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TKSAT-1: TUPAK KATARI Communications Satellite ProgramTELEPORT Know-How and Technology Transfer ProgramSelf-Assessment Tests - June 15, 2013

1

Self-Assessment TestBASIC SCIENCES

Part I: Physics

QUESTION 1: ANSWER (D)According to Newtons law of universal gravitation and his second law of motion, for a free falling

object with mass the following relationship holds: = 2 = = 2 (1)

On Earths surface: = and = . Using these equalities in equation (1) and solving for : = 2 = 2Then, introduce this expression for in equation (1):

= 22 Finally:

= 2

= + = + 2

= + 3

=

2

=

A)

B)

C)

D)

E)

1. Find the acceleration of an object due to Earths gravity as afunction of:

: Earths radius; and : distance from Earths center to that object.


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2. Which of the following statement about satellite elliptical orbit geometry is true?A) The perigee is the point on the orbit where the satellite is closest to Earth.B) The apogee is the point on the orbit where the satellite is furthest from Earth.C) The semi-major axis is the distance from the center of the ellipse to the apogee or

perigee.

D) The eccentricity is the distance from the center of the ellipse to one focus divided bythe semi-major axis.

E) All of the above

QUESTION 2: ANSWER (E)All listed statements are true.


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QUESTION 3: ANSWER (C)The centrifugal acceleration of the satellite is given by:

= = 2 If we combine the above expression with Newtons law of universal gravitation, we have:

2 = 2

Finally:

=

= 2

= = = 1

2 None of the above

A)

B)

C)

D)

E)

3. A satellite in an orbit that is always the same heightabove Earth moves in uniform circular motion with

radius . If Earths mass is , then the speed ofthat satellite orbiting Earth is:


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QUESTION 4: ANSWER (A)The forces acting on the system are:

4. Find the acceleration in an Atwoods machine where onemass is three times the other.

A) /2B) 2/3C) 3/5D) 3/4E) 4g/5

T

T

3

3 = 3 = 1

2

For the heavier mass: 3 = 3For the lighter mass: = Here we have: = + Introduce this in the equation of the heavier mass:

Finally:


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5

5. A layer of a liquid of type A with density floats on top of a layer of a liquid of type B withdensity 4. A block floats at the interface of both liquids with 1/3 of its volume in liquid A and2/3 of its volume in liquid B, as the following figure depicts. What is the density of the block interms of

?

A) 2B) 3C) 4D) 5E) None of the above

QUESTION 5: ANSWER (B)By Archimedes:

= = + 4 = 13 + 4 23Then:

= 13

+ 4 23

Finally: = 3

bLiquid ALiquid B


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6

6. A positive electric charge is uniformlydistributed along a thin wire ring with radius .Find the magnitude of the electric field strength

on the axis of the ring as a function of the

distance from the center of the ring.

QUESTION 6: ANSWER (A)The field intensity

at any point

on the axis of the ring can be found by the geometric sum of the

intensities created by small isolated elements of the charged ring. When adding up the intensity vectorsat point it is only necessary to consider the components directed along the axis of the ring. Due tosymmetry, the vector components of the intensity that areperpendicularto the axis of the ring will add

up to zero. Thus, the field intensity at point is:

= 140

(2 + 2)3 2

=

1

40

(2 + 2)2 3

= 140 (2 + 2)1 2

= 140 (2 + 2)1 2

= 140 (2 + 2)2 3

A)

B)

C)

D)

E)

= 1

40 2 + 2 cos = 140 2 + 2 2 + 2

=1

40 (2 + 2)3 2 Then:


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7. The electric charges and + are separated by a distance as the following figure depicts. Findthe approximate magnitude and direction of the electric field generated by the two charges at a

point P on the -axis that is located at a distance from the -axis.

QUESTION 7: ANSWER (E)

For values , we have:

1

40 3

= 140 22 ; Dir: + = 140 22 ; Dir: + = 140 2

2 ; Dir: = 1

40 3 ; Dir: + = 140 3 ; Dir:

A)

B)

C)

D)

E) +

2

, 0

2

, 0

(0, )

+

2, 0

2, 0

(0, )

cos = 22 + 2 4

= 2 cos = 2 140

2 + 2 4

2

2 + 2 4

As the figure depicts the direction of the electric field is ,then only options (C) and (E) are initially valid. The

magnitude of the electric field at point is = 2 cos =2 cos , where = are the magnitudes of theelectric field at point generated by charges and ,respectively. is the angle between the -axis and the linethat passes through point

and charge +

, thus:

And:


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8. Indicate which of the following circuitsare high-pass filters.

