What Is the RMS Amplitude of a Sinusoid in a Noisy Environment?

[TomUIC]

Q: A single sinusoidal signal is found in a large amount of noise. If the RMS value of the noise is 0.5 volts and the SNR is 10 dB, what is the RMS amplitude of the sinusoid?

I've related the question to the following equation: SNR=20log(singal/noise)dB

My answer is 1.58 volts, BUT I'm not entirely sure if that is what they are asking for. Just looking for a little guidance on the question.

 

[NascentOxygen]

Hi TomUIC. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

20.log(1.58/0.5) = 10 dB so your answer looks right.

 

[Windadct]

Hmmmm ... the answer is exactly 10 ... homework?

 

[TomUIC]

Hmmmm ... the answer is exactly 10 ... homework?

Yes homework, what relationship am I missing? I'm new to this area so any help is appreciated.

 

[rezeew]

I would like to clarify that the RMS amplitude of a sinusoid refers to the root-mean-square amplitude, which is the effective amplitude of the signal. This is calculated by taking the square root of the mean of the squared amplitudes of the signal over a given time period.

In this scenario, the RMS amplitude of the sinusoid can be calculated by first converting the SNR from decibels to a ratio. Using the given equation, we can rearrange it to solve for the signal-to-noise ratio (SNR) as follows:

SNR = 10^(SNR/20) = 10^(10/20) = 10^(0.5) = 3.16

Now, we can use this ratio to calculate the RMS amplitude of the sinusoid by multiplying it by the RMS value of the noise (0.5 volts).

RMS amplitude of sinusoid = 3.16 * 0.5 volts = 1.58 volts

Therefore, the RMS amplitude of the sinusoid in this scenario is 1.58 volts. It is important to note that this value represents the effective amplitude of the sinusoid in the presence of noise and not the actual peak amplitude of the sinusoid.