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Q1. | Fourier transform F(jω) of an arbitrary signal has the property |

A. | F(jω) = F(- jω) [Wrong Answer] |

B. | F(jω) = - F(- jω) [Correct Answer] |

C. | F(jω) = F*(- jω) [Wrong Answer] |

D. | F(jω) = - F*(jω) [Wrong Answer] |

View Answer
Explanation:-
Answer : BDiscuss it below :!! OOPS Login [Click here] is required to post your answer/resultHelp other students, write article, leave your comments |

**Also Read Similar Questions Below :**

⇒ Short-circuited stubs are preferred to open circuited stubs because the latter are

more difficult to make and connect

incapable of giving a full range of reactances

made of transmission line with a different characteristic impedance

liable to radiate

⇒ In the circuit of figure. The current supplied by battery at

*t*= &infin

0

0.208333333333333

less than 5 A

very large

⇒ Narrow band FM signal can be represented as

A cos (2p

*f*) - β

_{c}t*A*sin (2p

*f*) sin (2p

_{c}t*f*)

_{m}tA cos (2p

*f*) - β

_{m}t*A*sin (2p

*f*) sin (2p

_{c}t*f*)

_{m}tA cos (2p

*f*) + β

_{c}t*A*sin (2p

*f*) sin (2p

_{c}t*f*)

_{m}tA cos (2p

*f*) + β

_{m}t*A*sin (2p

*f*) sin (2p

_{m}t*f*)

_{c}t⇒ If an amplifier with gain of - 100 and feedback of β = - 0.1 has a gain change of 20% due to temperature the change in gain of the feedback amplifier will be

0.0002

0.02

0.001

0.0001

⇒ Which of the following statement is incorrect?

UJT is a 3-terminal diode having two unequally doped regions

A UJT can be triggered by any one of its three terminals

The bi-directional switching characteristics of a diac are symmetrical

As compared to JFETs, MOSFETs generate less noise in high gain amplifiers

⇒ In the given figure shows the equivalent circuit of a magic tee. If all ports are mat

*n*

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*n*

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*n*

_{H}= 0.707

*n*

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*n*

_{H}= 0.707

*n*

_{H}= 0.707,

*n*

_{H}= 2

⇒ Two voltages are -5 V and -10 V. In positive logic

-5 V is 1 and -10 V is 0

-10 V is 1 and -5 V is 0

-5 V is 1 in some circuits and 0 in others

-10 V is 1 in some circuits and 0 in others

⇒ Auto correlation for t = 0 is equal to

average power S of waveform

square of average power S of waveform

half of average power S of waveform

square root of average power S of waveform

⇒ A transmitted using AM has in modulated carrier output power of 10 KW and can be modulated to a maximum depth of 90% by a sinusoidal modulating voltage without causing overloading. Find the value to which an modulated carrier power can be increased without resulting in overloading if the maximum permitted modulating index is 40%

14 KW

12.96 KW

26.96 KW

2.96 KW

⇒ In fibre optics the use is made of thin glass fibres for efficient transmission of light.

TRUE

FALSE

⇒ A wave is frequency modulated with an index of 0.1 and its frequency is multiplied by 8 times, the modulation index is

not changed

decreased 8 times

increased 4 times

increased 8 times

⇒ Given two numbers A and B in sign magnitude representation in an eight bit format A = 00011110 and B = 10011100 then A ⊕ B gives

10000010

11111

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⇒ Companding is used

to overcome quantising noise in PCM

in PWM receivers to reduce impulse noise

to protect small signals in PCM from quantising noise

none of the above

⇒ When a single phase dual converter is operating in circulating current mode

both converters are working as rectifiers simultaneously

one converter is always working as rectifier and the other as inverter

only one converter is working at one time

both converters are either working as rectifiers or as inverters simultaneously

⇒ Consider the z-transform X(

*z*) = 5

*z*

^{2}+ 4

*z*

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*z*| < ∞ The inverse transform

*x*[

*n*] is

5δ[

*n*+ 2] + 3δ[

*n*] + 4δ[

*n*- 1]

5δ[

*n*- 2] + 3δ[

*n*] + 4δ[

*n*+ 1]

5

*u*[

*n*+ 2] + 3

*u*[

*n*] + 4

*u*[

*n*- 1]

5

*u*[

*n*- 2] + 3

*u*[

*n*] + 4

*u*[

*n*+ 1]

⇒ Trinitron is a monochrome picture tube.

TRUE

FALSE

⇒ Which of the following is not treated as hexadecimal constant by assembler in 8085?

45 H

6 AFH

234

64 H

⇒ In fabricating silicon BJT in ICs by the epitaxial process, the number of diffusions used is usually

2

3

4

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⇒ Consider the differential amplifier shown in figure. (Assume β is very hi

2 mA

1 mA

0.5 mA

3 mA

⇒ The complement of Boolean expression AB (B C + AC) is

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A B + (B + C)(A + C)

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none of the above

⇒ Find resistance R

_{B}to bring transistor to threshold of saturation V

_{CB}= 0, V

_{BE}= 0.7 V, a =

1.5 kΩ

15 kΩ

10 kΩ

1 kΩ

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rating about 1 A or less

rating of 20 A or less than 20 A

rating 100 A or less

⇒ If B

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B

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⇒ If the output current wave shape of class C circuit has a period of 1 μs and a pulse width of 0.006 μs, the duty cycle is

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0.00006

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⇒ What is an advantage of an electronic voltmeter over a non-electronic voltmeter?

Low power consumption

Low input impedance

The ability to measure wide ranges of voltages and resistance

Large portability

⇒ The waveform shown in the given Figure can be written

*v*(

*t*) =

*u*(

*t*- 1) +

*u*(

*t*- 2) +

*u*(

*t*- 3) +

*u*(

*t*-4)

*v*(

*t*) =

*u*(

*t*- 1) +

*u*(

*t*- 2) +

*u*(

*t*- 3) +

*u*(

*t*-4) -

*u*(

*t*- 5)

*v*(

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*u*(

*t*- 1) +

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*t*- 2) +

*u*(

*t*- 3) +

*u*(

*t*-4) - 4

*u*(

*t*- 5)

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*t*) =

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*t*- 1) +

*u*(

*t*- 2) +

*u*(

*t*- 3) +

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*t*-4) + 4

*u*(

*t*- 5)

⇒ An ideal op-amp can amplify dc signals

TRUE

FALSE

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High input impedance and low output impedance

Low input impedance and high output impedance

High voltage gain

High current gain

⇒

Match the following:

List I (Logic family) | List II (Characteristic) | ||
---|---|---|---|

A. | TTL | 1. | Maximum power consumption |

B. | ECL | 2. | Highest packing density |

C. | NMOS | 3. | Least power consumption |

D. | CMOS | 4. | Saturated Logic |

A-1, B-4, C-2, D-3

A-1, B-4, C-3, D-2

A-4, B-1, C-2, D-3

A-4, B-1, C-3, D-2

⇒ Saturation region of a JFET is also known as

source region

analog region

pinch off region

ohmic region