Question

The table below shows the average values of the power spectral density (PSD) of an EEG signal in different frequency bands.

Frequency bands (Hertz)	Average PSD (microvolts2/Hertz)
0.4 - 4.0	3
4.0 - 8.0	4
8.0 - 13.0	8
13.0 - 30.0	5
30.0 - 100.0	3

The PSD of the EEG signal has a noise floor of 2 microvolts²/Hertz. The signal to noise ratio for the alpha-band, expressed in decibels, is

• A. 1.76
• B. 3.01
• C. 6.02
• D. 3.98

Hint:

Alpha band is 8-13 Hz; SNR(dB) = 10 log10(signal PSD / noise floor) = 10 log10(8/2) = 6.02 dB.

Answer 1

The Correct Option is C

Step 1: Identify the alpha band.
EEG signals are grouped into standard frequency bands. The alpha band lies between 8 Hz and 13 Hz.
From the table, the average PSD for the 8.0-13.0 Hz band is \(8\) microvolts\(^2\)/Hertz.

Step 2: State the noise floor.
The question gives the noise floor of the EEG signal as \(2\) microvolts\(^2\)/Hertz. This is the background noise power level at every frequency, including inside the alpha band.

Step 3: Set up the SNR formula.
Signal to noise ratio (SNR) compares signal power to noise power. In decibels, for power quantities,
\[ SNR_{dB} = 10\log_{10}\left(\frac{P_{signal}}{P_{noise}}\right) \]
Here \(P_{signal}\) is the average PSD recorded in the alpha band, \(8\) microvolts\(^2\)/Hertz, and \(P_{noise}\) is the noise floor, \(2\) microvolts\(^2\)/Hertz.

Step 4: Compute the ratio.
\[ \frac{P_{signal}}{P_{noise}} = \frac{8}{2} = 4 \]

Step 5: Convert to decibels.
\(\log_{10}(4) = \log_{10}(2^2) = 2\log_{10}(2) = 2 \times 0.301 = 0.602\).
\[ SNR_{dB} = 10\log_{10}(4) = 10 \times 0.602 = 6.02 \text{ dB} \]

Step 6: Eliminate the wrong options.
Option (A) \(1.76\) does not correspond to any clean power ratio from this data.
Option (B) \(3.01\) is \(10\log_{10}(2)\), the dB value for a ratio of \(2\), not \(4\).
Option (D) \(3.98\) is close to the ratio \(4\) itself, not its dB value, so it confuses the ratio with the answer.

Final Answer:
The SNR for the alpha band is \(6.02\) dB.
\[ \boxed{6.02 \text{ dB}} \]