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Signal demodulation

67 bytes added, 19:03, 3 November 2017
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<math display="block">i_i(t) = I_i\sin(\omega_i t + \varphi_{I_i})</math>
where <math>I_i</math> is the amplitude of the <math> i_{th}</math>emitting coil waveform, <math> \omega_i</math> is the excitation frequency and <math> \varphi_i$ </math> is the current phase relative to an arbitrary reference. Summing all N current waveforms results in:
<math display="block">i(t) = \sum^N_{i=1} I_i\sin(\omega_i t + \varphi_{I_{i}})</math>
<math display="block">v(t) = \sum^N_{i=1} V_i\sin(\omega_i t + \varphi_{V_{i}})</math>
where $<math>V_i$ </math> is the amplitude of the induced voltage component and <math> \varphi_{V_i}$ </math> is the associated phase. Each frequency component of the voltage signal is extracted using two reference signals:
<math display="block">Y_i = \sin(\omega_{ri} t)</math>
<math display="block">X_i = \cos(\omega_{ri} t)</math>
where $<math>\omega_{ri}$ </math> is the frequency of the simulated reference signal. This demodulation results in the amplitudes and phases of all the frequency components relative to the simulated reference signal as follows:
<math display="block">\mathbf{V} = [V_1,V_2...V_n]</math>
<math display="block">\mathbf{I} = [I_1,I_2...I_n]</math>
The sign of this phase information indicates the axial orientation of the electromagnetic sensor with respect to the magnetic field.
With a simulated reference signal it can be difficult to lock the frequency to the signal source without the use of phase locking techniques. In our system this often results in a small mismatch in frequency since the simulated reference signal for a particular coil <math> \omega_{ri}</math> is slightly different from the frequency to be demodulated <math> \omega_i</math>since they are not locked. This results in a <math> \Delta\omega=\omega_{ri} - \omega_r$ </math> causing a low frequency oscillation in the demodulated signal, which would not be present in synchronous demodulation.
To demonstrate this, consider a single frequency where the coil current and sensor voltage waveforms are given by:
<math display="block">X(t) = \cos(\omega_r t)</math>
Starting with the sensor voltage $<math>v(t)$</math>, we multiply by the reference signals just as in the synchronous case to produce:
<math display="block">v(t)Y(t) = \frac{V}{2}[\cos((\omega-\omega_r) + \varphi_V) - \cos((\omega+\omega_r) + \varphi_V)]</math>
<math display="block">v(t)X(t) = \frac{V}{2}[\sin((\omega-\omega_r) + \varphi_V) - \sin((\omega+\omega_r) + \varphi_V)]</math>
<math display="block">\Delta\omega = \omega-\omega_r</math>
These signals are close to DC since they oscillate at a frequency of <math> \Delta\omega$</math>. The amplitude <math> V$ </math> can be determined as in the synchronous case using:
<math display="block">V = 2\sqrt{(v_x^2 + v_y^2)}</math>
The phase can be determined using:
<math display="block">\gamma_V(t) = \arctan\frac{v_x}{v_y} = \Delta\omega t +\varphi_V</math>
Its clear that the phase has a time dependency. This is due to the frequency mismatch of the carrier and reference signals. In order to remove this dependency, the same demodulation procedure above is performed to the coil current waveform i(t) to produce <math> I</math> and <math> \gamma_I$</math>. Subtracting the phase component <math> \gamma_V</math> from <math> \gamma_I</math> gives the constant relative phase angle between the two waveforms:
<math display="block">\gamma_V - \gamma_I = \varphi_V-\varphi_I</math>

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