
= 1
10
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
0
2𝜋𝜋
10
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
0
4𝜋𝜋
10
��
=5/10 = 0.5 d
1
= 1
10
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
1
2𝜋𝜋
10
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
1
4𝜋𝜋
10
��
= 0.3236 d
2
= 1
10
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
2
2𝜋𝜋
10
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
2
4𝜋𝜋
10
��
= 0 ….. x(k) d(n) |d(n)| ∠
d(n) 1.0000 0.5000 0.5000 0 1.0000 0.3236 + 0.0000i 0.3236 0 1.0000 -0.0000 - 0.0000i 0.0000 0 0 -0.1236 - 0.0000i 0.1236 -3.1416 0 0.0000 - 0.0000i 0.0000 0 0 0.1000 0.1000 0 0 0 - 0.0000i 0.0000 0 0 -0.1236 - 0.0000i 0.1236 -3.1416 1.0000 0.0000 - 0.0000i 0.0000 0 1.0000 0.3236 + 0.0000i 0.3236 0 -20
-15
-10
-5
0
5
10
15
20
0
0.2
0.4
0.6
0.8
1
0
1
2
3
4
5
6
7
8
9
0
0.2
0.4
0.6
0.8
n
|d
n
|
Magnitude
15 (b) 𝑑𝑑
𝑗𝑗
=
1
𝑁𝑁
�
𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑗𝑗
𝑗𝑗
1
+𝑁𝑁−1
𝑗𝑗=𝑗𝑗
1
=
1
10
�
1
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
𝑗𝑗
4
𝑗𝑗=0
=
1
10
�𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
0
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
1
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
2
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
3
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
4
)
�
=
1
10
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
2
)
�𝑒𝑒
𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
2
)
+
𝑒𝑒
𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
1
)
+ 1 +
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
1
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
2
)
�
= 𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
2
)
1
10
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �𝑛𝑛
2𝜋𝜋
10
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �𝑛𝑛
4𝜋𝜋
10
��
d
0
= 1
10
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
0
2𝜋𝜋
10
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
0
4𝜋𝜋
10
��
=5/10 = 0.5 d
1
= 𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
(
2
)
1
10
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
1
2𝜋𝜋
10
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
1
4𝜋𝜋
10
��
= 0.3236e
-j1.2655 d
2
= 1
10
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
2
2𝜋𝜋
10
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
2
4𝜋𝜋
10
��
= 0 ….. x(k) d(n) |d(n)| ∠
d(n) 1.0000 0.5000 0.5000 0 1.0000 0.1000 - 0.3078i 0.3236 -1.2566 1.0000 -0.0000 - 0.0000i 0.0000 0 1.0000 0.1000 - 0.0727i 0.1236 -0.6283 1.0000 0.0000 - 0.0000i 0.0000 0 0 0.1000 + 0.0000i 0.1000 0 0 0.0000 - 0.0000i 0.0000 0 0 0.1000 + 0.0727i 0.1236 0.6283 0 0.0000 - 0.0000i 0.0000 0 0
1
2
3
4
5
6
7
8
9
-4
-3
-2
-1
0
n
∠
d
n
Phase
-20
-15
-10
-5
0
5
10
15
20
0
0.2
0.4
0.6
0.8
1
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16 0 0.1000 + 0.3078i 0.3236 1.2566 (c) N = 12. 𝑑𝑑
𝑗𝑗
=
1
𝑁𝑁
�
𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑗𝑗
𝑗𝑗
1
+𝑁𝑁−1
𝑗𝑗=𝑗𝑗
1
=
1
12
�
1
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
𝑗𝑗
3
𝑗𝑗=−3
=
1
12
�𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
−3
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
−2
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
−1
)
+ 1 +
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
1
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
2
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
3
�
= 1
12
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �𝑛𝑛
2𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �𝑛𝑛
4𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �𝑛𝑛
6𝜋𝜋
12
��
d
0
= 1
12
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
0
2𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
0
4𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
0
6𝜋𝜋
12
��
=7/12 = 0.5833 d
1
= 1
12
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
1
2𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
1
4𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
1
6𝜋𝜋
12
��
= 0.110 0
1
2
3
4
5
6
7
8
9
0
0.2
0.4
0.6
0.8
n
|d
n
|
Magnitude
0
1
2
3
4
5
6
7
8
9
-2
-1
0
1
2
n
∠
d
n
Phase
-20
-15
-10
-5
0
5
10
15
20
0
0.2
0.4
0.6
0.8
1
17 d
2
= 1
12
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
2
2𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
2
4𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
2
6𝜋𝜋
12
��
= -0.0833 ….. x(k) d(n) |d(n)| ∠
d(n) 1.0000 0.5833 0.5833 0 1.0000 0.3110 + 0.0000i 0.3110 0 1.0000 -0.0833 + 0.0000i 0.0833 3.1416 1.0000 -0.0833 - 0.0000i 0.0833 -3.1416 0 0.0833 - 0.0000i 0.0833 0 0 0.0223 + 0.0000i 0.0223 0 0 -0.0833 + 0.0000i 0.0833 3.1416 0 0.0223 - 0.0000i 0.0223 0 0 0.0833 - 0.0000i 0.0833 0 1.0000 -0.0833 + 0.0000i 0.0833 3.1416 1.0000 -0.0833 - 0.0000i 0.0833 -3.1416 1.0000 0.3110 + 0.0000i 0.3110 0 (d) N = 12 0
2
4
6
8
10
12
0
0.2
0.4
0.6
0.8
n
|d
n
|
Magnitude
0
2
4
6
8
10
12
-4
-2
0
2
4
n
∠
d
n
Phase
18 𝑑𝑑
𝑗𝑗
=
1
𝑁𝑁
�
𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑗𝑗
𝑗𝑗
1
+𝑁𝑁−1
𝑗𝑗=𝑗𝑗
1
=
1
12
�
1
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
12
𝑗𝑗
6
𝑗𝑗=0
=
1
12
�𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
0
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
1
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
2
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
3
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
4
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
5
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
6
)
�
=
1
12
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
3
)
�𝑒𝑒
𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
3
)
+
𝑒𝑒
𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
2
)
+
𝑒𝑒
𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
1
)
+ 1 +
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
1
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
2
)
+
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
3
)
�
= 𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
(
3
)
1
12
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �𝑛𝑛
2𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �𝑛𝑛
4𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �𝑛𝑛
6𝜋𝜋
12
��
d
0
= 1
12
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
0
2𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
0
4𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
0
6𝜋𝜋
12
��
=7/12 = 0.5833 d
1
= 𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
12
(
3
)
1
12
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
1
2𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
1
4𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
1
6𝜋𝜋
12
��
= -0.3110
d
2
= 𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
12
(
3
)
1
12
�
1 + 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
2
2𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
2
4𝜋𝜋
12
�
+ 2
𝑐𝑐𝑐𝑐𝑐𝑐 �
2
6𝜋𝜋
12
��
= 0.0833 ….. x(k) d(n) |d(n)| ∠
d(n) 1.0000 0.5833 0.5833 0 1.0000 0.0000 - 0.3110i 0.3110 -1.5708 1.0000 0.0833 + 0.0000i 0.0833 0 1.0000 0.0000 - 0.0833i 0.0833 -1.5708 1.0000 0.0833 - 0.0000i 0.0833 0 1.0000 0.0000 - 0.0223i 0.0223 -1.5708 1.0000 0.0833 - 0.0000i 0.0833 0 0 0.0000 + 0.0223i 0.0223 1.5708 0 0.0833 + 0.0000i 0.0833 0 0 0.0000 + 0.0833i 0.0833 1.5708 0 0.0833 + 0.0000i 0.0833 0 0 -0.0000 + 0.3110i 0.3110 1.5708 -20
-15
-10
-5
0
5
10
15
20
0
0.2
0.4
0.6
0.8
1
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19 6. For the following periodic signals, find the coefficients d
n
of the discrete-time Fourier series representation, and plot |d
n
| and ∠
d
n
. (a) N = 5. 𝑑𝑑
𝑗𝑗
=
1
𝑁𝑁
�
𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑗𝑗
𝑗𝑗
1
+𝑁𝑁−1
𝑗𝑗=𝑗𝑗
1
=
1
5
�
0
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
12
0
+ 1
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
12
1
+ 2
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
12
2
+ 3
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
12
3
+ 4
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
12
4
�
d
0
= 1
5
�
0
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
5
0
+ 1
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
5
1
+ 2
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
5
2
+ 3
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
5
3
+ 4
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
5
4
�
=
1
5
(0 + 1 + 2 + 3 +
4) = 2
0
2
4
6
8
10
12
0
0.2
0.4
0.6
0.8
n
|d
n
|
Magnitude
0
2
4
6
8
10
12
-2
-1
0
1
2
n
∠
d
n
Phase
-10
-8
-6
-4
-2
0
2
4
6
8
10
0
1
2
3
4
5
20 d
1
= 1
5
�
0
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
5
0
+ 1
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
