frequency of a discrete signal and differentiators 1

[cmb042]

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

 

[Noway2]

A couple of points:

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.

 

[electricpete]

Noway hit the nail on the head. You have a harmonic non-sinusoidal waveform. What you calculated is the fundamental frequency (within a factor of 2). Since the waveform is not sinusoidal there will be harmonics. In general waveforms with sharp corners have lots of high order harmonics.

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[cmb042]

Ok so I need the frequency spectrum, thank you very much. I think I was trying to get away with just using the Nyquist rate but reading up on it I think I see where I went wrong. The Nyquist rate only applies if the original function is continuous.

 

[benta]

"randomly spaced pulses"
How will you relate this to a "frequency"?

Benta.

 

[cmb042]

The way I understand it, it relates to a frequency as a rate of change. The capacitor does not really care if it is periodic, only how fast it is changing. The formulas I have show you how to calculate corner or cuttoff frequency. So if I find the major frequency components of the signal, hopefully they don't drift too much, then I can tune the circuit to attenuate the rest. I hope I have it right.

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.

 

[Noway2]

If I may inquire, what is your objective? It looks like you are running into issues with some related concepts and "we" may be able to help you with your approach.

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.

 

[cmb042]

Thanks for you replys Noway2.

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.

 

[Skogsgurra]

It is only natural that you find a spread low frquency spectrum. Most of the energy in your signal is, as you say, in the 10 to 200 Hz region. Plus probably (because of your saw-tooth wave) odd harmonics up to tens of kHz. Your fast edges do also produce a spectrum, but you will need to set your oscilloscope timebase accordingly short. The FFT doesn't show lines any higher than the Nyquist frequency and that is dependent of the timebase setting.

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

http://www.gke.org

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