WIDEBAND (BROADBAND) TECHNOLOGY

[mikebg]

LOW DUTY CYCLE

Hi Mike,
nothing fancy,just simple,free and useful :http://www.zeitnitz.de/Christian/scope_en
("... free of charge for private and non-commercial use in educational institutions ...")

I use the same software by Zeitnitz. The difference is in frequency tab title only. The title of mine is "Frequency" instead "Frequency analysis" in post #6. Note what happens at low duty cycle:

[ODM]

Is the low-level non-harmonic content in these Zeitnitz spectral pictures some folded aliasing, or some other electronic/calculation artefact? It is present on the sound-card-software oscilloscope especially on high harmonic content waveforms, but the "proper" Agilent scope (#10) doesn't show any with its broad bandwidth and input band limiting.

[Bill512]

I use the same software by Zeitnitz. The difference is in frequency tab title only. The title of mine is "Frequency" instead "Frequency analysis" in post #6. Note what happens at low duty cycle:

Mike, the envelope frequency is always there, just in an other position.
If your main period is T, and the pulse on time is c, then the envelope has a freq of 1/c.
In 50% duty cycle the envelope is the even harmonics.
In a 6% duty cycle, is 25kHz.

[mikebg]

Mike, the envelope frequency is always there, just in an other position.
If your main period is T, and the pulse on time is c, then the envelope has a freq of 1/c.
In 50% duty cycle the envelope is the even harmonics.
In a 6% duty cycle, is 25kHz.

If you use the term "envelope" instead "fundamental frequency", this is not true. The fundamental frequency is always 1 / T independent on duty cycle. For example, if you have a PI machine radiating 3000 TX pulses per second, the period T=333us. In frequency domain that means:
This is a wideband metal detector with fundamental frequency 3kHz and target is illuminated with 3kHz, 6kHz, 9kHz etc. Duty cycle changes amplitudes of fundamental and harmonics, but not their frequencies.
If "envelope" means spectral density, here we have a discrete spectrum and this term is not suitable.

[Bill512]

If you use the term "envelope" instead "fundamental frequency", this is not true. The fundamental frequency is always 1 / T independent on duty cycle. For example, if you have a PI machine radiating 3000 TX pulses per second, the period T=333us. In frequency domain that means:
This is a wideband metal detector with fundamental frequency 3kHz and target is illuminated with 3kHz, 6kHz, 9kHz etc. Duty cycle changes amplitudes of fundamental and harmonics, but not their frequencies.
If "envelope" means spectral density, here we have a discrete spectrum and this term is not suitable.

Envelope has nothing to do with the fundamental frequency,or the value of the harmonics.
What I mean with the term "envelope frequency" is that the amplitude of the harmonics is a periodic function of time!
from a different point of view, is a superposition of two frequencies.
For low duty cycle looks like AM modulation signal (after rectification).
and finally for extreme low duty cycles, the envelope goes to infinite and you have the delta function.

[Davor]

The fundamental frequency is always 1 / T independent on duty cycle.

Unless you have a Dirac repeating at 3kHz rate. It would assume some kind of spectrum only if you have some targets to filter it ... assuming targets behave as 1st order LPF.

[ODM]

Still, I'd like to point out the hazard of conclusions based on spectrograms from that program - as they seem to be pretty close to the sort of artefacts you would get from aliasing. This would probably be hazardous to "ideal" purely mathematical analysis.

[mikebg]

RAMP RESPONSE

Unless you have a Dirac repeating at 3kHz rate. It would assume some kind of spectrum only if you have some targets to filter it ... assuming targets behave as 1st order LPF.

RAMP RESPONSE

Davor, Your TGT is LPF but My TGT is HPF.
See that My TGT is more suitable for your reinvention - metal detector operating in LF radio frequency band:

http://www.geotech1.com/forums/showt...162#post103162

At My TGT, the amplitude increases with 20dB/dec in LF radio frequency band.

Here are shown the positive and the negative ramp functions. The positive ramp means that TX current increases linear from minus infinity to plus infinity. The negative ramp represents TX current decreasing linear from plus infinity to minus infinity. The simulation shows that conductivity has no ramp response. Why ?

[Davor]

Davor, Your TGT is LPF but My TGT is HPF.