5
1
+ 2
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
5
2
+ 3
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
5
3
+ 4
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
5
4
�
= -0.5+j0.6882
d
2
= 1
5
�
0
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
5
0
+ 1
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
5
1
+ 2
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
5
2
+ 3
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
5
3
+ 4
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
5
4
�
= -0.5+j0.1625 ….. x(k) d(n) |d(n)| ∠
d(n) 0 2.0000 2.0000 0 1.0000 -0.5000 + 0.6882i 0.8507 2.1991 2.0000 -0.5000 + 0.1625i 0.5257 2.8274 3.0000 -0.5000 - 0.1625i 0.5257 -2.8274 4.0000 -0.5000 - 0.6882i 0.8507 -2.1991 (b) N = 10. 0
0.5
1
1.5
2
2.5
3
3.5
4
0
0.5
1
1.5
2
n
|d
n
|
Magnitude
0
0.5
1
1.5
2
2.5
3
3.5
4
-4
-2
0
2
4
n
∠
d
n
Phase
-20
-15
-10
-5
0
5
10
15
20
0
1
2
3
4
5
21 𝑑𝑑
𝑗𝑗
=
1
𝑁𝑁
�
𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑗𝑗
𝑗𝑗
1
+𝑁𝑁−1
𝑗𝑗=𝑗𝑗
1
=
1
10
�
0
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
0
+ 1
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
1
+ 2
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
2
+ 3
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
3
+ 4
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
4
�
d
0
= 1
10
�
0
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
0
+ 1
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
1
+ 2
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
2
+ 3
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
3
+ 4
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
4
�
=
1
10
(0 + 1 + 2 + 3 +
4) = 1
d
1
= 1
10
�
0
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
0
+ 1
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
1
+ 2
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
2
+ 3
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
3
+ 4
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
4
�
= -0.2736-j0.7694
d
2
= 1
10
�
0
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
0
+ 1
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
1
+ 2
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
2
+ 3
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
3
+ 4
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
4
�
= -0.25+j0.3441 ….. x(k) d(n) |d(n)| ∠
d(n) 0 1.0000 1.0000 0 1.0000 -0.2736 - 0.7694i 0.8166 -1.9125 2.0000 -0.2500 + 0.3441i 0.4253 2.1991 3.0000 0.1736 - 0.1816i 0.2513 -0.8080 4.0000 -0.2500 + 0.0812i 0.2629 2.8274 0 0.2000 + 0.0000i 0.2000 0 0 -0.2500 - 0.0812i 0.2629 -2.8274 0 0.1736 + 0.1816i 0.2513 0.8080 0 -0.2500 - 0.3441i 0.4253 -2.1991 0 -0.2736 + 0.7694i 0.8166 1.9125 7. For the following periodic signals, find the coefficients d
n
of the discrete-time Fourier series representation, and plot |d
n
| and ∠
d
n
. 0
1
2
3
4
5
6
7
8
9
0
0.2
0.4
0.6
0.8
1
n
|d
n
|
Magnitude
0
1
2
3
4
5
6
7
8
9
-4
-2
0
2
4
n
∠
d
n
Phase
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22 (a) Period = N = 10, one period of x(k) for 0 ≤ k ≤ 9 is given by
𝑥𝑥
(
𝑘𝑘
) = (0.8)
𝑗𝑗
, 0
≤ 𝑘𝑘 ≤
9
𝑑𝑑
𝑗𝑗
=
1
𝑁𝑁
�
𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑗𝑗
𝑗𝑗
1
+𝑁𝑁−1
𝑗𝑗=𝑗𝑗
1
=
1
10
�
1
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
0
+ 0.8
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
1
+ 0. 8
2
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
2
+
. . . . .
+0. 8
9
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
9
�
d
0
= 1
10
�
1
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
0
+ 0.8
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
1
+ 0. 8
2
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
2
+
. . . . .
+0. 8
9
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
9
�
=
1
10
1−0
.
8
10
1−0
.
8
=
0.4463
d
1
= 1
10
�
1
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
0
+ 0.8
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
1
+ 0. 8
2
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
2
+
. . . . .
+0. 8
9
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
9
�
= 0.0911-j0.1215
d
2
= 1
10
�
1
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
0
+ 0.8
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
1
+ 0. 8
2
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
2
+
. . . . .
