# Signal to Noise Ratio Calculation in Decibels: How to Simplify with Decibels

• Comp Sci
• Jason-Li
In summary: Thanks so much for your help!Yes, the formula for F is not stated in decibels. To divide decibel quantities we subtract, not divide.In summary, the voltage required on the input of the amplifier to give the maximum output of 85 dBμV is 5 volts. The signal level from the aerial is 5 dBmV and the input noise level is 20 dBμV, so the signal-to-noise ratio on the output of the amplifier is 45 dB.
Jason-Li
Homework Statement
(a) Determine the voltage required on the input of the amplifier to give
the maximum output of 85 dBμV.
(b) If the signal level from the aerial is 5 dBmV and the input noise level
is 20 dBμV, calculate the signal-to-noise ratio on the output of the
amplifier.

Values:
Bandwidth 40–862 MHz
Gain 20 dB
Noise Figure 6 dB
Max. Output 85 dBV
Input Impedance 75Ω
Output Impedance 75Ω
Relevant Equations
SNR=20Log(Vs/Vn)
SNR=10Log(Ps/Pn)
So pretty confident I understand part (a) however for part (b) I'm not sure if I have carried it out correctly if someone could give me a pointer?

(b)
5dBmV Input as a voltage:
5=20Log(V/1mV)
V=105/10
V=1.77827941mV

Then the noise level is 20dbμV so changing to a voltage:
20=20*Log(V/1μV)
V=1020/20
V=10μV

Then I can find the SNR of the input using:
SNRIn=20*Log(Vp/Vn)
SNRIn=20*Log((1.77827941*10-3)/(10*10-6))
SNRIn=45dB

Then as I have the Noise figure from the data as F=6dB I can do the following:
F=SNRIn/SNROut
SNROut=SNRIn/F
SNROut=45*6
SNROut=270dB

This seems far too high? I have found websites that say the formula is actually F = SNRIn-SNROut however the learning materials say:

I think you are correct as far as part (b). Then you know the SNR at the input, and the SNR at the output will be 6dB worse. So SNRout = 45 - 6 = 39dB.

Jason-Li
tech99 said:
I think you are correct as far as part (b). Then you know the SNR at the input, and the SNR at the output will be 6dB worse. So SNRout = 45 - 6 = 39dB.
Hi thanks for the reply and assurance, do you have any idea why my learning materials is citing the noise figure equation incorrectly?

Yes, the formula for F is not stated in decibels. To divide decibel quantities we subtract, not divide.

Merlin3189 and Jason-Li
Jason-Li said:
Homework Statement:: (a) Determine the voltage required on the input of the amplifier to give
the maximum output of 85 dBμV.
(b) If the signal level from the aerial is 5 dBmV and the input noise level
is 20 dBμV, calculate the signal-to-noise ratio on the output of the
amplifier.

Values:
Bandwidth 40–862 MHz
Gain 20 dB
Noise Figure 6 dB
Max. Output 85 dBV
Input Impedance 75Ω
Output Impedance 75Ω
Relevant Equations:: SNR=20Log(Vs/Vn)
SNR=10Log(Ps/Pn)

So pretty confident I understand part (a) however for part (b) I'm not sure if I have carried it out correctly if someone could give me a pointer?

(b)
5dBmV Input as a voltage:
5=20Log(V/1mV)
V=105/10
V=1.77827941mV

Then the noise level is 20dbμV so changing to a voltage:
20=20*Log(V/1μV)
V=1020/20
V=10μV

Then I can find the SNR of the input using:
SNRIn=20*Log(Vp/Vn)
SNRIn=20*Log((1.77827941*10-3)/(10*10-6))
SNRIn=45dB

Then as I have the Noise figure from the data as F=6dB I can do the following:
F=SNRIn/SNROut
SNROut=SNRIn/F
SNROut=45*6
SNROut=270dB

This seems far too high? I have found websites that say the formula is actually F = SNRIn-SNROut however the learning materials say:

View attachment 290575

Hi, I am struggling with part A to this question, could you assist? I am not sure where to start.

In Part (a) the output of the amplifier is 85 dBuV and the gain of the amplifier is 20 dB. How many dBuV aat the input?

Do I need to convert dB into Voltage for both gain and Vo and then just use a transposed ratio for gain to calculate Vi?

No you can stay in dB entirely. Do not over think this one!

With gain just being a ratio, is it as simple as G = output / input... transformed giving 85 / 20 = Vi (however needing to subtract rather than divide) 85 - 20 = Vi?

Last edited:
When we multiply or divide quantities using decibels we simply add or subtract. So for instance, if I have a signal of 10 dBV and I amplify it by 6 dB I end up with a signal of 10 + 6 = 16 dBV.
The whole point of decibels is to make gains and losses simple to work out without the need to carry out multiplication or division.
It is only if voltages or powers are actually added or subtracted that we have to come out of the decibel system.

## 1. What is signal to noise ratio (SNR)?

Signal to noise ratio (SNR) is a measure of the strength of a signal compared to the level of background noise present. It is typically expressed in decibels (dB) and is used to assess the quality of a signal in various fields such as telecommunications, electronics, and audio engineering.

## 2. How is SNR calculated?

SNR is calculated by dividing the power of the signal by the power of the noise. This can be done using the following formula: SNR = 10log10(Psignal/Pnoise), where Psignal is the power of the signal and Pnoise is the power of the noise.

## 3. What is a good SNR value?

A good SNR value depends on the specific application and the level of noise that is acceptable. In general, a higher SNR value indicates a stronger and more reliable signal. For example, a SNR of 20 dB is considered good for audio signals, while a SNR of 40 dB or higher is considered excellent for wireless communications.

## 4. How can SNR be improved?

SNR can be improved by increasing the power of the signal, reducing the level of noise, or a combination of both. This can be achieved through various methods such as using higher quality components, reducing interference, and implementing signal processing techniques.

## 5. What are the limitations of SNR?

SNR is a useful metric for evaluating the quality of a signal, but it does have some limitations. It does not take into account the type of noise present, which can affect the perceived quality of the signal. Additionally, SNR does not consider other factors such as distortion and frequency response, which can also impact the overall signal quality.

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