assignment 1 signals and systems student: ali abbasi (id
TRANSCRIPT
Signals and Systems
Laplace Transform and Its Applications
Student: Ali Abbasi (ID: 9531503)
Course Instructors: Dr. M Shahraki, Dr. M Mehrjoo
Course Number: 20 14 255
Faculty of Electrical and Computer Engineering,
University of Sistan and Baluchestan, Zahedan, Iran
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2/16
Frequency Range
Note: This slide has been used in assignment 9 to present applications of Fourier analysis.
3/16
Audio Systems with Multiple Speakers
Picture Source: https://www.bhphotovideo.com/explora/amp/audio/tips-and-solutions/what-about-all-those-speaker-specs
Note: This slide has been used in assignment 9 to present applications of Fourier analysis.
Filters Circuits
Picture Source: Santiago, J. (2013). Circuit Analysis for Dummies: John Wiley & Sons. (with modifications)
4/16
Low-pass Filter Analysis
5/16Picture Source: Santiago, J. (2013). Circuit Analysis for Dummies: John Wiley & Sons. (with modifications)
π£ππ π‘ = π£π (π‘) + π£ππ’π‘(π‘)
π£ππ π‘ = π πΆππ£ππ’π‘(π‘)
ππ‘+1
πΆΰΆ±ββ
π‘
πΆππ£ππ’π‘(π)
ππ‘ππ
π£ππ π‘ = π π(π‘) +1
πΆΰΆ±ββ
π‘
π(π) ππ
π£ππ π‘ = π πΆππ£ππ’π‘(π‘)
ππ‘+ π£ππ’π‘(π‘)
π£π (π‘)
π£ππ’π‘(π‘)π£ππ(π‘)
π
πΆ π΄ππ ππππ‘πππ ππππππ‘ππππ πππ π§πππ.
Low-pass Filter Analysis (cont.)
6/16
π£ππ π‘ = π πΆππ£ππ’π‘(π‘)
ππ‘+ π£ππ’π‘(π‘)
π£ππ’π‘ π‘ = π£ππ(π‘)(1 β πβ1π πΆ
π‘)
πππ π = π πΆπ πππ’π‘(π ) + πππ’π‘(π )
πππ π = πππ’π‘(π )(π πΆπ + 1)
πππ’π‘ π =1
π πΆ(
1
π +1π πΆ
)πππ(π )
πΏππππππ
πΏππππππβ1
Devices in S-Domain
7/16
π«πππππ π»πππ β π«πππππ
πΊ β π«πππππ
π½ππππππ πͺπππππππ°ππππ ππππ(ππππ πππππππππππ
ππππ πππππ)
πΌππ π£π(π‘) ππ(π ) β β
πΌπΆπ ππ(π‘) β πΌπ(π ) β
ππΆππ π£2 π‘ = ππ£1 π‘ π2 π = ππ1 π β β
ππΆπΆπ π2 π‘ = ππ£1 π‘ β πΌ2 π = ππ1 π β
πΆπΆππ π£2 π‘ = ππ1 π‘ π2 π = ππΌ1 π β β
πΆπΆπΆπ π2 π‘ = π½π1 π‘ β πΌ2 π = π½πΌ1 π β
π ππ ππ π‘ππ π£π π‘ = π ππ (π‘) ππ π = π πΌπ (π )πΌπ π =
1
π ππ (π )
ππ π = π
πΆππππππ‘πππ£πΆ π‘ = ΰΆ±
0
π‘
ππΆ(π) ππ ππΆ π =1
π πΆπΌπΆ π +
π£π(0)
π
πΌπΆ π = π πΆ ππΆ π β πΆπ£π(0) ππΆ π =1
π πΆ
πΌπππ’ππ‘πππ£πΏ π‘ = πΏ
πππΏ(π‘)
ππ‘
ππΏ π = π πΏπΌπΏ π β πΏππΏ(0)πΌπΏ π =
1
π πΏππΏ π +
ππΏ(0)
π
ππΏ π = π πΏ
S-Domain ThΓ©veninβs and Nortonβs Equivalent for Passive Elements
8/16Picture Source: Santiago, J. (2013). Circuit Analysis for Dummies: John Wiley & Sons. (with modifications)
Band-pass Filter: S-Domain Analysis
9/16Picture Source: Santiago, J. (2013). Circuit Analysis for Dummies: John Wiley & Sons. (with modifications)
πππ(π ) πππ’π‘(π )π 1
π πΆ
π πΏ
πππ’π‘ π =π
π πΏ +1π πΆ
+ π πππ π
π π =πππ’π‘(π )
πππ(π )=
π
πΏ
π
π 2 +π πΏ
π +1πΏπΆ
π΄ππ ππππ‘πππ ππππππ‘ππππ πππ π§πππ.
Band-pass Filter: S-Domain Analysis (cont.)
10/16
1
πΏπΆβ π2 = Β±
π
πΏπ β π2 Β±
π
πΏπ β
1
πΏπΆ= 0 β
ππΆ1 = βπ
2πΏ+
π
2πΏ
2
+1
πΏπΆ
ππΆ2 = +π
2πΏ+
π
2πΏ
2
+1
πΏπΆ
1
πΏπΆβ π2 = 0 β π0 =
1
πΏπΆ
π΅πππ_π€πππ‘β = ππΆ2 β ππΆ1 =π
πΏπππ π_ππππ‘ππ =
π0
π΅πππππππ‘β=1/ πΏπΆ
π /πΏ=1
π
πΏ
πΆ
π = ππ β π ππ =πππ’π‘(ππ)
πππ(ππ)=
π
πΏ
ππ
ππ 2 +π πΏ
π +1πΏπΆ
=π
πΏ
ππ
1πΏπΆ
β π2 +π πΏ
ππ
Band-pass Filter as a System
11/16
β’ For a band-pass filter (all initial conditions are zero):
π π =π΄02ππ0π
π 2 + 2ππ0π + π02 =
π΄0π0π π
π 2 +π0π π + π0
2=
π΄0π΅π
π 2 + π΅π + π02
π΄0:πππππππ πΊππππ0: π ππ πππππ‘ πΉππππ’ππππ¦π: π·ππππππ πΆππππππππππ‘π: ππ’ππππ‘π¦ πΉπππ‘ππ
Control Systems and S-Domain
12/16Picture Source: βApplications of Laplace Transform in Control Systems.β, βMobile Tutorβ channel on YouTube.
Control Systems and S-Domain (cont.)
13/16
β’ Gain Factor K:
β’ Nyquist Diagram
π π =1
π΄π 2 + π΅π + πΆ
β’ Help in solving differential equations of higher orders and evaluating system output.
Control Systems and S-Domain (cont.)
14/16
Picture Source: 2010218 (Linear System and Control) at University of Sistan and Baluchestan, Semester 3981, Midterm Exam, by Dr.
Saeed Tavakoli Afshari.
Control Systems and S-Domain (cont.)
15/16
β’ Faster forecasting of the outputs of a system.
Picture Source: (1) https://en.m.wikipedia.org/wiki/Aircraft_flight_control_system (2) https://lancmoms.com/adaptive-cruise-control/