+0. 8
9
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
9
�
= 0.0587-j0.0593 ….. x(k) d(n) |d(n)| ∠
d(n) 1.0000 0.4463 0.4463 0 0.8000 0.0911 - 0.1215i 0.1518 -0.9271 0.6400 0.0587 - 0.0593i 0.0834 -0.7907 0.5120 0.0522 - 0.0318i 0.0611 -0.5478 0.4096 0.0501 - 0.0143i 0.0521 -0.2781 0.3277 0.0496 - 0.0000i 0.0496 0 0.2621 0.0501 + 0.0143i 0.0521 0.2781 0.2097 0.0522 + 0.0318i 0.0611 0.5478 0.1678 0.0587 + 0.0593i 0.0834 0.7907 0.1342 0.0911 + 0.1215i 0.1518 0.9271 0
1
2
3
4
5
6
7
8
9
0
0.1
0.2
0.3
0.4
0.5
n
|d
n
|
Magnitude
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23 (b) Period = N = 10, one period of x(k) for 0 ≤ k ≤ 9 is given by
𝑥𝑥
(
𝑘𝑘
) =
𝑘𝑘
(0.8)
𝑗𝑗
, 0
≤ 𝑘𝑘 ≤
9
𝑑𝑑
𝑗𝑗
=
1
𝑁𝑁
�
𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑗𝑗
𝑗𝑗
1
+𝑁𝑁−1
𝑗𝑗=𝑗𝑗
1
=
1
10
�
0
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
0
+ 0.8
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
1
+ 2 × 0. 8
2
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
2
+
. . . . .
+9 × 0. 8
9
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
9
�
d
0
= 1
10
�
0
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
0
+ 0.8
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
1
+ 2 × 0. 8
2
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
2
+
. . . . .
+9 × 0. 8
9
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
9
�
= 1.2484 d
1
= 1
10
�
0
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
0
+ 0.8
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
1
+ 2 × 0. 8
2
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
2
+
. . . . .
+9 × 0. 8
9
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
9
�
= -
0.2730+j0.0196
d
2
= 1
10
�
0
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
0
+ 0.8
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
1
+ 2 × 0. 8
2
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
2
+
. . . . .
+9 × 0. 8
9
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
9
�
= -
0.13+j0.0527 ….. x(k) d(n) |d(n)| ∠
d(n) 0 1.2484 1.2484 0 0.8000 -0.2730 + 0.0196i 0.2737 3.0700 1.2800 -0.1300 + 0.0527i 0.1403 2.7567 1.5360 -0.0958 + 0.0329i 0.1013 2.8106 1.6384 -0.0845 + 0.0155i 0.0859 2.9608 1.6384 -0.0817 - 0.0000i 0.0817 -3.1416 1.5729 -0.0845 - 0.0155i 0.0859 -2.9608 1.4680 -0.0958 - 0.0329i 0.1013 -2.8106 1.3422 -0.1300 - 0.0527i 0.1403 -2.7567 1.2080 -0.2730 - 0.0196i 0.2737 -3.0700 0
1
2
3
4
5
6
7
8
9
-1
-0.5
0
0.5
1
n
∠
d
n
Phase
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24 8. For the following periodic signals, find the coefficients d
n
of the discrete-time Fourier series representation, and plot |d
n
| and ∠
d
n
. (a) Period = N = 10, one period of x(k) f
or 0 ≤ k ≤ 9 is given by
𝑥𝑥
(
𝑘𝑘
) =
�
𝑘𝑘
, 0
≤ 𝑘𝑘 ≤
4
5,5
≤ 𝑘𝑘 ≤
9
𝑑𝑑
𝑗𝑗
=
1
𝑁𝑁
�
𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑗𝑗
𝑗𝑗
1
+𝑁𝑁−1
𝑗𝑗=𝑗𝑗
1
=
1
10
�
0
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
0
+ 1
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
1
+ 2
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
2
+ 3
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
3
+ 4
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
4
+ 5
