Instrumentation Question

Major Exam Alert!!!

I don't like how this question is posed. My dislike might be why you are having trouble answering the question. You cannot tell the signal to noise ratio because you cannot calculate the true, random noise components from the information provided. None of the resistors are known. What the question is actually asking for is the output power ratio of the desired 5 Hz signal and the interfering 100 Hz signal. One must assume that none of the circuitry contributes any additional noise. This is, of course, not what happens in the real world. However, initially assuming a noiseless circuit is where one starts a circuit analysis. Additionally the CMRR of a differential amplifier will not really be a flat response. One must also then assume that it will have the same effect at all given frequencies.

Those are the two assumptions you must make to answer this question. I hope this helps.

thank you for the reply...

I have the final answers, I can obtain the part a but I get stock at part b.

the final answer of part b, the one about the SNR at the output of the filter should be 160 dB and I keep getting 120 dB.

do u know why?

thank u

OK, I think I see where you're confusing which tree is which in this forest of numbers.

The 120 dB you calculated is the rejection transfer ratio or attenuation in signal level for any interfering 100 Hz signal. What is the the initial signal to interference ratio?

the SNR at the input is 20 log (1x10^-3/100x10^-6)= 20dB

OK, That puts the SNR at what level after the CMRR amplifier, prior to the low pass filter?

[I was hoping you'd like to catch me in a deliberate error on my part. ]

Excellent leading questions and instruction methods. KUDOS!

I still will not do others homework. However, I will help others who are obviously doing their homework to complete their homework.

this is not a homework...this is simply a question that I cannot find answer to ;) so don't worry, in this case you have been very committed to your believes :)

OK so do you see yet where the extra 40 dB of attenuation comes from?

nope...and I simply hope I don't get it in the exam and if u know and aren't helping....well than at least u should be happy that u didn't do my homework for me ;)

cheers

You are so close to getting this, don't give up. I'll give you a few more pushes.

OK, what does a low pass filter do?

Is the 5 Hz and 100 Hz signals above or below the corner frequency?

What is the significance that your filter has two poles?

believe I have spent so much time on this one question that I came to a point to be absolutely disappointed in myself...a very bad feeling!

OK, what does a low pass filter do?

-a low-pass filter, let the low frequency signals pass and cuts all the high frequencies (lower and higher than cut-off freq)

Is the 5 Hz and 100 Hz signals above or below the corner frequency?

-the corner frequency is defined to be 10 Hz so it is above 5 Hz and below 100 Hz

What is the significance that your filter has two poles?

-well for a one pole filter the cutoff frequency is defined as the -6dB or 0.5 times the magnitude of the signal....but I don't see how these are going to give me answer. I went 10000 different ways...NOT WORKING

-well for a one pole filter the cutoff frequency is defined as the -6dB or 0.5 times the magnitude of the signal....but I don't see how these are going to give me answer. I went 10000 different ways...NOT WORKING

Here's the problem.

You have some basic misunderstandings about analog filters. To properly design an analog filter should be a separate class in your curriculum as the link should imply. However a few of the critical points you misunderstand are in the introduction of just three pages in the link provided. The corner or cutoff frequency is defined as the -3 dB point or half power level. The voltage magnitude will be 1/√2 less. This definition applies regardless of the number of poles your filter has. What the number of poles does effect is the slope of the out of band frequency response. For every octave out of band from the cutoff frequency the signal will drop 6 dB from the in band amplitude for each pole. So a 10 Hz low pass single pole filter will attenuate a 20 Hz signal by -6 dB and a 40 Hz signal by by -12 dB. Another way to describe a slope of 6 dB per octave per pole is 20 dB per decade per pole.

Hopefully this answers your question.

thank you very much...

I am just coming back from my exam and it went ok...I didn't get this question though but it was a big relief going to the exam having a clear idea about this. I had to read it 10 times to understand what is going on really, but it is my brain's problem of putting the story together!

thank u

cheers

You are very welcome.

As our CR4 FAQ states, this site is not a homework cheating site. People who bring obvious homework problems will not be treated kindly by many here, including myself. However, when I see somebody doing the work (as you demonstrated) but just missing something somewhere I will help a student to find their way through a maze. Often this will take several iterations to identify the root of the confusion. The drawback of doing this remotely via a forum is the time it takes will then be multiplied. Finding a person you can sit with will always speed this up. Look for the honors society at your school. They will often have a fellow student or teacher that can help. The IEEE honor society is Eta Kappa Nu (HKN), look for them.

I'm glad I could help, welcome to CR4.

And to think that some of us actually used a slide rule and the back of an envelope for these calculations.

The whole purpose of these exercises is to gain some idea of how the concept works. (Insert favorite deity here) forbid that we have to do more than punch numbers on a calculator... (Don't they call those people accountants?)

Once upon a time... I was badly in need of a calculator, but mine (and all the others in the study) had vanished. Either my wife or daughter had liberated them. BUT!! I went digging in a box and found my slide rule from college days and solved my problem. Luckily, neither wife or daughter know how to use such a critter.

Bill