1
Solution Problem 1 (5 points)
The first step is to assign clockwise mesh currents I1 and I2 to the left and right loops respectively.
Using KVL and mesh current analysis we get,
−10∠30◦ + 20I1 + j20(I1 − I2 ) = 0,
(1)
j20(I2 − I1 ) + I2 (20 − j20) = 0.
(2)
Solving for I1 and I2 and simplifying we get,
I2 = jI1 ,
(3)
I1 = 0.2236∠3.43◦ ,
(4)
I2 = 0.2236∠93.43◦
(5)
We use the mesh equality IL = I1 − I2 to get,
2
IL = 0.3162∠ − 41.57◦ ,
(6)
VR = 20I2 = 4.472∠93.43◦ .
(7)
Solution Problem 2 (5 points)
Because we have an independent source, we zero it to get the Thevenin equivalent impedance. Zeroing a
current sources is equivalent to replacing it with an open circuit wire. Therefore, the impedances on the left
branch of the circuit become in parallel with the impedances on the right branch. Solving for impedance
parallel impedances give us:
Zt =
(1 − j2)(1 − j2)
= 0.5 − j = 1.118∠−63.43◦
(1 − j2) + (1 − j2)
(8)
We notice that the current source is already in Norton configuration therefore:
In = 2∠0◦
(9)
The Thevenin voltage is the Norton current multiplied by the Thevenin impedance, therefore
Vt = In Zt = 2∠00 1.118∠−63.43◦ = 1 − 2i = 2.236∠−63.43◦
3
(10)
Solution Problem 3 (5 points)
We first notice that the circuit is a voltage divider in Phasor space. Therefore, it is straightforward to
formulate the transfer function as:
R2
H(f ) =
.
(11)
R1 + R2 + j2πf L
Using the equality fb ,
1
2π R
L
1 +R2
we get:
H(f ) =
1
R2
R1 +R2
.
+ j ffb
(12)
The magnitude and phase are:
R2
|H(f )| = q R1 +R2 ,
2
1 + ffb
∠H(f ) = tan
0
R2
R1 +R2
(13)
!
− tan
f
fb
1
!
= −tan
f
fb
.
(14)
We write the value of the norm in Decibels using the values R1 = R2 = 1Ω as:
|H(f )| = 20 log
R2
R1 + R2
f 2
=
− 10 log 1 +
fb
f 2
6 dB − 10 log 1 +
fb
Finally, the bode plot is:
(15)