LIU University – School of engineering
SOLUTION-Final Exam –Fall 2011 - 2012 Electric Circuits II (EENG300) Instructors: Dr. Hassan Bazzi, Dr. Ali Osheiba, Dr. Adnan Harb, Dr. Marc Mannah, Dr. Ziad Noun, Dr. Ahmad Haddad, Mr. Rodrigue Elias, Dr. Hussein Kassem
Please circle your instructor name. Date: Monday, February 6th 2012 Time: 09:00 – 11:00 Student Name:
Student ID:
Section:
Solve the four questions in the booklet: Marking Scheme: Questions
Weight
Question 1
25%
Question 2
25%
Question 3
30%
Question 4
20%
Mark
Final Mark
Conditions: 1. Closed Book Examination. 2. Do not take the staple out. The exam booklet must remain intact. 3. Programmable calculators are allowed. 4. This exam contains 14 pages including this page.
Good Luck
EENG300 – Final EXAM
Fall 2011-2012
Page 1 of 14
LIU University – School of engineering
Question 1– 25 points Given the Balanced three-phase circuit shown below.
4 j 3 Ω
Vca
j 36 Ω
j 36 Ω
12 j 12 Ω
Vab 12 j 12 Ω
Vbc
4 j 3 Ω
j 36 Ω
12 j 12 Ω
4 j 3 Ω
The three phase source has its voltage ܸ = 240√3∠ 30°ܸݏ ݉ݎ The load consists of a combination of a Δ-connected capacitors bank and a Y-Connected load. a) Draw the a-phase equivalent circuit. b) Calculate the line current. c) Calculate the phase voltage at Y-connected load. d) Calculate the phase voltage at the Δ-Connected load. e) Calculate the phase current in the Δ-Connected load. f) Determine the total complex power at the sending end of the line. g) Determine the total complex power at the Y-connected load. h) What percentage of the total average power at the sending end of the line is delivered to the loads?
EENG300 – Final EXAM
Fall 2011-2012
Page 2 of 14
LIU University – School of engineering Solution: a) Draw the a-phase equivalent circuit.
4j3Ω
j 12 Ω
Van
12 j 12 Ω
3 pts
Where
ܸ =
b) Calculate the line current.
ೌ್ √ଷ
2 pts
∠ − 30° = 240 ∠ 0 °Vrms
The equivalent Load impedance is : ܼ = −12݆×
The line current is:
ܫ =
12 + 12݆ = 12 − 12݆W = 12√2∠ − 45 W −12݆+ 12 + 12݆
2 pts
240 ∠ 0 = 13.07∠ 29.36°ݏ ݉ݎܣ 4 + 3݆+ 12 − 12݆
2 pts
c) Calculate the phase voltage at Y-connected load.
ܸே = (12 − 12݆) × ܫ = (12 − 12݆) × 13.07∠ 29.36° = 221.8 ∠ − 15.64°ܸݏ ݉ݎ ܸே = 221.8 ∠ − 15.64°ܸݏ ݉ݎ
2 pts
d) Calculate the phase voltage at the Δ-Connected load
ܸ = ܸே √3∠30° = 384.18 ∠14.36°ܸݏ ݉ݎ
2 pts
e) Calculate the phase current in the Δ-connected load.
