Question

5. The following data were taken from a voltage measurement on a nosy system : 1.37, 1.84,1.47, 1.10, 1.73, 154, 1.08 Call in my). The standard deviation (6) of the data set is 0.271 mv. (a) Assuming the noise is random , what is the S/N rative (b) How many measurements need to be averaged in order to achieve S/N=10?

Answer 1

(a)

Assuming noise is random :

The signal to noise ratio will equal the mean divided by the standard deviation.

mean of these data : 10.13 /7 = 0.145 mV

Standard deviation : 0.271 mv

S/N ratio =  1.447 mV / 0.271 mv = 5.34

(b) Measurements would have to be averaged to increase the S/N to 10 :

Since,  signal to noise ratio is equal the mean divided by the standard deviation.

standard deviation is proportional to (1/N )^1/2 (N= no. of measurement ; N1 = 7)

So, signal to noise ratio increases with the square root of the number of measurements :

we have (S/N)₂ = (S/N)₁ * (N₂ /N₁ )^1/2

10 = 5.34 * (N₂ /7 )^1/2

(N₂ )^1/2 = 4.954

so, no. of Measurements would have to be averaged to increase the S/N to 10

N₂ ~ 25