there are many ways to reduce the signal-to-noise ratio in NMR.

here, we will show the simplest one – by measuring it many times.

the reason behind this is the different of statistic nature between signal and noise.

for signal, the correlation between each measurement is strong. or to say, if you measured the signal is around 10 this time, it will probable measure it around 10 next time.

however, the noise is different, it is random in nature. so, in average, it will equal to zero.

in mathematics, we use the standard deviation $\sigma$ and covariance $cov$ to show it.

$\sigma(S_T)=\sigma(\sum S_i) = \sqrt{ \sum ( \sigma( S_i)^2 + 2cov(S_i,S_j) ) }$

$\sigma(N_T)=\sigma(\sum N_i) = \sqrt{ \sum ( \sigma( N_i)^2 + 2cov(N_i,N_j) ) }$

since the signal should be correlated, $cov(S_i , S_j) = \sigma(S_i) \sigma(S_j)$ thus,

$\sigma(S_T) = n \sigma(S_i)$

but the noise should have no correlation, $cov(N_i, N_j) = 0$

$\sigma(N_T) = \sqrt{n} \sigma(N_i)$

Thus, the signal-to-noise level will be

$\sigma( S/N _T ) = \sqrt{n} \sigma (S/N)$

For 100 times measurements, the signal-to-noise ratio will be 10 times stronger.