CN102169273A - Method for carrierless demodulation of output signal of asymmetrical 3*3 coupler - Google Patents

Abstract

The invention discloses a method for carrierless demodulation of an output signal of an asymmetrical 3*3 coupler. By the method, the output signal of the asymmetrical 3*3 coupler of an optical fibre acceleration sensor can be demodulated into a signal which is in a linear relationship with an acceleration signal. The method comprises the following steps of: firstly, obtaining the output signal of the asymmetrical 3*3 coupler by an M-Z interference system which consists of a light source, a 1*2 coupler, a sensor, a 3*3 coupler and a photoelectric detector, and demodulating the signal; secondly, removing a direct current component, carrying out core demodulation to obtain a result N, eliminating influence of an alternating current gain on a demodulation result to obtain a result D, and dividing the N by the D to obtain a result Z; and carrying out integration on the Z by an integrator to obtain a final phase signal. Compared with the conventional method for the carrierless demodulation of the output signal of the asymmetrical 3*3 coupler, the method provided by the invention has the advantages that: a requirement on the symmetry of a light path of a system is reduced; process is easier to implement; the more common problem of demodulation of the output signal of the asymmetrical 3*3 coupler is solved; and the method has quite high real significance and a use value.

Description

Carrier-free demodulation method for output signal of asymmetric 3 x 3 coupler

Technical Field

The invention belongs to the technical field of optical fiber sensing, and particularly relates to a carrier-free demodulation method for an output signal of an asymmetric 3 x 3 coupler.

Background

The interference type optical sensor can achieve high sensitivity and can be used for detecting weak physical quantity. In practical applications, the measured signal must be extracted automatically and linearly by appropriate signal processing techniques. At present, the phase detection methods which are applied more are phase generation carrier algorithms, including a homodyne method, a synthetic heterodyne method, an alternating current phase tracking method and the like. In comparison, the output signal of the 3 × 3 coupler method is reliable and stable, and carrier modulation is not required to be introduced into the optical path part, so that the frequency of the system modulated signal is reduced, the sampling frequency of the system can be reduced, and the design of a digital demodulation system is facilitated; meanwhile, the optical path part is relatively simple, and the research and development of the integrated optical sensor are facilitated.

The existing symmetrical 3 x 3 coupler output carrier-free demodulation method is carried out on the basis of the symmetrical design of a sensing system. However, due to the reasons of technology and the like, the sensing system is difficult to achieve ideal strict symmetry, so the existing symmetrical 3 × 3 coupler output carrier-free demodulation method has high requirements on the symmetry of the 3 × 3 coupler and the performances of the PIN and the preamplifier, and is not beneficial to wide application in practice

Disclosure of Invention

The invention aims to solve the problems that the existing symmetrical 3 x 3 coupler output carrier-free demodulation method has high requirements on the symmetry of the 3 x 3 coupler and the performance of PIN and a preamplifier, and is not beneficial to popularization and application, and provides a carrier-free demodulation method for an output signal of an asymmetrical 3 x 3 coupler.

The invention provides a demodulation method of an output signal of an asymmetric 3 x 3 coupler, which comprises the following steps:

1 st, asymmetric 3 x 3 coupler output signal u₁，u₂，u₃Obtained by

Obtaining an asymmetric 3 x 3 coupler output signal by an M-Z interferometric system comprising a light source, a 1 x 2 coupler, a sensor, a 3 x 3 coupler, and a photodetector, comprising:

1.1, generating a coherent light beam by a light source;

1.2, dividing light generated by a light source into two beams by a 1 x 2 coupler, wherein one beam is used as reference light and sent to a reference arm, and the other beam is used as signal light and sent to a signal arm;

1.3, the signal arm is deformed by the conversion of the external signal to be detected, so that the light in the signal arm is subjected to phase change and is subjected to phase difference with the light in the reference arm;

1.4, the two beams of light with the phase difference generated in the step 1.3 pass through a 3 × 3 coupler to generate three phase modulation light signals, and the three phase modulation light signals are received by three photodetectors and converted into three electrical signals u₁，u₂，u₃

u₁＝C₁+B₁ cosφ

u₂＝C₂+B₂ cos(φ-2π/3-δ₁)

u₃＝C₃+B₃ cos(φ+2π/3+δ₂)

Wherein, C_i，(i＝1，2，3)，δ₁，δ₂Is a constant related to the optical signal, C₁，C₂，C₃Respectively the DC component, delta, of the three signals₁And delta₂Is phase, B_i(i ═ 1, 2, 3) is the amplification factor of the preamplifier, and C is₁≠C₂≠C₃，B₁≠B₂≠B₃，δ₁≠δ₂；

2 nd, asymmetric 3 x 3 coupler output signal u₁，u₂，u₃Demodulation of

2.1 removing DC component

Using the formula a_i＝3u_i-k_iu_cI-1, 2, 3 removes the corresponding asymmetry3 x 3 coupler output signal u₁，u₂，u₃D.c. component in (a) to obtain a₁，a₂，a₃Wherein u is_cIs u₁，u₂，u₃Sum u of three signals_c＝u₁+u₂+u₃，k₁，k₂，k₃Respectively as follows:

${\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">k</figure-callout>}}_1=\frac{3C_1}{{\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">C</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="C"\; filenames="HDA0000055526010000012.png,HDA0000055526010000013.png"\; state="\{\{state\}\}">C</figure-callout>}}_2+{\mathrm{<figure-callout\; id="3"\; label="C"\; filenames="HDA0000055526010000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_3}$

${\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="HDA0000055526010000012.png,HDA0000055526010000013.png"\; state="\{\{state\}\}">k</figure-callout>}}_2=\frac{3C_2}{{\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">C</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="C"\; filenames="HDA0000055526010000012.png,HDA0000055526010000013.png"\; state="\{\{state\}\}">C</figure-callout>}}_2+{\mathrm{<figure-callout\; id="3"\; label="C"\; filenames="HDA0000055526010000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_3}$

