Converting dBV to dBSPL

Question

I am calibrating my microphone system consisting of a microphine whose signal is amplified and then converted from analog to digital with an ADC. The ADC converts the data to a 16-bit integer.

The calibration microphone produces 94 dBSPL at 1 kHz. I am sampling with 192 kHz, then plotting the FFT of that signal to find the given dBV at 1kHz.

First, converting the unit-free signal to V using:

calibration_94dB_1kHz = signal * (1 / 32768)

Then applying a window:

N = len(calibration_94dB_1kHz)
s = signal.windows.flattop(N) * calibration_94dB_1kHz

Then calculating the one-sided FFT with:

X = 2 * np.abs(np.fft.rfft(s))/N  # normalize the fft by the window-length and multiply by 2 since we are discarding the negative frequencies of the FFT
X_freq = np.fft.rfftfreq(N, 1.0 / 192000)

The data needs then to be plotted logarithmic so:

X_dbV = 20*np.log10(X)  # get fft in dBV

When plotting the data with:

axs[1, 0].clear()
axs[1, 0].semilogx(X_freq, X_dbV, color=colors[1], label='Frequency response |S(jw)| detrended and windowed')
axs[1, 0].grid(True)
axs[1, 0].legend(loc='upper left')

I get a peak at approximately 1 kHz that is -48.3 dBV. How can I convert this value to dBSPL if the mic has a sensitivity of -38 dBV +- 1, which is tested from the manufacturer with 1 kHz at 94 dBSPL?

EDIT: Here is the updated code:

calibration_94dB_1kHz_detrended = signal.detrend(data=calibration_94dB_1kHz, type='constant')
s = calibration_94dB_1kHz_detrended
Gain = 1 + 100 / 4.7
calibration_in_rms_Vpp = s / (2*np.sqrt(2))  # assuming p-p sinusoidal waveform
sensitivity = -38  # dBV

# Apply window
N = len(calibration_in_rms_Vpp)
s = signal.windows.flattop(N) * calibration_in_rms_Vpp
X = 2 * np.abs(np.fft.rfft(s))/N  # normalize the fft by the window-length and multiply by 2 since we are discarding the negative frequencies of the FFT
X_freq = np.fft.rfftfreq(N, 1.0 / 192000)
X_dbV = 20*np.log10(X)  # dBVrms
X_dbSPL = X_dbV - Gain - sensitivity + 94  # dBSPL

Answer 1

I'm not sure what to make of "The calibration microphone produces 94 dBSPL at 1 kHz." The microphone is not being used as a projector, thus it will produce some sort of voltage output, not a sound pressure output as a projector would.

Sound-Pressure Level (SPL) in air, is normally given in \$dB \; re \; 20\; \mu Pa\$, also written as \$dB/\!\!/20\; \mu Pa\$ in some literature.

Microphone sensitivity is normally given in \$ dB \; re \; 1V/20 \mu Pa \$, also can be written as \$ dBV \; re \;20 \mu Pa\$. The sensitivity number will look something like -45 dBV re 20 µPa (note: the sensitivity number is normally a negative number).

The 94 dB thing you refer to relates SPL to \$ 1 \; Pa \$, i.e. \$ 93.98 = {20 log({1 \over 20e-6})} \$.
You do not need to include the number 94 in your calculations unless you're trying to redefine SPL relative to one Pascal.

When measuring SPL at a single frequency it is customary to use \$V_{rms}\$.

$$ (SPL\_in\_dB/\!\!/20 \mu Pa) = \\ {(Output\_Voltage\_in\_dBV) - (Amplifier\_Gain\_in\_dB) - (microphone\_sensitivity\_in\_dBV/\!\!/20 \mu Pa)} $$

When measuring the dB value in the FFT, verify the reported number by inputting a reference signal at a known amplitude (rms value) and see that the FFT reports the voltage you expect, i.e., just do a simple amplitude measurement.