 Engineering
EE 353 PSUPSMC Fundamental Period Aperiodic & Real Valued Functions Exercises

EE 353

Pennsylvania State University - Penn State Main Campus

EE

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Solution:
1.
𝑓(𝑡) =

1 1
− cos(20𝜋𝑡 + 90𝑜 ) + cos(4𝜋𝑡 + 60𝑜 ) + sin(28𝜋 − 60𝑜 )
2 2

𝑓(𝑡) =

1 1
𝜋
𝜋
𝜋
− cos (20𝜋𝑡 + ) + cos (4𝜋𝑡 + ) + sin (28𝜋 − )
2 2
2
3
3

LCM for the time constants we have
𝑇0 = 𝐿𝐶𝑀 (

2𝜋 2𝜋 2𝜋
1 1 1
7 35 5
35 1
,
,
) = 𝐿𝐶𝑀 ( , , ) = ( , , ) =
=
20𝜋 4𝜋 28𝜋
10 2 14
70 70 70
70 2
𝑡ℎ𝑒 𝑠𝑖𝑔𝑛𝑎𝑙 𝑖𝑠 𝑝𝑒𝑟𝑖𝑜𝑑𝑖𝑐

2.
𝑓(𝑡) = sin(6𝑡) + cos (3𝜋𝑡)
2𝜋 2𝜋
𝜋 2
2𝜋
, ) = 𝐿𝐶𝑀 ( , ) =
6 3𝜋
3 3
3
𝑡ℎ𝑒 𝑠𝑖𝑔𝑛𝑎𝑙 𝑖𝑠 𝑝𝑒𝑟𝑖𝑜𝑑𝑖𝑐
The signal is periodic, since it is Fourier series,
2𝜋
𝑇0 =
𝑤𝑜
𝑇0 = 𝐿𝐶𝑀 (

3.

Solution
1.
𝑥𝑛 = cos(𝑛𝑤𝑜 𝑡) 𝑎𝑛𝑑 𝑥𝑚 = cos(𝑚𝑤𝑜 𝑡)
𝑇𝑜
2

𝑇
− 0
2

cos(𝑛𝑤𝑜 𝑡) cos(𝑚𝑤𝑜 𝑡) 𝑑𝑡

Since cosine is even
𝑇𝑜
2

𝑇
− 0
2

𝑇𝑜
2

cos(𝑛𝑤𝑜 𝑡) cos(𝑚𝑤𝑜 𝑡) 𝑑𝑡 = 2 ∫ cos(𝑛𝑤𝑜 𝑡) cos(𝑚𝑤𝑜 𝑡) 𝑑𝑡
0

For 𝑛 ≠ 𝑚
𝑇𝑜
2

𝑇𝑜
2

2 ∫ cos(𝑛𝑤𝑜 𝑡) cos(𝑚𝑤𝑜 𝑡) 𝑑𝑡 = ∫ cos((𝑚 + 𝑛)𝑤𝑜 𝑡) + cos((𝑚 − 𝑛)𝑤𝑜 𝑡) 𝑑𝑡
0

0

(𝑚 + 𝑛) 𝑤𝑜 𝑇0
(𝑚 − 𝑛) 𝑤𝑜 𝑇0
) sin (
)
2
2
+
(𝑚 + 𝑛)𝑤𝑜
(𝑚 − 𝑛)𝑤𝑜

sin (
=

For 𝑚 = 𝑛 ≠ 0
𝑇𝑜
2

𝑇𝑜
2

2 ∫ cos(𝑛𝑤𝑜 𝑡) cos(𝑛𝑤𝑜 𝑡) 𝑑𝑡 = 2 ∫
0

cos 2(𝑛𝑤𝑜 𝑡)

𝑇𝑜
2

𝑑𝑡 = ∫ (1 + cos (2𝑛𝑤0 𝑡)𝑑𝑡

0

=

0

𝑇0
1
𝑛𝑤𝑜 𝑇0
+
sin (
)
2 𝑛𝑤0
2

2.
𝑥𝑛 = sin(𝑛𝑤𝑜 𝑡) 𝑎𝑛𝑑 𝑥𝑚 = sin(𝑚𝑤𝑜 𝑡)
𝑇𝑜
2

𝑇
− 0
2

cos(𝑛𝑤𝑜 𝑡) cos(𝑚𝑤𝑜 𝑡) 𝑑𝑡

Since sine is odd
𝑇𝑜
2

𝑇
− 0
2

𝑇𝑜
2

sin(𝑛𝑤𝑜 𝑡) sin(𝑚𝑤𝑜 𝑡) 𝑑𝑡 = 2 ∫ sin(𝑛𝑤𝑜 𝑡) sin(𝑚𝑤𝑜 𝑡) 𝑑𝑡
0

For 𝑛 ≠ 𝑚
𝑇𝑜
2

2∫

𝑇
− 0
2

𝑇𝑜
2

sin(𝑛𝑤𝑜 𝑡) sin(𝑚𝑤𝑜 𝑡) 𝑑𝑡 = ∫ cos((𝑚 − 𝑛)𝑤𝑜 𝑡) − cos((𝑚 + 𝑛)𝑤𝑜 𝑡) 𝑑𝑡
0

(𝑚 − 𝑛) 𝑤𝑜 𝑇0
(𝑚 + 𝑛) 𝑤𝑜 𝑇0
) sin (
)
2
2

(𝑚 − 𝑛)𝑤𝑜
(𝑚 + 𝑛)𝑤𝑜

sin (
=
For 𝑚 = 𝑛 ≠ 0
𝑇𝑜
2

2∫

𝑇
− 0
2

𝑇𝑜
2

sin(𝑛𝑤𝑜 𝑡) sin(𝑚𝑤𝑜 𝑡) 𝑑𝑡 = 2 ∫

sin2 (𝑛𝑤𝑜 𝑡)

𝑑𝑡 = ∫ (1 − sis (2𝑛𝑤0 𝑡)𝑑𝑡

0

=

𝑇𝑜
2

0

𝑇0
1
𝑛𝑤𝑜 𝑇0

sin (
)
2 𝑛𝑤0
2

3.
𝑥𝑛 = cos(𝑛𝑤𝑜 𝑡) 𝑎𝑛𝑑 𝑥𝑚 = sin(𝑚𝑤𝑜 𝑡)
𝑇𝑜
2

𝑇
− 0
2

cos(𝑛𝑤𝑜 𝑡) sin(𝑚𝑤𝑜 𝑡) 𝑑𝑡

𝑡ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑜𝑑𝑑 𝑎𝑛𝑑 𝑒𝑣𝑒𝑛 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑔𝑖𝑣𝑒𝑠 𝑜𝑑𝑑 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙

𝑇𝑜
2

𝑇
− 0
2

cos(𝑛𝑤𝑜 𝑡) sin(𝑚𝑤𝑜 𝑡) 𝑑𝑡 = 0

Thus, the two function are orthogonal on the interval −

𝑇0
2

𝑡𝑜

𝑇𝑜
2

4.
𝑥𝑛 (𝑡) = 𝑒 𝑗𝑛𝑤0 𝑡 𝑎𝑛𝑑 𝑥𝑚 (𝑡) = 𝑒 𝑗𝑚𝑤𝑜 �... brianreyes958 (4351)
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