Signals and systems electrical engineering miscellaneous Difficult Questions and Answers | Page - 19

Signals and systems electrical engineering miscellaneous

Let x(t) = rect t - 1 (where rect (x) = 1
2

for - 1 ≤ x ≤ 1 and zero otherwise).
2 2

Then if since (x) = sin (πx) ,
πx

then Fourier Transform of x(t) + x(– t) will be given by

1. sinc ω
  2π
1. 2sinc ω
  2π
1. 2sinc ω cos ω
  2π 2
1. sinc ω sin ω
  2π 2

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rect(x) =1 for - 1 ≤ x ≤ 1 .......(1)
2 2

Given x(t) = rect t - 1
2

Simplifying x(t) with the help of equation (1).
∴ x(t) = 1, 0 ≤ t ≤ 1

Correct Option: C

rect(x) =1 for - 1 ≤ x ≤ 1 .......(1)
2 2

Given x(t) = rect t - 1
2

Simplifying x(t) with the help of equation (1).
∴ x(t) = 1, 0 ≤ t ≤ 1

If X(z) = z eith |z| > a,
(z - a)²

then residue of X(z)z^{n – 1} at z = a for n ≥ 0 will be

1. a^{n – 1}
1. aⁿ
1. naⁿ
1. na^{n – 1}

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X(z) = z with |z| > a
(z - a)²

Let f(z) = X(z) z^{n – 1} = zⁿ with |z| > a
(z - a)²

X(z) = z with |z| > a
(z - a)²

Let f(z) = X(z) z^{n – 1} = zⁿ with |z| > a
(z - a)²

Correct Option: D

X(z) = z with |z| > a
(z - a)²

Let f(z) = X(z) z^{n – 1} = zⁿ with |z| > a
(z - a)²

H(z) is a transfer function of a real system. When a signal x[n] = (t + j)ⁿ is the input to such a system, the output is zero. Further, the Region Of Convergence (ROC) of
1 - 1 z^-1
2

H(z) is the entire Z-plane (except z = 0). It can then be inferred that H(z) can have a minimum of

1. one pole and one zero
1. one pole and two zero
1. two poles and one zero
1. two poles and two zeros

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It will have minimum 1 pole and 2 zeros.
ROC = 1 - 1 z^-1 H(z) = (2z - 1) H(z)
2 2z

∴ H(z) will be in the form of zⁿ ,
(2z - 1)

Hence for minimum realization, n = 2
H(z) = z² ,
(2z - 1)

Correct Option: B

It will have minimum 1 pole and 2 zeros.
ROC = 1 - 1 z^-1 H(z) = (2z - 1) H(z)
2 2z

∴ H(z) will be in the form of zⁿ ,
(2z - 1)

Hence for minimum realization, n = 2
H(z) = z² ,
(2z - 1)

Let x(t) be a periodic signal with time period T. Let y(t) = x(t – t₀) + x(t + t₀) for some t₀.
The fourier Series coefficients of y(t) are denoted by b. If b_k = 0 for all odd k, then t₀ can be equal to

1. T/8
1. T/4
1. T/2
1. 2T

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y(t)= x(t – t₀) + x (t + t₀)
Since y(t) is periodic with period T, then x(t – t₀) and x(t + t₀) will also be periodic with T
Now, b_k = a_ke^-jkω₀t₀ + a_ke^jkω₀t₀
Where, a_k is fourier series coefficient of equal x(t),
b_k = a_k[e^-jkω₀t₀ + e^jkω₀t₀]
= 2a_k cos kω₀t₀
Given that, b_k = 0 for odd k
then, kω₀t₀ = k(π/2),
where k = 2m + 1, m is an integer
⇒ t₀ = T ω₀ = 2π
4 T

Correct Option: B

y(t)= x(t – t₀) + x (t + t₀)
Since y(t) is periodic with period T, then x(t – t₀) and x(t + t₀) will also be periodic with T
Now, b_k = a_ke^-jkω₀t₀ + a_ke^jkω₀t₀
Where, a_k is fourier series coefficient of equal x(t),
b_k = a_k[e^-jkω₀t₀ + e^jkω₀t₀]
= 2a_k cos kω₀t₀
Given that, b_k = 0 for odd k
then, kω₀t₀ = k(π/2),
where k = 2m + 1, m is an integer
⇒ t₀ = T ω₀ = 2π
4 T

A system with input x(t) and output y(t) is defined by the input-output relation

The system will be

1. causal, time-invariant and unstable
1. causal, time-invariant and stable
1. non-causal, time-invariant and unstable
1. non-causal, time-variant and unstable

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It is obvious that system will be time variant, unstable and non-causal.

Correct Option: D

It is obvious that system will be time variant, unstable and non-causal.