Signal to Noise Ratio: Questions for my multimedia and signals class

[Whoohw]

Question one

Homework Statement

x[n] = 5sin(0.22*pi)
The root mean square (RMS) value eof a sin wave is known to be A/root(2) where A is the amplitude. If the noise in a system is known to have an RMS value of 0.5, what si the signal to noise ration (in dB) of the x[n] in this syetem.

Homework Equations

RMS = A/root(2)
SNR = (Asignal/Anoise)^2
SNRdB = 10log10(SNR)

The Attempt at a Solution

Well, i figured that "A" for the RMS given (0.5) is root(2)/2

However, i do not know that which A is Asignal and which is Anoise (root(2)/2 or 5*sin(.22*pi) or just 5).

 

[Mark44]

Question one

Homework Statement

x[n] = 5sin(0.22*pi)

Shouldn't the formula above have n on the right side? Otherwise, all your values of x[n] are exactly the same.

The root mean square (RMS) value eof a sin wave is known to be A/root(2) where A is the amplitude. If the noise in a system is known to have an RMS value of 0.5, what si the signal to noise ration (in dB) of the x[n] in this syetem.

Homework Equations

RMS = A/root(2)
SNR = (Asignal/Anoise)^2
SNRdB = 10log10(SNR)

The Attempt at a Solution

Well, i figured that "A" for the RMS given (0.5) is root(2)/2

However, i do not know that which A is Asignal and which is Anoise (root(2)/2 or 5*sin(.22*pi) or just 5).

 

[Whoohw]

Thats what I keep thinking, but i that isn't what the problem has in the notation.

 

[Mark44]

Then I'd be willing to bet that it's a typo. Otherwise you don't have a sine wave. In that case x[n] is equal to about 3.1871 for all n. IOW, it's constant.

 

[DeeNos]

I would suggest that you clarify with your instructor or classmates which value represents the signal and which represents the noise in this specific problem. It is important to have a clear understanding of the variables before attempting to solve the problem. Once the values are clarified, you can use the equations provided to calculate the SNR in dB. Keep in mind that the SNR represents the ratio of the signal to the noise, so a higher value indicates a stronger signal compared to the noise.