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Justify/contradict following statements: Reduction in spatial resolution results in checker board degradation.

Mumbai University > Electronics > Sem 7 > Digital image processing

Marks: 5 M

Year: Dec 2013

1 Answer
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Higher the spatial resolution of the image, greater is the sampling rate i.e. the lower is the image area Δx Δy represented by each sampled point. Similarly, higher the grey level resolutions more are the number of quantized levels. Hence spatial resolution gives us an induction of the sampling while grey level resolution gives quantization.

Spatial resolution  Sampling

Grey level resolution Quantization

We have already stated that an image can be considered as a 2-D array. Image f(x,y) is arranged in the form of NxM array.

$$f (x, y) = \left[ \begin{array}{cccc} f(0,0) & f(0,1) & ..... & f(0, M -1) \\ f(1, 0) & f(1, 1) & ..... & f(1, M -1) \\ - \\ - \\ - \\ f(N - 1, 0) & f(N -1, 1) & & f(N-1, M-1) \end{array} \right]$$

Hence every image that is seen on the monitor is actually this matrix. Each element of the matrix is called a pixel. Whenever we see image on the computer it is actually the matrix which consists of N x M pixels each pixel is considered to be a sample. Hence more the pixels, more the samples, higher the sampling rate and hence better the spatial resolution. The value of each pixel is known as grey level. The computer understands only ones and zeros. Hence these grey levels need to be represented in the form of zero sand ones. If we have two bits to represent grey levels (2^2) can be defined viz. 00,01,10,11 where 00 is black, 11 is white and the other two are shades of grey. Similarly, if we have 8 bits to represent the grey levels, we will have 256 shades i.e. grey levels. Hence more the bits, more are the grey levels and better is the total clarity. The total size of the image is N x M x m, where m is the number of bits used.

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