ܫ =
ܸ 384∠14.36° = = 10.667∠104.36°ݏ ݉ݎܣ −36݆ −36݆
3 pts
f) Determine the total complex power at the sending end of the line ∗ ) = −3 × 240 ∠ 0 ° × 13.07∠ − 29.36° = −8201.7 + ݆4613.87ܸܣ ܵ௧௧ = −3(ܸ. ܫ
3 pts
EENG300 – Final EXAM
Fall 2011-2012
Page 3 of 14
LIU University – School of engineering g) Determine the total complex power at the Y - connected load
ܵି = 3 ∗
|ܸே |ଶ |221.8|ଶ = 3× = 6149.405 + ݆6149.405ܣܸܭ ܼ 12 − ݆12
4 pts
h) What percentage of the total average power at the sending end of the line is delivered to the loads? Since the Δ-connected load is purely capacitive, it will not contribute in the average power, so the total average power at the loads is that of the Y-connected load: %=ݎ݁ݓ
EENG300 – Final EXAM
ܲ 6149.405 × 100 = × 100 = 75% ܲ 8201.7
Fall 2011-2012
2 pts
Page 4 of 14
LIU University – School of engineering
Question 2– 25 points Consider the circuit shown below:
R 30 kΩ
vi
C 100 pF
L 10 mH
vo
(a) Show (via a qualitative analysis) the type of this filter. (b) Determine the magnitude and the phase of the transfer function ܸ(ఠ ) ܸ(ఠ )
= )݆߱(ܪ
(c) At what condition of H(jω) we determine the center frequency? -
Deduce the center frequency of the filter.
(d) What is the bandwidth of the filter? (e) At what condition of H(jω) we determine the cutoff frequency? -
Find the 2 cutoff frequencies of the circuit.
(f) What is the quality factor of the circuit? (g) Draw an approximate plot of the filter response. (h) Suppose a resistor R L is inserted in parallel to the LC combination at the output, What would be its effect on the following quantities (no calculation required): o
H_max
o
Center frequency ω o
o
Quality factor Q
Solution
Vi (s)
EENG300 – Final EXAM
R
1 cs
Fall 2011-2012
sL
Vo (s)
Page 5 of 14
LIU University – School of engineering
Vi (j )
R
1 j ωc
j L
Vo (j )
(a) Qualitative study: In analyzing the circuit qualitatively we visualize vi as a sinusoidal voltage and we seek the steady-state nature of the output voltage vo . -
At zero frequency the inductor represents a short circuit at the output, hence vo 0
2 pts
when ω 0 . -
At infinite frequency the capacitor represents short circuit, hence vo 0 when ω .
-
The impedance of the parallel combination of L and C is
Z(LC)
jwl 1 w2 LC
will be infinite and thus open circuit at the resonant frequency ݓ =
1 pts
1
√ܥܮ
and hence the output voltage will be maximum at ω ω o .
∴ At frequencies on either side of ωo the amplitude of the output voltage will be nonzero, thus the circuit behaves like a band- filter.
1 pts
(b) The transfer function: 1 ܮݏቀ ቁ ܮݏ ܥݏ ܼ= = ଶ 1 ܥܮ ݏ+ 1 ܮݏ+ ܥݏ ܮݏ ܸ ܼ ܮݏ ݏଶܥܮ+ 1 = )ݏ(ܪ = = = ଶܥܮ+ 1) ܮݏ ܸ݅ ܼ+ ܴ ܮݏ+ ܴ(ݏ +ܴ ଶ ܥܮ ݏ+ 1 ݏ ܴܥ = 1 1 ݏଶ + ቀ ቁݏ+ ܴܥ ܥܮ
EENG300 – Final EXAM
Fall 2011-2012
2 pts
Page 6 of 14
LIU University – School of engineering Which have the following form: = )ݏ(ܪ
ݏଶ
ߚݏ + ߚݏ+ ߱ ଶ
1 pts
∴ it is a band- filter of transfer function H(jω)
1 ܴܥ = )݆߱( ܪ 1 1 −߱ ଶ + ݆߱ ቀ ቁ+ ܴܥ ܥܮ ߱ ܴܥ |= |)݆߱(ܪ ଶ ଶ ට ቀ 1 − ߱ ଶቁ + ቀ ߱ ቁ ܥܮ ܴܥ ݆߱
