A 10 ns rise time does not have a 100 MHz bandwidth.

And I don't think a lot of the effects you mentioned - resistive losses, skin effect, dielectric losses - are much in play at the frequencies involved here - 30 KHz and 10 ns edge times.

You really can't have a 10 ns edge without a 100 MHz bandwidth, SteveSh. The argument is simple: if you need to go from low to high in that time, by adding up sinusoidals of increasing frequency, you need at least one that had 10 ns between its minimum and maximum, which is half the period. Inverse of 20 ns is 50 MHz.

Now, you need an edge, not a periodic function. So, you need to add something twice as fast oscillating as the first tone to convert the previous high and the following lows to something somewhat constant. (If you are following this from an approximation theory perspective, you will have recognized the Chebychev polynomial approach there.)

sina bala - All you really need to get a good 10 ns edge (10%-90%) is 35 MHz BW.

Ah that's the old lecroy rule of thumb, right? 35%! Yeah that's just bogus, it assumes that you define bandwidth for an oscilloscope RC filter with relatively relaxed transition, not the actual 3dB bandwidth! (So it is useful for selecting the right bandwidth for your analog scope frontend, but doesn't really describe the signal)

No. It starts with a square wave signal with 0 ns rise & fall times. We know that in order to accurately represent this we need the fundamental plus an infinite number of harmonics. Now we start eliminating higher harmonics, turn that frequency domain signal back into a time domain signal, plot that signal and measure the 10%-90% rise time of the signal.

Ah, we can't even reconstruct the exact step with an infinite amount of discrete harmonics, Gibbs won't allow us to. That's the fascinating thing about Gibbs phenomenon. The series didn't uniforms converge within the space of Fourier serieses to the approximated function, the step (which doesn't mean it doesn't have a Fourier transform).

But that nitpicking from my side aside:

A sine of frequency 1/(2π) goes from 0 to 0.8 in arcsrin(0.8)≈0.937. We need to go from -0.8 to 0 to +0.8 (to cross the middle 90% of the sine' rising edge), meaning we need about 1.874 (rad). We set that time to be 10 ns, so for the period T of the time scaled sine it holds that 10 ns/T = 1.874/(2π) —> f=1/T=1.874/(2π • 10 ns) = 29.9 MHz

Well, we were both wrong, but you were wrong by 5% and I was by orders of magnitude!

So why am I stubborn?

Argh on Android my keyboard hides the entry box of the chat, this is terrible

I'm stubborn because, yes, the edge speed sufficient to make something else that fast just indeed not very high

But the frequency components needed to not make it a sine, but keep it at a high level, are inherently multiples of the fundamental that w

One works need. Sadly, adding the harmonica needed to reduce the "pay junk" ripple

Pay junk = post-jump

These harmonics ,

As per Gibbs thrown, none the machine of the Fourier series cost to the center of the jump; they don't only make the thing after the jump more plateau-like

They also make the jump stripper

Steeper

My point was that you don't have to treat this signal/interface like a super high-speed one. 35 MHz BW for this particular interface (30 KHz PWM signal & 10 ns rise/fall times for a switching power supply) should be fine.

So, this really becomes a bit of an ugly optimization problem, where the question of " how much bandwidth does a jump to a plateau need" is defined both by the speed of the edge as well as the acceptable ripple

@nick I think that is the point

Neither me nor Steve would recommend a raise time that day

Fast

@SteveSh I see. I answered "if you really had arise time this day, and needed the plateau afterwards to be free if ringing, here's an educated guess on needed bandwidth", you answer, "if I need a 30 kHz PWM...

@Nick you're preaching to the choir. We're not the asker

Smells like an XY problem.

Again, why are you telling us this?

Op will not read this chat!

@sinabala I'm not preaching to the choir. I'm joining the choir.

Problem is that the person who should here the song will not hear it :) I think Steve and I both agree with you, and you should instead tell the asker!

Only reason I can think of for the 10 ns rise time number is that OP is planning on using this signal to drive the gate of a switching MOSFET,and he thinks he will get lower switching losses with 10 ns than with, say 50 ns rise time. But this is just a WAG.

@SteveSh If that's the situation, then the gate driver should be on the far end of the cable next to the MOSFET.

On closer look, the O.P. already wrote that he'll have a gate driver (UCC27517A) on the far end of the cable.