Signal and systems miscellaneous Easy Questions and Answers | Page - 16

Signal and systems miscellaneous

Relation between a and b when a random variable has exponential pdf
f_x(x) = ae^{– b|x|}

1. ab = 1
1. 2a = b
1. a = 2
  b
1. ab = 2

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Given F_x(x) = ae^{– b|x|}
F_x(x)= ^∞ F_x(x)dx
_{– ∞}

1 = ^∞ ae^{– b|x|} .dx
_{– ∞}

or 1 = ⁰ ae^{+ bx} dx + ^∞ ae^{– bx}.dx
_{– ∞} ₀

or 1 = ae^bx + bx ⁰ + ae^{– bx} ^∞
b _{– ∞} – b ₀

or 1 = a – ae^{– b∞} + a e^{– b∞} + a · 1
b b – b a

or 1 = 2a
b

or 2a = b

Correct Option: B

Given F_x(x) = ae^{– b|x|}
F_x(x)= ^∞ F_x(x)dx
_{– ∞}

1 = ^∞ ae^{– b|x|} .dx
_{– ∞}

or 1 = ⁰ ae^{+ bx} dx + ^∞ ae^{– bx}.dx
_{– ∞} ₀

or 1 = ae^bx + bx ⁰ + ae^{– bx} ^∞
b _{– ∞} – b ₀

or 1 = a – ae^{– b∞} + a e^{– b∞} + a · 1
b b – b a

or 1 = 2a
b

or 2a = b

Value of p(x > a/2) in fig.

1. 1
  2
1. 1
  3
1. 1
  4
1. 1
  8

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P X > a = ^∞ f_x(x)dx
2 _a/2

1 = ^∞ – bx + 1 dx
_a/2 a

= – b x² + bx ^a
a 2 _a/2

= – b a² + ba + b (a/2)² – ba
a 2 a a 2

= – ab + ab + ab – ab
a 8 2

= ab = 1
8 8

Correct Option: D

P X > a = ^∞ f_x(x)dx
2 _a/2

1 = ^∞ – bx + 1 dx
_a/2 a

= – b x² + bx ^a
a 2 _a/2

= – b a² + ba + b (a/2)² – ba
a 2 a a 2

= – ab + ab + ab – ab
a 8 2

= ab = 1
8 8

The relation between a and b for pdf shown below—

1. ab = 1
1. ab = 0
1. ab = 2
1. ab = 3

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Fx(x) = b x + b; x < 0
a
– b x + b; x > 0
a

Now, Fx(x) = ^∞ f x(x)dωx
_{– ∞}

1 = ⁰ b x + b dx + b x + b dx
_{– a} a a

or 1 = b x² + bx ⁰ + – b x² + bx ^a
a 2 _{– a} a 2 ₀

or 1 = –ba² + ab – b a² + ab
2a a 2

or 1 = – ab + ab – ab + ab
2 2

or 1 = ab

Correct Option: A

Fx(x) = b x + b; x < 0
a
– b x + b; x > 0
a

Now, Fx(x) = ^∞ f x(x)dωx
_{– ∞}

1 = ⁰ b x + b dx + b x + b dx
_{– a} a a

or 1 = b x² + bx ⁰ + – b x² + bx ^a
a 2 _{– a} a 2 ₀

or 1 = –ba² + ab – b a² + ab
2a a 2

or 1 = – ab + ab – ab + ab
2 2

or 1 = ab

The value of p(2 < x < 3) in
fx(x) = a(x – 1); 1 ≤ x ≤ 4
0; otherwise

1. 1
  2
1. 1
  4
1. 1
  3
1. 1
  5

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P(2 < x < 3) = ³ a(x – 1)dx
₂

= ³ 2 · (x – 1)dx
₂ 9

= 2 x² – x ³
9 2 ₂

= 1
3

Correct Option: C

P(2 < x < 3) = ³ a(x – 1)dx
₂

= ³ 2 · (x – 1)dx
₂ 9

= 2 x² – x ³
9 2 ₂

= 1
3

The pdf for a random variable x is given
fx(x) = a(x – 1); 1 ≤ x ≤ 4
0; otherwise

then a is—

1. 1
  9
1. – 1
  9
1. 2
  9
1. – 2
  9

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Fx(n) = ^∞ Fx(x) dω
_{– ∞}

1 = ⁴ a(x – 1)dω
₁

1 = a x² – x ⁴
2 ₁

1 = 9 a
2

or a = 9
2

Correct Option: C

Fx(n) = ^∞ Fx(x) dω
_{– ∞}

1 = ⁴ a(x – 1)dω
₁

1 = a x² – x ⁴
2 ₁

1 = 9 a
2

or a = 9
2