${\mathrm{<figure-callout\; id="3"\; label="k"\; filenames="HDA0000055526010000022.png"\; state="\{\{state\}\}">k</figure-callout>}}_3=\frac{3C_3}{{\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">C</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="C"\; filenames="HDA0000055526010000012.png,HDA0000055526010000013.png"\; state="\{\{state\}\}">C</figure-callout>}}_2+{\mathrm{<figure-callout\; id="3"\; label="C"\; filenames="HDA0000055526010000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_3}$

C₁，C₂，C₃same as in 1.4;

2.2, removing the signal a of the DC component in the step 2.1₁，a₂，a₃Performing core demodulation

For three paths of signals a with direct current components removed₁，a₂，a₃Time differentiation，

i is 1, 2, 3, and three electric signals d are generated₁，d₂，d₃And making a difference between two of them, then making a difference with a₁，a₂，a₃Multiplying respectively to obtain:

N＝a₁(d₃-d₂)+a₂(d₁-d₃)+a₃(d₂-d₁)；

2.3 eliminating the influence of AC gain on the demodulation result

In the asymmetric case, the sum of the squares of the AGC block inputs is

<math><mrow><msubsup><mi>a</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>2</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>3</mn><mn>2</mn></msubsup><mo>=</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><mo>&CenterDot;</mo><mo>[</mo><msubsup><mi>B</mi><mn>1</mn><mn>1</mn></msubsup><mo>+</mo><msubsup><mi>B</mi><mn>2</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>B</mi><mn>3</mn><mn>2</mn></msubsup><mo>+</mo><msub><mi>W</mi><mn>2</mn></msub><mi>cos</mi><mrow><mo>(</mo><mn>2</mn><mi>&phi;</mi><mo>-</mo><msub><mi>&Lambda;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>]</mo></mrow></math>

In the formula W₂And Λ₂And an AC gain B₁，B₂，B₃And delta₁And delta₂It is related. When B is present₁，B₂，B₃When the following formula is satisfied, W₂Is equal to 0, thereby

The sum of the two is a constant,

<math><mrow><mfrac><msub><mi>B</mi><mn>2</mn></msub><msub><mi>B</mi><mn>1</mn></msub></mfrac><mo>=</mo><msub><mi>&xi;</mi><mn>1</mn></msub><mo>=</mo><msup><mrow><mo>[</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>+</mo><mi>sin</mi><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub></mrow><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>-</mo><mi>sin</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></mfrac><mo>]</mo></mrow><mrow><mn>1</mn><mo>/</mo><mn>2</mn></mrow></msup></mrow></math>

<math><mrow><mfrac><msub><mi>B</mi><mn>3</mn></msub><msub><mi>B</mi><mn>1</mn></msub></mfrac><mo>=</mo><msub><mi>&xi;</mi><mn>2</mn></msub><mo>=</mo><msup><mrow><mo>[</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mi>sin</mi><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub></mrow><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>-</mo><mi>sin</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></mfrac><mo>]</mo></mrow><mrow><mn>1</mn><mo>/</mo><mn>2</mn></mrow></msup></mrow></math>

in the AGC block of FIG. 1, the gain is adjusted by adjusting the amplifier coefficient g₁，g₂，g₃To pair

Is adjusted so that

<math><mrow><msub><mi>g</mi><mn>2</mn></msub><msubsup><mi>B</mi><mn>2</mn><mn>2</mn></msubsup><mo>/</mo><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>=</mo><msubsup><mi>&xi;</mi><mn>1</mn><mn>2</mn></msubsup></mrow></math>

<math><mrow><msub><mi>g</mi><mn>3</mn></msub><msubsup><mi>B</mi><mn>3</mn><mn>2</mn></msubsup><mo>/</mo><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>=</mo><msubsup><mi>&xi;</mi><mn>2</mn><mn>2</mn></msubsup></mrow></math>

The output result of the AGC block is

<math><mrow><mi>D</mi><mo>=</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><mo>&CenterDot;</mo><mo>[</mo><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msub><mi>g</mi><mn>2</mn></msub><msubsup><mi>B</mi><mn>2</mn><mn>2</mn></msubsup><mo>+</mo><msub><mi>g</mi><mn>3</mn></msub><msubsup><mi>B</mi><mn>3</mn><mn>2</mn></msubsup><mo>]</mo></mrow></math>

<math><mrow><mo>=</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>[</mo><mn>1</mn><mo>+</mo><msubsup><mi>&xi;</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>&xi;</mi><mn>2</mn><mn>2</mn></msubsup><mo>]</mo></mrow></math>

And 2.4, dividing the output result N of the step 2.2 and the output result D of the step 2.3 to obtain a result Z, and integrating the result Z by an integrator to obtain a final phase signal.

Demodulation principle of the invention

First, with the M-Z interference system shown in fig. 5, the phase of the light in the signal arm changes due to the change in the external signal (being measured), thereby creating a phase difference with the light in the reference arm. Two beams of coherent light enter a 3 x 3 coupler, generate interference phenomenon in the coupler, are output from three output ends of the 3 x 3 coupler, are respectively received by three connected photoelectric detectors, and convert optical signals into corresponding three paths of electric signals u proportional to light intensity₁，u₂，u₃For subsequent processing by the demodulation circuit.

