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