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
5
+ 5
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
6
+5
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
7
+ 5
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
8
+ 5
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
9
�
d
0
= 1
10
�
0
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
0
+ 1
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
1
+ 2
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
2
+ 3
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
3
+ 4
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
4
+ 5
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
5
+ 5
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
6
+5
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
7
+ 5
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
8
+ 5
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
9
�
=(0+1+2+3+4+5+5+5+5+5)/10=35/10=3.5 d
1
=
1
10
�
0
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
0
+ 1
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
1
+ 2
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
2
+ 3
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
3
+ 4
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
4
+ 5
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
5
+ 5
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
6
+5
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
7
+ 5
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
8
+ 5
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
9
�
0
1
2
3
4
5
6
7
8
9
0
0.2
0.4
0.6
0.8
1
1.2
1.4
n
|d
n
|
Magnitude
0
1
2
3
4
5
6
7
8
9
-4
-2
0
2
4
n
∠
d
n
Phase
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25 = -0.7736+j0.7694 d
2
=
1
10
�
0
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
0
+ 1
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
1
+ 2
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
2
+ 3
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
3
+ 4
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
4
+ 5
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
5
+ 5
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
6
+5
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
7
+ 5
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
8
+ 5
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
9
�
= -0.25+j0.3441 ….. x(k) d(n) |d(n)| ∠
d(n) 0 3.5000 3.5000 0 1.0000 -0.7736 + 0.7694i 1.0911 2.3589 2.0000 -0.2500 + 0.3441i 0.4253 2.1991 3.0000 -0.3264 + 0.1816i 0.3735 2.6338 4.0000 -0.2500 + 0.0812i 0.2629 2.8274 5.0000 -0.3000 - 0.0000i 0.3000 -3.1416 5.0000 -0.2500 - 0.0812i 0.2629 -2.8274 5.0000 -0.3264 - 0.1816i 0.3735 -2.6338 5.0000 -0.2500 - 0.3441i 0.4253 -2.1991 5.0000 -0.7736 - 0.7694i 1.0911 -2.3589 0
1
2
3
4
5
6
7
8
9
0
0.5
1
1.5
2
2.5
3
3.5
n
|d
n
|
Magnitude
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26 (b) Period = N = 10, one period of x(k) for 0 ≤ k ≤ 9 is given by
𝑥𝑥
(
𝑘𝑘
) =
�
𝑘𝑘
, 0
≤ 𝑘𝑘 ≤
3
4,4
≤ 𝑘𝑘 ≤
6
10
− 𝑘𝑘
, 7
≤ 𝑘𝑘 ≤
9
𝑑𝑑
𝑗𝑗
=
1
𝑁𝑁
�
𝑥𝑥
(
𝑘𝑘
)
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
𝑁𝑁
𝑗𝑗
𝑗𝑗
1
+𝑁𝑁−1
𝑗𝑗=𝑗𝑗
1
=
