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.