# Pre Assessment Test - Basics Of Signals And Systems

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| By DEVIE P
DEVIE P, Assistant Professor
Devie, a seasoned finance professional specializing in fintech and innovation. Known for driving strategic financial initiatives and fostering industry collaboration.
Quizzes Created: 1 | Total Attempts: 155
Questions: 11 | Attempts: 155

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Questions and Answers
• 1.

### All the real time natural signals are of which type?

• A.

Discrete

• B.

Continous

• C.

Digital

Correct Answer
B. Continous
Explanation
Real time natural signals are analog which are always continous

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• 2.

### Is analog signals are more flexible for easy transmission?

• A.

True

• B.

False

Correct Answer
B. False
Explanation
Digital signals are more flexible to transmit than analog signals

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• 3.

### Select the conditions for the existence of Initial value theorem.

• A.

If time t approaches to infinity then the function f(t) should exists.

• B.

If time t approaches to zero then the function f(t) should vanish.

• C.

If time t approaches to zero then the function f(t) should exists.

• D.

If time t approaches to infinity then the function f(t) should vanish.

Correct Answer
C. If time t approaches to zero then the function f(t) should exists.
Explanation
The initial value theorem states that if a function f(t) is Laplace transformable and has a finite Laplace transform F(s), then the limit of f(t) as t approaches zero exists and is equal to the initial value of F(s) at s=0. Therefore, the condition for the existence of the initial value theorem is that if time t approaches zero, then the function f(t) should exist.

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• 4.

### In the following conditions, for which Final value doesn’t exist.

• A.

If the function sF(s)has the infinite pole

• B.

If the function sF(s)has pole on origin

• C.

If the function sF(s)has no zeros

• D.

If the function sF(s)has pole at unity

Correct Answer
B. If the function sF(s)has pole on origin
Explanation
If the function sF(s) has a pole on the origin, it means that the denominator of the function has a factor of (s-0), which simplifies to just s. This indicates that the function has a singularity at s=0, causing it to blow up to infinity at that point. As a result, the final value of the function does not exist.

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• 5.

### Expand the series ->(1 + x)^−1

• A.

1 + x + x^2 + x^3 + x^4 + . . .

• B.

1 − x + 2x^2 − 3x^3 + 4x^4 − . . .

• C.

1 − x + x^2 − x^3 + x^4 − . . .

Correct Answer
C. 1 − x + x^2 − x^3 + x^4 − . . .
Explanation
The given series is an expansion of the expression (1 + x)^-1. This can be derived using the formula for the sum of an infinite geometric series, which is a/(1 - r), where a is the first term and r is the common ratio. In this case, the first term is 1 and the common ratio is -x. Thus, the series can be written as 1 + x + x^2 + x^3 + x^4 + ... which simplifies to 1 − x + x^2 − x^3 + x^4 − ...

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• 6.

### Expand the below binomial series->(1 + x)^n

• A.

1 + nx +[n(n − 1)/2!] x^2 +[n(n − 1)(n − 2)/3!] x^3 + . . .

• B.

1 + nx +2[n(n − 1)/2!] x^2 +3[n(n − 1)(n − 2)/3!] x^3 + . . .

• C.

1 - nx +[n(n − 1)/2!] x^2 +[n(n − 1)(n − 2)/3!] x^3 + . . .

• D.

1 - nx +[n(n − 1)/2!] x^2 -[n(n − 1)(n − 2)/3!] x^3 - . . .

Correct Answer
A. 1 + nx +[n(n − 1)/2!] x^2 +[n(n − 1)(n − 2)/3!] x^3 + . . .
Explanation
The given answer is the correct expansion of the binomial series (1 + x)^n. It represents the terms of the series where each term is obtained by raising x to a power and multiplying it by the corresponding coefficient. The coefficients are derived from the binomial coefficient formula, which involves the values of n and the power of x in each term. The given answer correctly follows the pattern of increasing powers of x and the corresponding coefficients based on the binomial coefficient formula.

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• 7.

### High Pass Filter allows the

• A.

Frequency lesser than infinity

• B.

Frequency less than cut-off frequency

• C.

Frequency higher than infinity

• D.

Frequency higher than cut-off frequency

Correct Answer
D. Frequency higher than cut-off frequency
Explanation
A high pass filter allows frequencies that are higher than the cut-off frequency to pass through while attenuating or blocking frequencies that are lower than the cut-off frequency. This means that any signal with a frequency higher than the cut-off frequency will be allowed to pass through the filter with minimal loss or distortion.

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• 8.

### Transfer function of the system is defined as the

• A.

Ratio of laplace transform of input to output response

• B.

Ratio of laplace transform of input to output response with zero initial condition

• C.

Ratio of laplace transform of output to input response with zero initial condition

• D.

Ratio of laplace transform of output to input response

Correct Answer
C. Ratio of laplace transform of output to input response with zero initial condition
Explanation
The transfer function of a system is defined as the ratio of the Laplace transform of the output response to the Laplace transform of the input. The inclusion of "with zero initial condition" means that the system is assumed to start from a state where there is no initial energy or activity. This assumption allows for a more accurate representation of the system's behavior in response to the input. Therefore, the correct answer is the ratio of the Laplace transform of the output to the input response with zero initial condition.

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• 9.

### Low pass filter eliminates frequencies ______________ than the cut-off frequency

• A.

Higher

• B.

Lower

Correct Answer
A. Higher
Explanation
A low pass filter eliminates frequencies that are higher than the cut-off frequency. This means that any frequencies above the cut-off frequency are attenuated or reduced in amplitude, while frequencies below the cut-off frequency are allowed to pass through with minimal attenuation. Therefore, the correct answer is "higher."

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• Current Version
• Mar 20, 2023
Quiz Edited by
ProProfs Editorial Team
• Dec 18, 2016
Quiz Created by
DEVIE P

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