Signal to Noise Ratio: in Probability of Detection

Status
Not open for further replies.
T

tabraiz

Full Member level 3
Joined
Nov 23, 2007
Messages
165
Helped
14
Reputation
28
Reaction score
14
Trophy points
1,298
Location
USA
Activity points
2,008
In calculating PD, for a n-sample case we assume that the noise will be reduced by increasing the samples, Mathematically this is right n(SNR) but how is that conceptually right ???
:roll:
 

Hi,

SNR is conceptually defined by the equation : SNR = 6.02N + 1.76 dB.

Meaning- higher the resolution (or sampling bits), higher is the S to N ratio- meaning higer is the numerator part of the SNR (that is Signal sampling quality). Hence the Noise will be reduced with increase of sampling bits. From the same equation it has been found that each extra bit of sampling will result drop in in noise power by 6dB.

At the same time please note that Noise power is independent of sampling frequency. Meaning it does not mean that high speed ADC devices will have better noise performance.

Hope this helps.
 
But this is signal power-to-quantization noise power ratio not SNR (short of Signal-to-Noise Ratio). The former measures the performance of the quantizer, while the later measures the performance of the receiver.
 

Hi David,

I believe we are discussing about a quantizer. Please correct me if I am wrong. The question is to see why noise power will reduce by increasing the sample bits (quantization)...
 

Yes, I know. But the term used is not the most correct one in my opinion.

Simply put, when we have more bits/sample, this means that we have more levels for the quantizer. Imagine you have the real line part from -10 to 10, and you divide it to 2 equal segments, and each segment is represented by a point. This will have large quantization error, because say the representatives are -5 and 5, then any number between 0 and 10 will be taken as 5, and any number between -10 and 0 will be taken as -5. So, the maximum error here is 5. On the other hand, if we divide this line to 8 equally segments, the maximum error is reduced to 1. Asymptotically, as the number of levels (segments) goes to infinity the quantization error goes to zero.
 
Reactions: cks3976

    cks3976

    Points: 2
    Helpful Answer Positive Rating
Hi tabraiz,

it is not clear what do you mean as "n-sample case". Can you clarify?
Nevertheless, i guess that it has to do with averaging signals or random variables.
In that case, signals are added coherently (always the same value), but noise contributions are added noncoherently (with random variations).
regards

Z
 

Dear Friends,
I appreciate your response.
We are discussing a problem related with Communication Receiver or a simple radar receiver.
 

the reduction of time interval between samples raises the possible bandwidth of the signals being digitized and reduces quantization noise.
Averaging can be used to improve SNR from random noise by root of N sample ratio . Depends if noise is due to Rayleigh Fading or Rician fading(coherent noise) vs Rnd Noise.
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…