Circuit I Circuit II Circuit III Circuit IV

QUESTION 8: ANSWER (A)For high frequencies (

) the high-pass filters make

. For the above circuits we have:

I. = + , then for high frequencies we have . This is a high-pass filter.II. = + = 1+1, then for high frequencies we have . This is a high-pass filter.

III. =+ =

1+1, then for high frequencies we have 0. This is a low-pass filter.IV. = +, then for high frequencies we have 0. This is a low-pass filter.

A) I and IIB) II and IIIC) III and IVD) II and IVE) I and III


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9

9. Each of the 8 resistors of the circuit depicted in the following figure has a resistance value equal to. Find the total resistance between terminals A and B as a function of.

QUESTION 9: ANSWER (D)By symmetry the equivalent resistance of the circuit is equal to three resistors + = 2connected in parallel. Then:

1 = 12 + 12 + 12 = 32 = 2

3

3

212

1

3

2

3

4

3

A)

B)

C)

D)

E)


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10

Self-Assessment TestBASIC SCIENCES

Part II: Mathematics

10. The rectangle is inscribed in a circle as depicted inthe figure. If = and = 2, find the length of the arc in terms of.

A) 2 B)

3

C) 2 3 D) 3 2 E) None of the above

QUESTION 10: ANSWER (C)The midpoint of the diagonal of the inscribed rectangle coincides with the center of the circle.Thus, the diameter of the circle is equal to 2 and its radius is equal to . Here, all three sides of thetriangle

are equal to

and the angle

is equal to 60. Then the angle

is equal to 120

which is 1/3 (i.e., 120/360) of the complete angle that the arc makes in the circle. Finally, thelength of the arc will be the same fraction of the circumference of the circle: = 13 2 = 2 3 .


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11

11. In the rectangle shown in the figure thearea of the triangle is 100. If = 5and = 4, find the area of the rectangle.


QUESTION 11: ANSWER (B)From the figure we have:

=

+

12

(5 + 4) = 100 + 12 (4)This leads to = 40/. Then the area of the rectangle is:

= 2 = 2 12

(9) 40 = 360


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12

12. Find the simplified value of the expression = cos( ) + sin + 2.A)

= cos

2

B) = 1C) = 1D) = sin 2 E) = sin()

QUESTION 12: ANSWER (E)Expanding the expression we have:cos( ) = (cos )(cos) + (sin )(sin ) = (cos )(1) + (sin )(0) = cos sin + 2 = (sin) cos2 + (cos ) sin2 = (sin)(0) + (cos )(1) = cos Then: = cos + cos = 0From the available options only answer (E) is valid since sin() = 0


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13. Find the value of the following expression:

2 + 33

1 =

QUESTION 13: ANSWER (A)With the simple change of variable = 2 + 3, we have = 2. Then: = 12Back to the integral:

2 + 33

1 = 1

2 =3+3

=1+3 =1

2 log 412

=

1

2 (log 12 log4) =1

2 log

12

4 = log3 = log3

A) log3B) log(3)C) log(

)

D)

log(3)E) None of the above


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14. Find the area of the region delimited by thecoordinate axes and the line tangent to the curveof the function (see below) at the point (0, ).

