ECE 301 Zdybel Lab 5 Worksheet
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Jonathan Zdybel
ECE 301 - Foundation of Electrical Circuits II
11/05/2023
Lab 5 - RFID Card Transient Response Analysis
Introduction
This assignment aims to simulate an RFID scanning device that will be used to scan
tags, match parts and their known specifications, and then ship the “parts” to the correct
destinations. This circuit (Fig 1) will be designed, simulated, and run through Multisim,
next, data analysis, specifically on finding the nominal damping coefficient, will be
performed through MATLAB.
Fig. 1. Implementation of the RFID
transceiver and tag in Multisim. The
values of L
s
, L
p
, and the coil coupling
coefficient is entered as properties
of the ideal transformer T1. In this
case, L
s
= L
p
= 1 mH, and the
coupling coefficient = 1
In the case of the RLC RFID tag simulation, the shape of the RFID signal envelope is
known to be a decaying exponential (Fig 2) this will make it easier to determine the
correct coefficient pertaining to each box.
Fig 2.
Using the equation above, constants A, B, and α must be determined from the initial and
final conditions of the data. The value of α can be extracted by using the coordinates of
two peaks of the voltage response. This is the purpose of the MATLAB portion of the
assignment. After acquiring vast swaths of data, we must comb through and calculate
the decaying coefficient to determine the boxes correctly, thus sending the “parts” to the
correct locations.
,
Simulation
Fig. 3. RFID transceiver attached to the subcircuit representing RFID card 1.
After creating and running the circuit shown in Fig 3, replacing the RFID cards, and
rerunning the experiment, we will have four data files from the oscilloscope. There was
a lot of data in these files, but we specifically need the decaying coefficient part of the
signal. To determine this, we created the MATLAB file below
This code imported the files that were generated by the circuit next, it ran the data
through a function which calculated the decaying coefficient by finding the maximum
value along each column. Then multiply coordinates corresponding to the maximum
values in the second column of the data. Through the subplot function, we were able to
plot the signals and produce Fig 4.
Fig 4.
Conclusion
In conclusion, we can see that nearly all the measured α values have a corresponding
nominal α, which is linked to a location. There is a considerable standard deviation of
±65
between some measured values and their expected values; this can be chalked up
to human error.
RFID number
measured α
nominal α
box destination
1
-463
500
Havertown
2
-379
400
Huston
3
-535
600
Boston
4
-292
300
Chicago
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Related Questions
Circuits Lab Simulation | Schoc x
6 https://iti-submission-google.a x
->
A ti-submission-google.app.schoology.com/assignment/student/59013660
I Netflix M maaoun4@hawkm..
HFC Current Students..
Home | Schoology M Inbox (2,780) - 20.
Q Student Portal
D DocHul
ps
Course: Student R..
Maya Aoun - Circuits Lab Simulation - 10382745 * O
File Edit View Insert Format Tools Add-ons Help
Last edit was made 8 minutes ago by Zainab Chebib
Comic San.
B IU
A
CD 田回▼
E - E
100%
Normal text
12
2 3 4 . I .S 6 .I 7
4. r in Tne Dianks Dasea on Tne Two measuring aevices beiow.
Voltmeter
Ammeter
a. A voltmeter is a device that measures
b. An ammeter is a device that measures
Part B: A Simple Circuit
5. First, start by building a simple circuit using a battery, switch, connecting wires and one light
bulb.
6. Based on the circuit you built in the last question, figure #_
representation of a simple circuit.
is the correct picture
3
7. Fill in the blanks using the switch feature of the circuit.
a. When the…
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TV
Series Configuration
Parallal Configuration
a lf the two surge protectors are arranged in series, what is the probability that a voltage surge will not damage the television?
Do not round.)
b. If the two surge protectors are arranged in parallel, what is the probability that a voltage surge will not damage the television?
(Do not round.)
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O The parallel arangment provides better protection because it has a lower probability of protection.
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INSTRUCTIONS:
Under DC conditions, in steady state, find the current i and the voltage Vo.
the exercise of the image, it is in Spanish but it is easy to understand.
Note:Please describe what method/application you are using in each step. (so I can understand it)
(and that the result is good, of course).
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classify.
Y1 (n) = 5x(n) + 2x²(n)
y2(n) = x(n – 1) – 2x(n – 1)
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However, keep the lamps LIspice parameters fixed (i.e. do not change the attributes of the lamp).
3. Using the RC circuit Power your Neon Lamp so that it turns on every 0.1 seconds (10 Hz frequency). The time for which it remains on is not critical, but should obviously be much less than 0.1 seconds.
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A.
Pz
P
MM-
www
P3
-mu-
AMB
Py
m
In the electrical circuit shown above, the values p1 =0.915, p2=0.626, p3=0.762, p4=
0.820, p5 =0.927 are the probabilities that each of the components of the circuit,
independently, will work. Determine the probability that the circuit operates between
points A and B.
Express your result as a number between zero and one, with four decimal places.
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A. What is the dependent source control parameter i∆ in terms of the node voltages?
B. Owing to the voltage source between nodes a and b, what is the constraint equation that relates va and vb? (Other than numeric values, your answer should contain only the terms vb and vb.)
C. What is the KCL equation for the supernode? (Your equation may contain both va and vb. Or, you are free to use the constraint equation to write the equation in terms of a single node voltage.) Using this equation or the ones previous, what is vb (give a numeric value.) What is i∆ (provide a numerical value).
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current.
21
-X(t)
K=1/C
jo
F(t)
r=1/R
m
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Use the equations (1) to (5) to find the value for R1 AND R2.
tc = 0.693(R1 + R2)(C1)
.(1)
td = 0.693(R2)(C1)
.(2)
f=1/T1/(tc + td)..
(3)
Duty cycle (tc)/T
=
.(4)
(5)
T = tc + td
A. For frequencies = 15 KHz, 10 KHz, and 5 KHz with a 70% duty cycle.
B. For frequency = 10 kHz, generate the waveform for the following duty cycles:
90%, 70%, 50%, 30%, and 10%.
V1
10V
R1
200
A1
VCC
ន
RST
OUT
R2:
DIS
00
THR
TRI
CON
GHD
C1
9.4nF
C2
9.4nF
XSC1
Tektronix
555 VIRTUAL
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Assignment # 04
Chapter 4, Norman.S.Nise
What are these responses...?
Undamped
Under-damped
Critically damped
Overdamped
Undamped
2.0
1.8
1.6
1.4
Under-
damped
Critically
damped
1.2
1.0
0.8
0,6
Overdamped
0.4
0.2
0.5
1.5
2.5
3.
3.5
4.
FIGURE 4.10 Step responses for second-order system damping cases
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1.
How can first or second-order circuits be identified by inspecting the circuits?
2.
How can an underdamped second order system be identified?
3.
What form does the complementary solution of an underdamped second order system take?
4.
How can a critically damped system be identified?
5.
What form does the complementary solution of a critically damped second order system take?
6.
How can an overdamped system be identified?
7.
What form does the complementary solution of an overdamped second order system take?
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Recommended textbooks for you
- Delmar's Standard Textbook Of ElectricityElectrical EngineeringISBN:9781337900348Author:Stephen L. HermanPublisher:Cengage Learning

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