Finding SNR when only SNRdB and Bandwidth is known...

[audio_mosquito]

Given: Bandwidth = 200KHz, SNRdB = 4

What is the theoretical capacity of the channel?

capacity = bandwidth x log2 (1 + SNR)

I need to find SNR when only SNRdB is known, but how do I do that?

Thanks in advance.

Helpful formulas below.
=================

SNR = average signal power / average noise power

SNRdB = 10 log10 SNR

 

[avaserfi]

Given: Bandwidth = 200KHz, SNRdB = 4

What is the theoretical capacity of the channel?

capacity = bandwidth x log2 (1 + SNR)

I need to find SNR when only SNRdB is known, but how do I do that?

Thanks in advance.

Helpful formulas below.
=================

SNR = average signal power / average noise power

SNRdB = 10 log10 SNR

Using your formulas you would have:

SNRdB (4) = 10 ln(SNR)
4/10 = ln SNR - the opposite function of Log is e^ so:
e^.4 = SNR
SNR = 1.49

Pretty sure my math is right there. Just plug 1.49 into the other formula and it should work out. Someone please correct me if I am wrong.

Finding capacity:

capacity = bandwidth x log2 (1 + SNR)
C = 200 x [log(1.49)/log(2)] (the latter is how you find log base 2 of a specific number)
C = 115.42

Let me know if specific clarification is needed or if I made a mistake.

edit: I made a mistake in the math and fixed it.

 

[audio_mosquito]

Thanks a lot avaserfi, your logic seems correct to me as I was thinking similarly myself, but wasn't positive.

If you don't mind, let me know if the following logic makes sense to you...

A signal with 200 milliwatts power passes through 10 devices, each with an average noise of 2 microwatts.

snr = 200 / 2 = 100

SNR db = 10 log10 100 = 20

I'm just wondering if I need to do anything differently with the calculations since the signal is passing through ten devices.

Thanks again.

 

[avaserfi]

I don't think passing through multiple devices will make a difference in calculations, but I am unsure to be honest.

One thing to note log10 is log base 10 which is the natural log (ln on a calculator this is not the same as hitting the log button). So there is a slight problem with your calculations. ln 100 = 2.30 so SNRdB = 23.0.

 

[Adam]

One thing to note log10 is log base 10 which is the natural log (ln on a calculator this is not the same as hitting the log button). So there is a slight problem with your calculations. ln 100 = 2.30 so SNRdB = 23.0.

I think that you just mistyped on this, Andrew. Log base 10 is not the natural log. Also, ln(100) = 4.6.

Sorry if I'm misunderstanding and butting in.

 

[avaserfi]

I think that you just mistyped on this, Andrew. Log base 10 is not the natural log. Also, ln(100) = 4.6.

Sorry if I'm misunderstanding and butting in.

You are dead on. For some reason I remembered learning that in highschool, but I guess I remembered wrong.

Here are corrected calculations:

SNRdB (4) = 10 log(SNR)
4/10 = log SNR - the opposite function of Log is 10^ so:
10^.4 = SNR
SNR = 2.51

Finding capacity:

capacity = bandwidth x log2 (1 + SNR)
C = 200 x [log(3.51)/log(2)] (the latter is how you find log base 2 of a specific number)
C = 362.29KHz

 

[Adam]

My calculation of capacity is slight different.

SNR = 2.51 (I get the same answer as you)

From the given equation for capacity:

capacity = bandwidth x log2 (1 + SNR)​

I get:

capacity = 200 kHz x log2(1 + 2.51)​

capacity = 200 kHz x log2(3.51)​

capacity = 200 kHz x 1.81 (log calculated using Excel and verified as 2^1.81 = 3.51)​

capacity = 362 kHz​

Please check that, as it differs from yours.

EDIT: I just noticed where we differ. You took log2(2.51) instead of log2(1+2.51). That's all.

 

[avaserfi]

My calculation of capacity is slight different.

SNR = 2.51 (I get the same answer as you)

From the given equation for capacity:

capacity = bandwidth x log2 (1 + SNR)​

I get:

capacity = 200 kHz x log2(1 + 2.51)​

capacity = 200 kHz x log2(3.51)​

capacity = 200 kHz x 1.81 (log calculated using Excel and verified as 2^1.81 = 3.51)​

capacity = 362 kHz​

Please check that, as it differs from yours.

