Exam-solution-final-fall2011-2012_eeng300.pdf 41351w

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.

4j3Ω

 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 ܵ௧௢௧௔௟ = −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

ି ଶ଴ூభ 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