Exam-solution-Final-fall2011-2012_EENG300.pdf

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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 Mark

Question 1 25% Question 2 25% Question 3 30% Question 4 20% 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.

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Question 1

– 25 points

Given the Balanced three-phase circuit shown below.

4 j 3 Ω ca

V

ab

V

bc

V

j 36 Ω  j 36 Ω  j 36 Ω  4 j 3 Ω 4 j 3 Ω 12 j 12 Ω 12 j 12 Ω 12 j 12 Ω

The three phase source has its voltage ܸ௔௕= 240√3∠ 30°ܸݎ݉ ݏ

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?

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Solution:

a) Draw the a-phase equivalent circuit.

4 j 3 Ω j 12 Ω  12 j 12 Ω an V Where ܸ௔௡=௏_√ଷೌ್∠ − 30° = 240 ∠ 0 °Vrms

b) Calculate the line current.

The equivalent Load impedance is :

ܼ௘௤= −12݆×_{−12݆+ 12 + 12݆}12 + 12݆ = 12 − 12݆W= 12√2∠ − 45W

The line current is:

ܫ_௔஺=_{4 + 3݆+ 12 − 12݆= 13.07∠ 29.36°ܣݎ݉ ݏ}240 ∠ 0 c) Calculate the phase voltage at Y-connected load.

ܸ஺ே = (12 − 12݆) × ܫ௔஺= (12 − 12݆) × 13.07∠ 29.36° = 221.8 ∠ − 15.64°ܸݎ݉ ݏ

ܸ஺ே= 221.8 ∠ − 15.64°ܸݎ݉ ݏ

d) Calculate the phase voltage at the Δ-Connected load

ܸ஺஻ = ܸ஺ே√3∠30° = 384.18 ∠14.36°ܸݎ݉ ݏ

e) Calculate the phase current in the Δ-connected load.

ܫ_஺஻=_−36݆=ܸ஺஻ 384∠14.36°_−36݆ = 10.667∠104.36°ܣݎ݉ ݏ f) Determine the total complex power at the sending end of the line

ܵ௧௢௧௔௟= −3(ܸ௔௡. ܫ௔஺∗ ) = −3 × 240 ∠ 0 ° × 13.07∠ − 29.36° = −8201.7 + ݆4613.87ܸܣ

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g) Determine the total complex power at the Y - connected load ܵ௒ି௅ = 3 ∗|ܸ஺ே| ଶ ܼ௒ = 3 × |221.8|ଶ 12 − ݆12 = 6149.405 + ݆6149.405ܭܸܣ

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:

%݌݋ݓ݁ݎ=ܲ_ܲ௅ ௚× 100 = 6149.405 8201.7 × 100 = 75%

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Question

2– 25 points

Consider the circuit shown below:

i

v

o

v

L

C

R 30 kΩ



pF

100

10 mH

(a) Show (via a qualitative analysis) the type of this filter.

(b) Determine the magnitude and the phase of the transfer function ܪ(݆߱) =ܸ_ܸ௢(ఠ )

௜(ఠ )

(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 RLis 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 ωo o Quality factor Q Solution

(s)

V

_o

sL

cs

1 R

i V (s)

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o V (j ) j L

R

i V (j )

ωc

j

1

(a) Qualitative study:

In analyzing the circuit qualitatively we visualize

v

_i as a sinusoidal voltage and we seek the steady-state nature of the output voltage

v

_o.

- At zero frequency the inductor represents a short circuit at the output, hence

v

_o



0

when

ω 0



.

- At infinite frequency the capacitor represents short circuit, hence

v

_o



0

whenω . - The impedance of the parallel combination of L and C is

(LC) 2

Z

1 jwl

w LC





will be infinite and thus open circuit at the resonant frequency ݓ_௢= 1

√ܮܥ

and hence the output voltage will be maximum atω ω _o.

∴ At frequencies on either side of

ω

_othe amplitude of the output voltage will be nonzero, thus the circuit behaves like a band-pass filter.

