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Signal and Systems
Chapter 1
Islam K. Sharawneh
islam.Sharawneh@ptuk.edu.ps
DT Complex Exponential Signal
• 𝑥 𝑛 = 𝐶𝑒𝛼𝑛 , C 𝑎𝑛𝑑 𝛼 are complex numbers
• 1: if 𝐶 = 1 𝑎𝑛𝑑 𝛼 = 𝑗𝜔0 then 𝑥 𝑛 = 𝑒𝑗𝜔0𝑛
This 𝑒𝑗𝜔0𝑛 is not periodic for all 𝜔0
• 𝑥 𝑛 + 𝑁 = 𝑒𝑗𝜔0(𝑛+𝑁)
= 𝑒𝑗𝜔0𝑛
𝑒𝑗𝜔0𝑁
. If 𝜔0𝑁 = 2𝜋𝑚, this implies
that 𝑥 𝑛 = 𝑥 𝑛 + 𝑁 . 𝑚, 𝑛 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 and 𝜔0 =
2𝜋𝑚
𝑁
or
𝜔0
2𝜋
=
𝑚
𝑁
is
a “rational number”
DT Complex Exponential Signal
• 𝑥 𝑛 = 𝑒𝑗𝜔0𝑛
is periodic, then we have
𝜔0
2𝜋
=
𝑚
𝑁
under the assumption that
N and m have no factors in common, 𝑁 =
2𝜋𝑚
𝜔0
, 𝜔0≠ 0.
• The fundamental frequency =
2𝜋
𝑁
=
𝜔0
𝑚
• CT: 𝑥 𝑡 = 𝑒𝑗𝜔0𝑡
are distinct signals for distinct 𝜔
• DT: 𝑥 𝑛 = 𝑒𝑗𝜔0𝑛
= 𝑒𝑗𝜔0(𝑛+𝑁)
= 𝑒
𝑗𝜔0(𝑛+
2𝜋𝑚
𝜔0
)
= 𝑒𝑗𝜔0𝑛
. 𝑒𝑗𝜔02𝜋𝑚
Periodic complex exponential signals with angular frequencies 𝜔0, 𝜔0± 2𝜋,
𝜔0± 4𝜋, … are identical signals. That reason we consider a frequency
interval of length 2𝜋. E.g. −𝜋 ≤ 𝜔0 < 𝜋 ; 0 ≤ 𝜔0 < 2𝜋.
Other Signals: Unit Rectangular Pulse
Other Signals: Unit Ramp
Other Signals: Unit Parabolic
Other Signals: Unit Signum
Other Signals: Unit Sinc
Other Signals: Unit Area Triangular
Example on using unit impulse and unit step functions
• 𝑦1 𝑡 = 𝑢 𝑡 − 𝑢 𝑡 − 1
• 𝑦2 𝑡 =
𝑑𝑦(𝑡)
𝑑𝑡
• 𝑦3 𝑡 = 𝑢 𝑡 + 1 − 2𝑢 𝑡 + 𝑢 𝑡 − 1
Extra Questions
• Q1: Determine the average power of the following signals
1. 𝑥 𝑡 = 𝐴𝑒𝑗(𝜔0+𝜃)
2. 𝑥 𝑡 = 𝐴𝑐𝑜𝑠(𝜔0 + 𝜃)
Q2: If 𝑥 𝑛 = cos
𝜋𝑛2
8
, show that 𝑥 𝑛 is periodic with fundamental
period 𝑁 = 8.
Q3: Express the following as a sum of weighted delayed impulses
𝑥 𝑛 = {0,1,-2,3,-2,1}
(0 sample starts at n=0)
Extra Questions
• Q4: Sketch the following functions
• 1. 𝑥1 𝑡 = ൝
1 −
𝑡
12
, 0 < 𝑡 ≤ 12
0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
• 2. 𝑥2 𝑡 = 𝑥1 1 − 𝑡 𝑢 𝑡 + 1 − 𝑢 𝑡 − 2
• 3. 𝑥3 𝑡 = 𝑥1 1 − 𝑡 𝑢 𝑡 + 1 − 𝑢 2 − 3𝑡
Q5: 𝑢 𝑡 is a unit impulse function, show that 𝑠𝑛𝑔 𝑡 = 𝑢 𝑡 − 𝑢 −𝑡
Q6: If x t = cos 2𝜋𝑓𝑡 is a continuous time signal, and x[n] = cos 2𝜋𝐹𝑛
is a discrete time signal which is the sampled version of x t at sampling rate
𝐹𝑠 =
1
𝑇𝑠
, show that the sampling rate 𝐹𝑠 =
𝑓
𝐹
.