1
10
�
0
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
0
+ 1
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
1
+ 2
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
2
+ 3
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
3
+ 4
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
4
+ 4
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
5
+ 4
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
6
+3
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
7
+ 2
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
8
+ 1
𝑒𝑒
−𝑗𝑗𝑗𝑗
2𝜋𝜋
10
9
�
d
0
= 1
10
�
0
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
0
+ 1
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
1
+ 2
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
2
+ 3
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
3
+ 4
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
4
+ 4
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
5
+ 4
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
6
+3
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
7
+ 2
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
8
+ 1
𝑒𝑒
−𝑗𝑗0
2𝜋𝜋
10
9
�
= 2.4 d
1
= 1
10
�
0
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
0
+ 1
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
1
+ 2
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
2
+ 3
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
3
+ 4
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
4
+ 4
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
5
+ 4
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
6
+3
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
7
+ 2
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
8
+ 1
𝑒𝑒
−𝑗𝑗1
2𝜋𝜋
10
9
�
= -
0.9472
d
2
= 1
10
�
0
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
0
+ 1
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
1
+ 2
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
2
+ 3
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
3
+ 4
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
4
+ 4
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
5
+ 4
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
6
+3
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
7
+ 2
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
8
+ 1
𝑒𝑒
−𝑗𝑗2
2𝜋𝜋
10
9
�
= -
0.1 ….. x(k) d(n) |d(n)| ∠
d(n) 0 2.4000 2.4000 0 1.0000 -0.9472 - 0.0000i 0.9472 -3.1416 0
1
2
3
4
5
6
7
8
9
-4
-3
-2
-1
0
1
2
3
n
∠
d
n
Phase
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27 2.0000 -0.1000 - 0.0000i 0.1000 -3.1416 3.0000 -0.0528 - 0.0000i 0.0528 -3.1416 4.0000 -0.1000 - 0.0000i 0.1000 -3.1416 4.0000 -0.0000 + 0.0000i 0.0000 0 4.0000 -0.1000 0.1000 3.1416 3.0000 -0.0528 + 0.0000i 0.0528 3.1416 2.0000 -0.1000 - 0.0000i 0.1000 -3.1416 1.0000 -0.9472 - 0.0000i 0.9472 -3.1416 9. Let the input data be x(k) = [1 3 2 5 9 6 7 4 ]
t
. This data is applied to 8-point (N = 8) radix-
2 FFT calculator. Find the values everywhere for
(a) decimation in time radix-2 FFT x(0) = 1, x(1) = 3, x(2) = 2, x(3) = 5, x(4) = 9, x(5) = 6, x(6) = 7, x(7) = 4 0
1
2
3
4
5
6
7
8
9
0
0.5
1
1.5
2
2.5
n
|d
n
|
Magnitude
