Answer

Correct option: (ii) Extract meaningful insights

Explanation

Data processing converts raw acquired data into useful information that can be interpreted or used for decision-making.

Typical processing operations include:

• Data cleaning and removal of invalid values.
• Noise reduction and filtering.
• Calibration and scaling.
• Averaging and statistical calculations.
• Feature extraction.
• Conversion into engineering units.
• Preparation of data for visualization or control.

For example, raw vibration samples can be processed to calculate RMS vibration, peak amplitude or frequency components, helping identify a machine fault.

Therefore, the correct answer is (ii).

Answer

Correct option: (iv) Transformer

Explanation

Embedded systems commonly use different types of memory and storage devices, including:

• EEPROM: Non-volatile memory used for storing configuration data and calibration values.
• SRAM: Fast volatile memory used for variables and temporary runtime data.
• SD card: Removable non-volatile storage suitable for large amounts of logged data.

A transformer is an electrical power/isolating component. It is not a data-storage device.

Therefore, the correct answer is (iv).

Solution: Complete Data Processing Cycle

Definition

Data processing converts raw acquired samples into meaningful information that can be stored, displayed, transmitted or used for control.

Processing Cycle

Data Acquisition
      ↓
Data Validation
      ↓
Pre-processing / Filtering
      ↓
Calibration & Scaling
      ↓
Feature Extraction / Computation
      ↓
Analysis
      ↓
Storage / Display / Communication / Control

1. Data Acquisition

Sensors and ADCs generate digital samples at a selected sampling rate.

2. Data Validation

The system checks for missing, out-of-range or corrupted values.

3. Pre-processing

Noise may be reduced using averaging or digital filters. Signals can also be normalized or resampled when appropriate.

4. Calibration and Scaling

ADC codes are converted into engineering units such as °C, V, Pa or rpm.

5. Feature Extraction

Useful quantities are calculated from raw samples, such as peak, RMS, mean, frequency or heart rate.

For a set of $N$ samples $x_i$, the mean is:

$$\bar{x}=\frac{1}{N}\sum_{i=1}^{N}x_i$$

6. Analysis

Statistical, frequency-domain or trend analysis can be applied depending on the application.

7. Storage and Output

Processed or raw data can be saved to flash, EEPROM, SD card or a computer and displayed or transmitted.

Real-Time Considerations

The processing time must be sufficiently small compared with the sampling interval. Buffering and interrupts/DMA can help prevent sample loss.

Conclusion

A complete data-processing cycle transforms raw sensor samples into reliable, useful information through validation, filtering, calibration, computation and analysis.

Solution: Data Storage Options and Data Logging

Data Storage Options

Storage	Nature	Typical use
SRAM	Volatile, fast	Temporary buffers
EEPROM	Non-volatile	Calibration/configuration data
Internal Flash	Non-volatile	Firmware and small datasets
External Flash	Non-volatile	Larger embedded storage
SD card	Removable, high capacity	Continuous data logging
USB/Computer	External	Long-term transfer/storage

Data Logging Process

Sensor → ADC → MCU → RAM Buffer → Storage Interface → File/Memory

Steps

1. Initialize the sensor and ADC.
2. Configure the sampling timer.
3. Acquire a sample at every sampling instant.
4. Convert the ADC code into the required format.
5. Add a timestamp or channel identifier when necessary.
6. Store samples in a RAM buffer.
7. When the buffer is sufficiently full, write a block to non-volatile storage.
8. Repeat until logging ends.

Why Buffering is Important

Storage devices such as SD cards may have variable write latency. A RAM buffer allows acquisition to continue while data is being written in blocks.

Example Pseudocode

while (logging)
{
    sample = read_adc();
    buffer[index++] = sample;

    if (index == BUFFER_SIZE)
    {
        storage_write(buffer, BUFFER_SIZE);
        index = 0;
    }
}

A real implementation should also handle write failures, buffer overflow, timestamps, file-system errors and safe shutdown.

Storage Selection

• Use EEPROM for small calibration/configuration data.
• Use Flash for firmware and moderate non-volatile data.
• Use SD cards for large continuous logs.
• Use computer/USB/network storage when very large datasets must be transferred or archived.

Conclusion

Efficient data logging depends on selecting suitable storage, using buffering and ensuring that the sustained storage throughput is greater than the incoming data rate.

Solution: Vibration Data Storage Calculation

Given:

• Sampling rate = 10 kSamples/s = $10,000$ samples/s
• Sample size = 16 bits = 2 bytes/sample
• Duration = 5 minutes = 300 s

(i) Minimum Memory Required

Data rate:

$$10,000\times16=160,000\text{ bits/s}$$

Convert to bytes/s:

$$\frac{160,000}{8}=20,000\text{ bytes/s}$$

For 300 seconds:

$$20,000\times300=6,000,000\text{ bytes}$$

Therefore:

$$\boxed{6,000,000\text{ bytes}\approx6\text{ MB}}$$

Using binary units, this is approximately:

$$\frac{6,000,000}{1,048,576}\approx5.72\text{ MiB}$$

So at least 6 MB decimal storage is required, excluding filesystem overhead and metadata.