II, demodulation principle:

1. removing DC part of three-way output of coupler

In the asymmetric case, the 3 × 3 coupler in the M-Z interferometric system shown in FIG. 5 is used to obtain a signal with a DC component C₁，C₂，C₃Three output signals of

u₁＝C₁+B₁ cosφ

u₂＝C₂+B₂ cos(φ-2π/3-δ₁)

u₃＝C₃+B₃ cos(φ+2π/3+δ₂)

Wherein, C_i，B_i，δ₁，δ₂(i ═ 1, 2, 3) is a constant as described above. C₁≠C₂≠C₃，B₁≠B₂≠B₃，δ₁≠δ₂。

The sum of the three output signals is

u₁+u₂+u₃＝{C₁+C₂+C₃+W₁ cos[φ(t)-Δ₁]}

Wherein,

${\mathrm{<figure-callout\; id="1"\; label="W"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">W</figure-callout>}}_1=\sqrt{{\mathrm{<figure-callout\; id="1"\; label="U"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">U</figure-callout>}}_1^2+{\mathrm{<figure-callout\; id="1"\; label="V"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">V</figure-callout>}}_1^2},$ <math><mrow><msub><mi>&Delta;</mi><mn>1</mn></msub><mo>=</mo><mi>arctan</mi><mfrac><msub><mi>V</mi><mn>1</mn></msub><msub><mi>U</mi><mn>1</mn></msub></mfrac><mo>+</mo><msub><mi>&kappa;</mi><mn>1</mn></msub><mi>&pi;</mi></mrow></math>

<math><mrow><msub><mi>U</mi><mn>1</mn></msub><mo>=</mo><msub><mi>B</mi><mn>1</mn></msub><mo>-</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mrow><mo>(</mo><msub><mi>B</mi><mn>2</mn></msub><mi>cos</mi><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><msub><mi>B</mi><mn>3</mn></msub><mi>cos</mi><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>-</mo><mfrac><msqrt><mn>3</mn></msqrt><mn>2</mn></mfrac><mrow><mo>(</mo><msub><mi>B</mi><mn>2</mn></msub><mi>sin</mi><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><msub><mi>B</mi><mn>3</mn></msub><mi>sin</mi><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></math>

<math><mrow><msub><mi>V</mi><mn>1</mn></msub><mo>=</mo><mfrac><msqrt><mn>3</mn></msqrt><mn>2</mn></mfrac><mrow><mo>(</mo><msub><mi>B</mi><mn>2</mn></msub><mi>cos</mi><msub><mi>&delta;</mi><mn>1</mn></msub><mo>-</mo><msub><mi>B</mi><mn>3</mn></msub><mi>cos</mi><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>-</mo><mfrac><mn>1</mn><mn>2</mn></mfrac><mrow><mo>(</mo><msub><mi>B</mi><mn>2</mn></msub><mi>sin</mi><msub><mi>&delta;</mi><mn>1</mn></msub><mo>-</mo><msub><mi>B</mi><mn>3</mn></msub><mi>sin</mi><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></math>

<math><mrow><msub><mi>&kappa;</mi><mn>1</mn></msub><mo>=</mo><mo></mo><mfenced open='{' close=''><mtable><mtr><mtd><mn>0</mn><mo>,</mo><msub><mi>U</mi><mn>1</mn></msub><mo>></mo><mn>0</mn></mtd></mtr><mtr><mtd><mn>1</mn><mo>,</mo><msub><mi>U</mi><mn>1</mn></msub><mo>&lt;</mo><mn>0</mn></mtd></mtr></mtable></mfenced></mrow></math>

by analysis, it was found that B is present₁，B₂，B₃When the following formula is satisfied, W₁And 0, so that the sum of the three outputs is constant.

<math><mrow><mfrac><msub><mi>B</mi><mn>2</mn></msub><msub><mi>B</mi><mn>1</mn></msub></mfrac><mo>=</mo><msub><mi>&eta;</mi><mn>1</mn></msub><mo>=</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><msub><mi>&delta;</mi><mn>2</mn></msub><mo>-</mo><mi>sin</mi><msub><mi>&delta;</mi><mn>2</mn></msub></mrow><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mrow><mo>(</mo><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>+</mo><mi>sin</mi><mrow><mo>(</mo><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></mfrac></mrow></math>

<math><mrow><mfrac><msub><mi>B</mi><mn>3</mn></msub><msub><mi>B</mi><mn>1</mn></msub></mfrac><mo>=</mo><msub><mi>&eta;</mi><mn>2</mn></msub><mo>=</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><msub><mi>&delta;</mi><mn>1</mn></msub><mo>-</mo><mi>sin</mi><msub><mi>&delta;</mi><mn>1</mn></msub></mrow><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mrow><mo>(</mo><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>+</mo><mi>sin</mi><mrow><mo>(</mo><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></mfrac></mrow></math>

B can be controlled by adjusting the amplification factor of the preamplifier₁，B₂，B₃Satisfy the above formula

B₂＝η₁B₁

B₃＝η₂B₁

Thereby obtaining: u. of_c＝u₁+u₂+u₃＝C₁+C₂+C₃

In order to remove the DC component, so that

3C₁-k₁(C₁+C₂+C₃)＝0

3C₂-k₁(C₁+C₂+C₃)＝0

3C₃-k₁(C₁+C₂+C₃)＝0

Adjusting the coefficients k of the three amplifiers in fig. 1, respectively₁，k₂，k₃So that

${\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">k</figure-callout>}}_1=\frac{3C_1}{{\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">C</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="C"\; filenames="HDA0000055526010000012.png,HDA0000055526010000013.png"\; state="\{\{state\}\}">C</figure-callout>}}_2+{\mathrm{<figure-callout\; id="3"\; label="C"\; filenames="HDA0000055526010000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_3}$

${\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="HDA0000055526010000012.png,HDA0000055526010000013.png"\; state="\{\{state\}\}">k</figure-callout>}}_2=\frac{3C_2}{{\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">C</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="C"\; filenames="HDA0000055526010000012.png,HDA0000055526010000013.png"\; state="\{\{state\}\}">C</figure-callout>}}_2+{\mathrm{<figure-callout\; id="3"\; label="C"\; filenames="HDA0000055526010000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_3}$