(b) The transfer function:

ܼ= ݏܮቀ1ݏܥቁ ݏܮ+ 1_ݏܥ= ݏܮ ݏଶ_{ܮܥ+ 1} ܪ(ݏ) =ܸ݋_ܸ݅=_{ܼ+ ܴ =}ܼ ݏܮ ݏଶ_{ܮܥ+ 1} ݏܮ ݏଶ_{ܮܥ+ 1 + ܴ} = _{ݏܮ+ ܴ(ݏ}ݏܮ_ଶ_{ܮܥ+ 1)} = ݏ ܴܥ ݏଶ _{+ ቀ 1} ܴܥቁݏ+ܮܥ1

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Which have the following form:

ܪ(ݏ) =_ݏ_ଶ_{+ ߚݏ+ ߱}ߚݏ

௢ଶ

∴ it is a band-pass filter of transfer function H(jω) ܪ(݆߱) = ݆߱ 1ܴܥ −߱ଶ_{+ ݆߱ ቀ 1} ܴܥቁ+ܮܥ1 |ܪ(݆߱)| = ߱ ܴܥ ටቀ1_{ܮܥ− ߱}ଶ_ቁଶ_{+ ቀ߱} ܴܥቁ ଶ And ߔ(݆߱) = ߨ/2 − tanିଵ_ቌ ߱ ܴܥ 1 ܮܥ − ߱ଶ ቍ

(c) The center frequency is found at H(jωo) = Hmax= 1

- The center frequency is simply deduced from the expression of the transfer function to be: o ₃ ₁₂ 1 1 ω 1000 krad/s LC 10x10 x100x10    

(d) The bandwidth of the filter from the expression of the transfer function:

12 3 1 10 β 333.33 krad/s RC 30 x 10 x100   

(e) The cutoff frequency is found at

ܪ(݆߱௖) =ܪ݉ ܽݔ

√2 =

1 √2

- The cutoff frequencies can be written in terms of

ω

_oandβ as follows:

2 2 c1 o β β ω ω 847 krad/s 2 2      _{ }     , and

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1 pts

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2 2 c2 o β β ω ω 1180.455 krad/s 2 2     _{ }     or : 2 c1 1 1 1 ω 2RC 2RC LC      _ _    _{, and} 2 c2 1 1 1 ω 2RC 2RC LC     _ _   

(f) The quality factor Q is

o

ω 1000

Q 3

β 333.33

  

(g) Filter response plot:

   __{, R/s} 70.71 o

H(j ω ) 1



o H(j ω ) 2 o ω c1 ω ω_c2 ω rad/s 0.0 0.0



(h) If a resistor RLis added in || to the output :

- Hmaxwill be reduced since the max output voltage Vin in our case will be divided over R

and RL(voltage divider) ,

ܪ୫ ୟ୶=_{ܴ + ܴ}ܴ௅ ௅

- 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 Hmaxand Hmax/√2.

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Question 3

– 30 points For the circuit shown below:

1

V

40 



1

I

2

V



2

I

a- Determine the expression for the input impedance Zin= V1/I1when no load is connected

at the 2nd_port.

b- Determine the expression for the voltage gain AV=V2/V1.

c- Determine the expressions for VTHand ZTHof 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:

 

h

1K

- 2

10 200 S









 







e- Calculate the max-power.

Hint:

1 11 1 12 2

V  h I  h V_, I2  h I21 1 h V22 2

Solution

(a) From the circuit we have:

1 11 1 12 2

V  h I  h V (1)

2 21 1 22 2

I  h I  h V (2)

60 = 40ܫଵ+ ܸଵ (3)

When no load is connected at the 2nd_{port I}₂_=0.

From (2), _ܸ_ଶ_{= −}௛మభ ௛మమܫଵ

Substitute in (1) _{=> ܸ}_ଵ_{= ℎ}_ଵଵ_{+ ℎ}_ଵଶ_ቀ−௛మభ ௛_మమቁܫଵ

This leads to: _ܼ_௜௡ ₌ ௏భ

ூ_భ= ℎଵଵ– ቀ ௛_భమ௛_మభ ௛_మమ ቁ= ௱௛ ௛_మమ

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(b) The voltage gain

ܣ_௏=ܸ_ܸଶ

ଵ=

ܸଶ

ܫ_ଵ.ܸܫଵ_ଵ= −ℎℎଶଵ_ଶଶ.ℎ߂݄ = −ଶଶ ℎ߂݄ଶଵ

(c) To find

Z

_Th, remove the 60-V voltage source at the input port and find 2 Th 2

V

Z

I



at the output port, as shown:

From the circuit equations 3 becomes:

ܸଵൌ െͶͲܫଵ (3)

Substituting into (1) we obtains

12 2 1 11 1 12 2 1 11

h

-40I

h I

h V

I

40 h

V







 



(4) But from (2) 2 22 1 21 2

h

I

h

V

I





(2)

Substituting (4) into (2), we have

12 2 11 22 21 12 22 2 22 2 21 2 11 11

h

h h

40 h

I

h

40 h

40 V

V





V









Therefore, as 2 Th 2

V

Z

I



11 Th 11 22 21 12 22 h 40 Z h h h h 40 h    

To find

V

_Th, we find the open-circuit voltage

V

₂.