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signal and system chapter1-part1.pdf

  • 1. Signal and Systems Chapter 1 Islam K. Sharawneh islam.Sharawneh@ptuk.edu.ps
  • 2. DT Complex Exponential Signal • 𝑥 𝑛 = 𝐶𝑒𝛼𝑛 , C 𝑎𝑛𝑑 𝛼 are complex numbers • 1: if 𝐶 = 1 𝑎𝑛𝑑 𝛼 = 𝑗𝜔0 then 𝑥 𝑛 = 𝑒𝑗𝜔0𝑛 This 𝑒𝑗𝜔0𝑛 is not periodic for all 𝜔0 • 𝑥 𝑛 + 𝑁 = 𝑒𝑗𝜔0(𝑛+𝑁) = 𝑒𝑗𝜔0𝑛 𝑒𝑗𝜔0𝑁 . If 𝜔0𝑁 = 2𝜋𝑚, this implies that 𝑥 𝑛 = 𝑥 𝑛 + 𝑁 . 𝑚, 𝑛 𝑎𝑟𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 and 𝜔0 = 2𝜋𝑚 𝑁 or 𝜔0 2𝜋 = 𝑚 𝑁 is a “rational number”
  • 3. DT Complex Exponential Signal • 𝑥 𝑛 = 𝑒𝑗𝜔0𝑛 is periodic, then we have 𝜔0 2𝜋 = 𝑚 𝑁 under the assumption that N and m have no factors in common, 𝑁 = 2𝜋𝑚 𝜔0 , 𝜔0≠ 0. • The fundamental frequency = 2𝜋 𝑁 = 𝜔0 𝑚 • CT: 𝑥 𝑡 = 𝑒𝑗𝜔0𝑡 are distinct signals for distinct 𝜔 • DT: 𝑥 𝑛 = 𝑒𝑗𝜔0𝑛 = 𝑒𝑗𝜔0(𝑛+𝑁) = 𝑒 𝑗𝜔0(𝑛+ 2𝜋𝑚 𝜔0 ) = 𝑒𝑗𝜔0𝑛 . 𝑒𝑗𝜔02𝜋𝑚 Periodic complex exponential signals with angular frequencies 𝜔0, 𝜔0± 2𝜋, 𝜔0± 4𝜋, … are identical signals. That reason we consider a frequency interval of length 2𝜋. E.g. −𝜋 ≤ 𝜔0 < 𝜋 ; 0 ≤ 𝜔0 < 2𝜋.
  • 4. Other Signals: Unit Rectangular Pulse
  • 9. Other Signals: Unit Area Triangular
  • 10. Example on using unit impulse and unit step functions • 𝑦1 𝑡 = 𝑢 𝑡 − 𝑢 𝑡 − 1 • 𝑦2 𝑡 = 𝑑𝑦(𝑡) 𝑑𝑡 • 𝑦3 𝑡 = 𝑢 𝑡 + 1 − 2𝑢 𝑡 + 𝑢 𝑡 − 1
  • 11. Extra Questions • Q1: Determine the average power of the following signals 1. 𝑥 𝑡 = 𝐴𝑒𝑗(𝜔0+𝜃) 2. 𝑥 𝑡 = 𝐴𝑐𝑜𝑠(𝜔0 + 𝜃) Q2: If 𝑥 𝑛 = cos 𝜋𝑛2 8 , show that 𝑥 𝑛 is periodic with fundamental period 𝑁 = 8. Q3: Express the following as a sum of weighted delayed impulses 𝑥 𝑛 = {0,1,-2,3,-2,1} (0 sample starts at n=0)
  • 12. Extra Questions • Q4: Sketch the following functions • 1. 𝑥1 𝑡 = ൝ 1 − 𝑡 12 , 0 < 𝑡 ≤ 12 0, 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 • 2. 𝑥2 𝑡 = 𝑥1 1 − 𝑡 𝑢 𝑡 + 1 − 𝑢 𝑡 − 2 • 3. 𝑥3 𝑡 = 𝑥1 1 − 𝑡 𝑢 𝑡 + 1 − 𝑢 2 − 3𝑡 Q5: 𝑢 𝑡 is a unit impulse function, show that 𝑠𝑛𝑔 𝑡 = 𝑢 𝑡 − 𝑢 −𝑡 Q6: If x t = cos 2𝜋𝑓𝑡 is a continuous time signal, and x[n] = cos 2𝜋𝐹𝑛 is a discrete time signal which is the sampled version of x t at sampling rate 𝐹𝑠 = 1 𝑇𝑠 , show that the sampling rate 𝐹𝑠 = 𝑓 𝐹 .