Tufts University
School of Engineering
Department of Electrical and Computer Engineering
ES3 - Introduction to Electrical Circuits
Fall 2007 Lab Section: Tuesday 2:30 4:30
Experiment 2 Measurement of Time-Varying Signals
Name: Submitted to: Exper
Tufts University
School of Engineering
Department of Electrical and Computer Engineering
ES3 - Introduction to Electrical Circuits
Fall 2007 Lab Section: Tuesday 2:30 4:30 Experiment 5 Interfacing Circuits to Computers:
Analog-to-Digital Conversion
Tufts University
School of Engineering
Department of Electrical and Computer Engineering
ES3 - Introduction to Electrical Circuits
Fall 2007 Lab Section: Tuesday 2:30 4:30 Experiment 4 Amplification, Impedance, and Frequency Response
Name: Submitt
Tufts University
School of Engineering
Department of Electrical and Computer Engineering
ES3 - Introduction to Electrical Circuits
Fall 2007 Lab Section: Tuesday 2:30 4:30
(I did a make up lab on Friday, October 26 instead of my normal Tuesday sect
Tufts University
School of Engineering
Department of Electrical and Computer Engineering
ES3 - Introduction to Electrical Circuits
Fall 2007 Lab Section: Tuesday 2:30 4:30
Experiment 1 DC Measurements
Name: Submitted to: Experiment Performed: Exp
CHAPTER 11
Solutions for Exercises
E11.1 (a) A noninverting amplifier has positive gain. Thus v o (t ) Avv i (t ) 50v i (t ) 5.0 sin(2000 t ) (b) An inverting amplifier has negative gain. Thus v o (t ) A v i (t ) 50v i (t ) 5.0 sin( 2000 t ) v E11.2
CHAPTER 7
Solutions for Exercises
E7.1 (a) For the whole part, we have: Quotient Remainders 23/2 11 1 11/2 5 1 5/2 2 1 2/2 1 0 1/2 0 1 Reading the remainders in reverse order, we obtain: 2310 = 101112 For the fractional part we have 2 0.75 = 1 + 0.5
CHAPTER 6
Solutions for Exercises
E6.1 (a) The frequency of v i n (t ) frequency H (f )
2 cos(2
2000t ) is 2000 Hz. For this
2 60 . Thus, Vout
and we have v out (t ) frequency H (f ) v out (t ) 0. E6.2
4 cos(2
2000t cos(2
H (f )Vn i
60 ). 30
CHAPTER 5
Exercises
E5.1 (a) We are given v (t )
150 cos(200 t 30 ) . The angular frequency is
the coefficient of t so we have 200 f /2 100 Hz T 1 /f Vrms Vm / 2 150 / 2 106.1 V
radian/s . Then 10 ms
Furthermore, v(t) attains a positive peak when
CHAPTER 4
Solutions for Exercises
E4.1 The voltage across the circuit is given by Equation 4.8: v C (t ) Vi exp( t / RC )
in which Vi is the initial voltage. A t the time t1% for which the voltage reaches 1% of the initial value, we have 0.01 exp( t
CHAPTER 3
Solutions for Exercises
E3.1
v (t )
q (t ) / C 10 6 sin(105t ) /(2 10 6 ) 0.5 sin(105t ) V dv i (t ) C (2 10 6 )(0.5 105 ) cos(105t ) 0.1 cos(105t ) A dt
E3.2
Because the capacitor voltage is zero at t = 0, the charge on the capacitor i
CHAPTER 2
Solutions for 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 ser
CHAPTER 1
Solutions for Exercises
E1.1 E1.2 E1.3 Charge = Current Time = (2 A) (10 s) = 20 C
0.01 200cos(200t ) 2cos(200t ) A
i (t )
dq (t ) dt
d (0.01sin(20 0t) dt
Because i2 has a positive value, positive charge moves in the same direction as t