From (3), we can get V3 =
Hence, y12 =
I1
V2
=
V1 = 0
15I 1
, substitute it into (4), we get V2 = 57I 1 .
2
I
1
, y 22 = 2
57
V2
=
V1 = 0
I1
8V2
=
V1 = 0
1
.
456
Observe that the determinant of the Y matrix in (i) (left circuit) is not zero, hence its
z
s
+1
Y =
4
0.25 s(s + 4 )
s+3
3
1
4
s+4 4
1
3
1
+
+
s + 3 s + 4 4
(b)
Input impedance seen at port 1
y y
Yin (s ) = y11 12 21
y 22 + YL
Z in (s ) =
1
5s + 16
=
Yin (s + 4 )2
(c)
Since we are finding steady-state magnitude of gain, we can find the tran
(b)
(i) (Left circuit)
V1 = I 1 * 16 + (I 1 + I 2 ) * 2 s + 8 I 1
4
V2 = I 2 * s + (I 1 + I 2 ) * 2 s + 8 I 1
So
2s I
V1 2s + 24
1
V = 2 s + 8 2s + 4 I
2
s 2
1
2s V
I 1 2 s + 24
4 1
=
I
+
+
2
s
8
2
s
V
2
s 2
4
1 2s +
2s V1
=
s
(2 s + 8) 2 s
1
0
31.291 0.6038
Z 1 = 10 3
, Z 2 = 10 3
3
31.875 0.625
0.5 10 10
Z in 2 (s ) = z 2,11
z 2,12 z 2, 21
z 2, 22 + Z L
= (31.291) 10 3
(0.638)(31.875) 10 3
0.625 + 0.008
= 886.3791
Z in (s ) = z1,11
= 10 3
z1,12 z1, 21
(
)
(
)
z1, 22 + 10 3 | Z in 2
ECE5020 Some Hmwrk 4 Solutions
2. and 3. Problems 4.10, 4.12 description are in attached pages:
4. Problem using Figure 4.16 (3rd ed.) or Figure 4.29 (4th ed.)
see slide.
a). Change the final load capacitance from 20 to 72. Calculate h and gate size for e
Our groups robot testing was a painstaking and long process with a lot of problems. At
first, our robot did not even work, with the motor control only starting in response to
sound instead of stopping. Besides this major problem, the wheels would run fast
Challenge #1
Problem: One of the most common chemical reactors found around us is the car engine.
It takes gasoline and reacts it with air to release chemical energy, which in turn is
transformed into mechanical energy. Using the concepts of material bala
Page 1 of 12
The Sound Controlled Robot
ES140 EE Module 3
Wednesday, December 6
Table of Contents
1. Process
of
Assembly:
Electronic.3
2. Process
of
Assembly:
Mechanical.6
3. Robot
Testing
4. Troubleshooting
5. The
Robot
and
Adjustments.7
Electronic
Probl
Fall 2006
ES 140 CEE Module
ES 140
Civil Engineering Module
Group Project
A project modified from the "Quakes and Shakes" program sponsored by the National Science
Foundation and the University of Notre Dame, Earthquake Research Institute.
Buildings and E
Wind Load Assignment
Fall, 2004
ES140 CEE Module
You may work with your group members to solve this problem
This problem requires you to combine the skills you developed in Module 1 with the
concepts of equivalent forces and moments that you have learned
CEE Module
Wind Load Assignment
1 of 1
Part 1
Part 3
The total gravity weight (dead load) of the building will be the weight per floor, 1000 kips, multiplied
by the number of floors, 80, (the building is 800 ft tall and the floors are each 10 ft tall, the
Get all ingredients in terms of IPs and constants
rr
1
The integral we have to perform is E(r) = 4 0 dq |rrqq|3 . We must express all
of its ingredients entirely in terms of our sweeping parameter zq and constants:
Dierential of charge: dq = 0 dlq = 0 d
Problem 2: Spiraling Around the Earth
(a) One of our Tips & Tricks suggests that the best coordinate system to use
is almost always the one that best matches the region of integration. Clearly,
thats spherical coordinates in this problem. Even though were
6. Determine the voltage gain, input resistance and output resistance (both are
indicated in the graph) of the amplifier circuit shown below.
o =80, VA =, VT =25mV.