${\mathrm{<figure-callout\; id="3"\; label="k"\; filenames="HDA0000055526010000022.png"\; state="\{\{state\}\}">k</figure-callout>}}_3=\frac{3C_3}{{\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">C</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="C"\; filenames="HDA0000055526010000012.png,HDA0000055526010000013.png"\; state="\{\{state\}\}">C</figure-callout>}}_2+{\mathrm{<figure-callout\; id="3"\; label="C"\; filenames="HDA0000055526010000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_3}$

Thereby utilizing 3u_i-k_iu_cRemoving corresponding u₁，u₂，u₃The direct current component in the three paths of the non-direct current signals a is obtained₁，a₂，a₃。

2. Performing core demodulation on signal with DC component removed

The signal without the DC component is sent to the core demodulation module of FIG. 1 for processing

N＝a₁(d₃-d₂)+a₂(d₁-d₃)+a₃(d₂-d₁)

In the formula $d_i=\frac{{\mathrm{da}}_i}{\mathrm{dt}},i=\mathrm{1,2,3}$

The expression of N thus output is

<math><mrow><mi>N</mi><mo>=</mo><mo>-</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><mo>&CenterDot;</mo><mfrac><mi>d&phi;</mi><mi>dt</mi></mfrac><mo>&CenterDot;</mo><mo>{</mo><msqrt><mn>3</mn></msqrt><mo>[</mo><msub><mi>B</mi><mn>1</mn></msub><msub><mi>B</mi><mn>2</mn></msub><mi>cos</mi><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><msub><mi>B</mi><mn>1</mn></msub><msub><mi>B</mi><mn>3</mn></msub><mi>cos</mi><msub><mi>&delta;</mi><mn>2</mn></msub><mo>+</mo><msub><mi>B</mi><mn>2</mn></msub><msub><mi>B</mi><mn>3</mn></msub><mi>cos</mi><mrow><mo>(</mo><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>]</mo></mrow></math>

<math><mrow><mo>-</mo><msub><mi>B</mi><mn>1</mn></msub><msub><mi>B</mi><mn>2</mn></msub><mi>sin</mi><msub><mi>&delta;</mi><mn>1</mn></msub><mo>-</mo><msub><mi>B</mi><mn>1</mn></msub><msub><mi>B</mi><mn>3</mn></msub><mi>sin</mi><msub><mi>&delta;</mi><mn>2</mn></msub><mo>+</mo><msub><mi>B</mi><mn>2</mn></msub><msub><mi>B</mi><mn>3</mn></msub><mi>sin</mi><mrow><mo>(</mo><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>}</mo></mrow></math>

And can obtain

<math><mrow><mi>N</mi><mo>=</mo><mo>-</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><mo>&CenterDot;</mo><mfrac><mi>d&phi;</mi><mi>dt</mi></mfrac><mo>&CenterDot;</mo><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mi>Q</mi></mrow></math>

Wherein <math><mrow><mi>Q</mi><mo>=</mo><msqrt><mn>3</mn></msqrt><mo>[</mo><msub><mi>&eta;</mi><mn>1</mn></msub><mi>cos</mi><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><msub><mi>&eta;</mi><mn>2</mn></msub><mi>cos</mi><msub><mi>&delta;</mi><mn>2</mn></msub><mo>+</mo><msub><mi>&eta;</mi><mn>1</mn></msub><msub><mi>&eta;</mi><mn>2</mn></msub><mi>cos</mi><mrow><mo>(</mo><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>]</mo></mrow></math>

<math><mrow><mo>-</mo><msub><mi>&eta;</mi><mn>1</mn></msub><mi>sin</mi><msub><mi>&delta;</mi><mn>1</mn></msub><mo>-</mo><msub><mi>&eta;</mi><mn>2</mn></msub><mi>sin</mi><msub><mi>&delta;</mi><mn>2</mn></msub><mo>+</mo><msub><mi>&eta;</mi><mn>1</mn></msub><msub><mi>&eta;</mi><mn>2</mn></msub><mi>sin</mi><mrow><mo>(</mo><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></math>

It can be seen from the above equation that for asymmetric input, as long as the dc component can be completely removed, the output of the core demodulation module is still in the form of the product of the differential of the phase difference and a constant, and after integration, the demodulation result that is linear with the measured signal can be obtained. But the above equation still includes correlation with the coupler output

The term, needs to be removed by the AGC block.

3. Removing the effect of asymmetric input on automatic gain control circuit

In the asymmetric case, the sum of the squares of the AGC block inputs is

<math><mrow><msubsup><mi>a</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>2</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>3</mn><mn>2</mn></msubsup><mo>=</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><mo>&CenterDot;</mo><mo>[</mo><msubsup><mi>B</mi><mn>1</mn><mn>1</mn></msubsup><mo>+</mo><msubsup><mi>B</mi><mn>2</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>B</mi><mn>3</mn><mn>2</mn></msubsup><mo>+</mo><msub><mi>W</mi><mn>2</mn></msub><mi>cos</mi><mrow><mo>(</mo><mn>2</mn><mi>&phi;</mi><mo>-</mo><msub><mi>&Lambda;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>]</mo></mrow></math>

In the formula W₂And Λ₂With AC gain and delta₁And delta₂It is related. When B is present₁，B₂，B₃When the following formula is satisfied, W₂Is equal to 0, thereby

The sum of the two is a constant,

<math><mrow><mfrac><msub><mi>B</mi><mn>2</mn></msub><msub><mi>B</mi><mn>1</mn></msub></mfrac><mo>=</mo><msub><mi>&xi;</mi><mn>1</mn></msub><mo>=</mo><msup><mrow><mo>[</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>+</mo><mi>sin</mi><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub></mrow><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>-</mo><mi>sin</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></mfrac><mo>]</mo></mrow><mrow><mn>1</mn><mo>/</mo><mn>2</mn></mrow></msup></mrow></math>