You are right of course. I thought of HPF, but fingers were faster

[mikebg]

THE SECOND DERIVATIVE

What happens at ramp function of TX current?

The EMV (electromotive voltage) that causes eddy current in a target, is proportional to the derivative of TX current, ie the rate at which magnetic flux changes.
Derivative in TD (time domain) means multiplication with frequency in FD (frequency domain). For EMV induced in RX coil, we need again multiplication with frequency. That means the signal from a conductive target is proportional to second derivative of frequency squared.

Note what happens when TX current changes as ramp function. In this case, a DC voltage causes eddy current. The target operates as a permanent magnet.
It is seen that the linear change of TX current does not create TGT signal at conductive target because the second derivative d2B/dt2 is zero. The AIR signal is also a DC voltage because it is proportional to first derivative.

[tony_av]

Nice joke

[mikebg]

Nice joke

CORRECTION
The third sentence of post #25 instead
"That means the signal from a conductive target is proportional to second derivative of frequency squared." should be:
"That means the signal from a conductive target is proportional to second derivative or frequency squared."

[Tinkerer]

What happens at ramp function of TX current?

The EMV (electromotive voltage) that causes eddy current in a target, is proportional to the derivative of TX current, ie the rate at which magnetic flux changes.
Derivative in TD (time domain) means multiplication with frequency in FD (frequency domain). For EMV induced in RX coil, we need again multiplication with frequency. That means the signal from a conductive target is proportional to second derivative of frequency squared.

Note what happens when TX current changes as ramp function. In this case, a DC voltage causes eddy current. The target operates as a permanent magnet.
It is seen that the linear change of TX current does not create TGT signal at conductive target because the second derivative d2B/dt2 is zero. The AIR signal is also a DC voltage because it is proportional to first derivative.

Professor,
if you are saying that the linear ramp of the triangular TX wave form does not create a signal you are wrong. The words linear ramp imply a linear rate of change in the current. Therefore there there are eddy currents generated.
Further more, you state your self, that there is a linear change of TX current, therefore a rate of change, therefore there are eddy currents generated.

We, the students, ask you to please revise your statements.

Tinkerer

[Qiaozhi]

Professor,
if you are saying that the linear ramp of the triangular TX wave form does not create a signal you are wrong. The words linear ramp imply a linear rate of change in the current. Therefore there there are eddy currents generated.
Further more, you state your self, that there is a linear change of TX current, therefore a rate of change, therefore there are eddy currents generated.

We, the students, ask you to please revise your statements.

Tinkerer

Mikebg is talking about the second derivative of the signal. i.e. the rate of change of the rate of change.
In this case, for the linear slope of the triangular waveform, the second derivative is zero.

[mikebg]

Professor,
if you are saying that the linear ramp of the triangular TX wave form does not create a signal you are wrong. The words linear ramp imply a linear rate of change in the current. Therefore there there are eddy currents generated.
Further more, you state your self, that there is a linear change of TX current, therefore a rate of change, therefore there are eddy currents generated.

We, the students, ask you to please revise your statements.

Tinkerer

Thinker,
You have no reason to call me "Professor", because I'm just a former ham with primitive knowledge for design TXs, RXs and antennas. However I have reasons to call you "Thinker" because you think what I'm saying:

1. You think that ramp functon is part of triangular TX wave. I do not think so. I showed the frequency spectrum of triangular wave in other thread:

http://www.geotech1.com/forums/showt...202#post164202

http://www.geotech1.com/forums/showt...377#post164377

The spectrum of sawtooth wave also is shown here - in this thread, but it also is not ramp function:

http://www.geotech1.com/forums/showt...687#post166687

2. You think that ramp function does not create eddy current. I do not think so. Even connected a battery in eddy current loop to illustrate that eddy current created by source "e" is DC.
http://www.geotech1.com/forums/showt...399#post167399

Participants of this forum are not my students because they (and you) are much more educated than me. Radio clubs in Bulgaria do not issue diplomas. They are trained hams to make primitive analysis in order not make mistakes at design.

Thinker, please think whether your draft contains something stupid before to post it in a worlds forum! Stupid is to think that triangular wave and sawtooth wave contains ramp function. The ramp function is a mathematical abstract for analysis because starts from infinity and goes to infinity. However analisis shows that d2i/dt2 - second derivative of TX current creates signal.
Professor Mike