HW # 1: 1.20, 1.25, 3.2, 3.9, 3.26, 3.33 - Due Sept. 9, 2011 Solutions.pdf · 2013-03-03 · HW #...
Transcript of HW # 1: 1.20, 1.25, 3.2, 3.9, 3.26, 3.33 - Due Sept. 9, 2011 Solutions.pdf · 2013-03-03 · HW #...
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HW # 1: 1.20, 1.25, 3.2, 3.9, 3.26, 3.33 - Due Sept. 9, 2011
Solutions
1.20 Find complex numbers t = z1 + z2 and s = z1 – z2, both in polar form, for each of the following
pairs:
a) z1 = 2 + j3 and z2 = 1 – j2
b) z1 = 3 and z2 = -j3
c) z1 = 430° and z2 = 3-30°
d) z1 = 330° and z2 = 3-150°
a) z1 + z2 = 2 + j3 + 1 – j2 = 3 + j = √ ( ) = √ 18.43°
z1 – z2 = 2 + j3 – 1 + j2 = 1 + j5 = √ ( )= √ 78.7°
b) z1 + z2 = 3 – j3 = √ ( ) = 4.24-45°
z1 - z2 = 3 + j3 = √ ( ) = 4.2445°
c) z1 + z2 = 330° + 3-30° = ( ) ( )
√ (
)=5.190
z1 - z2 = 330° - 3-30°= ( ) ( )
√ ( )=390
d) z1 + z2 =330° + 3-150°= ( ) ( )
=0
z1 - z2 =330° - 3-150°= ( ) ( )
√ (
)=630
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1.25 A voltage source is given by ( ) ( ) is connected to a series RC
load as shown is Fig. 1-20. If R = 1 MΩ and C = 200pF, obtain an expression for ( ), the voltage
across the capacitor.
Essentially, this is a voltage divider, for which
, where Rpartial is the amount
of resistance taking place after the point of the Vout measurement. However, since this uses a
capacitor, one must use complex resistance values (also known as impedance), which will be
noted here with a Z.
From Table 1-5,
√( ) (
)
√( )
( )
√( ) (
( ))|
( ) ( )
3.2 Given vectors , show that C is
perpendicular to both A and B.
Two vectors are perpendicular if their dot product is 0.
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3.9 Find an expression for the unit vector directed toward the origin from an arbitrary point on the
line described by x = 1 and z = -3.
The unit vector to be found is depicted, in red, by this figure:
At any point along this line, a vector pointing away from the origin is given by the coordinates of
the point:
.
To reverse the direction, and thus point towards the origin, one needs only to take the negative of
each vector component:
To make this a unit vector, divide by the magnitude:
√ ( )
√ ( )
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3.26 Find the volumes described by
a) ⁄
b) ⁄
a) This can be solved by algebra. One need only to take the volume of a cylinder of radius 5
and height 2 and subtract from it the volume of a cylinder of radius 2 and height 2, which
gives ( ) . Additionally, a radial variance from ⁄
corresponds to only one quadrant, which means only one fourth of this volume should be
counted. ⁄ ⁄
b) With density assumed to be 1, the integral in spherical coordinates takes the form:
∫ ∫ ∫
⁄
∫ ∫
|
⁄
∫ ∫
⁄
∫ |
⁄
∫
|
3.33 Transform the vector into cylindrical coordinates
and then evaluate it at ( ⁄ ⁄ ).
From table 3-2: and
√ ( ⁄ ) .
So, ( ) ( )
( ) ( )
( ( ( ⁄ )) ( ( ⁄ )))
( ( ( ⁄ )) ( ( ⁄ ))
( ( ⁄ )))
Since ⁄ and ⁄ , all terms cancel out except the