Energy within an electric field

[BvU]

I have the oscilloscope data. I cropped the data so I have one wavelength of the pulse.

Can you explain what this means ?

Did I miss the answer ?

Yes the pulse is very similar to your image but not exact.

Your reluctance to show the image is almost commendable.

So on the scope you almost see

Re-read post #20: $x = \pi\$ corresponds to t = 75 $\mu$s, so $a = 75 \mu s /\pi\$ $$

10 \,\text{kV} \ \int_{-\infty}^\infty \operatorname {sinc} \left (t\over a\right ) \, dt= 0.75\,{\text Vs} $$

What would you like a picture of? The setup is a pulse source connected to a capacitor and the voltage drop across the capacitor is measured.

Your reluctance to show a schematic of the setup is also almost commendable.
Fortunately you dedicate a few words to it, so we can try to erverse engineer what we are talking and confusing each other about since Thursday:

Oscilloscope 1MΩ impedance, Scope 900MΩ impedance. Not sure of the impedance of the pulse generating device but the resistance is approx. 9kΩ.

So what is it ? 1MΩ or 900MΩ ? I hope the latter...
Now let's try to reproduce our excavations and draw the circuit:

which, according to your description, might be equivalent to

which means the capacitor discharges over the 9 k$\Omega$ in about 70 ns. [edit: 0.7 $\mu$s]

I can't for the life of me understand how the pulse generator can come up with such a weird pulse shape, but other than that it looks like we have a bad case of impedance mismatching with ringing as a consequence.

Without further context in more detail it's hard to say anything sensible about this.

$\$

 

[Bhope69199]

Yes your interpretation is correct! Apologies I meant probe impedance is 900MΩ (scope impedance is 1MΩ).

What further context would you need?

From the posts I believe I have found out the energy within the capacitor at a specific time during the pulse. Now what I would like to understand is how to calculate the energy delivered by the pulse generator over 1 second.

 

[BvU]

What further context would you need?

Brand name and type number of the pulse generator, so I can look up the specs
And what's on the scope if you remove the 80 pF capacitor (if the probe can stand the pulse ...)

From the posts I believe I have found out the energy within the capacitor at a specific time during the pulse.

Freedom of religion, but I have hard time believing it...

Now what I would like to understand is how to calculate the energy delivered by the pulse generator over 1 second.

Basically zero: whatever is pumped into the capacitor is almost immediately dissipated again by the 9 k$\Omega$ .

But you could do $(\int E(t)\,dt )/(\int \,dt )\$
IF (big IF) indeed the voltage over the capacitor is $V(t) = \operatorname {sinc} {\pi t \over 75 \mu{\text s}}$ then $E= {1\over 2} \int CV^2\, dt$ (over 1 pulse) times 4 pulses/s

$\$

 

[Bhope69199]

Great, thanks.

Brand name and type number of the pulse generator, so I can look up the specs
And what's on the scope if you remove the 80 pF capacitor (if the probe can stand the pulse ...)
$\$

It is a bespoke made generator so no specs. Open circuit the pulse is the same shape but about 20% larger in both amplitude and width.

Basically zero: whatever is pumped into the capacitor is almost immediately dissipated again by the 9 k$\Omega$ .
$\$

OK but even if it is dissipated into the 9kΩ there would have been a draw from the power supply surely?. So how do we calculate that energy coming from the power supply?

But you could do $(\int E(t)\,dt )/(\int \,dt )\$
IF (big IF) indeed the voltage over the capacitor is $V(t) = \operatorname {sinc} {\pi t \over 75 \mu{\text s}}$ then $E= {1\over 2} \int CV^2\, dt$ (over 1 pulse) times 4 pulses/s

$\$

Isn't it times 250 pulses in a second? 250Hz, is a pulse every 4ms isn't it?

 

[BvU]

Of course. I mixed up the 250 and the 4. So: yes.

Open circuit the pulse is the same shape but about 20% larger in both amplitude and width.

Most peculiar. I know a capacitor causes lag in a signal. But I never expect lead ...
Or are the peaks shifted in time ?

 

[Bhope69199]

Of course. I mixed up the 250 and the 4. So: yes.Most peculiar. I know a capacitor causes lag in a signal. But I never expect lead ...
Or are the peaks shifted in time ?

The peaks are actually shifted in time.