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Channel Astronomy (Brief Introduction): Astronomical Signal Channel Processing Mode
DOI: 10.12677/AAS.2019.72003, PDF, HTML, XML, 下载: 6,268  浏览: 12,591

Abstract: Based on the modern communication model, this paper deals with the channel interaction of as-tronomical signals, especially the signals from distant extragalactic galaxies and supernovae, and draws different conclusions from current astronomy, which are more consistent with the observa-tions of astronomical phenomena such as cosmic red shift and supernova signals. The conclusion of the channel mode of processing of astronomical signals is that galaxies do not need to retreat, and the universe does not have a big explosion. The cosmic red shift is only caused by the propagation of light wave signals in space channels.

1. 天文信号处理模式发展过程

Figure 1. Ptolemy (geocentric theory), Copernicus (heliocentric theory), Hubble (Big Bang) stood on the earth, observing and interpreting astronomical information

${V}_{f}={H}_{C}×D$ (1)

2. 信道天文学：天文信号的信道模式处理

Figure 2. Processing astronomical signals using modern communication models

$Y\left(x,t\right)=H\left(x,t\right)*s\left(t\right)+n\left(t\right)$ (2)

$s\left(t\right)=\frac{y\left(x,t\right)-n\left(t\right)}{H\left(x,t\right)}$ (3)

3. 宇宙红移的信道模式处理

$s\left(t\right)=A\mathrm{sin}\varpi t$ (4)

$I\left(x\right)=I\left({x}_{0}\right){e}^{-2\alpha x}$ (5)

$A\left(x\right)=A\left({x}_{0}\right){e}^{-\alpha x}$ (6)

${f}_{o}={f}_{s}{e}^{-{\alpha }_{f}x}$${\lambda }_{o}={\lambda }_{s}{e}^{-{\alpha }_{f}x}$ (7)

$Z=\frac{{\lambda }_{o}-{\lambda }_{s}}{{\lambda }_{s}}$ (8)

$Z=\frac{{\lambda }_{o}-{\lambda }_{s}}{{\lambda }_{s}}={e}^{{\alpha }_{f}x}-1$ (9)

$Z={e}^{{\alpha }_{f}x}-1\approx {\alpha }_{f}x$ (10)

1 Mpc = 106 pc = 3.2616 * 106 l.y. = 3.2616 * 106 * 0.94607 * 1016 m = 3.08568 * 1022 m；

H1 m = 0.8 * 10−26/m (11)

$Z=\frac{{\lambda }_{o}-{\lambda }_{i}}{{\lambda }_{i}}\approx \frac{{f}_{i}-{f}_{o}}{{f}_{i}}$ (12)

$\nabla f={f}_{i}-{f}_{o}\approx {f}_{i}*Z=Z*D*{f}_{i}$ (13)

$\Delta \theta =2\text{π}*\Delta t*\Delta f=2\text{π}*\Delta t*Z*{f}_{i}=2\text{π}*\Delta t*HD*{f}_{i}$ (14)

$\Delta \theta$ = 360˚ * 86400 s * 0.8 * 10−26/m * 2 * 104 m * 1015 Hz/s = 5˚(15)

4. 超新星信号的信道模式处理

${a}_{n}\frac{{d}^{n}y\left(t\right)}{d{t}^{n}}+\cdots +{a}_{0}y={b}_{m}\frac{{d}^{m}x}{d{t}^{m}}+\cdots +{b}_{0}x$ (16)

$\delta \left(t\right)=\left[\begin{array}{l}\infty ,t=0\\ 0,t\ne 0\end{array}\text{\hspace{0.17em}}与\text{\hspace{0.17em}}\underset{-\infty }{\overset{+\infty }{\int }}\delta \left(t\right)\text{d}t=A$ (17)

Figure 3. Experimental diagram for measuring frequency attenuation

Figure 4. Supernova signal observations (copy reference)

${a}_{2}\frac{\text{d}{S}_{o}^{2}}{\text{d}{t}^{2}}+{a}_{1}\frac{\text{d}{S}_{o}\left(t\right)}{\text{d}t}+{a}_{0}{S}_{o}\left(t\right)={b}_{0}{S}_{i}\left(t\right)$ (18)

$H\left(S\right)=\frac{{b}_{0}}{{a}_{2}{s}^{2}+{a}_{1}s+{a}_{0}}$ (19)

1) 临界阻尼状态响应：

${S}_{o}\left(t\right)=\frac{{b}_{0}}{{a}_{2}}*t*{e}^{-\sqrt{\frac{{a}_{0}}{{a}_{2}}}*t}$ (20)

2) 过阻尼状态

${S}_{o}\left(t\right)=\frac{{\varpi }_{0}k}{2\sqrt{{\xi }^{2}-1}}\left[{\text{e}}^{-\left(\xi +\sqrt{{\xi }^{2}-1}\right){\varpi }_{0}t}-{\text{e}}^{{}^{-\left(\xi -\sqrt{{\xi }^{2}-1}\right){\varpi }_{0}t}}\right]$ (21)

3) 欠阻尼状态

${S}_{o}\left(t\right)=\frac{{\varpi }_{0}k}{2\sqrt{1-{\xi }^{2}}}\left[{\text{e}}^{-\xi {\varpi }_{0}t}\mathrm{sin}\left(\sqrt{1-{\xi }^{2}}{\varpi }_{0}t\right)\right]$ (22)

${S}_{o}\left(t\right)=0.41t*{\text{e}}^{-0.05977t}$ (23)

Figure 5. Fitting with the second order channel critical damping impulse response function

2011年诺贝尔获奖者成果：发现超新星亮度比预期的低、红移比预期的大

4.1. 亮度比预期低的信道解释

$M-m=-2.5\mathrm{log}\left(\frac{{F}_{10}}{{F}_{D}}\right)=5-5\mathrm{log}d\left(pc\right)$ (24)

${S}_{o}\left(t\right)=\frac{{b}_{0}}{{a}_{2}}*t*{e}^{-\sqrt{\frac{{a}_{0}}{{a}_{2}}}*t}$

Figure 6. Impact response of pulse function to channel

4.2. 红移比预期的大

${\lambda }_{o}={\lambda }_{s}{e}^{{\alpha }_{f}d}$

$Z=\frac{{\lambda }_{o}-{\lambda }_{s}}{{\lambda }_{s}}={e}^{{\alpha }_{f}d}-1\approx {\alpha }_{f}d$

${S}_{o}\left(t\right)\frac{{b}_{0}}{{a}_{2}}*t*{e}^{-\sqrt{\frac{{a}_{0}}{{a}_{2}}}*t}$

5. 信道对波的弥散作用

Figure 7. Dispersion diagram of wave pulses by channel

6. 结论

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