Solved Problems on Chapter (1)

[1] Consider waveform shown in figure-1: 𝒈(𝒕)

a) Classify g(t).
𝟓
b) Find signal size (power or energy).
c) Find 𝑮(𝒇) 20 𝒕
d) Convert g(t) into periodic signal then sketch 0 10

amplitude and phase of its spectrum

−𝟓
Solution: 𝑭𝒊𝒈. 𝟏

𝑎) 𝑔(𝑡) is continuous – deterministic – digital – Aperiodic

10 20

𝑏) 𝐸𝑔 = ∫ (5)2 𝑑𝑡 + ∫ (−5)2 𝑑𝑡 = 2 × (25 × 10) = 500 𝐽𝑜𝑢𝑙𝑒

0 10

𝑡−5 𝑡 − 15
𝑐) 𝑔(𝑡) = 5𝑟𝑒𝑐𝑡 ( ) − 5𝑟𝑒𝑐𝑡 ( )
10 10

𝐺(𝑓) = 50𝑠𝑖𝑛𝑐(10𝑓)𝑒 −10𝜋𝑓 − 50𝑠𝑖𝑛𝑐(10𝑓)𝑒 −30𝜋𝑓

𝒈(𝒕)
𝟓

10 20 30 40 50 60
0 𝒕

−𝟓

𝑇𝑜 = 20

2𝜋 𝜋
𝜔𝑜 = =
20 10

Since g(t) is symmetric about x- axis, ∴ 𝑎𝑜 = 0

 And since g(t) is symmetric about the origin (odd function), ∴ 𝑎𝑛 = 0

10 20
2
𝑏𝑛 = [∫ 5 sin(𝑛𝜔𝑜 𝑡) 𝑑𝑡 − ∫ 5 sin(𝑛𝜔𝑜 𝑡) 𝑑𝑡]
20
0 10

1 cos(𝑛𝜔𝑜 𝑡) 0 1 cos(𝑛𝜔𝑜 𝑡) 20
= ( | )+ ( | )
2𝑛𝜋 10 2𝑛𝜋 10

2
1 𝑓𝑜𝑟 𝑜𝑑𝑑 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑛
= (1 − cos(𝑛𝜋) + 1 − cos(𝑛𝜋)) = { 𝑛𝜋
2𝑛𝜋
0 𝑓𝑜𝑟 𝑒𝑣𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑛

𝑎𝑜 = 0 𝑓𝑜𝑟 𝑛 = 0
2
𝐴𝑚𝑝𝑙𝑖𝑡𝑢𝑑𝑢𝑒 𝑠𝑝𝑒𝑐𝑡𝑟𝑎𝑙 𝑑𝑒𝑛𝑠𝑖𝑡𝑦: 𝐶𝑛 = { 𝑓𝑜𝑟 𝑜𝑑𝑑 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑛
𝑛𝜋
0 𝑓𝑜𝑟 𝑒𝑣𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑛

−𝜋 𝑓𝑜𝑟 𝑜𝑑𝑑 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑛

−1
𝑏𝑛
𝑃ℎ𝑎𝑠𝑒 𝑠𝑝𝑒𝑐𝑡𝑟𝑎𝑙 𝑑𝑒𝑛𝑠𝑖𝑡𝑦: 𝜃𝑛 = −𝑡𝑎𝑛 ( )={
𝑎𝑛
0 𝑓𝑜𝑟 𝑒𝑣𝑒𝑛 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑛

𝑪𝒏 𝜽𝒏

1 2 3 4 5
𝒏

0 1 4 5 𝒏
2 3
[2] Sketch waveform and spectrum of following signals:
𝒂) 𝒙(𝒕) = 𝟒 𝒔𝒊𝒏𝒄𝟐 (𝟐𝟎𝟎𝒕) 𝒄𝒐𝒔(𝟓𝟎𝟎𝟎𝝅𝒕)
𝒃) 𝒈(𝒕) = 𝟐𝟎 𝒓𝒆𝒄𝒕(𝟎. 𝟏𝒕 − 𝟓𝟎) + 𝟐𝟎 𝒓𝒆𝒄𝒕(𝟎. 𝟏𝒕 + 𝟓𝟎)
𝒄) 𝒚(𝒕) = 𝟖 + 𝟏𝟎 𝒄𝒐𝒔(𝟏𝟎𝟎𝟎𝟎𝝅𝒕)
Solution:

𝒙(𝒕)
4
1 a) 𝒄𝒐𝒔(𝟓𝟎𝟎𝟎𝝅𝒕)

0.5

0 𝒕 (ms)
−𝟏𝟎𝟎 −𝟓𝟎 𝟓𝟎 𝟏𝟎0

-0.5

-1 -4
-20 -15 -10 -5 0 5 10 15 20

X(f)

0.02

−2700 −2500 −2300 2300 2500 2700

0 𝒇 (Hz)

b)

g(t)

20

𝒕
−505 −500 −495 495 500 505
0
 𝑮(𝒇)
4
1 𝒄𝒐𝒔(𝟏𝟎𝟎𝟎𝝅𝒇)

0.5

0 𝒇 (Hz)
−𝟎. 𝟐 −𝟎. 𝟏 𝟎. 𝟏 𝟎. 𝟐

-0.5

-1
-20 -15 -10 -5 0 5 10 15 20

𝒚(𝒕)
c)
1
18
0.8

0.6

0.4

0.2

-0.2

-0.4 𝒕
-0.6

-0.8

-1
0 10 20
−2 30 40 50 60

𝒀(𝒇)

𝒇 (Hz)
−𝟓𝟎𝟎𝟎 0 𝟓𝟎𝟎𝟎
[3] Calculate power of following signals:
𝝅
𝒂) 𝟓 𝐜𝐨𝐬(𝟓𝟎𝟎𝟎𝝅𝒕 + 𝟑 )
𝟓𝝅
𝒃) 𝟔 𝐜𝐨𝐬(𝟒𝟎𝟎𝝅𝒕) + 𝟏𝟎 𝐜𝐨𝐬(𝟓𝟎𝟎𝟎𝝅𝒕 + )
𝟔
𝝅 𝟓𝝅
c) 𝟏𝟎 𝐬𝐢𝐧 (𝟐𝟎𝟎𝟎𝝅𝒕 + 𝟐 ) + 𝟕 𝐜𝐨𝐬(𝟐𝟎𝟎𝟎𝝅𝒕 + )
𝟔

Solution:
(5)2
𝑎) 𝑃𝑥 = = 12.5 V 2 / sec
2
(6)2 (10)2
𝑏) 𝑃𝑥 = + = 68 V 2 / sec
2 2
𝜋 5𝜋 𝜋
𝑐) 𝑥(𝑡) = 10 𝑠𝑖𝑛 (2000𝜋𝑡 + ) + 7 𝑠𝑖𝑛(2000𝜋𝑡 + + )
2 6 2
𝜋 4𝜋
= 10 𝑠𝑖𝑛 (2000𝜋𝑡 + ) + 7 𝑠𝑖𝑛(2000𝜋𝑡 + )
2 3
(10)2 (7)2 4𝜋 𝜋
𝑃𝑥 = + + (10) × (7) × cos ( − ) = 13.8782 V 2 / sec
2 2 3 2