detector_exp
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EE 412 Communications Laboratory AM Detection
Theory
A detector is designed to present a 50 ohm load to the RF transmission medium, and to generate a low
frequency output which represents the envelope (contour of positive peaks) o
Module 1, Video 2: The Laplace Transform
Timothy A. Wilson
Department of Electrical, Computer, Software, and Systems Engineering
Embry-Riddle Aeronautical University
EE 300, Fall 2016
T. Wilson (ERAU)
The Laplace Transform
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This Video
1
Laplace Tran
GEORGIA INSTITUTE OF TECHNOLOGY
SCHOOL of ELECTRICAL and COMPUTER ENGINEERING
ECE 2026
Summer 2015
Lab #7: Frequency Response of FIR Filters
Date: 89 July 2015
Pre-Lab: You should read the Pre-Lab section of the lab and do all the exercises in the Pre-Lab
P2.39: Given D = 3ax +2xyay +8x2y3az C/m2, (a) determine the charge density at the point
P(1,1,1). Find the total flux through the surface of a cube with 0.0 x 2.0m, 0.0 y 2.0m
and 0.0 z 2.0m by evaluating (b) the left side of the divergence theorem and (
P2.71: A parallel plate capacitor with a 1.0 m2 surface area for each plate, a 2.0 mm plate
separation, and a dielectric with relative permittivity of 1200 has a 12. V potential difference
across the plates. (a) What is the minimum allowed dielectric stre
P2.71: A parallel plate capacitor with a 1.0 m2 surface area for each plate, a 2.0 mm plate
separation, and a dielectric with relative permittivity of 1200 has a 12. V potential difference
across the plates. (a) What is the minimum allowed dielectric stre
P2.19: In free space, there is a point charge Q = 8.0 nC at (-2.0,0,0)m, a line charge L = 10
nC/m at y = -9.0m, x = 0m, and a sheet charge s = 12. nC/m2 at z = -2.0m. Determine E at the
origin.
The situation is represented by Figure P2.19, and the total
P2.39: Given D = 3ax +2xyay +8x2y3az C/m2, (a) determine the charge density at the point
P(1,1,1). Find the total flux through the surface of a cube with 0.0 x 2.0m, 0.0 y 2.0m
and 0.0 z 2.0m by evaluating (b) the left side of the divergence theorem and (
P2.29: Given D = 2a + sin az C/m2, find the electric flux passing through the surface
defined by 2.0 m, 90. 180, and z = 4.0 m.
Dg S, dS d d a z
d
4
2
2
2 a sin a z g d d a z d
sin d 6C
P2.32: Given a 2.00 cm radius solid wire centered on the z-axis
P2.57: A material has 12.0 V/m ax field intensity with permittivity 194.5 pF/m. Determine the
electric flux density.
D E 194.5 x1012 F
m
C
12V m FV 2.3 nC a
m
2
x
P2.63: For z 0, r1 = 9.0 and for z > 0, r2 = 4.0. If E1 makes a 30 angle with a normal to t
P2.57: A material has 12.0 V/m ax field intensity with permittivity 194.5 pF/m. Determine the
electric flux density.
D E 194.5 x1012 F
m
C
12V m FV 2.3 nC a
m
2
x
P2.63: For z 0, r1 = 9.0 and for z > 0, r2 = 4.0. If E1 makes a 30 angle with a normal to t
P2.51: Two spherical conductive shells of radius a and b (b > a) are separated by a material with
conductivity . Find an expression for the resistance between the two spheres.
First find E for a < r < b, assuming +Q at r = a and Q at r = b. From Gausss la
Laboratory 3
Introduction to complex numbers and complex exponentials in MATLAB.
Laboratory Objectives
To learn how to manipulate complex numbers in MATLAB.
To learn how to plot complex numbers in MATLAB.
To learn how to add sinusoidal waveforms using
Laboratory 4
Introduction to Sampling, Aliasing, and Reconstruction with Digital Images.
Laboratory Objectives
To learn how to load and display images in MATLAB.
To learn how to synthesize test images in MATLAB.
To learn how to sample and reconstruct i
Laboratory 4
Introduction to Sampling, Aliasing, and Reconstruction with Digital Images.
Laboratory Objectives
To learn how to load and display images in MATLAB.
To learn how to synthesize test images in MATLAB.
To learn how to sample and reconstruct i
Lesson 11 Solutions
5.9 (a)
1. By superposition we can express the input sequence as a linear combination of two distinct sequences
[] = 1 [] + 2 []. Taking the upper path in Fig. 5-17 gives
[] = 1 [] + 2 []
while the lower path gives
[] = ( 1 [] + 2 [])
Lesson 10 Solutions
5.7 The impulse response is computed from
[] = [ ] = cfw_
=0
for = 0,1,2,
0
otherwise
Because the delta function is non-zero only when n = k the impulse response is just the the sequence of
difference equation coefficients.
= cfw_3,
Lesson 17 Solutions
7.3 (a) By definition of the z-transform
() []
then performing an inverse z-transform of () gives the impulse response
[] = [] + 5[ 1] 3[ 2] + 2.5[ 3] + 4[ 8]
Recalling that the input response results from inputting [] = []to yield []
Lesson 19 Solutions
7.11(a)
() = 1 + 3 =
3 + 1
3
3
This has zeros at = 1 = (2+1)/3 for = 0,1,2 so = /3 , , 5/3, = /3 ,
or = +/3, + , +5/3, = /3 , + so zeros are located at = /3 ,
Note that the zeros not on the real axis appear as complex conjugates.
(b)
Lesson 17 Solutions
7.3 (a) By definition of the z-transform
() []
then performing an inverse z-transform of () gives the impulse response
[] = [] + 5[ 1] 3[ 2] + 2.5[ 3] + 4[ 8]
Recalling that the input response results from inputting [] = []to yield []
Lesson 19 Solutions
7.11(a)
() = 1 + 3 =
3 + 1
3
3
This has zeros at = 1 = (2+1)/3 for = 0,1,2 so = /3 , , 5/3, = /3 ,
or = +/3, + , +5/3, = /3 , + so zeros are located at = /3 ,
Note that the zeros not on the real axis appear as complex conjugates.
(b)