Here are some considerations for observing with the radio telescope:

**What is going to be the typical integration time per point?**

To calculate a reasonable integration time, we need to know a but about noise, as we hope to have an integration time long enough that our signal-to-noise ratio (SNR) is quite high.

**The root-mean-square error \(\sigma\) is given by:**

\[\sigma = \frac{T_{\text{sys}}}{\sqrt{\tau \Delta \nu}}\]

Where \( T_{\text{sys}}\) is the noise of each sample, \(\tau\) is the integration time, and \(\Delta \nu\) is the channel width.

We will define the signal strength to be \(T_A\). Therefore, the SNR is defined as:

\[\text{SNR} = \frac{T_A}{\sigma}\]

If we solve this for \(\sigma\) and plug it into the equation we found above, then solve for \(\tau\), we get:

\[\tau = \frac{T_{\text{sys}}^2 (\text{SNR})^2}{T_A^2 \Delta \nu}\]

We are told that \(T_{\text{sys}}\) is about 500 K and that the average signal peak will be about 2 K. Furthermore, we will consider using the 0.5 MHz wide channel. As for the SNR, let's calculate the integration time that corresponds to an average error of 1%, meaning SNR=100.

\[\tau = \frac{(500 \text{ K})^2 (100)^2}{(2\text{ K})^2 (0.5\times 10^6 \text{ s}^{-1})} = 1250 \text{ s} \]

This means that a good integration time would be about

*21 minutes*.

**Over what range of longitude do you plan to observe?**

According to the lab manual, we will be observing at Galactic longitudes of 10° to 70°.

**How many positions do you plan to observe?**

The manual seems to suggest that we will be observing four giant molecular clouds, so we will be observing at

*four*positions.

**At what LST are you going to start observing? At what EST?**

We will probably begin our observations

**1:00pm EST on (for example) November 10. This corresponds to about 15:34 LST.**

**Are you going to position switch or frequency switch to "flat field" the spectrum?**

We want the flat-field correction to correspond to the sensitivity of the CCD at the specific frequency at which we are observing. As a result, we do not want to switch the frequency to flat field our images. Instead, we can perform a "position switch" to do this.

Above, we started Problem 1 on Worksheet 9.2, so let's finish it:

**(b) Find an equation for the velocity resolution \(\Delta v\) km/s that corresponds to a channel width of \(\Delta \nu\) MHz at the frequency of \(^{12} CO\), \(\nu = 115.271 GHz\). Write this in the form:**

\[\Delta v = \text{____} \text{km/s} \left(\frac{\Delta \nu}{1 \text{ MHz}}\right)\]

To do this, we can use the Doppler equation, which tells us:

\[\Delta v = c \frac{\Delta \nu}{\nu}\]

To find the equation in the desired form, we simply need to multiply the speed of light \(c\) (\(3\times 10^5 \text{ km/s}\)) by \(10^6\), which corresponds to the MHz term we put in the denominator. This gives us:

\[\boxed{\Delta v = 3.8\times 10^{11} \text{km/s}\left(\frac{\Delta \nu}{1 \text{ MHz}}\right)}\]

I've shown solutions very similar to (a) and (c) above.

(b) is way off. It should just be ~2.6 km/s. You calculate \( \Delta v \) for a MHz and then scale it by \( \nu \) in MHz.

ReplyDelete4.5