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At the input port, From (3)

1 1

V



60 40 I



(5) Substituting (5) into (1) we have

1 11 1 12 2

60 40 I





h I



h V

Or 11 1 12 2

60 (40





h ) I



h V

(6) At the output, 2

I



0

(7)

Substituting (7) into (2) we have

22 21 1 22 2 1 2 21

h

0 h I

h V

I

-

V

h









(8)

Substituting (8) into (6), one obtains

22 11 12 2 21

h

60 (40

h )

h

V

h





 

_



_





Or 21 Th 2 11 22 21 12 12 21 11 22 22 60 h 60 V V (40 h ) h / h h h h h h -40 h       

(d) Substituting the values of the h parameters;

Th 3 -6 -6

1000 40

1040

Z

51.46 Ω

10 x 200 x10

20 40 x 200 x10

20.21 





And Th

60 x 10

V

29.69V

20.21 

 



Since ZTHis purely resistive, for max-average power ZL= RL=RTH = 51.46 Ω

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(e) The max-power is : ܲ݉ ܽݔ= ¼ܸ_்ܴுଶ ௅ = 1 4(−29.69) ଶ 51.46 = 4.28ܹ

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Question 4

– 20 points

Determine the y-parameters for the two-port shown in the figure below:

2i



4 8 2

i

Solution

To get

y

₁₁and

y

₂₁, consider the circuit in which port2 is short-circuited:

KCL at node 1, ܫଵ=ܸ₂௢+ܸ₄௢൅ ʹܫଵൌ൐ ܫଵ= −͵ܸ₄௢

ܸ

ଵ

ൌ ͺܫ

ଵ

൅ ܸ

௢

ൌ ͺܫ

ଵ

−

ସ_ଷ

ܫ

ଵ

=

ଶ଴_ଷ

ܫ

ଵ

=>

ܻ

ଵଵ

=

_ܸ

ܫ

ଵ ଵ

=

3

20 ൌ ͲǤͳͷܵǤ

ܫ

ଶ

ൌ െʹܫ

ଵ

–

௏_ସ೚

ൌ െʹܫ

ଵ

–

ି ర యூభ ସ

= −

ହூభ ଷ And so ܻଶଵ=_ܸܫଶ ଵ= ܫଶ ܫଵ. ܫଵ ܸଵ= − 5 3. 3 20= − 5 20ൌ െͲǤʹͷܵǤ

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Similarly, we get

y

₁₂ and

y

₂₂, consider the figure where port 1 is short circuited: KCL at node 1 ܫଵ=ܸ₂௢൅ ʹܫଵ+ܸ௢െ ܸ₄ ଶ=>ܸ₄ଶ=͵ܸ₄௢൅ ܫଵ (1) KCL at node 2 ܫଶ=ܸଶെ ܸ₄ ௢െ ʹܫଵ (2) Where _ܸ_௢_{ൌ െͺ ܫ}_ଵ (3) So, sub (3) in (1) : ܸଶ 4 = ͵ሺെͺ ܫଵ) 4 ൅ ܫଵൌ െͷܫଵൌ൐ ܻଵଶ= ܫଵ ܸଶ= − 1 20ൌ െͲǤͲͷܵ Sub (3) in (2) :

ܫ

ଶ

=

௏_ସమ

–

௏_ସ೚

–ʹܫ

ଵ

=

ିଶ଴ூ_ସ భ

–

ି଼ூ_ସభ

–ʹܫ

ଵ

ൌ െͷܫ

ଵ Finally,

ܻ

ଶଶ

=

_௏ூమ_మ

=

_ூூమ_భ

.

_௏ூభ_మ

= (−5)(−0.05) = 0.

25 ܵ

∴ the Y parameter matrix:

 

y

0.15 -0.05

S

-0.25

0.25 



 







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2 pts

1 pts

2 pts