= 2 + +

QUESTION 14: ANSWER (B)The derivative of the function is: = = 2 + By the pointslope formula the equation of the tangent line at (0, ) is:

= (0) ( 0) = (2(0) + ) = + The region is a triangle with vertices at the points (0,0), (0, ) and ( , 0). The area can becalculated by doing:

= 12

() = 2

2

A) 2 2 B) 2 2 C)

2

2

D) 22 E) None of the above


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2sgn( 2)

44 15. Find the value of the integral

A) 14B) 16C) 18D) 20E) 22

QUESTION 15: ANSWER (D)Notice that

2sgn(2) = | 2|. This is the v-like curve of || shifted by 2 to the right. Then the area of theresulting two triangles can be calculated:

12 (6)(6) + 12 (2)(2) = 20


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16. Consider the sequence where (1) = 1 and () = ( 1) + 17 for all integers n>1. Here, thevalue of(43) is:

A) 3B) 4C) 5D) 6E) 7

QUESTION 16: ANSWER (E)The general expression for the above sequence is: () = (1) + ( 1) 17Then:

(43) = 1 + (43 1) 17 = 1 + 6 = 7


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17. In the domain of real numbers consider and . Then the expression 12 ( + | |)is equal to:

A) The difference between and B) The average of and C) The maximum of and D) The minimum of and E) None of the above

QUESTION 17: ANSWER (D)If > , then | | = , then 12 ( + | |) = If

>

, then |

| =

, then

1

2(

+

|

|) =

If = , then | | = 0, then 12 ( + | |) = = Therefore, the expression

12 ( + | |) is equal to the minimum of and


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18

18. Find the minimum distance between the surface of the sphere ( 3)2 + ( 2)2 + ( 1)2 = 9and the surface of the sphere ( + 3)2 + 2 + ( + 2)2 = 4


QUESTION 18: ANSWER (A)The distance between the centers of the spheres is:

= (3 3)2 + (2 0)2 + (1 2)2 = 36 + 4 + 9 = 7Then, the minimum distance between the surfaces of both spheres is equal to the above distance minus

the radii of the two spheres: 7 9 4 = 2


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Self-Assessment TestAPPLIED SCIENCES

Part I: Electronics

QUESTION 1: ANSWER (A)In the figure: 1 = 2, y 3 = 2 + 1 4 . Replace these values in the following system of equations:

1. Consider the circuit of the figure with a voltage-controlledcurrent source. Find the value of voltage 2 between theterminals of the 4 resistor.

A) V2 = 4 voltsB) V2 = 2 voltsC)

V2 = 2 voltsD) V2 = 4 volts

E) None of the above

1 = 2(1 2) = 2(2 2)1 = 22 + 43 = 22 + 4 2 + 14

2 = 4 2 + 14

= 4 0 + 44

= 4 voltsSolving the above system we have: 2 = 0, 1 = 4.Then:1

2 3


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3. Simplify the function(, ) = A) B)

C) D) E) None of the above

QUESTION 3: ANSWER (D)By Boolean algebra:

(, ) = =

(

)

= = ()() = Using the De Morgan law:

(, ) = = =


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4. Find the simplified expression for the mintermsof the Karnaugh map shown in the following

figure.

A) 2 0 B) 3 1 C) 32 1D) 2 1 0 E) 21 0

QUESTION 4: ANSWER (A)The cells of the Karnaugh map are connected in a toroid-like shape. Thus, the four corners form a

region

of 2 2 cells where:

0 does not change and it is equal to 0 in all region ; so its complement, 0 , is included in therepresentation of the simplified expression.

1 changes (taking values 0 and 1) and must be excluded.2 does not change and it is equal to 0 in all region ; so its complement, 2 , is included in the

representation of the simplified expression.

3 changes (taking values 0 and 1) and must be excluded.Therefore, the simplified expression is: = 2 0


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QUESTION 5: ANSWER (B)Only the 2's and 1s complements are defined for binary numbers, thus option (A) is discarded.

Notice also that the least significant bit (LSB) of is straightforwardly copied to the LSB of. This rulesout option (D) because the 1s complement inverts all bits, and discards also option (C) because a

multiplication by 2 will always set the LSB equal to 0.

To get the 2's complement of a binary number we can start at the LSB and copy all the 0's (i.e., going

from the LSB to the MSB) until we find the first 1, at that point we copy that 1 and then invert the

remaining bits. This conversion is performed by the combinational circuit of the figure and thus the

answer is option (B).

5. The circuit of the figure converts the inputs into the outputs in order to show:

A)

The 9's complement ofB) The 2's complement ofC) The value of multiplied by 2D) The 1's complement ofE) None of the above


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6. The switch of the figure is closed at = 0. Find the expressions for the current and the energystored in the inductor .