<math><mrow><mfrac><msub><mi>B</mi><mn>3</mn></msub><msub><mi>B</mi><mn>1</mn></msub></mfrac><mo>=</mo><msub><mi>&xi;</mi><mn>2</mn></msub><mo>=</mo><msup><mrow><mo>[</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mi>sin</mi><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub></mrow><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>-</mo><mi>sin</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></mfrac><mo>]</mo></mrow><mrow><mn>1</mn><mo>/</mo><mn>2</mn></mrow></msup></mrow></math>

in the AGC block of FIG. 1, the gain is adjusted by adjusting the amplifier coefficient g₁，g₂，g₃To pair

Is adjusted so that

<math><mrow><msub><mi>g</mi><mn>2</mn></msub><msubsup><mi>B</mi><mn>2</mn><mn>2</mn></msubsup><mo>/</mo><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>=</mo><msubsup><mi>&xi;</mi><mn>1</mn><mn>2</mn></msubsup></mrow></math>

<math><mrow><msub><mi>g</mi><mn>3</mn></msub><mo>=</mo><msubsup><mi>B</mi><mn>3</mn><mn>2</mn></msubsup><mo>/</mo><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>=</mo><msubsup><mi>&xi;</mi><mn>2</mn><mn>2</mn></msubsup></mrow></math>

The output result of the AGC circuit is

<math><mrow><mi>D</mi><mo>=</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><mo>&CenterDot;</mo><mo>[</mo><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msub><mi>g</mi><mn>2</mn></msub><msubsup><mi>B</mi><mn>2</mn><mn>2</mn></msubsup><mo>+</mo><msub><mi>g</mi><mn>3</mn></msub><msubsup><mi>B</mi><mn>3</mn><mn>2</mn></msubsup><mo>]</mo></mrow></math>

<math><mrow><mo>=</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>[</mo><mn>1</mn><mo>+</mo><msubsup><mi>&xi;</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>&xi;</mi><mn>2</mn><mn>2</mn></msubsup><mo>]</mo></mrow></math>

4. After N and D are divided, negated and integrated, the demodulation result is obtained

<math><mrow><mi>V</mi><mrow><mo>(</mo><mi>t</mi><mo>)</mo></mrow><mo>=</mo><mo>&Integral;</mo><mo>-</mo><mfrac><mi>N</mi><mi>D</mi></mfrac><mi>dt</mi><mo>=</mo><mfrac><mi>Q</mi><mrow><msub><mi>g</mi><mn>1</mn></msub><mrow><mo>(</mo><mn>1</mn><mo>+</mo><msubsup><mi>&xi;</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>&xi;</mi><mn>2</mn><mn>2</mn></msubsup><mo>)</mo></mrow></mrow></mfrac><mo>&CenterDot;</mo><mi>&phi;</mi><mrow><mo>(</mo><mi>t</mi><mo>)</mo></mrow></mrow></math>

From the above equation, the output result of the asymmetric 3 × 3 coupler output carrier-free demodulation method shown in fig. 1 has a linear relationship with the optical phase difference caused by the detected signal, and the linear demodulation of the asymmetric 3 × 3 coupler output can be realized.

The invention has the advantages and beneficial effects that:

the method of the invention does not need to load high-frequency carrier waves, allows symmetry deviation on the optical path design, adopts the SOPC technology of high integration and parallel processing to realize the digitization of the demodulation system, and effectively solves the demodulation problem of the output signals of the asymmetrical 3X 3 coupler.

Drawings

FIG. 1: the invention discloses a schematic block diagram of an asymmetric 3 x 3 coupler output signal carrier-free demodulation method.

FIG. 2: the demodulation method of the invention is utilized to carry out the simulation result of the demodulation when the measured signal is 10 Hz. (in the vertical order: the signal to be measured, the demodulation result, 3 modulated signals)

FIG. 3: the demodulation method of the invention is utilized to carry out the simulation result of the demodulation when the measured signal is 100 Hz. (in the vertical order: the signal to be measured, the demodulation result, 3 modulated signals)

FIG. 4: the demodulation method of the invention is utilized to carry out the simulation result of the demodulation when the measured signal is 1 KHz. (in the vertical order: the signal to be measured, the demodulation result, 3 modulated signals)

FIG. 5: a block diagram of an M-Z interference system for obtaining three paths of modulated signals; 1 is a 1 × 2 coupler; 2 is a signal arm; 3 is a reference arm; 4 is an external signal; 5 is a 3 x 3 coupler.

FIG. 6: is an acceleration signal processing system SOPC architecture block diagram.

Detailed Description

The schematic block diagram of the non-carrier demodulation method of the output signal of the asymmetric 3 × 3 coupler of the present invention is shown in fig. 1, and is divided into three parts, namely, dc component removal, core demodulation and Automatic Gain Control (AGC). The present invention is described in further detail below with reference to the schematic block diagrams.

FIG. 5 is a schematic diagram of an M-Z fiber optic interferometer with a 1 × 2 coupler at the input and a 3 × 3 coupler at the output. Laser emitted by a laser light source is divided into two beams through a 1 x 2 coupler 1 to enter a reference arm 3 and a signal arm 2 respectively, a sensor on the signal arm senses the change of an external signal 4 to enable the signal arm to deform so as to enable the optical phase in the signal arm to change, and the optical phase of the reference arm is kept unchanged. Two beams of light in the signal arm and the reference arm generate phase difference, the two beams of light enter the 3 multiplied by 3 coupler 5, three beams of light are separated out and enter three light detectors respectively, optical signals in the optical fibers are converted into electric signals, and the electric signals are demodulated by programming of a following digital signal processing method.

Fig. 1 is a schematic block diagram of asymmetric 3 × 3 digital signal demodulation, which is divided into three parts, i.e., dc component removal, core demodulation, and Automatic Gain Control (AGC). The demodulator adopts a digital demodulation method, and adopts the SOPC technology to design the whole demodulator system according to the algorithm functional block diagram, and the SOPC architecture is shown in figure 6. The generating system on chip is configured with the SOPC Builder integrated in Quartu II by Altera corporation. And generating a 32-bit soft core Nios II embedded processor by using the SOPC Builder, and compiling the system together with Quartu II design software.