(ii) Suitable Interface

The raw required throughput is:

$$20,000\text{ bytes/s}=20\text{ kB/s}$$

or:

$$160\text{ kbps}=0.16\text{ Mbps}$$

SPI

SPI is the best choice among the given options for a typical embedded data logger because it provides high throughput with simple hardware and is widely supported by SD cards and external memory.

Even a modest SPI clock of 1 MHz can theoretically transfer:

$$1\text{ Mbit/s}=125\text{ kB/s}$$

which is well above the required 20 kB/s, leaving margin for protocol and storage overhead.

I²C

I²C is convenient for sensors and EEPROMs, but its lower typical speed makes it less attractive for continuous high-rate storage.

Parallel

Parallel interfaces can provide high throughput but require more pins and hardware complexity.

Final Answer

• Minimum data storage: $6,000,000$ bytes ≈ 6 MB
• Raw data rate: 0.16 Mbps = 20 kB/s
• Recommended interface: SPI, because it easily provides sufficient bandwidth with low pin count and is well suited to SD-card/external-memory logging.

Solution: Multichannel DAQ Data Rate and SD-Card Framing

Given:

• Number of channels = 4
• Sampling rate per channel = 25 kS/s
• ADC resolution = 12 bits/sample
• SPI rate to SD card = 12 Mbps
• Sector size = 512 bytes

(i) Raw Data Rate

Total samples per second across all channels:

$$4\times25,000=100,000\text{ samples/s}$$

Each sample contains 12 bits, so:

$$R_{raw}=100,000\times12=1,200,000\text{ bits/s}$$

Therefore:

$$\boxed{R_{raw}=1.2\text{ Mbps}}$$

In bytes/s:

$$\frac{1,200,000}{8}=150,000\text{ bytes/s}$$

So:

$$\boxed{R_{raw}=0.15\text{ MB/s}}$$

Account for packing

Because SD-card writes are byte-addressable, 12-bit samples cannot simply occupy 12 bits independently unless the firmware packs them.

A convenient packing scheme stores two 12-bit samples in 3 bytes:

$$2\times12=24\text{ bits}=3\text{ bytes}$$

Thus efficient packing produces exactly the theoretical raw rate of 0.15 MB/s, excluding filesystem and protocol overhead.

The available SPI bandwidth is:

$$12\text{ Mbps}=1.5\text{ MB/s}$$

Therefore the nominal raw data rate is only:

$$\frac{0.15}{1.5}=10\%$$

of the stated SPI link capacity, leaving substantial bandwidth margin.

(ii) 512-Byte Sector Framing

A 512-byte sector contains:

$$512\times8=4096\text{ bits}$$

With 12-bit samples, the maximum number of complete samples is:

$$\left\lfloor\frac{4096}{12}\right\rfloor=341\text{ samples}$$

341 samples use:

$$341\times12=4092\text{ bits}=511.5\text{ bytes}$$

So one sector can contain 341 complete 12-bit samples if the packing is bit-level, leaving:

$$4096-4092=4\text{ bits}$$

unused.

However, a cleaner byte-aligned scheme is to pack 340 samples:

$$340\times12=4080\text{ bits}=510\text{ bytes}$$

leaving exactly 2 bytes for metadata, a timestamp, frame counter, CRC, or padding.

Recommended framing

Use 340 samples per 512-byte sector and reserve 2 bytes for metadata, or use a 341-sample bit-packed payload with 4 unused bits plus a defined header/trailer.

At 100,000 samples/s:

$$\frac{100,000}{340}=294.12\text{ sectors/s}$$

Thus the system needs approximately:

$$\boxed{295\text{ sectors/s}}$$

with 340 samples/sector, ignoring additional metadata.

For maximum packing efficiency using 341 samples/sector:

$$\frac{100,000}{341}=293.26\text{ sectors/s}$$

so approximately 294 sectors/s are required.

Final Answer

• Raw data rate: 1.2 Mbps = 0.15 MB/s
• SPI capacity: 12 Mbps = 1.5 MB/s
• Maximum complete 12-bit samples in 512 bytes: 341 samples
• Efficient practical framing: 340 samples + 2 bytes metadata/padding
• Sectors/s: approximately 295 sectors/s for 340-sample frames, or 294 sectors/s with 341-sample bit packing.

The stated SPI bandwidth is comfortably sufficient for the raw acquisition rate.