ECE205 Circuits & Systems
Winter 2023-24
ECE205 Circuits & Systems
Final Exam
02/19/24, 1:00 p.m., Moench B105
NAME: ___________________________ CM: ___________
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You must show work to receive partial and full credit
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Put a box around your final answer and it must include units, if necessary
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You can leave answers as fractions, under radicals, with tan-1(x) or in integrals unless told
to solve
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You may use a calculator for simple calculations but not for integrals, differential equations
or systems of equations, computers are not allowed
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No cellphones allowed during the exam, put it in your backpack or the front of the room
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No drinks allowed on the table during the exam, put it in the front of the room
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You must use a pencil to complete the test
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A formula sheet is provided at the end of the exam
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Time allowed: 180 minutes
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You must do 8 of the following 10 questions, the first 3 are required. Make sure to mark
what you want graded. If you don’t mark any, the first 8 will be graded
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Grade
This
1
X
Steady-State Frequency
Response
6
System Properties
2
X
Bode Plots
7
Impulse and Step Response
3
X
Filters
8
Convolution
9
Laplace Circuit Analysis
10
Feedback Control System
4
5
Awarded
Points
Topic
First-Order
Circuits/Systems
Second-Order
Circuits/Systems
#
Total
ECE205 Final (W2324) - Review Sheet
Grade
This
Awarded
Points
Topic
Total
Page 1 of 5
ECE205 Circuits & Systems
Winter 2023-24
Helpful Hints
Details:
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The exam has 10 questions, you must complete 8 of them. The first three are required.
The problems are worth 25 points each
The exam is 180 minutes (3 hours)
Topics include
o First-Order Systems
o Second-Order Systems
o System Properties
o Impulse and Step Response
o Convolution
o Laplace Circuit Analysis
o Feedback Control Systems
o Steady-State Frequency Response