The SOPC architecture shown in fig. 6 includes an AD converter, a digital processing chip, a DA converter, an ethernet interface. The analog electric signals are collected and converted into digital signals through an AD converter, a digital processing chip writes operation logic according to an asymmetric 3 x 3 signal demodulation principle to demodulate the digital signals converted by the AD, and the demodulated digital signals are restored into analog signals through a DA converter to be output, or demodulated results are directly output through an Ethernet.

The specific process of the output signal demodulation method of the asymmetric 3 x 3 coupler provided by the invention is as follows:

1 st, asymmetric 3 x 3 coupler output signal u₁，u₂，u₃Obtained by

Obtaining an asymmetric 3 x 3 coupler output signal by an M-Z interferometric system comprising a light source, a 1 x 2 coupler, a sensor, a 3 x 3 coupler, and a photodetector, comprising:

1.1, generating a coherent light beam by a light source;

1.2, dividing light generated by a light source into two beams by a 1 x 2 coupler, wherein one beam is used as reference light and sent to a reference arm, and the other beam is used as signal light and sent to a signal arm;

1.3, the signal arm is deformed by the conversion of the external signal to be detected, so that the light in the signal arm is subjected to phase change and is subjected to phase difference with the light in the reference arm;

1.4, the two beams of light with the phase difference generated in the step 1.3 pass through a 3 × 3 coupler to generate three phase modulation light signals, and the three phase modulation light signals are received by three photodetectors and converted into three electrical signals u₁，u₂，u₃

u₁＝C₁+B₁ cosφ

u₂＝C₂+B₂ cos(φ-2π/3-δ₁)

u₃＝C₃+B₃ cos(φ+2π/3+δ₂)

Wherein, C_i，(i＝1，2，3)，δ₁，δ₂Is a constant related to the optical signal, C₁，C₂，C₃Respectively the DC component, delta, of the three signals₁And delta₂Is phase, B_i(i ═ 1, 2, 3) is the amplification factor of the preamplifier, and C is₁≠C₂≠C₃，B₁≠B₂≠B₃，δ₁≠δ₂；

2 nd asymmetric 3 x 3 couplingOutput signal u of the device₁，u₂，u₃Demodulation of

2.1 removing DC component

Using the formula a_i＝3u_i-k_iu_cI-1, 2, 3 removing the corresponding asymmetry 3 × 3 coupler output signal u₁，u₂，u₃D.c. component in (a) to obtain a₁，a₂，a₃Wherein u is_cIs u₁，u₂，u₃Sum u of three signals_c＝u₁+u₂+u₃，k₁，k₂，k₃Respectively as follows:

${\mathrm{<figure-callout\; id="1"\; label="k"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">k</figure-callout>}}_1=\frac{3C_1}{{\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">C</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="C"\; filenames="HDA0000055526010000012.png,HDA0000055526010000013.png"\; state="\{\{state\}\}">C</figure-callout>}}_2+{\mathrm{<figure-callout\; id="3"\; label="C"\; filenames="HDA0000055526010000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_3}$

${\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="HDA0000055526010000012.png,HDA0000055526010000013.png"\; state="\{\{state\}\}">k</figure-callout>}}_2=\frac{3C_2}{{\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">C</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="C"\; filenames="HDA0000055526010000012.png,HDA0000055526010000013.png"\; state="\{\{state\}\}">C</figure-callout>}}_2+{\mathrm{<figure-callout\; id="3"\; label="C"\; filenames="HDA0000055526010000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_3}$

${\mathrm{<figure-callout\; id="3"\; label="k"\; filenames="HDA0000055526010000022.png"\; state="\{\{state\}\}">k</figure-callout>}}_3=\frac{3C_3}{{\mathrm{<figure-callout\; id="1"\; label="C"\; filenames="HDA0000055526010000012.png"\; state="\{\{state\}\}">C</figure-callout>}}_1+{\mathrm{<figure-callout\; id="2"\; label="C"\; filenames="HDA0000055526010000012.png,HDA0000055526010000013.png"\; state="\{\{state\}\}">C</figure-callout>}}_2+{\mathrm{<figure-callout\; id="3"\; label="C"\; filenames="HDA0000055526010000022.png"\; state="\{\{state\}\}">C</figure-callout>}}_3}$

C₁，C₂，C₃same as in 1.4;

2.2, removing the signal a of the DC component in the step 2.1₁，a₂，a₃Performing core demodulation

For three paths of signals a with direct current components removed₁，a₂，a₃The time differential is obtained by the calculation of the time differential,

i is 1, 2, 3, and three electric signals d are generated₁，d₂，d₃And making a difference between two of them, then making a difference with a₁，a₂，a₃Multiplying respectively to obtain:

N＝a₁(d₃-d₂)+a₂(d₁-d₃)+a₃(d₂-d₁)；

2.3 eliminating the influence of AC gain on the demodulation result

In the asymmetric case, the sum of the squares of the AGC block inputs is

<math><mrow><msubsup><mi>a</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>2</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>3</mn><mn>2</mn></msubsup><mo>=</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><mo>&CenterDot;</mo><mo>[</mo><msubsup><mi>B</mi><mn>1</mn><mn>1</mn></msubsup><mo>+</mo><msubsup><mi>B</mi><mn>2</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>B</mi><mn>3</mn><mn>2</mn></msubsup><mo>+</mo><msub><mi>W</mi><mn>2</mn></msub><mi>cos</mi><mrow><mo>(</mo><mn>2</mn><mi>&phi;</mi><mo>-</mo><msub><mi>&Lambda;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>]</mo></mrow></math>

