network profile
| bio | website | |
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| location | ||
| age | ||
| visits | member for | 1 year, 4 months |
| seen | Mar 22 '12 at 13:32 | |
| stats | profile views | 5 |
I am now a ninja.
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Mar 22 |
awarded | Supporter |
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Jan 28 |
awarded | Scholar |
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Jan 28 |
accepted | Signal-to-noise ratio of the difference between two signals |
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Jan 28 |
comment |
Signal-to-noise ratio of the difference between two signals From this, and using Henden's formula, I can obtain the error in magnitudes: $-2.5 \log {1\pm\frac{1}{300}} = 0.0036$. Hopefully I'm not wrong when I consider this error (in Henden's words) to be our noise component. It is at this point where I'm stuck. In order to use your formula, would the signal, $S_1^2$, be still 8.02? |
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Jan 28 |
comment |
Signal-to-noise ratio of the difference between two signals Thanks a whole lot, twistor59. Although your answer is exactly what I had been looking for, I am not quite sure how it could be applied in our case. I have updated my question in order to include the formula that we use, given the SNR, to compute the error in magnitudes. The thing is that our data set does not distinguishes between signal and noise components — we have a magnitude, say 8.02, and an reliable estimation of its SNR, say 300. [Continues in next comment] |
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Jan 28 |
revised |
Signal-to-noise ratio of the difference between two signals Added the formula to compute the error in magnitudes, given the SNR |
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Jan 26 |
awarded | Editor |
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Jan 26 |
revised |
Signal-to-noise ratio of the difference between two signals Missing links fixed. |
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Jan 20 |
awarded | Student |
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Jan 20 |
asked | Signal-to-noise ratio of the difference between two signals |