0
1
2
3
4
5
6
7
8
9
-4
-2
0
2
4
n
∠
d
n
Phase
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28 x(0) x(4) x(2) x(6) x(1) x(5) x(3) x(7) 𝑊𝑊
8
1
=
𝑒𝑒
−𝑗𝑗
2𝜋𝜋
8
1
=
1
√2
− 𝑗𝑗
1
√2
= 0.7071
− 𝑗𝑗
0.7071
, 𝑊𝑊
8
2
=
𝑒𝑒
−𝑗𝑗
2𝜋𝜋
8
2
=
−𝑗𝑗
𝑊𝑊
8
3
=
𝑒𝑒
−𝑗𝑗
2𝜋𝜋
8
3
=
−
1
√
2
− 𝑗𝑗
1
√
2
=
−
0.7071
− 𝑗𝑗
0.7071
B
it reversed order:
x(0) = 1, x(4) = 9 , x(2) = 2, x(6) = 7, x(1) = 3, x(5) = 6, x(3) = 5 , x(7) = 4 X1(0) =x(0)+x(4)=10, X1(1) = x(0)-x(4)=-8, X1(2) =x(2)+x(6)=9, X1(3) = (x(2)-x(6))
𝑊𝑊
8
2
= j5, X1(4) =x(1)+x(5)=9, X1(5) =x(1)-x(5)= -3, X1(6) =x(3)+x(7)= 9, X1(7) = (x(3)-x(7)) 𝑊𝑊
8
2
= -j. X2(0)=X1(0)+X1(2)=19, X2(1)=X1(1)+X1(3)=-8+j5, X2(2)=X1(0)-X1(2)=1, X2(3)=X1(1)-X1(3)=-8-j5,X2(4)=X1(4)+X1(6)=18, X2(5)=(X1(5)+X1(7)) 𝑊𝑊
8
1
=-
2.8284+j1.4142, X2(6)=(X1(4)-X1(6))
𝑊𝑊
8
2
=0, X2(7)=(X1(5)-X1(7))
𝑊𝑊
8
3
=2.8284+j1.4142 X(0)=X2(0)+X2(4) =19+18=37 X(1)=X2(1)+X2(5)=(-8+j5)+( -2.8284+j1.4142)= -10.8284 + j6.4142 -1 -1 -1 -1 -1 -1 -1 -1 -1 2
W
N
2
W
N
1
W
N
-1 -1 2
W
N
3
W
N
X(0) X(1) X(2) X(3) X(4) X(5) X(6) X(7) X1(0) X1(1) X1(2) X1(3) X1(4) X1(5) X1(6) X1(7) X2(0) X2(1) X2(2) X2(3) X2(4) X2(5) X2(6) X2(7) -1
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29 X(2)=X2(2)+X2(6)=(1)+( 0)= 1 X(3)=X2(3)+X2(7)= (-8+j5)+( -2.8284+j1.4142)= -5.1716 - j3.5858 X(4)=X2(0)-X2(4)=(19)-( 18)= 1 X(5)=X2(1)-X2(5)=(-8+j5)-( -2.8284+j1.4142)= -5.1716 + j3.5858 X(6)=X2(2)-X2(6)=(1)-( 0)= 1 X(7)=X2(3)-X2(7)= (-8+j5)-( -2.8284+j1.4142)= -10.8284 - j6.4142 (b) decimation in frequency radix-2 FFT x(0) x(1) x(2) x(3) x(4) x(5) x(6) x(7) x(0) = 1, x(1) = 3, x(2) = 2, x(3) = 5, x(4) = 9, x(5) = 6, x(6) = 7, x(7) = 4 X1(0) =x(0)+x(4)=10, X1(1) =x(1)+x(5)=9, X1(2) =x(2)+x(6)=9, X1(3)=x(3)+x(7)=9, X1(4)=x(0)-x(4)=-8, X1(5)=(x(1)-x(5))
𝑊𝑊
8
1
= -2.1213+j2.1213, X1(6)=(x(2)-x(6))
𝑊𝑊
8
2
= j5, X1(7) = (x(3)-x(7)) 𝑊𝑊
8
3
= -0.7071-j0.7071 X2(0)=X1(0)+X1(2)=19, X2(1)=X1(1)+X1(3)=18, X2(2)= X1(0)-X1(2)=1, X2(3)= (X1(1)-X1(3))
𝑊𝑊
8
2
=0 X2(4)=X1(4)+X1(6)=-8+j5, X2(5)=X1(5)+X1(7) =-2.8284+j1.4142, -1 -1 -1 -1 -1 -1 -1 -1 -1 2
W
N
2
W
N
1
W
N
-1 -1 2
W
N
3
W
N
X(0) X(4) X(2) X(6) X(1) X(5) X(3) X(7) X
1
(0) X
1
(1) X
1
(2) X
1
(3) X
1
(4) X
1
(5) X
1
(6) X
1
(7) X
2
(0) X
2
(1) X
2
(2) X
2
(3) X
2
(4) X
2
(5) X
2
(6) X
2
(7) -1
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30 X2(6)=X1(4)-X1(6)=-8-j5, X2(7)=(X1(5)-X1(7))
𝑊𝑊
8
2
=2.8284+j1.4142 X(0)=X2(0)+X2(1) =19+18=37 X(4)=X2(0)-X2(1)=(19)-( 18)= 1 X(2)=X2(2)+X2(3)=(1)+( 0)= 1 X(6)=X2(2)-X2(3)=(1)-( 0)= 1 X(1)=X2(4)+X2(5)=(-8+j5)+( -2.8284+j1.4142)= -10.8284 + j6.4142 X(5)=X2(4)-X2(5)=(-8+j5)-( -2.8284+j1.4142)= -5.1716 + j3.5858 X(3)=X2(6)+X2(7)= (-8+j5)+( -2.8284+j1.4142)= -5.1716 - j3.5858 X(7)=X2(6)-X2(7)= (-8+j5)-( -2.8284+j1.4142)= -10.8284 - j6.4142 Applying bit reverse order, we get
X(0)=X2(0)+X2(1) =19+18=37 X(1)=X2(4)+X2(5)=(-8+j5)+( -2.8284+j1.4142)= -10.8284 + j6.4142 X(2)=X2(2)+X2(3)=(1)+( 0)= 1 X(3)=X2(6)+X2(7)= (-8+j5)+( -2.8284+j1.4142)= -5.1716 - j3.5858 X(4)=X2(0)-X2(1)=(19)-( 18)= 1 X(5)=X2(4)-X2(5)=(-8+j5)-( -2.8284+j1.4142)= -5.1716 + j3.5858 X(6)=X2(2)-X2(3)=(1)-( 0)= 1 X(7)=X2(6)-X2(7)= (-8+j5)-( -2.8284+j1.4142)= -10.8284 - j6.4142 37.0000 -10.8284 + 6.4142i 1.0000 -5.1716 - 3.5858i 1.0000 -5.1716 + 3.5858i 1.0000 -10.8284 - 6.4142i 10. Let the transform X(n) be 43.0000 6.0711 - 0.5858i 5.0000 + 6.0000i -8.0711 + 3.4142i 7.0000 -8.0711 - 3.4142i 5.0000 - 6.0000i 6.0711 + 0.5858i This data is applied to 8-point (N = 8) radix-