(1) Av 1; Rin =204.5k; Rout 37
(2) Av 100; Rin =20k; Rout 37
(3) Av -100; Rin =2k; Rout 2.
The rst two integrals are zero since the integrands are oscillating functions with
periods 2 and respectively. The third integral is
2
1
sin q dq =
2
2
2
0
0
1
1
q sin(2q )
1 cos(2q ) dq =
2
2
2
=
0
Hence,
0 R 2
y
E=
4 0 (R2 + z 2 )3/2
Problem 5: Field of
Phys 225 - Homework 10 Solutions
Problem 1: Div, Curl, and Maxwell in cylindrical
(a) To nd the volume charge density we need to compute E. Lets compute
rst out
1
1
)= 0
out = 0 Eout = 0
)=0
(s
(
s s 2 0 s
s s 2 0
We can see that out = 0, so there is no
added together around a ring (the integral over q from 0 to 2). We are left with
only the z component:
E(z)|x=y=0
a2
=
z
4 0
sin q (z a cos q )
dq
2
2
3/2
/2 (z + a 2az cos q )
dq
0
a2
=
z (2)
4 0
a2
=
20
2
z2
z2
a z cos q
a2 + z 2 2az cos q
/2
a+z
a
4. Determine the collector current, Ic, for the following BJT circuit.
F=99, VBE(on)=0.7V.
VCC=10V
RC=4.65k
(1) Ic =2mA
(2) Ic =1.98mA
(3) Ic =2.15mA
(4) Ic =0.02mA
(5) Ic =1.435mA
(6) None of the above
5. For the MOS circuit shown below, what is the outp
L
2
R
z(1 + cos ) s ds d dz
z dm = 0
L
R
0
0
2
s ds
= 0
0
2
L
(1 + cos ) d
zdz
L
0
L
z2
R
= 0 [ + sin ]2
0
2
2 L
R2
L2 L2
= 0 2
= 0
2
2
2
The total mass of the cylinder is
L
2
R
dm = 0
(1 + cos ) s ds d dz
L
0
R
= 20 L
0
2
s ds
0
2
(1 + cos ) d
0
R
[ + si
(f ) f (s, , z) = |z z + x s|
Since z and s are members of the same orthonormal coordinate system (cylindri
cal), we can apply Pythagoras. We simply have to convert that Cartesian x to
cylindrical coordinates to obtain the eld f (s, , z):
f (s, , z) =
z 2
The rst and last terms (the ones involving Es ) can be grouped:
Es Es
1
+
=
(s Es )
s
s
s s
Thus we get the formula from the formula sheet:
E =
1
1 E Ez
(s Es ) +
+
s s
s
z
Problem 2: Potential Simplication
The potential of the dipole eld is given by
V (
(d)
In part (a) we obtained these expressions for the spherical unit vectors in terms of
Cartesian unit vectors:
r = sin (cos x + sin y ) + cos z
= cos (cos x + sin y ) sin z
= sin x + cos y
Since x, y , z are independent of position, all the position-d
Problem 2: Unit Vector Transformations
3
(v ri ) ri , we can write
(a) Using the equation provided in the question v =
i=1
r = ( r x) x + ( r y ) y + ( r z ) z
= ( x) x + ( y ) y + ( z ) z
= ( x) x + ( y ) y + ( z ) z
From the 3D gur
Problem 4: Solenoidal Field
Using Newtons second law F = ma, we have:
F =qv B = q(s s + s + z z ) B0 z = qB0 (s + s s)
ma =m ( s2 ) s + (2s + s) + z z )
s
Comparing every component in F and in ma we get three equations of motion (one
for each compon