In the formula W₂And Λ₂And an AC gain B₁，B₂，B₃And delta₁And delta₂It is related. When B is present₁，B₂，B₃When the following formula is satisfied, W₂Is equal to 0, therebyThe sum of the two is a constant,

<math><mrow><mfrac><msub><mi>B</mi><mn>2</mn></msub><msub><mi>B</mi><mn>1</mn></msub></mfrac><mo>=</mo><msub><mi>&xi;</mi><mn>1</mn></msub><mo>=</mo><msup><mrow><mo>[</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>+</mo><mi>sin</mi><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub></mrow><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>-</mo><mi>sin</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></mfrac><mo>]</mo></mrow><mrow><mn>1</mn><mo>/</mo><mn>2</mn></mrow></msup></mrow></math>

<math><mrow><mfrac><msub><mi>B</mi><mn>3</mn></msub><msub><mi>B</mi><mn>1</mn></msub></mfrac><mo>=</mo><msub><mi>&xi;</mi><mn>2</mn></msub><mo>=</mo><msup><mrow><mo>[</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mi>sin</mi><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub></mrow><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>-</mo><mi>sin</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></mfrac><mo>]</mo></mrow><mrow><mn>1</mn><mo>/</mo><mn>2</mn></mrow></msup></mrow></math>

in the AGC block of FIG. 1, the gain is adjusted by adjusting the amplifier coefficient g₁，g₂，g₃To pair

Is adjusted so that

<math><mrow><msub><mi>g</mi><mn>2</mn></msub><msubsup><mi>B</mi><mn>2</mn><mn>2</mn></msubsup><mo>/</mo><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>=</mo><msubsup><mi>&xi;</mi><mn>1</mn><mn>2</mn></msubsup></mrow></math>

<math><mrow><msub><mi>g</mi><mn>3</mn></msub><msubsup><mi>B</mi><mn>3</mn><mn>2</mn></msubsup><mo>/</mo><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>=</mo><msubsup><mi>&xi;</mi><mn>2</mn><mn>2</mn></msubsup></mrow></math>

The output result of the AGC block is

<math><mrow><mi>D</mi><mo>=</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><mo>&CenterDot;</mo><mo>[</mo><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msub><mi>g</mi><mn>2</mn></msub><msubsup><mi>B</mi><mn>2</mn><mn>2</mn></msubsup><mo>+</mo><msub><mi>g</mi><mn>3</mn></msub><msubsup><mi>B</mi><mn>3</mn><mn>2</mn></msubsup><mo>]</mo></mrow></math>

<math><mrow><mo>=</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>[</mo><mn>1</mn><mo>+</mo><msubsup><mi>&xi;</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>&xi;</mi><mn>2</mn><mn>2</mn></msubsup><mo>]</mo></mrow></math>

And 2.4, dividing the output result N of the step 2.2 and the output result D of the step 2.3 to obtain a result Z, and integrating the result Z by an integrator to obtain a final phase signal.

According to the asymmetric 3 x 3 coupling demodulation theory, simulation software Simulink in MATLAB environment is used for carrying out mathematical modeling on the asymmetric 3 x 3 coupling output carrier-free demodulation method, and the demodulation theory under the condition of asymmetric input is verified. Parameters in the formula 1.4 in simulation are set as follows: delta₁＝0.147，δ₂＝0.209，C₁＝B₁＝1，C₂＝B₂＝0.8，C₃The values of the adjustment parameters can be calculated from these parameters according to the demodulation method described above, where B3 is 1.2.

FIGS. 2 to 4 are simulation results (in the order of up and down: measured signal, demodulation result, 3 channels of modulated signals) of the demodulation of measured signals of 10Hz, 100Hz and 1KHz, respectively, by the demodulation method of the present invention.

It can be seen from fig. 2 to 4 that the demodulation method can realize demodulation in the frequency range of 0Hz to 1KHz, and maintain good phase consistency despite slight distortion in the case that the measured signal is 1 KHz. The slight demodulation distortion when the detected signal is higher is related to the sampling frequency, and the undistorted demodulation signal can be obtained by only increasing the sampling frequency

It should be noted that the present invention provides a method for implementing asymmetric 3 × 3 signal demodulation by SOPC technology, but is not limited thereto, and the asymmetric 3 × 3 signal demodulation may also be implemented by using technologies such as ARM or DSP according to the algorithm of the present invention.

Claims (1)

1. A method for carrier-less demodulation of an output signal from an asymmetric 3 x 3 coupler, the method comprising:

1 st, asymmetric 3 x 3 coupler output signal u₁，u₂，u₃Obtained by

Obtaining an asymmetric 3 x 3 coupler output signal by an M-Z interferometric system comprising a light source, a 1 x 2 coupler, a sensor, a 3 x 3 coupler, and a photodetector, comprising:

1.1, generating a coherent light beam by a light source;

1.2, dividing light generated by a light source into two beams by a 1 x 2 coupler, wherein one beam is used as reference light and sent to a reference arm, and the other beam is used as signal light and sent to a signal arm;

1.3, the external signal to be detected causes the signal arm to deform, so that the light in the signal arm changes phase and generates a phase difference with the light in the reference arm;

1.4, the two beams of light with the phase difference generated in the step 1.3 pass through a 3 × 3 coupler to generate three phase modulation light signals, and the three phase modulation light signals are received by three photodetectors and converted into three electrical signals u₁，u₂，u₃

u₁＝C₁+B₁ cosφ

u₂＝C₂+B₂ cos(φ-2π/3-δ₁)

u₃＝C₃+B₃ cos(φ+2π/3+δ₂)

Wherein, C_i，i＝1，2，3，δ₁，δ₂Constant in relation to the optical signal, C₁，C₂，C₃Respectively the direct current quantity of the three signals, phi is the phase of the light changing along with the time, delta₁And delta₂To deviate the phase, B_iI is 1, 2, 3 is the amplification factor of the preamplifier, and C₁≠C₂≠C₃，B₁≠B₂≠B₃，δ₁≠δ₂；

