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.