# How does one compute frequency bandwidth based on FWHM

In summary, the authors report a full width at half maximum (FWHM) of 2 km/s for the observed OH transition emission frequency of 1720 MHz in the local standard of rest (LSR). However, this is wider than the expected thermal line width of 0.5 to 0.7 km/s, leading the authors to suggest that it is a blend of several maser spots along the line of sight. They also observe another OH line at different LSR velocities, indicating relative bulk movement or masing properties of objects in the line of sight.
In the following text the authors state:

"The full width at half maximum (FWHM) we measure is about 2 km/s"

and this is in the local standard of rest (LSR), ##v_{LSR}##. I have seen basic doppler shift equations to convert from these velocities to frequency (shown below in link). Although if I consider the bandwidth to be 2 km/s and the frequency of 1720 MHz to correspond to ~-45 km/s, then the frequency bandwidth would be on the order of 100 MHz which doesn't seem to be quite right intuitively as masers are typically much narrower. Also, in the paper, they state they observe another OH line at ~30 km/s so I don't believe my above interpretation is correct.

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So, 1720 MHz is the OH transition emission frequency of interest. the ##v_{LSR}## of ~ -45km/s is the observed redshift of that frequency due to the relative motion of the intervening OH cloud, and the 2km/s bandwidth is the FWHM line width, which is rather more than the 0.5 to 0.7 km/s expected thermal line width, and which leads the authors to suggest that "we are most likely seeing a blend of several maser spots along the line of sight."

As you state, another OH line at ##v_{LSR}## ~ -30km/s was also discovered in observations of other PSR's, and also due to OH cloud(s) between those PSR's and earth.

EDIT: So, the 1720 MHz frequency, and the 2 km/s bandwidth, and the ##v_{LSR}## of ~-45 km/s or ~30 km/s are unrelated to each other, apart from their being indicators of relative bulk movement or of masing properties of objects in the line of sight of the observations.

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## 1. How do you calculate frequency bandwidth based on FWHM?

The frequency bandwidth can be calculated by multiplying the full width at half maximum (FWHM) by 2. This is because the FWHM represents the range of frequencies where the signal amplitude is greater than half its maximum value.

## 2. What is FWHM and why is it important in computing frequency bandwidth?

FWHM stands for full width at half maximum and it is a measure of the width of a frequency peak or signal. It is important in computing frequency bandwidth because it represents the range of frequencies where the signal is significant and contributes to the overall bandwidth.

## 3. Can frequency bandwidth be calculated using other parameters besides FWHM?

Yes, frequency bandwidth can also be calculated using other parameters such as the 3 dB bandwidth or the -3 dB bandwidth. These parameters represent the range of frequencies where the signal amplitude is 3 dB or -3 dB lower than its maximum value, respectively.

## 4. How do you measure FWHM from a frequency spectrum?

To measure FWHM from a frequency spectrum, you would need to identify the peak of the signal and then measure the width of the peak at half its maximum amplitude. This can be done using a software tool or by manually measuring the peak width on a printed spectrum.

## 5. Is there a specific unit for frequency bandwidth?

Frequency bandwidth is typically measured in hertz (Hz) or kilohertz (kHz). However, it can also be measured in other units such as megahertz (MHz) or gigahertz (GHz) depending on the frequency range being considered.

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