Given a linear Vin, what frequency to assume as input to a differentiator?

[atferrari]

I intend to design a differentiator stage. I measured "all" possible input signals, so to speak, getting the highest rate of a linear signal (steadily increasing) of 265mV in 60 seconds.

If I do 265/60 = 4,42 mV/sec, could I consider my input signal as a triangle of 4,42mV pk - pk / freq= 0,5 HZ?

 

[unclejed613]

i'm having difficulty understanding where you get 0.5hz from a 60 second ramp. frequency is 1/t where t is the period of a repetitive signal. if you input a steady slope into a differentiator, what you get out is a DC offset proportional to the dv/dt of the input. if you input a triangle wave you get a square wave out where the amplitude is determined by the slope of the triangle wave. if you input a square wave you get a spiked wave out because the dv/dt of the leading and trailing edges are very high and the dv/dt of the top and bottom are zero.

 

[gophert]

i'm having difficulty understanding where you get 0.5hz from a 60 second ramp. frequency is 1/t where t is the period of a repetitive signal. if you input a steady slope into a differentiator, what you get out is a DC offset proportional to the dv/dt of the input. if you input a triangle wave you get a square wave out where the amplitude is determined by the slope of the triangle wave. if you input a square wave you get a spiked wave out because the dv/dt of the leading and trailing edges are very high and the dv/dt of the top and bottom are zero.

Slope of the 0.5Hz triangle wave is 265mV/minute
So, at 0.5Hz, the voltage change over 1 sec rise is, as the The OP said, is 4.42mV and the fall time is 4.42mV.

 

[atferrari]

i'm having difficulty understanding where you get 0.5hz from a 60 second ramp. frequency is 1/t where t is the period of a repetitive signal. if you input a steady slope into a differentiator, what you get out is a DC offset proportional to the dv/dt of the input. if you input a triangle wave you get a square wave out where the amplitude is determined by the slope of the triangle wave. if you input a square wave you get a spiked wave out because the dv/dt of the leading and trailing edges are very high and the dv/dt of the top and bottom are zero.

Dear Uncle:
Have you read my post as a question? It is.

Given that everywhere where the differentiator is explained, a reference is made to frequency to decide on the design. I used the rate of change in one second. Not knowing any better I am asking if that could be the way.
If not, what?

 

[gophert]

Dear Uncle:
Have you read my post as a question? It is.

Given that everywhere where the differentiator is explained, a reference is made to frequency to decide on the design. I used the rate of change in one second. Not knowing any better I am asking if that could be the way.
If not, what?

It really depends on the noise that dilutes your signal. Noise levels of 4mV or more are common so you may want to use a time interval of 10 seconds so you see 40mV over the time period. Ultimately, you'll have to play with the system to make sure you are not making corrections to your system that are actually caused by noise and not from an input. If your system is very tightly designed and immune from noise, then 1 second or even 0.1 second could be possible. Also, you'll have to do what make sense to the system. Updating PID parameters of a ship's throttle at 0.1 second accuracy doesn't make sense but that is not enough for a fast line-following robot.

 

[atferrari]

It really depends on the noise that dilutes your signal. Noise levels of 4mV or more are common so you may want to use a time interval of 10 seconds so you see 40mV over the time period. Ultimately, you'll have to play with the system to make sure you are not making corrections to your system that are actually caused by noise and not from an input. If your system is very tightly designed and immune from noise, then 1 second or even 0.1 second could be possible. Also, you'll have to do what make sense to the system. Updating PID parameters of a ship's throttle at 0.1 second accuracy doesn't make sense but that is not enough for a fast line-following robot.

In fact gophert, my circuit is already built but while tuning I realized that I still have to struggle with noise, say, everywhere. Behavior of P & I parts seem reasonable but then discovered that D does not work. Literally 0 output. That is why I decided to revise my design.

For the bolded parts above, could you explain what do they mean? Where to apply that time interval? At lost here.

Just in case, note that this is purely analog. All TL07X opamps.

Thanks for your help!!

 

[unclejed613]

a reference is made to frequency to decide on the design.

i'm from the "old school" so forgive me if i'm not familiar with that design method... when i learned differentiators it was in terms of time constants (the flip side of frequency)... as an example, a common use of differentiators and integrators was in extracting sync pulses from broadcast video... the signal specs were published in terms of time measurements (milliseconds and microseconds) with the frequency often shown in brackets.... this is because engineers at the time designed differentiators and integrators in time constants (which is actually a very straightforward calculation of 0.69RC without the extra step of inverting for frequency 1/(0.69RC))...

it's an approach which makes even more sense today than it did then because microprocessors use time calculations to construct analog signals...

 

[atferrari]

i'm from the "old school" so forgive me if i'm not familiar with that design method... when i learned differentiators it was in terms of time constants (the flip side of frequency)... as an example, a common use of differentiators and integrators was in extracting sync pulses from broadcast video... the signal specs were published in terms of time measurements (milliseconds and microseconds) with the frequency often shown in brackets.... this is because engineers at the time designed differentiators and integrators in time constants (which is actually a very straightforward calculation of 0.69RC without the extra step of inverting for frequency 1/(0.69RC))...

it's an approach which makes even more sense today than it did then because microprocessors use time calculations to construct analog signals...

Thanks for returning Uncle.
I hardly could make any additional remark. With little actual experience on differentiator all what I have read in designing them is explained base on the frequency of the input.

Thanks for taking the time to assist.