o Bode Plots
o Filters
Study/Review:
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Study Guide Week 1 - 10
Course Notes Ch. 1 - 8
Homework 1 - 10
Exams/Old Exams 1 - 3
Quizzes/Old Quizzes 1-10
View Videos Online
Advice:
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Get a good night’s sleep
Eat a good breakfast
Time yourself doing practice exams
Join a study group
Bring at least 2 pencils and erasers
Go to the learning center
Go to Percopo tutors
Go to HKN tutors – Moench D210
Go to a study session
Ask professor questions
ECE205 Final (W2324) - Review Sheet
Page 2 of 5
ECE205 Circuits & Systems
Winter 2023-24
Equation/Formula Sheet
First Order Circuits
𝑑𝑑𝑑𝑑(𝑡𝑡)
+ 𝑦𝑦(𝑡𝑡) =
𝑑𝑑𝑑𝑑
𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜(∞)
𝐾𝐾 =
𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖(∞)
τ
τ = RTHC
𝐾𝐾𝐾𝐾(𝑡𝑡)
τ = L/RTH
y(t) = KA + [y(to) - KA]e-(t-to)/τ
𝑡𝑡
𝑦𝑦(𝑡𝑡) = 𝑦𝑦(𝑡𝑡𝑜𝑜 )𝑒𝑒 −(φ(𝑡𝑡)−φ(𝑡𝑡𝑜𝑜 )) + � 𝑒𝑒 −�φ(𝑡𝑡)−φ(λ)� 𝑏𝑏𝑏𝑏(λ)𝑑𝑑λ
Second Order Circuits
𝑡𝑡𝑜𝑜
𝑑𝑑2 𝑦𝑦(𝑡𝑡)
𝑑𝑑𝑑𝑑(𝑡𝑡)
+
2
ζω
+ ω𝑛𝑛 2 𝑦𝑦(𝑡𝑡) = 𝐾𝐾 ω𝑛𝑛 2 𝑥𝑥(𝑡𝑡)
𝑛𝑛
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑 2
𝑇𝑇𝑝𝑝 =
π
ω𝑑𝑑
ζπ
−
�1−ζ2
𝑃𝑃𝑃𝑃 = 𝑒𝑒
𝑥𝑥 100%
4
= 4τ
𝑇𝑇𝑠𝑠 =
𝑇𝑇𝑝𝑝 =
ζω𝑛𝑛
𝜋𝜋
𝜔𝜔𝑑𝑑
,
𝜔𝜔𝑑𝑑 = 𝜔𝜔𝑛𝑛 �1 − 𝜁𝜁 2
System Types
Type
overdamped
Damping ratio
ζ>1
critically damped
ζ=1
Underdamped
ζ<1
Complete Response
𝑦𝑦(𝑡𝑡) = 𝐾𝐾𝐾𝐾 + 𝑐𝑐1 𝑒𝑒 𝑟𝑟1 𝑡𝑡 + 𝑐𝑐2 𝑒𝑒 𝑟𝑟2 𝑡𝑡
𝑦𝑦(𝑡𝑡) = 𝐾𝐾𝐾𝐾 + 𝑐𝑐1 𝑒𝑒 𝑟𝑟𝑟𝑟 + 𝑐𝑐2 𝑡𝑡𝑒𝑒 𝑟𝑟𝑟𝑟
𝑦𝑦(𝑡𝑡) = 𝐾𝐾𝐾𝐾 + 𝑐𝑐𝑒𝑒 −σ𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠(ω𝑑𝑑 𝑡𝑡 + θ)
ω𝑑𝑑 = ω𝑛𝑛 �1 − ζ2
Undamped
r = -σ ± jωd= −ζω𝑛𝑛 ±ω𝑛𝑛 �ζ2 − 1
𝑦𝑦(𝑡𝑡) = 𝐾𝐾𝐾𝐾 + 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐(ω𝑛𝑛 𝑡𝑡 + θ)
r = ± jωn
ζ=0
Feedback Systems
Closed loop transfer function:
PID Controller:
Steady-state Error:
Steady-state Output:
Static gain:
𝑇𝑇(𝑠𝑠) =
𝑌𝑌(𝑠𝑠)
𝑅𝑅(𝑠𝑠)
=
𝐺𝐺𝑝𝑝𝑝𝑝 𝐺𝐺𝑐𝑐 𝐺𝐺𝑝𝑝
1+𝐺𝐺𝑐𝑐 𝐺𝐺𝑝𝑝 𝐻𝐻
𝐺𝐺𝐶𝐶 (𝑠𝑠) = 𝑘𝑘𝑝𝑝 + 𝑘𝑘𝑠𝑠𝑖𝑖 + 𝑘𝑘𝑑𝑑 𝑠𝑠
𝑒𝑒𝑠𝑠𝑠𝑠 = lim [𝑥𝑥(𝑡𝑡) − 𝑦𝑦(𝑡𝑡)] = lim 𝑠𝑠[𝑋𝑋(𝑠𝑠) − 𝑌𝑌(𝑠𝑠)] = 𝐴𝐴[1 − 𝐻𝐻(0)]
𝑡𝑡→∞
𝑠𝑠→0
𝑦𝑦𝑠𝑠𝑠𝑠 = lim [𝑦𝑦(𝑡𝑡)] = lim[𝑠𝑠𝑠𝑠(𝑠𝑠)] = lim [𝐴𝐴𝐴𝐴(𝑠𝑠)] = 𝐴𝐴𝐴𝐴(0)
𝐾𝐾 =
ECE205 Final (W2324) - Review Sheet
𝑡𝑡→∞
𝑦𝑦𝑠𝑠𝑠𝑠 (𝑡𝑡)
𝑥𝑥𝑠𝑠𝑠𝑠 (𝑡𝑡)
= 𝐻𝐻(0)
𝑠𝑠→0
Page 3 of 5
𝑠𝑠→0
ECE205 Circuits & Systems
Winter 2023-24
Laplace Transforms
𝐿𝐿{δ(𝑡𝑡)} = 1
𝐿𝐿{𝑢𝑢(𝑡𝑡)} =
1
𝑠𝑠
𝐿𝐿{𝑡𝑡𝑡𝑡(𝑡𝑡)} =
𝐿𝐿 �
1
𝑠𝑠 2
𝑡𝑡 (𝑚𝑚−1)
1
𝑢𝑢(𝑡𝑡)� = 𝑚𝑚
(𝑚𝑚 − 1)!
𝑠𝑠
1
𝑠𝑠 + 𝑎𝑎
1
𝐿𝐿{𝑡𝑡𝑒𝑒 −𝑎𝑎𝑎𝑎 𝑢𝑢(𝑡𝑡)} =
(𝑠𝑠 + 𝑎𝑎)2
𝐿𝐿{𝑒𝑒 −𝑎𝑎𝑎𝑎 𝑢𝑢(𝑡𝑡)} =
𝑡𝑡 (𝑚𝑚−1) −𝑎𝑎𝑎𝑎
1
𝐿𝐿 �
𝑒𝑒 𝑢𝑢(𝑡𝑡)� =
(𝑠𝑠 + 𝑎𝑎)𝑚𝑚
(𝑚𝑚 − 1)!