And
ߔ (݆߱) =
ߨ/2 − tan ିଵ ቌ
߱ ܴܥ
1 − ߱ଶ ܥܮ
(c) The center frequency is found at H(jω o) = Hmax = 1 -
ቍ
2 pts
2 pts
1 pts
The center frequency is simply deduced from the expression of the transfer function to be:
ωo
1 1 1000 krad/s 3 LC 10x10 x100x10 12
1 pts
(d) The bandwidth of the filter from the expression of the transfer function:
β
1 1012 333.33 krad/s RC 30 x 103 x100
2 pts
(e) The cutoff frequency is found at ݆߱( ܪ) = -
ݔܽ ݉ܪ √2
=
1
√2
1 pts
The cutoff frequencies can be written in of ωo and β as follows:
2
β β ωc1 ωo 2 847 krad/s , 2 2
1 pts
and
EENG300 – Final EXAM
Fall 2011-2012
Page 7 of 14
LIU University – School of engineering 2
ωc2
β β ωo 2 1180.455 krad/s 2 2
1 pts
or : 2
ωc1
2
1 1 1 1 1 1 ωc2 2RC 2RC 2RC LC , and 2RC LC
(f) The quality factor Q is
Q
ωo 1000 3 β 333.33
2 pts
(g) Filter response plot:
H(j ωo ) 1
H(j ω o ) 2
70.71
2 pts
0.0 0.0
ωc1
ωo
, R/s
ω c2
ω rad/s
(h) If a resistor RL is added in || to the output : -
Hmax will be reduced since the max output voltage Vin in our case will be divided over R and RL (voltage divider) ,
1 pts
ܴ ܪ୫ ୟ୶ = ܴ + ܴ
1 pts
-
The center frequency will retain its value.
-
The quality factor will decrease since the bandwidth of the filter with RL will increase due to the decrease of both Hmax and Hmax/√2.
1 pts
EENG300 – Final EXAM
Fall 2011-2012
Page 8 of 14
LIU University – School of engineering
Question 3 – 30 points For the circuit shown below:
40
I1
I2
V1
V2
a- Determine the expression for the input impedance Z in = V1/I1 when no load is connected at the 2nd port. b- Determine the expression for the voltage gain A V=V2/V1 . c- Determine the expressions for V TH and ZTH of Thevenin’s equivalent seen at the output port. d- Calculate the values of the load impedance ZL that when connected to the output will provide max-average power transferred to it:
1K
-2 200 S
h 10
e- Calculate the max-power.
Hint: V1 h11 I1 h12 V2 , I2 h 21 I1 h 22 V2
Solution (a) From the circuit we have:
V1 h11 I1 h12 V2
(1)
I2 h 21 I1 h 22 V2
(2)
60 = 40ܫଵ + ܸଵ
(3)
2 pts
When no load is connected at the 2nd port I2 =0. From (2), Substitute in (1) This leads to:
EENG300 – Final EXAM
ܸଶ = −
మభ
ܫ మమ ଵ
=> ܸଵ = ℎଵଵ + ℎଵଶ ቀ− ܼ =
భ ூభ
మభ మమ
ቁܫଵ
మభ
= ℎଵଵ– ቀ భమ
మమ
௱
ቁ=
Fall 2011-2012
2 pts
మమ
Page 9 of 14
LIU University – School of engineering (b) The voltage gain ܣ =
ܸଶ ܸଶ ܫଵ ℎଶଵ ℎଶଶ ℎ ଶଵ = . = − . = − ܸଵ ܫଵ ܸଵ ℎଶଶ ߂݄ ߂݄
2 pts
(c) To find ZTh , remove the 60-V voltage source at the input port and find ZTh
V2 at the I2
output port, as shown:
1 pts
From the circuit equations 3 becomes: ܸଵ ൌ െ ͶͲܫଵ
(3)
2 pts
Substituting into (1) we obtains
-40I1 h11 I1 h12 V2
I1
h12V2 40 h11
(4)
2 pts
But from (2)
I 2 h 21 I1 h 22 V2
(2)
Substituting (4) into (2), we have
I2 h 22V2 h 21 Therefore, as ZTh Z Th
h12V2 h h h 21 h12 40 h 22 11 22 V2 h11 40 h11 40
V2 I2
h11 40 h11 h 22 h 21 h12 40 h 22
To find VTh , we find the open-circuit voltage V2 .
