Signal and systems miscellaneous Easy Questions and Answers | Page - 33

Signal and systems miscellaneous

A square wave is defined by
x(t) = A; 0 < t < T₀/2
– A; T₀/2 < t < T₀

It is periodically extended outside this interval. What is the general coefficient a_n in the Fourier expansion of the wave?

1. 0
1. 2A(1 – cos nπ)
  nπ
1. 2A(1 + cos nπ)
  nπ
1. 2A(1 – cos nπ)
  [(n + 1)π]

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Given that
x(t) = A; 0 < t < T₀/2
– A; T₀/2 < t < T₀

where T₀ is the fundamental period. If we plot this periodic square wave.

From the wave, it is clear that wave is an odd function, therefore a_n = 0.
Hence, alternative (A) is the correct choice.

Correct Option: A

Given that
x(t) = A; 0 < t < T₀/2
– A; T₀/2 < t < T₀

where T₀ is the fundamental period. If we plot this periodic square wave.

From the wave, it is clear that wave is an odd function, therefore a_n = 0.
Hence, alternative (A) is the correct choice.

(i) y[n] = x[2 – n] is non-causal
(ii) y[n] = x[n] cos ω₀ is causal
(iii) y[n] = sgn [x(n)] is non-causal
Which of the above statement is false?

1. 1
1. 2
1. 3
1. None of these

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(i) y[n] = x[2 – n] is non causal for any negative value of n.
(ii) y[n] = x[n] cos ω₀n is causal for any value of n.
(iii) y[n] = sgn [x(n)] is causal for any value of n.
Hence, alternative (C) is the correct choice.

Correct Option: C

(i) y[n] = x[2 – n] is non causal for any negative value of n.
(ii) y[n] = x[n] cos ω₀n is causal for any value of n.
(iii) y[n] = sgn [x(n)] is causal for any value of n.
Hence, alternative (C) is the correct choice.

The Fourier series expansion of a real periodic signal with fundamental frequency f 0 is given by
_{n = ∞}
g_P(t) = C_n.e^j2πnf₀t
^{n = – ∞}

It is given that C₃ = 3 + j5 then C_{– 3} is—

1. 5 + j3
1. – 3 – j5
1. – 5 + j3
1. 3 – j5

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If function is real (as given), then
x(t) = x*(t)
or x[t] = a_k = a*_{– k}
So, c_{– 3} = c₃* = 3 – j5
Hence, alternative (D) is the correct choice.

Correct Option: D

If function is real (as given), then
x(t) = x*(t)
or x[t] = a_k = a*_{– k}
So, c_{– 3} = c₃* = 3 – j5
Hence, alternative (D) is the correct choice.

For the signal x₁(t) = e^{– k₁t} u(t) and x₂(t) = e^{– k₂} u(t). their convolution will be—

1. [e^k₁t – e^{– k₂t}]
  [k₁ + k₂]
1. [e^k₁t – e^{– k₂t}]
  [k₂ – k₁]
1. [e^k₁t + e^{– k₂t}]
  [k₁ + k₂]
1. [e^k₁t + e^{– k₂t}]
  [k₂ + k₁]

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Given -
x₁(t) = e^k1^t u(t)
x₂(t) = e^–k2^t u(t)
x(t) = x₁(t)^* x₂(t) = x₁(s).x₂(s)
x₁(s) = 1
s – k₁

x₂(s) = 1
s – k₂

x₁(s).x₂(s) = 1 = x(s)
(s – k₁) (s + k₂)

By using partial fraction
x(s) = A + B
(s – k₁) (s + k₂)

x(s) = As + Ak₂ + Bs – Bk₂
(s – k₁) (s + k₂)

A + B = 0 or A = – B …(i)
Ak₂ – Bk₁ = 1 …(ii)
From equations (i) and (ii), we get
or
Ak₂ + Ak₁ = 1
or
A = 1
k₁ + k₂

and
B = - 1
k₁ + k₂

Now,
x(s) = 1 1 - 1
k₁ + k₂ s – k₁ s + k₁

By taking inverse Laplace transform
x(t) = 1 [e^k1^t– e– ^k2^t]
k₁ + k₂

Hence, alternative (A) is the correct choice.

Correct Option: B

Given -
x₁(t) = e^k1^t u(t)
x₂(t) = e^–k2^t u(t)
x(t) = x₁(t)^* x₂(t) = x₁(s).x₂(s)
x₁(s) = 1
s – k₁

x₂(s) = 1
s – k₂

x₁(s).x₂(s) = 1 = x(s)
(s – k₁) (s + k₂)

By using partial fraction
x(s) = A + B
(s – k₁) (s + k₂)

x(s) = As + Ak₂ + Bs – Bk₂
(s – k₁) (s + k₂)

A + B = 0 or A = – B …(i)
Ak₂ – Bk₁ = 1 …(ii)
From equations (i) and (ii), we get
or
Ak₂ + Ak₁ = 1
or
A = 1
k₁ + k₂

and
B = - 1
k₁ + k₂

Now,
x(s) = 1 1 - 1
k₁ + k₂ s – k₁ s + k₁

By taking inverse Laplace transform
x(t) = 1 [e^k1^t– e– ^k2^t]
k₁ + k₂

Hence, alternative (A) is the correct choice.

Which one of the following must be satisfied if a signal is to be periodic for – ∞ < t < ∞?

1. x(t + T₀) = x(t)
1. x(t + T₀) = dx(t)
  dt
1. x(t + T₀) = ^t₀ x(t)dt
  _t
1. x(t + T₀) = x(t) = kT₀

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NA

Correct Option: A

NA