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WIDEBAND (BROADBAND) TECHNOLOGY

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  • Qiaozhi
    replied
    Originally posted by Tinkerer View Post
    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.

    Leave a comment:


  • Tinkerer
    replied
    Originally posted by mikebg View Post
    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

    Leave a comment:


  • mikebg
    replied
    Originally posted by tony_av View Post
    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."

    Leave a comment:


  • tony_av
    replied
    Nice joke

    Leave a comment:


  • mikebg
    replied
    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.
    Attached Files

    Leave a comment:


  • Davor
    replied
    Originally posted by mikebg View Post
    Davor, Your TGT is LPF but My TGT is HPF.
    You are right of course. I thought of HPF, but fingers were faster

    Leave a comment:


  • mikebg
    replied
    RAMP RESPONSE

    Originally posted by Davor View Post
    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:

    General tech discussions on all types of metal detectors: VLF, 2-box, BFO, off-resonance, PLL, etc. Questions, ideas, and anything else that moves you.


    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 ?
    Attached Files

    Leave a comment:


  • ODM
    replied
    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.

    Leave a comment:


  • Davor
    replied
    Originally posted by mikebg View Post
    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.

    Leave a comment:


  • Bill512
    replied
    Originally posted by mikebg View Post
    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.

    Leave a comment:


  • mikebg
    replied
    Originally posted by Bill512 View Post
    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.

    Leave a comment:


  • Bill512
    replied
    Originally posted by mikebg View Post
    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.

    Leave a comment:


  • ODM
    replied
    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.

    Leave a comment:


  • mikebg
    replied
    LOW DUTY CYCLE

    Originally posted by Bill512 View Post
    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:
    Attached Files

    Leave a comment:


  • Bill512
    replied
    Originally posted by mikebg View Post
    I need software that can show me what happens in Frequency domain when amended parameter "b" (explained in post # 9).

    Can somebody show me a link where I can download for free a signal generator with Frequency analysis similar to that used by Bill512 in posts #5 and #6?

    In next posts I should show the frequency spectrum of a "ringing" TX pulsation (underdamped PI metal detector or without damping resistor).
    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 ...")

    Leave a comment:

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