QUESTION 6: ANSWER (C)With = 4 we use the following expression to find the value of current:

= 1 10sin3 = 104 13 cos3 + = 56 cos 3 + Considering that = 0 when = 0:

0 = 56

cos(3 0) + = 56

Then:

=

5

6cos 3

+

5

6=

5

6(1

cos3

)

For the energy stored in the inductor:

= 12

2 = 12

4 56

(1 cos3)2 = 2518

(1 cos3)2 = 2518

(cos3 1)2

A)

= 56 cos3 ; = 2518 (cos3)2B) = 2518 cos3 ; = 56 (cos 3)2C) = 56 (1 cos3); = 2518 (cos3 1)2D) = 56 (1 + c o s 3); = 2518 (1 + c o s 3)2E) None of the above


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Part II: Telecommunications

7. Indicate which of the following sequences corresponds to the HDB3 coding of the sequence of bits0101100000000100100001.

A) 0+0- +00000000- 00+0000-B) 0+0- +000+- 0- +- 00+0000-C) 0+0- +000+- 00- +00- +00+-D) 0+0- +00+- 0- 00- 00- +0+0-E) 0+0- +0000+000- 00+0000-

QUESTION 7: ANSWER (C)For the sequence of bits: 0101100000000100100001

The HDB3 coding is: 0+0-+000VB00V+00-B00V-

That finally is: 0+0-+000+-00-+00-+00+-


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8. What is the alarm signal that will be forwardly transmitted in an E1 frame if the signal at the inputnode is interrupted?

A) RAIB) REDC) AISD) DMAE) YELLOW

QUESTION 8: ANSWER (C)The signal transmitted is the Alarm Indication Signal (AIS).


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9. Which of the following statements is true for an antenna that has a 20 dBi gain?A) The antenna is omnidirectionalB) The antenna produces 100 times more power than an isotropic antenna in equal

conditionsC) The antenna produces 20 times more power than an isotropic antenna in equal conditionsD) The antenna produces 1000 times more power than an isotropic antenna in equal

conditions

E) The antenna amplifies by a factor of 20 the input RF energy

QUESTION 9: ANSWER (B)The equivalent gain in dBi (isotropic the forward gain of an antenna compared with the hypothetical

isotropic antenna) can be obtained with the following expression:

= 10log10 If = 20 dBi; then G, the power ratio compared with an isotropic antenna, must be equal to 100.


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10. Which of the following values represents more closely the practical spectral efficiency of QPSKmodulation?

A) 0.1 b/s/HzB) 10 b/s/HzC) 1.5 b/s/HzD) .0015 b/s/HzE) 150 b/s/Hz

QUESTION 10: ANSWER (C)The theoretical value of the spectral efficiency of QPSK is equal to 2/bits/Hz. Thus, option (C)

represents more closely a practical spectral efficiency of QPSK.


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11. Which of the following values is the closest to the required bandwidth to transmit an E1 streammodulated with QPSK and using a FEC of 7/8?

A) 1.5 MHzB) 2.0 MHzC) 2.6 MHzD) 4.0 MHzE) 1.0 MHz

QUESTION 11: ANSWER (A)With FEC the coding rate R is obtained by = /, then the total number of bits taking into accountthe redundancy bits is = /. The bit rate of the original baseband is measured in bits/s andthe symbol rate rate of the modulated carrier is quantified symbols/s or bauds according tothe following expression:

= log2 Where is the number of phases that the carrier can take. If we replace by when using FEC, wehave: =

log2 = log2 The formula for the required RF bandwidth to transmit a QPSK modulation indicates that 20% should

be added to the theoretical value: = 1.2 = 1.2 log2 For E1 = 2.048 Mb/s, for QPSK = 4, and with FEC 7/8 we have = 7/8. Replacing these valuesin the above expression gives: = 1.2 2.048 1067

8log2 4

1.4 MHz


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12. In a transmission system that uses 8-PSK modulation, given a bandwidth of 4 MHz, find the bitrate.

A) 6 MbpsB) 8 MbpsC) 10 MbpsD) 12 MbpsE) 14 Mbps

QUESTION 12: ANSWER (D)For PSK the baud rate is the same as the bandwidth, which means the baud rate is 4000000. But in 8-

PSK the bit rate is 3 times the baud rate, so the bit rate is 12 Mbps.