1 pts
2 pts
LIU University – School of engineering
1 pts
At the input port, From (3)
V1 60 40 I1 Substituting (5) into (1) we have
1 pts
(5)
60 40 I1 h11 I1 h12 V2
2 pts
Or
60 (40 h11 ) I1 h12 V2 At the output, I2 0
(6)
1 pts
(7)
Substituting (7) into (2) we have
0 h 21 I1 h 22 V2
I1 -
h 22 V2 h 21
(8)
2 pts
Substituting (8) into (6), one obtains
h 60 (40 h11 ) 22 h12 V2 h 21
2 pts
Or VTh V2
60 60 h 21 (40 h11 ) h 22 / h 21 h12 h12 h 21 h11 h 22 -40 h 22
1 pts
(d) Substituting the values of the h parameters;
ZTh
1000 40 1040 51.46 Ω -6 10 x 200 x10 20 40 x 200 x10 20.21
VTh
60 x 10 29.69V 20.21
And
3
-6
1 pts
1 pts
Since ZTH is purely resistive, for max-average power ZL = RL =RTH = 51.46 Ω
2 pts
LIU University – School of engineering
(e) The max-power is :
ܲ݉ ܽ¼ =ݔ
EENG300 – Final EXAM
ଶ ்ܸு 1 (−29.69)ଶ = = 4.28ܹ ܴ 4 51.46
Fall 2011-2012
2 pts
Page 12 of 14
LIU University – School of engineering
Question 4 – 20 points Determine the y-parameters for the two-port shown in the figure below:
2i
i
8
4
2
Solution To get y11 and y 21 , consider the circuit in which port2 is short-circuited:
1 pts
KCL at node 1, ܫଵ =
ܸ ܸ ͵ܸ + ʹܫଵ ൌ ܫଵ = − 2 4 4 ସ
ܸଵ ൌ ͺ ܫଵ ܸ ൌ ͺ ܫଵ − ܫଵ = ܻଵଵ =
And so
ܸଵ
ܫଶ ൌ െ ʹܫଵ– ܻଶଵ =
ܫଵ
ସ
=
3
20
ଷ
ଶ
ܫ ଷ ଵ
ൌ ͲǤͳͷܵǤ
ൌ െʹܫଵ–
ర య
ି ூభ ସ
=−
2 pts => ହூభ ଷ
ܫଶ ܫଶ ܫଵ 5 3 5 = . =− . =− ൌ െ ͲǤʹͷܵǤ ܸଵ ܫଵ ܸଵ 3 20 20
2 pts
2 pts 2 pts
LIU University – School of engineering Similarly, we get y12 and y 22 , consider the figure where port 1 is short circuited:
1 pts
KCL at node 1 ܫଵ = KCL at node 2
ܸ ܸ െ ܸଶ ܸଶ ͵ܸ ʹܫଵ + => = ܫଵ 2 4 4 4 ܫଶ =
Where
ܸ ൌ െ ͺ ܫଵ
So, sub (3) in (1) :
Sub (3) in (2) : Finally,
ܸଶ െ ܸ െ ʹܫଵ 4
(3)
(1)
2 pts
(2)
2 pts
1 pts
ܸଶ ͵ሺെ ͺ ܫଵ) ܫଵ 1 = ܫଵ ൌ െ ͷܫଵ ൌ ܻଵଶ = = − ൌ െ ͲǤͲͷܵ 4 4 ܸଶ 20
ܫଶ =
మ ସ
ܻଶଶ =
∴ the Y parameter matrix:
–
ூమ
మ
ସ
=
– ʹܫଵ = ூమ ூభ
.
ூభ మ
ି ଶூభ ସ
–
ି ଼ூభ ସ
– ʹܫଵ ൌ െ ͷܫଵ
= (−5)(−0.05) = 0.25 ܵ 0.15
y -0.25
-0.05 S 0.25
1 pts 2 pts 2 pts