2 IFFT calculator. Find the values everywhere for
(a) decimation in time radix-2 IFFT X(0) = 43, X(1) = 6.071-j0.5858, X(2) = 5+j6, X(3) = -8.0711+j3.4142, X(4) = 7, X(5) = -8.0711-j3.4142, X(6) = 5-j6, X(7) = 6.0711+j0.5858 X(0) x(0) X1(0) X2(0) 1/8
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31 X(4) X(2) X(6) X(1) X(5) X(3) X(7) 𝑊𝑊
8
−1
=
𝑒𝑒
𝑗𝑗
2𝜋𝜋
8
1
=
1
√2
+
𝑗𝑗
1
√2
= 0.7071 +
𝑗𝑗
0.7071
, 𝑊𝑊
8
−2
=
𝑒𝑒
𝑗𝑗
2𝜋𝜋
8
2
=
𝑗𝑗
𝑊𝑊
8
−3
=
𝑒𝑒
𝑗𝑗
2𝜋𝜋
8
3
=
−
1
√
2
+
𝑗𝑗
1
√
2
=
−
0.7071 +
𝑗𝑗
0.7071
X1(0)=X(0)+X(4)=50, X1(1)=X(0)-X(1)=36, X1(2)=X(2)+X(6)=10, X1(3)=(X(2)-X(6)) 𝑊𝑊
8
−2
=j12×j=-12, X1(4)=X(1)+X(5) =-2-j4, X1(5)=X(1)-X(5)=14.1421+j2.8284, X1(6)=X(3)+X(7)=-2+j4, X1(7)=(X(3)-X(7)) 𝑊𝑊
8
−2
=-2.8284-j14.1422 X2(0)=X1(0)+X1(2)=60, X2(1)=X1(1)+X1(3)=24, X2(2)=X1(0)-X1(2)=40, X2(3)=X1(1)-X1(3)=48, X2(4)=X1(4)+X1(6)=-4, X2(5)=(X1(5)+X1(7)) 𝑊𝑊
8
−1
=16 X2(6)=(X1(4)-X1(6)) 𝑊𝑊
8
−2
=8, X2(7)=(X1(5)-X1(7)) 𝑊𝑊
8
−3
=-24 x(0)=(X2(0)+X2(4))/8=7, x(1)=(X2(1)+X2(5))/8=5, x(2)=(X2(2)+X2(6))/8=6, x(3)=(X2(3)+X2(7))/8=3, x(4)=(X2(0)-X2(4))/8=8, x(5)=(X2(1)-X2(5))/8=1, x(6)=(X2(2)-X2(6))/8=4, x(7)=(X2(3)-X2(7))/8=9 (b) decimation in frequency radix-2 IFFT X(0) = 43, X(1) = 6.071-j0.5858, X(2) = 5+j6, X(3) = -8.0711+j3.4142, -1 -1 -1 -1 -1 -1 -1 -1 -1 2
W
N
−
2
W
N
−
1
W
N
−
-1 -1 2
W
N
−
3
W
N
−
x(1) x(2) x(3) x(4) x(5) x(6) x(7) X1(1) X1(2) X1(3) X1(4) X1(5) X1(6) X1(7) X2(1) X2(2) X2(3) X2(4) X2(5) X2(6) X2(7) -1 1/8 1/8 1/8 1/8 1/8 1/8 1/8
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32 X(4) = 7, X(5) = -8.0711-j3.4142, X(6) = 5-j6, X(7) = 6.0711+j0.5858 X(0) X(1) X(2) X(3) X(4) X(5) X(6) X(7) X1(0)=X(0)+X(4)=50, X1(1)= X(1)+X(5)=-2-j4, X1(2)= X(2)+X(6)=10, X1(3)= X(3)+X(7)=-2+j4, X1(4)=X(0)-X(4)=36, X1(5)= (X(1)-X(5)) 𝑊𝑊
8
−1
=8+j12, X1(6)= (X(2)-X(6)) 𝑊𝑊
8
−2
=-12, X1(7)= (X(3)-X(7)) 𝑊𝑊
8
−3
= 8-j12 X2(0)=X1(0)+X1(2)=60, X2(1)=X1(1)+X1(3)=-4, X2(2)=X1(0)-X1(2)=40, X2(3)=(X1(1)-X1(3)) 𝑊𝑊
8
−2
=8, X2(4)=X1(4)+X1(6)=24, X2(5)=X1(5)+X1(7)=16, X2(6)=X1(4)-X1(6)=48, X2(7)=(X1(5)-X1(7)) 𝑊𝑊
8
−2
=-24 x(0)=(X2(0)+X2(1))/8=7 x(4)=(X2(0)-X2(1))/8=8 x(2)=(X2(2)+X2(3))/8=6 x(6)=(X2(2)-X2(3))/8=4 x(1)=(X2(4)+X2(5))/8=5 x(5)=(X2(4)-X2(5))/8=1 x(3)=(X2(6)+X2(7))/8=3 -1 -1 -1 -1 -1 -1 -1 -1 -1 2
W
N
−
2
W
N
−
1
W
N
−
-1 -1 2
W
N
−
3
W
N
−
x(0) x(4) x(2) x(6) x(1) x(5) x(3) x(7) X
1
(0) X
1
(1) X
1
(2) X
1
(3) X
1
(4) X
1
(5) X
1
(6) X
1
(7) X
2
(0) X
2
(1) X
2
(2) X
2
(3) X
2
(4) X
2
(5) X
2
(6) X
2
(7) -1 1/8 1/8 1/8 1/8 1/8 1/8 1/8 1/8
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33 x(7)=(X2(6)-X2(7))/8=9 After the bit reverse order, we have
x(0)=(X2(0)+X2(1))/8=7 x(1)=(X2(4)+X2(5))/8=5 x(2)=(X2(2)+X2(3))/8=6 x(3)=(X2(6)+X2(7))/8=3 x(4)=(X2(0)-X2(1))/8=8 x(5)=(X2(4)-X2(5))/8=1 x(6)=(X2(2)-X2(3))/8=4 x(7)=(X2(6)-X2(7))/8=9
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