𝐿𝐿{𝑐𝑐𝑐𝑐𝑐𝑐(ω𝑜𝑜 𝑡𝑡)𝑢𝑢(𝑡𝑡)} =
𝐿𝐿{𝑠𝑠𝑠𝑠𝑠𝑠(ω𝑜𝑜 𝑡𝑡)𝑢𝑢(𝑡𝑡)} =
𝑠𝑠 2
𝑠𝑠
+ ω2𝑜𝑜
ω𝑜𝑜
𝑠𝑠 2 + ω2𝑜𝑜
𝐿𝐿{𝑒𝑒 −𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐(ω𝑜𝑜 𝑡𝑡)𝑢𝑢(𝑡𝑡)} =
𝐿𝐿{𝑒𝑒 −𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠(ω𝑜𝑜 𝑡𝑡)𝑢𝑢(𝑡𝑡)} =
𝐿𝐿 �
𝑠𝑠 + 𝑎𝑎
(𝑠𝑠 + 𝑎𝑎)2 + ω2𝑜𝑜
ω𝑜𝑜
(𝑠𝑠 + 𝑎𝑎)2 + ω2𝑜𝑜
𝑑𝑑𝑑𝑑(𝑡𝑡)
� = 𝑠𝑠𝑠𝑠(𝑠𝑠) − 𝑥𝑥(0− )
𝑑𝑑𝑑𝑑
𝐿𝐿 �
𝑑𝑑2 𝑥𝑥(𝑡𝑡)
� = 𝑠𝑠 2 𝑋𝑋(𝑠𝑠) − 𝑠𝑠𝑠𝑠(0− ) − 𝑥𝑥̇ (0− )
𝑑𝑑𝑡𝑡 2
𝐿𝐿{𝑥𝑥(𝑡𝑡 − 𝑎𝑎)} = 𝑒𝑒 −𝑎𝑎𝑎𝑎 𝑋𝑋(𝑠𝑠)
𝐿𝐿{𝑒𝑒 −𝑎𝑎𝑎𝑎 𝑥𝑥(𝑡𝑡)} = 𝑋𝑋(𝑠𝑠 + 𝑎𝑎)
𝑡𝑡
𝐿𝐿 �𝑥𝑥 � � , 𝑎𝑎 > 0� = 𝑎𝑎𝑎𝑎(𝑎𝑎𝑎𝑎)
𝑎𝑎
Initial Value Theorem: If 𝑥𝑥(𝑡𝑡) ↔ 𝑋𝑋(𝑠𝑠) 𝑙𝑙𝑙𝑙𝑙𝑙+𝑥𝑥(𝑡𝑡) = 𝑙𝑙𝑙𝑙𝑙𝑙 𝑠𝑠𝑠𝑠(𝑠𝑠)
𝑡𝑡→0
𝑠𝑠→∞
Final Value Theorem: If 𝑥𝑥(𝑡𝑡) ↔ 𝑋𝑋(𝑠𝑠) 𝑙𝑙𝑙𝑙𝑙𝑙 𝑥𝑥(𝑡𝑡) = 𝑙𝑙𝑙𝑙𝑙𝑙𝑠𝑠𝑠𝑠(𝑠𝑠)
𝑡𝑡→∞
ECE205 Final (W2324) - Review Sheet
𝑠𝑠→0
Page 4 of 5
ECE205 Circuits & Systems
Winter 2023-24
Bode Plots Summary
Constant
𝐻𝐻(𝑗𝑗ω) = 𝐾𝐾
Magnitude = 20 log10 𝐾𝐾
Phase = 0° (K>0)
Zero at the origin
180° (K<0)
𝐻𝐻(𝑗𝑗ω) = (𝑗𝑗ω)𝑛𝑛
Magnitude = 20n dB/dec crosses 0 dB at 1 rad/s
Phase = n90°
Pole at the origin
1
𝐻𝐻(𝑗𝑗ω) = (𝑗𝑗ω)𝑛𝑛
Magnitude = -20n dB/dec crosses 0 dB at 1 rad/s
Phase = -n90°
Simple Zero
𝐻𝐻(𝑗𝑗ω) = �1 +
𝑗𝑗𝑗𝑗 𝑛𝑛
𝑧𝑧
�
Magnitude = 0 dB for ω ≤ z
20n dB/dec for ω ≥ z
Phase = 0° for ω ≤ 0.1z
45n °/dec for 0.1z < ω ≤ 10z
90n° for ω > 10z
Simple Pole
𝐻𝐻(𝑗𝑗ω) =
1
�1+
𝑗𝑗𝑗𝑗 𝑛𝑛
�
𝑝𝑝
Magnitude = 0 dB for ω ≤ p -20n dB/dec for ω ≥ p
Phase = 0° for ω ≤ 0.1p
-90n° for ω > 10p
ECE205 Final (W2324) - Review Sheet
Page 5 of 5
-45n °/dec for 0.1p < ω ≤ 10p