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 #...
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
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.
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:
√ ( )
√ ( )
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