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Part III: Computer Systems

13. Consider the structure of the following tables:

Create a report that displays the lastname of the employee, name of the department and localization.

Which of the following SQL commands must be used in order to display the aforementioned

information?

A) SELECT lastname, name_department, localization_id FROM employee, departmentB) SELECT employee.lastname, department.name_department, department.localization_id

FROM employee e, department d WHERE e.department_id = d.department_id

C) SELECT e.lastname, d.name_department, d.localization_id FROM employee e,department d WHERE manager_id = manager_id

D) SELECT e.lastname, d.name_department, d.localization_id FROM employee e,department d WHERE e.department_id = d.department_id

E) None of the aboveQUESTION 13: ANSWER (D)The SQL command is:

SELECT e.lastname, d.name_department, d.localization_id FROM employee e, department d WHERE

e.department_id = d.department_id

departmentColumn name Type

department_id number

manager_id number

name_department varchar(25)

localization_id number

employeeColumn name Type

employee_id number

department_id number

manager_id number

lastname varchar(25)


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14. Which of the listed alternatives better describes the following IP address and mask?IP address: 192.192.192.56

Subnet mask: 255.255.255.240

A) This is a Class B address with a subnet mask of 12 bits, providing 14 valid subnets.B) The subnet mask is using 4 subnet bits that allow the creation of 14 usable subnets and 14

hosts per subnet.

C) This is a Class C address that uses 4 subnet bits that allow the creation of 16 usablesubnets and 14 hosts per subnet.

D) This is a Class C address that uses 4 subnet bits that allow the creation of 14 usablesubnets and 14 hosts per subnet.

E) None of the above

QUESTION 14: ANSWER (C)The address 192.192.192.56 with the subnet mask 255.255.255.240 corresponds to a Class C network

and is equivalent to 192.192.192.56/28; this means that there are 4 bits to determine the number of

usable hosts per network: 2 2 = 16 2 = 14. The number of usable networks is 2 = 16.


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15. What information is added during encapsulation at OSI Layer 3?A) Source and destination MACB) Source and destination application protocolC) Source and destination port numberD) Source and destination IP addressE) None of the above

QUESTION 15: ANSWER (D)


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16. A Stored Procedure:A) Is a sequence of SQL statements to perform a specific function.B) Is stored in compiled form in a database.C) Can be called from all client environments.D) All of the above are true statementsE) None of the above are true



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17. Indicate which of the following statements is true with regard to the management of systemsinformation in an RDBMS?

A) The information stored in an RDBMS can be accessed using SQL.B) An RDBMS stores database definition information in system-created tables.C) The information stored in an RDBMS habitually cannot be updated by a user.D) All of the above are true statementsE) None of the above are true



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18. If the following sequence of instructions is executed,what will be printed at the end?

QUESTION 18: ANSWER (D)Check the following table that lists down the details of the instructions that are executed and the valuesof all variables (each row shows the values at the end of the execution of the corresponding sequence of

steps):

Steps executed i n r a b1 1 6 0 0 1

2 to 7 2 6 1 1 12 to 7 3 6 2 1 22 to 7 4 6 3 2 32 to 7 5 6 5 3 5

2 6 6 8 5 88 6 6 8 5 8

1. Set a=0, b=1, i=1, n=6 and r=02. I f i


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Self-Assessment TestENGLISH PROFICIENCY

Part I: GrammarIn this section sentences or short conversations include blanks for which you must choose the correct

word from four choices. Your choice must complete the sentence both in terms of grammar and

meaning.

