frequency of a discrete signal and differentiators
frequency of a discrete signal and differentiators
(OP)
I have a input signal of randomly spaced negative sawtooth-like pulses of varying amplitude a few microseconds in duration. This is fed to a differentiator. I understand the capacitor and feedback resistor of the differentiator select the corner frequency of the amplifier circuit. What I am not sure of is what exactly is the frequency of my input.
I thought it was f=1/2*period but when I use that value, I get small resistances that does not jive with what I have seen in the lab. The period in this case is the duration of the pulse correct?
Also, is there any way to increase the gain of a differentiator? Can I make the output lets say 5 times the derivative without another opamp?
Thanks
I thought it was f=1/2*period but when I use that value, I get small resistances that does not jive with what I have seen in the lab. The period in this case is the duration of the pulse correct?
Also, is there any way to increase the gain of a differentiator? Can I make the output lets say 5 times the derivative without another opamp?
Thanks





RE: frequency of a discrete signal and differentiators
1 - Understand that a physical differentiator is a very noisy circuit and this can lead to MANY problems. It would be a better idea to do the differentiation in software, but even in this domain it can be a source of significant noise.
2 - Regarding the "frequency" of your input, there are several ways in which you could interpret the meaning of the frequency. I am assuming that you are refering to the signal bandwidth. Your sawtooth signal can be thought of as being comprised of a sum of sinewaves of the fundamental and its phase shifted harmonics. The easiest way to measure this would be with a spectrum analyzer or to perform an FFT calculation on the data.
2A - Frequency = 1/period, not 1/2*period. This definition would apply to the concept of the sawtooth signal being a singular event, or in relation to a singal tone (sinewave) of the composite signal.
RE: frequency of a discrete signal and differentiators
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RE: frequency of a discrete signal and differentiators
RE: frequency of a discrete signal and differentiators
How will you relate this to a "frequency"?
Benta.
RE: frequency of a discrete signal and differentiators
Now I do not have a spectrum analyzer at my disposal, but I may have a FFT function in my O-scope. If not maybe I can build a FFT circuit.
RE: frequency of a discrete signal and differentiators
Are you working with a purely analog circuit or does the signal get sampled by an ADC for use in a processor?
Regarding a comment from your original post, you asked about making the derivative 5x larger without a second opamp. The problem you will run into is that the differentiator is frequency selective and the gain will be too. Yes, it will have a "cutoff" freqeuncy - as it is a high pass filter in the frequency domain.
If the signal is processed by an ADC, the differentiation and the gain can be handled digitally which will eliminate a lot of the problems you are running into.
The nyquist rate will apply if you are sampling a countinous (analog) signal with an ADC. In order to prevent aliasing, it will be necessary to sample at a rate >2x the highest signal component. Depending on your application aliasing may or may not be a problem. If aliasing is a problem then performing the differentiation in the processor, which amounts to a series of subtractions and a scaling, will help in this regard.
Based on what I suspect is going on, I think a better approach would be to re-arrange your differentiator into a low pass (anti-aliasing) filter to limit the bandwidth then and process the signal digitally.
RE: frequency of a discrete signal and differentiators
I am working with a purely analog signal. There is no software. I have a sensor that operates at high voltages, so the capacitor of the differentator is more of a neccessity of DC isolation than a desire for a derivitive output. The sensor outputs 10-200 pulses a second. The output of the amplifier is input to a multivibrator that outputs larger longer square pules to a data collection device.
I have found the FFT function of my O-scope and to my shagrin it looks like the main components of my signal are low frenquency and spread across the spectrum. I had hoped to find a large tight high frenquency component like I remembered from my labs in college. But at least I have a starting point.
RE: frequency of a discrete signal and differentiators
If you want to see the spectrum of the fast edges, just trig on the edge and make the display show just the edge. Then go FFT to see the spectrum.
Gunnar Englund
www.gke.org
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