Detecting Cassiopeia A

[Andrew Thornett]

I suspect I didn't understand properly what the original Cass A thread was about but continuing down the line of detecting Cass A rather than resolving it, this is what ChatGPT advises - I am giving ChatGPT credit because this did come from ChatGPT and also because if it is wrong then I would like the criticism aimed elsewhere!

Andy

To estimate the integration time required to detect Cassiopeia A at 1420 MHz using a 1-meter dish, we need to consider the signal-to-noise ratio (SNR) and radiometer equation. Here's a breakdown:

---

Key Parameters

1. Flux Density of Cas A at 1420 MHz:
   Cas A is decreasing in brightness over time, but as of recent measurements:

   Sν≈1500 Jy (1 Jy = 10−26 W/m2/Hz)S_\nu \approx 1500\ \text{Jy} \ \text{(1 Jy = }10^{-26}\ \text{W/m}^2/\text{Hz)}

2. Dish Diameter: 1 meter
   Effective area (assuming efficiency ~70%):

   Aeff=η⋅π⋅(D2)2≈0.7⋅0.785⋅12≈0.55 m2A_{\text{eff}} = \eta \cdot \pi \cdot \left(\frac{D}{2}\right)^2 \approx 0.7 \cdot 0.785 \cdot 1^2 \approx 0.55\ \text{m}^2

3. System Temperature:
   At L-band, T_sys ≈ 100 K is typical for amateur setups (depends on LNA, environment, etc.)

4. Bandwidth:
   Let's assume a moderate bandwidth of 1 MHz for neutral hydrogen work.

5. Radiometer Equation:

   SNR=Sν⋅Aeffk⋅Tsys⋅2⋅Δν⋅t\text{SNR} = \frac{S_\nu \cdot A_{\text{eff}}}{k \cdot T_{\text{sys}}} \cdot \sqrt{2 \cdot \Delta \nu \cdot t}

   where:

• kk is Boltzmann’s constant =1.38×10−23= 1.38 \times 10^{-23}

• tt is integration time in seconds

---

Solving for Integration Time

We rearrange to solve for tt when a detectable SNR ~5 is desired:

t=(SNR⋅k⋅TsysSν⋅Aeff)2⋅12⋅Δνt = \left( \frac{SNR \cdot k \cdot T_{\text{sys}}}{S_\nu \cdot A_{\text{eff}}} \right)^2 \cdot \frac{1}{2 \cdot \Delta \nu}

Plugging in values:

• Sν=1500×10−26S_\nu = 1500 \times 10^{-26}

• Aeff=0.55A_{\text{eff}} = 0.55

• Tsys=100T_{\text{sys}} = 100

• Δν=1×106\Delta\nu = 1 \times 10^6

• SNR=5SNR = 5

t=(5⋅1.38×10−23⋅1001500×10−26⋅0.55)2⋅12⋅106t = \left( \frac{5 \cdot 1.38 \times 10^{-23} \cdot 100}{1500 \times 10^{-26} \cdot 0.55} \right)^2 \cdot \frac{1}{2 \cdot 10^6} t≈(8.36×10−21)2⋅12×106≈6.99×10−41⋅5×10−7=(clearly too small—check units)t \approx (8.36 \times 10^{-21})^2 \cdot \frac{1}{2 \times 10^6} \approx 6.99 \times 10^{-41} \cdot 5 \times 10^{-7} = \text{(clearly too small—check units)}

Let’s correct units and simplify numerically:

t=(5⋅1.38⋅1001500⋅0.55×10−23+26)2⋅12⋅106t = \left( \frac{5 \cdot 1.38 \cdot 100}{1500 \cdot 0.55} \times 10^{-23+26} \right)^2 \cdot \frac{1}{2 \cdot 10^6} t=(690825⋅103)2⋅12⋅106=(0.836⋅103)2⋅5⋅10−7t = \left( \frac{690}{825} \cdot 10^3 \right)^2 \cdot \frac{1}{2 \cdot 10^6} = (0.836 \cdot 10^3)^2 \cdot 5 \cdot 10^{-7} t=(836)2⋅5⋅10−7≈698000⋅5⋅10−7=0.35 secondst = (836)^2 \cdot 5 \cdot 10^{-7} \approx 698000 \cdot 5 \cdot 10^{-7} = 0.35\ \text{seconds}

---

Answer:

Approximately 0.35 seconds of integration time would be enough to detect Cas A at 5σ using a 1-meter dish with a 1 MHz bandwidth, assuming system temperature ~100 K and decent LNA.

---

Reality Check:

• This is an ideal estimate.

• In practice, interference, system noise variations, and calibration errors will require longer integration, typically:

• 1–10 minutes to clearly distinguish the signal in drift scan or total power mode.

• Cas A is also resolved out by some small dishes—beamwidth should encompass the full source.

[kb3puw]

I have been working on validating my raster study (you can see the details in the April 2025 SARA Journal).

To do the validation I have been using the AIFHi survey database configured for 15 degrees which is about what a 1 meter dish is.

Here is a comparison of my results (using my RAISE software).  My software allows me to gather data sets at the same coordinates

as my raster work from the Survey database and display them in a comparable manner.  I am not claiming anything other that I am

centered on CAS A evaluating 1.42 GHz with a small dish.

--
--
You received this message because you are subscribed to the Google
Groups "Society of Amateur Radio Astronomers" group.
To post to this group, send email to sara...@googlegroups.com
To unsubscribe from this group, send email to
sara-list-...@googlegroups.com
For more options, visit this group at
http://groups.google.com/group/sara-list?hl=en
---
You received this message because you are subscribed to the Google Groups "Society of Amateur Radio Astronomers" group.
To unsubscribe from this group and stop receiving emails from it, send an email to sara-list+...@googlegroups.com.
To view this discussion visit https://groups.google.com/d/msgid/sara-list/DB9PR06MB7564983F791FD221C3512FF6FF65A%40DB9PR06MB7564.eurprd06.prod.outlook.com.

[Anthony]

Well, with all the great conversation, it's been a while. Guess I'll turn my amateur RAS interferometer towards CASS A, it's been a while. 

Transit happens in 19 minutes, 25 seconds...

To view this discussion visit https://groups.google.com/d/msgid/sara-list/CAB-X21OURnksdifv9GFpAY18wHEbEEDGtkh86YWYpn-ZtDZHYQ%40mail.gmail.com.

[Anthony]

Calibrated my system a few days ago or was it yesterday using the Sun. 

Images attached...

On Mon, May 26, 2025 at 2:25 PM kb3puw <kb3...@gmail.com> wrote:

To view this discussion visit https://groups.google.com/d/msgid/sara-list/CAB-X21OURnksdifv9GFpAY18wHEbEEDGtkh86YWYpn-ZtDZHYQ%40mail.gmail.com.

[Anthony]

Oops, Taurus A, transit in 2 minutes 

On Mon, May 26, 2025 at 2:25 PM kb3puw <kb3...@gmail.com> wrote:

To view this discussion visit https://groups.google.com/d/msgid/sara-list/CAB-X21OURnksdifv9GFpAY18wHEbEEDGtkh86YWYpn-ZtDZHYQ%40mail.gmail.com.