You got me again. I am trying to multitask and not doing a good job today. Looking over my previous work I did log(2.51)/log(2) when it should have been 3.51. I just did log2(2.51) not log2(1+2.51).

Yet again you are keeping me honest . I have fixed my previous post.

 

[Adam]

You got me again. I am trying to multitask and not doing a good job today. Looking over my previous work I did log(2.51)/log(2) when it should have been 3.51. I just did log2(2.51) not log2(1+2.51).

Yet again you are keeping me honest . I have fixed my previous post.

Yeah, I noticed the difference about the same time that you did, I think. I put an edit in my post, but not fast enough.

You probably still have some euphoria from this morning's ascension to The Couch. No one will blame you for being a little off.

 

[audio_mosquito]

Yeah, 362Kbps is the theoretical highest capacity (bit rate) for the given bandwidth and noise. The formula is called the Noisy Channel: Shannon Capacity.

To calculate the theoretical maximum capacity or bit rate for a noiseless channel, you can use the Noiseless Channel: Nyquist Bit Rate.

BitRate = 2 x bandwidth x log2 L

L = number of signal levels used to represent data

 

[AcuDefTechGuy]

What is this, a physics forum or something?
What are you guys thinking?
I've taken Engineering physics I, Calculus I, II, III, Engineering Math I, and you guys are giving me really bad FLASHBACKS!!!!
Stop it.

 

[Adam]

...you guys are giving me really bad FLASHBACKS!!!!
Stop it.

Partial differential equations...

If that's not a horrible flashback, then it's an even more horrible foreshadow (assuming that you're still in engineering studies).

 

[AbyssalLoris]

For the signal passing through 10 devices, the calculation appears to be somewhat wrong. The signal power is 200mW and the noise is 2 micro watt. So the SNR by that calculation would be 200/2x10e-3 = 100000, and SNR (dB) would be 10log(100000) = 50.

Anyway, the question is not very clear at all. What does the signal passing through 10 devices mean? Does it pass in series or are the 10 devices in parallel? What happens to the signal as it passes through the device? Is it amplified, attenuated or preserved? This will matter depending on the point in the chain where you are calculating SNR, for example if you want SNR at the output.

 

[AcuDefTechGuy]

Partial differential equations...

If that's not a horrible flashback, then it's an even more horrible foreshadow (assuming that you're still in engineering studies).

Oh no, no, no. That was 15 years ago!
But I still have nightmares

 

[avaserfi]

Partial differential equations...

If that's not a horrible flashback, then it's an even more horrible foreshadow (assuming that you're still in engineering studies).

I am the one still taking math classes here , kind of, statistics mostly, but I have had my fair share of calculus ! Statistics sucks just as much though its just a different way of thinking about things.

 

[AcuDefTechGuy]

I am the one still taking math classes here , kind of, statistics mostly, but I have had my fair share of calculus ! Statistics sucks just as much though its just a different way of thinking about things.

When I was still in school, I was a great "A" student. But looking back, I just don't know the heck I did it. I hate school. I really do. It seems like the only kind of nightmares I have these days are of myself studying for an exam!
Okay, that's it. This thread is going to cause me to have nightmares now.

 

[audio_mosquito]

For the signal passing through 10 devices, the calculation appears to be somewhat wrong. The signal power is 200mW and the noise is 2 micro watt. So the SNR by that calculation would be 200/2x10e-3 = 100000, and SNR (dB) would be 10log(100000) = 50.

Anyway, the question is not very clear at all. What does the signal passing through 10 devices mean? Does it pass in series or are the 10 devices in parallel? What happens to the signal as it passes through the device? Is it amplified, attenuated or preserved? This will matter depending on the point in the chain where you are calculating SNR, for example if you want SNR at the output.

Yeah I forgot to convert 2 microwatt to milliwatt (2uW x 10^-3 = .002mW).

SNR = 200 / .002 = 100,000
SNRdB = 10log(100,000) = 50

Thanks for catching that.

Concerning the wording of the question, I agree more information is needed as the question is rather ambiguous without specifics.