# Digital Image Processing Mcqs

Q:

A) Mask that excludes diagonal neighbors has more sharpness than the masks that doesn’t | B) Mask that includes diagonal neighbors has more sharpness than the masks that doesn’t |

C) Both masks have same sharpness result | D) None of the mentioned |

Answer & Explanation
Answer: B) Mask that includes diagonal neighbors has more sharpness than the masks that doesn’t

Explanation: Including diagonal neighbor pixels enhances sharpness of the image. So, Mask that includes diagonal neighbors has more sharpness than the masks that doesn’t.

Explanation: Including diagonal neighbor pixels enhances sharpness of the image. So, Mask that includes diagonal neighbors has more sharpness than the masks that doesn’t.

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Q:

A) Because it contain high positive values | B) Because it contain high negative value |

C) Because it contain both positive and negative values | D) None of the mentioned |

Answer & Explanation
Answer: C) Because it contain both positive and negative values

Explanation: A Laplacian filtered image contain both positive and negative values of comparable magnitudes. So, scaling is necessary.

Explanation: A Laplacian filtered image contain both positive and negative values of comparable magnitudes. So, scaling is necessary.

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Q:

A) That the center spike would be negative | B) That the immediate neighbors of center spike would be positive. |

C) All of the mentioned | D) None of the mentioned |

Answer & Explanation
Answer: C) All of the mentioned

Explanation: For the above given enhanced image the Laplacian subtraction suggest that the center coefficient of Laplacian mask is negative and so the center spike is negative with its immediate neighbors being positive.

Explanation: For the above given enhanced image the Laplacian subtraction suggest that the center coefficient of Laplacian mask is negative and so the center spike is negative with its immediate neighbors being positive.

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Q:

A) Fourier transform pair notation | B) Laplacian |

C) Gradient | D) None of the mentioned |

Answer & Explanation
Answer: A) Fourier transform pair notation

Explanation: The Fourier transform of the Laplacian result in spatial domain is equivalent to multiplying the F(u, v) and H(u, v). This dual relationship is expressed as Fourier transform pair notation given by: ∇² f(x,y)-[(u – M/2)²+ (v – N/2)²]F(u,v), for an image of size M *N.

Explanation: The Fourier transform of the Laplacian result in spatial domain is equivalent to multiplying the F(u, v) and H(u, v). This dual relationship is expressed as Fourier transform pair notation given by: ∇² f(x,y)-[(u – M/2)²+ (v – N/2)²]F(u,v), for an image of size M *N.

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Q:

A) H(u, v)= -(u²+ v²) | B) H(u, v)= -(u²– v²) |

C) H(u, v)= -[(u – M/2)²+ (v – N/2)²]. | D) H(u, v)= -[(u – M/2)²– (v – N/2)²]. |

Answer & Explanation
Answer: C) H(u, v)= -[(u – M/2)²+ (v – N/2)²].

Explanation: The given operation f(x, y)(-1)x+y shifts the center transform so that (u, v)=(0, 0) is at point (M/2, N/2) and hence the filter is: H(u, v)= -[(u – M/2)²+ (v – N/2)²].

Explanation: The given operation f(x, y)(-1)x+y shifts the center transform so that (u, v)=(0, 0) is at point (M/2, N/2) and hence the filter is: H(u, v)= -[(u – M/2)²+ (v – N/2)²].

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Q:

A) (M -1, N -1) | B) (M/2, N/2) |

C) (M+1, N+1) | D) (0, 0) |

Answer & Explanation
Answer: B) (M/2, N/2)

Explanation: The given operation f(x, y)(-1)x+y shifts the center transform so that (u, v)=(0, 0) is at point (M/2, N/2) for F and f of same size M*N.

Explanation: The given operation f(x, y)(-1)x+y shifts the center transform so that (u, v)=(0, 0) is at point (M/2, N/2) for F and f of same size M*N.

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Q:

A) Resize the transform | B) Rotate the transform |

C) Shifts the center transform | D) All of the mentioned |

Answer & Explanation
Answer: C) Shifts the center transform

Explanation: The given operation f(x, y)(-1)x+y shifts the center transform so that (u, v)=(0,0) is at point (M/2, N/2) for F and f of same size M*N.

Explanation: The given operation f(x, y)(-1)x+y shifts the center transform so that (u, v)=(0,0) is at point (M/2, N/2) for F and f of same size M*N.

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Q:

A) H(u, v)= -(u²– v²) | B) H(u, v)= -(1) |

C) H(u, v)= -(u²+ v²) | D) none of the mentioned |

Answer & Explanation
Answer: C) H(u, v)= -(u²+ v²)

Explanation: Laplacian in frequency domain is: I[(∂² f(x,y))/∂x² +(∂² f(x,y))/∂y² ]= -(u²+v²)F(u,v), where ℑ is the Fourier transform operator and F(u, v) is Fourier transformed function of f(x, y) and -(u²+ v²) is the filter.

Explanation: Laplacian in frequency domain is: I[(∂² f(x,y))/∂x² +(∂² f(x,y))/∂y² ]= -(u²+v²)F(u,v), where ℑ is the Fourier transform operator and F(u, v) is Fourier transformed function of f(x, y) and -(u²+ v²) is the filter.

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