1. Have you seen the new restaurant in the center of town? No, but ______ my friends are going there tonight.

A) fewB) there are a fewC) lessD) a few of

2. It was _____ that most of the students passed.A) so easy testB) such easy testC) so easy a testD) so hard test

3. The doctor had more patients______ he had time to see them.A) asB) andC) thanD) even

4. The bridge has been flooded, so _____ you do, it will take two hours to get there.A) no matterB) whateverC) wheneverD) it doesnt matter

5. The small child crossed the busy street and, ______, was not harmed.A) amazed enoughB) enough amazingC) amazingly enoughD) enough amazement

6. Will Jay try out for Othello? __________ the role before, so Im sure he will.A) Hes playedB) His playingC) To playD) He will have played


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7. Find someone for the computer job quickly; it doesnt matter ______ in.A) what city they liveB) what city do they liveC) where they liveD) where do they live

8. Andrew was hoping ______ give the cat away, but Sylvias allergies required it.A) he mustntB) he wouldntC) not to have toD) having not to


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Part II: VocabularyThis section includes 8 sentences containing gaps. You must choose one word from four choices which

best completes the sentences in terms of accuracy and meaning.

9. Barbara was able to ______ her weekly income by taking wedding photos on the weekend.A) sum upB) supplyC) augmentD) supplicate

10. Tom works 60 hours a week! Hes really ______ to his job.A) closeB) devotedC) stuckD) pledged

11. In politics, Paula quickly learned never to_______ her opponents weaknesses.A) interplayB) put offC) underestimateD) undermine

12. Although she was _______ of a good education, Yvonne went on to succeed in business.A) refusedB) defeatedC) deprivedD) denied

13. This cream was ________ to make the skin look younger.A) formulatedB) institutedC) congealedD) inventive

14. The accused drug dealer was held in _____ until the trial.A) custodyB) ransomC) perpetuityD) arrangement

15. They got Dr. Roberts to _______ over the meeting of the animal rights committee.A) enforceB) captivateC) conductD) preside


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16. I couldnt understand your report, because your main points were too ________.A) ambiguousB) irreverentC) anonymousD) illegitimate


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Part III: ReadingThis section consists of the next short text followed by 4 multiple choice questions.

17. What do the great auroral displays follow?A) Big streams of electrically charged atoms which fall into the Earth's atmosphere.B) Solar flares full of electrons and protons.C) A short period of powerful cosmic flares near important sunspots.D) An interval of solar flares in the Earth's magnetic field.

18. The author mentions the electrons and protons to showA) that they cause sunspots.B) that they leave the Sun at a constant rate.C) they affect the Earths magnetic field.D) the Aurora cannot be seen from everywhere on Earth.

19. Which of the following is not true about the Northern lights?A) Sometimes they seem to be stationary.B) They are long vertical lines of green lights, colored red at the bottom.C) The solar flares give out extensive amounts of particles full of electricity.D) The light is the result of energy emitted from the Sun and Moon.

The Aurora, also known as the Northern Lights, has mystified people down through the ages. Written

records of so-called great auroral displays date back more than two thousand years.

Sometimes the Aurora over Alaska and other auroral zones is barely visible or appears colorless and

unmoving. But at other times the auroras can be incredibly bright, multi-hued and fast moving. Tallgreen curtains of lights, red tipped at their bottoms, stretch from horizon to horizon. They ripple andsway, fold and unfold, then suddenly disappear, only to reform in a new shape minutes later.

Auroral light is produced by a high-vacuum electrical discharge, and is powered by interactionsbetween the Sun and Earth. The rare great auroral displays follow one or two days of violent solarflares in the vicinity of the Suns major sunspots. These solar flares cast out vast streams ofelectrically charged particles, mostly electrons and protons, which stream into the Earths

atmosphere. The Earths magnetic field steers them away from tropical regions, towards the poles.When the particles strike the gases, the atoms and molecules of the Earth's upper atmosphere glow.

Although much is known about the Aurora, this fascinating phenomenon still withholds some of itsmysteries. Scientists, for example, do not yet understand why the Aurora is so highly structured. The

most beautiful Aurora is composed of thin sheets that stretch upward a hundred miles or more andextend across the sky from horizon to horizon. These arcs and bands whip and weave across the sky,and bright rays ripple along them at fantastic speed. Some of these intricate multicolored forms have

a thickness of only 100 meters, yet may be several hundred kilometers tall and well over 1000

kilometers in length.


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20. It can be inferred from the passage thatA) the Aurora has been a mystery for over two thousand years and will continue to be.B) the structure of the Aurora is not the only thing that scientists do not understand aboutit.C) the written records of the great auroral displays are incorrect.D) the Aurora is always visible from Earth.

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