Question

Please help Instrumental Analysis

Answer 1

I don't know how to do this, but I was told it was easy?

Answer 2

a) rms noise is just the squared root of average of the square of the points around the noise, so eyeballing it:

\(\sf rms ~noise=\sqrt{\dfrac{0.001^2+(-0.001)^2+0.001^2+(-0.001)^2}{4}}=0.001\)

b) the Signal-to-noise ratio is just what the name describes, \(\sf\dfrac{0.0036}{0.002}=1.8\)

c) the lowest low pass filter should be half of the width of the peak, so 5 ms. This is in order to be able to capture the signal, you need at least 3 points,

Answer 3

The thing I was getting confused on was that my professor gave me this:
\[rms=\frac{P-P_{nosie}}{6}\]
and I was supposed to get this from taking a ruler to the data given.
I was also given this:
\[\frac{S}{N}=\frac{S}{rms}\]
the signal of the peak should also be measured.
My book does it the way you have done it so I just don't understand where he is coming from.

Answer 4

idk either, he might be doing it differently for some reason.

the rms looks like he's the average of 6 points.

Answer 5

I'll show you the example he did? Maybe you can explain it to me?

Answer 6

The signal is centered around 1 from the baseline of zero. i'm not sure why he's dividing the noise by 5, he's probably averaging it (because with more measurements, you can reduce the noise) but i'm not sure why he chose 5.

Answer 7

so how exactly would I apply this to the problem I was given