How Does Frequency Affect C/N in Dish Antennas?

In summary: Maybe the dish profile is not accurate enough for 40 GHz. At this frequency, the profile would need to be within, say, an eighth of a wavelength, which at 40 Ghz is only a small distance. Lambda = 300/40 = 7.5mm, so Lamba/8 is approx 1 mm.
  • #1
ashah99
60
2
Thread moved from the technical forums to the schoolwork forums
Summary:: I am stuck on a problem on dish antenna C/N and it's relation to frequency. Please see below for thought process.

Problem Statement:

1634148927402.png
,
Link budget formula in dB form: Pr = Pt + Gt + Gr + 20*log10(lambda/(4*pi*distance)) - Losses [dB]
C/N = Pr - Pn [dB], where Pn = 10*log10(k*T*B)

I am stuck on coming up with the final answer. Without knowing all the details (RX/TX powers, antenna efficiency, additional losses, etc.), I can only conclude that some factors are equal, like noise power. The antenna gains Gt, Gr, and free space propagation losses all are frequency dependent and are proportional to f^2, so with 2 gain factors and 1 loss, my net C/N increase would be with respoect to 1 f^2. I'm just not sure how to proceed from here. any help is very much appreciated.
 
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  • #2
Antenna gain ∝ 10 log (pi D / lambda)^2
So gain rises with freq squared.
At 12 GHz:-
Increase in gain of Gt at 12 GHz = 20 log 12/6 = 6dB
Increase in gain of Gr at 12 GHz = 6dB
Path loss is also dependent on freq squared, so it will increase by 6dB at 12 GHz
Pr change = Gt + Gr - Lp = 6 + 6 - 6 = +6dB
CNR = 10 + 6 = 16dB
At 40GHz:-
Increase in Gt at 40 GHz = 20 log 40/6 = 16,5dB
Increase in gain of Gr = 16.5 dB
Path loss increase = 20 log 40/6 = 16.5 dB
Pr change = Gt+Gr -Lp = 16.5 + 16.5 - 16.5 = +16.5 dB
CNR = 10 + 16.5 = 26.5 dB
 
  • #3
tech99 said:
Antenna gain ∝ 10 log (pi D / lambda)^2
So gain rises with freq squared.
At 12 GHz:-
Increase in gain of Gt at 12 GHz = 20 log 12/6 = 6dB
Increase in gain of Gr at 12 GHz = 6dB
Path loss is also dependent on freq squared, so it will increase by 6dB at 12 GHz
Pr change = Gt + Gr - Lp = 6 + 6 - 6 = +6dB
CNR = 10 + 6 = 16dB
At 40GHz:-
Increase in Gt at 40 GHz = 20 log 40/6 = 16,5dB
Increase in gain of Gr = 16.5 dB
Path loss increase = 20 log 40/6 = 16.5 dB
Pr change = Gt+Gr -Lp = 16.5 + 16.5 - 16.5 = +16.5 dB
CNR = 10 + 16.5 = 26.5 dB

Hi tech99. Thank you for your reply and help. I'm a little confused on how you went
from
G_ant = 10 log (pi D / lambda)^2 or 20 log (pi D / lambda) = 20 log (pi (8 m) / [(3e8 m/s)/ 12e9 Hz)] ) to
20 log 12/6 above for the Increase in gain of Gt/Gr at 12 GHz.
Would you be willing to explain?
 
  • #4
If we take the log of a number squared, that is the same as twice the log of the number not squared.
10 log N^2 = 20 log N.
I think that is the confusion.
 
  • #5
tech99 said:
If we take the log of a number squared, that is the same as twice the log of the number not squared.
10 log N^2 = 20 log N.
I think that is the confusion.
Not really. I understand that. I’m having trouble understanding how you got the 20log(12/6) and 20log(40/6) parts above.
 
  • #6
The antenna power gain is proportional to frequency squared. If we increase the frequency from 6 to 12 GHz, the power gain rises by (12/6)^2 = 4. Or in decibal form Gain Ratio = 20 log 12/6.
The 40/6 mentioned is the frequency ratio between 6 GHz and 40 GHz.
 
  • #7
tech99 said:
The antenna power gain is proportional to frequency squared. If we increase the frequency from 6 to 12 GHz, the power gain rises by (12/6)^2 = 4. Or in decibal form Gain Ratio = 20 log 12/6.
The 40/6 mentioned is the frequency ratio between 6 GHz and 40 GHz.

I see that now. Thanks for the explanations; this cleared things up. Would you mind if I annoyed you with a follow-up question to this scenerio?

1634246464762.png

As to why this antenna will fail to deliver more power gain at the higher frequency, I can only think of atmospheric attentuation, mostly rain, but the problem says that the weather is clear. I'm not an expert at antenna design, do you have any thoughts?
 
  • #8
Maybe the dish profile is not accurate enough for 40 GHz. At this frequency, the profile would need to be within, say, an eighth of a wavelength, which at 40 Ghz is only a small distance. Lambda = 300/40 = 7.5mm, so Lamba/8 is approx 1 mm.
 

1. What is the C/N problem in dish antennas?

The C/N (carrier-to-noise) problem in dish antennas refers to the issue of signal interference and noise that can affect the quality of the received signal. This can result in a poor signal-to-noise ratio, making it difficult to accurately receive and decode the signal.

2. What causes the C/N problem in dish antennas?

The C/N problem can be caused by a variety of factors, including atmospheric conditions, interference from other signals, and equipment malfunction. It can also be affected by the design and construction of the dish antenna itself.

3. How does the C/N problem affect the performance of dish antennas?

The C/N problem can significantly impact the performance of dish antennas, as it can result in a weaker and less reliable signal. This can lead to errors in data transmission and reception, and can also affect the overall speed and efficiency of the antenna.

4. Are there any solutions to the C/N problem in dish antennas?

Yes, there are several solutions to the C/N problem, including using higher quality materials in the construction of the dish antenna, implementing signal filters to reduce interference, and adjusting the positioning or alignment of the antenna to optimize signal reception.

5. How can I prevent the C/N problem in my dish antenna?

To prevent the C/N problem in your dish antenna, it is important to regularly maintain and inspect your equipment for any potential issues. It is also helpful to research and invest in high-quality dish antennas that are designed to minimize signal interference and noise. Additionally, proper installation and positioning of the antenna can also help prevent the C/N problem.

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