2 nd, asymmetric 3 x 3 coupler output signal u₁，u₂，u₃Demodulation of

2.1 removing DC component

Using the formula a_i＝3u_i-k_iu_cI-1, 2, 3 removing the corresponding asymmetry 3 × 3 coupler output signal u₁，u₂，u₃D.c. component in (a) to obtain a₁，a₂，a₃Wherein u is_cIs u₁，u₂，u₃Sum u of three signals_c＝u₁+u₂+u₃，k₁，k₂，k₃Respectively as follows:

$k_1=\frac{3C_1}{C_1+C_2+C_3}$

$k_2=\frac{3C_2}{C_1+C_2+C_3}$

$k_3=\frac{{3C}_3}{C_1+C_2+C_3}$

C₁，C₂，C₃same as in 1.4;

2.2, removing the signal a of the DC component in the step 2.1₁，a₂，a₃Performing core demodulation

For three paths of signals a with direct current components removed₁，a₂，a₃The time differential is obtained by the calculation of the time differential,

i is 1, 2, 3, and three electric signals d are generated₁，d₂，d₃And making a difference between two of them, then making a difference with a₁，a₂，a₃Multiplying respectively to obtain:

N＝a₁(d₃-d₂)+a₂(d₁-d₃)+a₃(d₂-d₁)；

2.3 eliminating the influence of AC gain on the demodulation result

In the asymmetric case, the sum of the squares of the AGC block inputs is

<math><mrow><msubsup><mi>a</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>2</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>a</mi><mn>3</mn><mn>2</mn></msubsup><mo>=</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><mo>&CenterDot;</mo><mo>[</mo><msubsup><mi>B</mi><mn>1</mn><mn>1</mn></msubsup><mo>+</mo><msubsup><mi>B</mi><mn>2</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>B</mi><mn>3</mn><mn>2</mn></msubsup><mo>+</mo><msub><mi>W</mi><mn>2</mn></msub><mi>cos</mi><mrow><mo>(</mo><mn>2</mn><mi>&phi;</mi><mo>-</mo><msub><mi>&Lambda;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>]</mo></mrow></math>

In the formula W₂And Λ₂And an AC gain B₁，B₂，B₃And delta₁And delta₂(ii) related; when B is present₁，B₂，B₃When the following formula is satisfied, W₂Is equal to 0, thereby

The sum of the two is a constant,

<math><mrow><mfrac><msub><mi>B</mi><mn>2</mn></msub><msub><mi>B</mi><mn>1</mn></msub></mfrac><mo>=</mo><msub><mi>&xi;</mi><mn>1</mn></msub><mo>=</mo><msup><mrow><mo>[</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>+</mo><mi>sin</mi><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub></mrow><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>-</mo><mi>sin</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></mfrac><mo>]</mo></mrow><mrow><mn>1</mn><mo>/</mo><mn>2</mn></mrow></msup></mrow></math>

<math><mrow><mfrac><msub><mi>B</mi><mn>3</mn></msub><msub><mi>B</mi><mn>1</mn></msub></mfrac><mo>=</mo><msub><mi>&xi;</mi><mn>2</mn></msub><mo>=</mo><msup><mrow><mo>[</mo><mfrac><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mi>sin</mi><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub></mrow><mrow><msqrt><mn>3</mn></msqrt><mi>cos</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow><mo>-</mo><mi>sin</mi><mrow><mo>(</mo><mn>2</mn><msub><mi>&delta;</mi><mn>1</mn></msub><mo>+</mo><mn>2</mn><msub><mi>&delta;</mi><mn>2</mn></msub><mo>)</mo></mrow></mrow></mfrac><mo>]</mo></mrow><mrow><mn>1</mn><mo>/</mo><mn>2</mn></mrow></msup></mrow></math>

in the AGC block, by adjusting the amplifier coefficient g₁，g₂，g₃To pairIs adjusted so that

<math><mrow><msub><mi>g</mi><mn>2</mn></msub><msubsup><mi>B</mi><mn>2</mn><mn>2</mn></msubsup><mo>/</mo><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>=</mo><msubsup><mi>&xi;</mi><mn>1</mn><mn>2</mn></msubsup></mrow></math>

<math><mrow><msub><mi>g</mi><mn>3</mn></msub><msubsup><mi>B</mi><mn>3</mn><mn>2</mn></msubsup><mo>/</mo><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>=</mo><msubsup><mi>&xi;</mi><mn>2</mn><mn>2</mn></msubsup></mrow></math>

The output result of the AGC block is

<math><mrow><mi>D</mi><mo>=</mo><mfrac><mn>2</mn><mn>9</mn></mfrac><mo>&CenterDot;</mo><mo>[</mo><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msub><mi>g</mi><mn>2</mn></msub><msubsup><mi>B</mi><mn>2</mn><mn>2</mn></msubsup><mo>+</mo><msub><mi>g</mi><mn>3</mn></msub><msubsup><mi>B</mi><mn>3</mn><mn>2</mn></msubsup><mo>]</mo></mrow></math>

<math><mrow><mo>=</mo><mfrac><mn>9</mn><mn>2</mn></mfrac><msub><mi>g</mi><mn>1</mn></msub><msubsup><mi>B</mi><mn>1</mn><mn>2</mn></msubsup><mo>[</mo><mn>1</mn><mo>+</mo><msubsup><mi>&xi;</mi><mn>1</mn><mn>2</mn></msubsup><mo>+</mo><msubsup><mi>&xi;</mi><mn>2</mn><mn>2</mn></msubsup><mo>]</mo></mrow></math>

And 2.4, dividing the output result N of the step 2.2 and the output result D of the step 2.3 to obtain a result Z, and integrating the result Z by an integrator to obtain a final phase signal.

Images