Q1Digital Image Processing
Question
Q.1. (a) For a 24-bit color image of size 5 inches by 6 inches, scanning is done at the rate of 300 dots per inches. Calculate the total number of bits required to represent the image. How much time is required to transmit the image if the available data rate is 100 kbps? [8]
(b) Discuss down sampling (or subsampling) and upsampling of an image with suitable example. [8]
Answer
For a 5x6 inch 24-bit color image scanned at 300 dpi, the total pixel count is 2,700,000, requiring 64,800,000 bits (7.725 MB), which takes 648 seconds (10.8 minutes) to transmit at 100 kbps; down sampling and up sampling are explained with an example showing loss of detail on the round trip.
(a) Numerical: Bits Required and Transmission Time
We are given an image of physical size 5 inches by 6 inches, scanned at a resolution of 300 dots per inch (dpi), with 24 bits used per pixel (24-bit color). We must find the total number of bits required to represent the image and the time needed to transmit it at a data rate of 100 kbps.
Step 1: Compute pixel dimensions
The width in pixels is obtained by multiplying the physical width in inches by the scanning resolution in dots per inch, and similarly for height.
Step 2: Compute total number of pixels
Step 3: Compute total bits
Since each pixel is represented using 24 bits (for 24-bit color), the total number of bits is obtained by multiplying the total pixel count by 24.
Expressed in bytes, this is 64,800,000 / 8 = 8,100,000 bytes, which is approximately 7.725 MB (using 1 MB = 1,048,576 bytes), or about 8.1 MB if using the decimal definition of 1 MB = 1,000,000 bytes. The commonly quoted figure using binary MB conventions is approximately 7.725 MB.
Step 4: Compute transmission time
The available data rate is 100 kbps, i.e., 100,000 bits per second. The transmission time is obtained by dividing the total number of bits by the data rate in bits per second.
Converting 648 seconds into minutes: 648 / 60 = 10.8 minutes. Therefore, the total number of bits required to represent this image is 64,800,000 bits (approximately 7.725 MB), and the time required to transmit this image over a 100 kbps channel is 648 seconds, which is equal to 10.8 minutes.
- Width in pixels = 5 inches x 300 dpi = 1500 pixels
- Height in pixels = 6 inches x 300 dpi = 1800 pixels
- Total pixels = 1500 x 1800 = 2,700,000 pixels
- Total bits = 2,700,000 x 24 bits/pixel = 64,800,000 bits (approx. 7.725 MB)
- Transmission time = 64,800,000 bits / 100,000 bps = 648 seconds = 10.8 minutes
(b) Down Sampling and Up Sampling
Down sampling, also called subsampling, reduces the spatial resolution of an image by keeping only every Nth pixel, row, or column and discarding the rest, thereby reducing the data volume (an image down sampled by factor 2 in each dimension retains only one quarter of the original pixel count). It is used in building image pyramids, generating thumbnails, and reducing computation for coarse-to-fine processing pipelines. Before down sampling, the image must be passed through an anti-alias low-pass filter to remove high-frequency content that cannot be represented at the reduced sampling rate; skipping this step causes aliasing artifacts such as jagged edges or moire patterns to appear, as governed by the Nyquist sampling criterion.
Up sampling increases the spatial resolution of an image by inserting new pixels between existing ones, with the values of these new pixels estimated through interpolation, such as nearest-neighbor (copying the closest known pixel, fast but blocky), bilinear (weighted average of the four nearest pixels, smoother), or bicubic (weighted average of sixteen nearest pixels using a cubic kernel, smoothest and sharpest of the three). Crucially, up sampling can only estimate plausible values for the new pixels; it can never recover the true fine detail that was permanently discarded during an earlier down sampling step, so the up sampled result is always inherently smoother/blurrier than a genuine full-resolution capture.
As an example, consider a 4x4 image down sampled to 2x2 by retaining only pixels at even row and column indices; this discards 12 of the original 16 pixel values (75 percent of the data). Up sampling this 2x2 result back to 4x4 using bilinear interpolation inserts three new rows and columns whose values are computed as distance-weighted averages of the four surviving corner pixels; the reconstructed image resembles the coarse structure of the original but appears visibly blurred and permanently lacks the fine detail present in the rows and columns that were discarded during down sampling, demonstrating that the down-sample/up-sample round trip is fundamentally lossy.
It is worth connecting this concept back to the numerical calculation performed in part (a): the scanned 1500x1800 pixel, 24-bit image occupies 64,800,000 bits and takes 648 seconds to transmit at 100 kbps, which may be impractically slow for some applications such as live preview over a slow link. Down sampling that same image by a factor of 2 in each dimension (to 750x900 pixels) would reduce the total pixel count to one quarter (675,000 pixels), reducing the total bit count and transmission time proportionally to roughly 162 seconds, at the cost of losing three quarters of the spatial detail; this illustrates the practical trade-off that motivates down sampling in bandwidth- or storage-constrained transmission and storage scenarios, where a lower-resolution preview or thumbnail is transmitted quickly, and the full-resolution original is only transmitted afterward if genuinely required.
Nearest-neighbor, bilinear, and bicubic interpolation, used during up sampling, differ not only in visual smoothness but also in computational cost: nearest-neighbor requires no arithmetic beyond an index lookup and is therefore the fastest, bilinear requires a weighted sum of 4 neighboring pixels, and bicubic requires a weighted sum of 16 neighboring pixels using cubic polynomial weights, making it the most computationally expensive of the three but also the one producing the smoothest, least blocky results with the fewest visible artifacts along diagonal edges.
This calculation illustrates why uncompressed digital image and video data volumes can become extremely large very quickly - a single 5x6 inch photograph scanned at a modest 300 dpi already requires nearly 65 million bits (about 7.7 megabytes) to represent in raw, uncompressed 24-bit color form, and the resulting transmission time of nearly 11 minutes over a relatively slow 100 kbps link (typical of older dial-up or low-bandwidth mobile connections) directly motivates the widespread practical use of image compression techniques (both lossless and lossy, as discussed elsewhere in this paper) to reduce this data volume to a more manageable size for storage and transmission, particularly for higher-resolution images, video (many images per second), or bandwidth-constrained transmission channels such as low-bandwidth wireless or satellite links.
It is also worth noting that this calculation assumes a simple, uncompressed representation with a fixed number of bits per pixel (24 bits for full RGB color, 8 bits per color channel) - real imaging and scanning systems may use different color depths (such as 8-bit grayscale, 16-bit per channel for high dynamic range imaging, or various color subsampling schemes such as YCbCr 4:2:0 used in JPEG and video compression, which allocate fewer bits to the chrominance/color-difference channels than to luminance, exploiting the human visual system's lower sensitivity to color detail compared to brightness detail), each of which would change the total bit count and hence the transmission time calculation accordingly.