'Scope probe compensation.

Why does the HF compensation adjustment only move the end of the square wave edge, the beginning stays lovely and square? I would have thought that both "right angles" had the same frequency spectrum.

[Radio Wrangler]

Causality!

Some transfer functions with squarewave input will give pre-ripples leading up to an edge. This looks dangerously like a circuit predicting the coming of an edge. But the class of circuits which do this exhibit sufficient insertion delay that the onset of the pre-ripples is not earlier than the entry of the edge into the input of the circuit. The bulk of the edge only exits after the pre-ripples.

In the case of the scope probe, the high frequency content is the result of both right angles, and the step magnitude and only shows up afterwards. It's the Fourier transform of a Heaviside function. Symmetry is preserved by the equivalent action on the falling edge.

David

[trobbins]

A scope comp squarewave is polarised with one half of the waveform at ground.

[Radio Wrangler]

It would be just the same with a squarewave symmetrical about ground. As it is with a wave switching between ground and a calibration voltage, if set to overcompensate, the indication on the scope shows it overshooting below ground as well as above the calibration voltage.

Have a play with a general purpose lab pulse generator and a scope and you'll see that the compensation capacitors have no effect on the DC component.

David

[kalee20]

Originally Posted by trobbins

A scope comp squarewave is polarised with one half of the waveform at ground.

That doesn't make any difference- you'd see the same effect if the square wave transitioned between +5V and -5V, or +1V and +3V.

It's down to causality, as David said: the probe and 'scope doesn't 'know' that a transition is about to occur, so can't put in an anticipatory curve or overshoot before the event itself.

Many thanks, boomin’ obvious when told!