Office of Continuing and Distance Education
College of Engineering
301A Engr. Unit C
University Park, PA 16801
Phone: (814) 8657643
Fax: (814) 8653969
www.engr.psu.edu/cde
Office of Continuing & Distance Education
College of Engineering
The Pennsylvani
The Pennsylvania State University
Department of Electrical Engineering
EE 211/EE 212
Sample Midterm Exam #1A
Student Name:
PSU ID #:
This exam is intended to give you practice solving problems that are typically found on midterm
exams. There are 20 multi
The Pennsylvania State University
Department of Electrical Engineering
EE 211/EE 212
Sample Midterm Exam #2A
Student Name:
PSU ID #:
This exam is intended to give you practice solving problems that are typically found on midterm
exams. There are 20 multi
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Module #3: Clear and Muddiest Points
Also, it is pretty difficult memorizing all of the terms.
Is there a different between DO and oxygen concentration or are they same thing?
Seed and unseeded tests   when to use which
Meaning, why are additional vari
Circle or check offthe correct answem
There are 25 muhipiechoice problems at 4 points each.
Name:
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22% m = {)5 raﬂsec.
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CHAPTER 5
Exercises
E5.1
(a) We are given v (t ) 150 cos(200t 30 ) . The angular frequency is
the coefficient of t so we have 200 radian/s . Then
f / 2 100 Hz
T 1 /f 10 ms
Vrms Vm / 2 150 / 2 106 .1 V
Furthermore, v(t) attains a positive peak when the arg
CHAPTER 6
Exercises
E6.1
(a) The frequency of v in (t ) 2 cos(2 2000t ) is 2000 Hz. For this
frequency H (f ) 260 . Thus, Vout H (f )V 260 20 460
in
and we have v out (t ) 4 cos(2 2000t 60 ).
(b) The frequency of v in (t ) cos(2 3000t 20 ) is 3000 Hz. Fo
CHAPTER 4
Exercises
E4.1
The voltage across the circuit is given by Equation 4.8:
vC (t ) Vi exp( t / RC )
in which Vi is the initial voltage. At the time t1% for which the voltage
reaches 1% of the initial value, we have
0.01 exp( t1% / RC )
Taking the n
CHAPTER 3
Exercises
E3.1
v (t ) q (t ) / C 10 6 sin(10 5t ) /(2 10 6 ) 0.5 sin(10 5t ) V
dv
i (t ) C
(2 10 6 )( 0.5 10 5 ) cos(10 5t ) 0.1 cos(10 5t ) A
dt
E3.2
Because the capacitor voltage is zero at t = 0, the charge on the
capacitor is zero at t = 0.
CHAPTER 2
Exercises
E2.1
(a) R2, R3, and R4 are in parallel. Furthermore R1 is in series with the
combination of the other resistors. Thus we have:
1
Req R1
3
1 / R2 1 / R3 1 / R4
(b) R3 and R4 are in parallel. Furthermore, R2 is in series with the
comb
CHAPTER 1
Exercises
E1.1
Charge = Current Time = (2 A) (10 s) = 20 C
E1.2
i (t )
E1.3
Because i2 has a positive value, positive charge moves in the same
direction as the reference. Thus, positive charge moves downward in
element C.
dq (t ) d
(0.01sin(20
Mesh Analysis Algorithm
Type II interior branch current sources
An interior current source cannot be accommodated in
the basic mesh analysis algorithm, because there is no vi
relationship with which to express the voltage across the
current source when w
Idealized Model Simplifications
Two assumptions that greatly simplify closedloop op amp
circuit analysis can be made if
Negative feedback causes the voltages at the inverting and noninverting terminals to be driven to (virtually) the same value
v v+ virt
Operational Amplifiers
1
EE 211/EE 212
Basic Information
Inputoutput Voltage Relationship
Openloop vs. Closedloop (Feedback) Operation
Standard Behavioral Op Amp Model
Idealized Op Amp Model
Canonical Op Amp Circuits
Other Op Amp Circuits
Multistage A
NonInverting Amplifier
Find the inputoutput relationship of
the noninverting amplifier. Determine
the lower limit on the value of the gain
(vout/vin). Assume an idealized op amp. R1
+
vin

23
op amps
R2
v
v+
+

EE 211/EE 212
Jeffrey Mayer 20032006,
(Inverting) LevelShifting Amplifier
31
Similar topology to subtracting
amplifier except that one of the
inputs is a DC constant
+
va(t)

op amps
R2
R1
+
~

+
vb

+

EE 211/EE 212
Jeffrey Mayer 20032006, David Salvia 20062013, Svetla Jivkova 2010,
Saturation
EE 211/EE 212
The inputoutput relationships for the canonical amplifiers are valid
only as long as the op amp is operated in its linear region (i.e. output
voltage remains out of saturation.)
When output voltage hits VSat clipping
Saturation v
The Pennsylvania State University
Department of Electrical Engineering
EE 211/EE 212
Sample Midterm Exam #1B
Student Name:
PSU ID #:
This exam is intended to give you practice solving problems that are typically found on midterm exams. There are 20 multip
CHAPTER 8
Exercises
E8.1
The number of bits in the memory addresses is the same as the address
bus width, which is 20. Thus, the number of unique addresses is 220 =
1,048,576 = 1024 1024 = 1024K.
E8.2
(8 bits/byte) (64 Kbytes) = 8 64 1024 = 524,288 bits
E
EE211Spring2016Recit#201/29/2016page1of9Student
Name: _
Student Name: _
Student Name: _
EE 211 Recitation #2
Friday 01/29/2016
Spring 2016
 GRADED GROUP WORK OF UP TO THREE STUDENTS,
 ONE COMPLETED REPORT PER GROUP , ONE GROUP GRADE!
Before you start wo
The Pennsylvania State University
Department of Electrical Engineering
EE 211/EE 212
Sample Final Exam B
Student Name:
PSU ID #:
This exam is intended to give you practice solving problems that are typically found on final
exams. There are 20 multiplecho
The Pennsylvania State University
Department of Electrical Engineering
EE 211/EE 212
Sample Final Exam A
Student Name:
PSU ID #:
This exam is intended to give you practice solving problems that are typically found on final
exams. There are 20 multiplecho
Saturation
EE 211/EE 212
The inputoutput relationships for the canonical amplifiers are valid
only as long as the op amp is operated in its linear region (i.e. output
voltage remains out of saturation.)
When output voltage hits VSat
clipping
Saturation v
What is light?
Radiation at different wavelengths
Light is a type of _, which is capable
of exciting the human _ system.
%
%
%
%
%
%
%
%
%
%
%
%
380
480
580
680
780
%
%
%
%
%
%
%
%
%
%
%
%
380
480
Wavelength (nm)
580
680
1
Spectral power distribution
780
4/5/2017
Building Energy Codes
Building energy codes limit the amount of lighting power
(watts) that is permitted in a building (by building type alone
or by a sum of power allotted by space types).
Energy use is further controlled through mandatory light