There is no one to one relationship between rise time and bandwidth. A slew rate limiter is a non-linear filter, so can't directly be characterized as a low pass filter with some obvious rolloff frequency. Think of it in the time domain, and you can see that a slew rate limit effects signals proportional to amplitude. A 5 Vpp signal limited to 5 V/µs can't have a period shorter than 2 µs, at which point it degenerates to a 500 kHz triangle wave. However, if the amplitude only needed to be 1 Vpp, then the limit is a 2.5 MHz triangle wave. Since the concept of bandwidth get less clear when a non-linear filter is envolved, you can at best talk about it approximately.

Your answer can also vary greatly depending on what exactly "rise time" is. This is a term that should never be used without some qualification. Even a simple R-C filter has ambiguous rise time. Its step response is a exponential with no place being a clear "end". It's rise time is therefore infinite. Without a threshold of how close to the end you need to be to considered to have risen, the term "rise time" is meaningless. This is why you need to either talk about rise time *to a specific fraction of the final value*, or slew rate.

The equation you site is therefore just plain wrong, at least without a set of qualifications. Perhaps those are found on the page you got it from, but quoting it out of contect makes it wrong. Your question is unaswerable in its current form.

## Added:

You now say the real issue is limiting high frequencies from sharp edges so that parts of the signal don't get into the frequency range where your wire becomes a transmission line. This has little directly to do with rise time. Since the real issue is frequency content, deal with that directly. The simplest way is probably a R-C low pass filter. Set it to roll off above the highest frequency of interest in the signal, and well below the frequency at which your system can no longer be considered lumped. If there is no frequency space between these, then you can't what you want. In that case you need to use a lower bandwidth signal, a shorter wire, or deal with the transmission line aspects of the wire.

In your case, you say the highest frequency of interest is 30 MHz, so adjust the filter to that or a little higher, let's say 50 MHz since that will leave your desired signal pretty much intact. The wavelength of 50 MHz is 6 meters in free space. You didn't say what impedence your transmission line is, but let's figure propagation will be half the speed of light, which leaves 3 meter wavelength on the wire. To be pretty safe just ignoring transmission line issues, you want the wire to be 1/10 wavelength or less, which is 300 mm or about a foot. So if the wire is a foot or less in length, then you can add a simple R-C filter at 50 MHz and forget about it.

Transmission line effects don't just suddenly appear at some magic wavelength relative to the wire length, so how long is too long is a gray area. Up to 1/4 wavelength can often be short enough. If it is "long", then the best thing is to use a impedence controlled driver and a terminator at the other end. However, that is cumbersome and also attenuates the signal by half. You either deal with the lower amplitude at the receiver, or boost it at the transmitter before it gets divided by the driving impedence and the transmission line characteristic impedence.

A simpler solution that may take some experimental tweaking, is to simply put a small resistor in series with the driver and be done with it. That will form a low pass filter with the capacitance of the cable and whatever other stray capacitance is around. It's not as predictable as a deliberate R-C, but much simpler and often good enough.

Yes, there is such a effect. A ideal transmission line is modeled with lots of little series inductors and parallel capacitors. See this answer by Phil.

In such a ideal transmission line, frequencies above a certain amount are removed and the remainder of the step propagates forever unchanged. This approximation is usually good enough for "short" transmission lines.

Real "long" transmission lines differ in that the series resistance matters, which is ignored in the ideal model presented above. This series resistance effectively adds some low pass filtering. Since the resistance accumulates with length, the resulting filter becomes ever lower in frequency. The more low pass filtered edge at the end of a long transmission line therefore looks more spread out since ever more high frequencies are removed over the length of the transmission line.

## Best Answer

The boundary between lumped and distributed systems is not clear-cut but there are some commonly used values. For distributed systems transmission line theory is required.

The distinction is usually made based on the effective length of a signal or the feature of a signal like an edge. So it's important to consider the rise- and fall time of a signal and not the frequency. Nevertheless the frequency imposes an upper limit on the risetime.

In air a signal travels with about 85ps/in (~ 33ps/cm). The propagation delay depends on the dielectric constant, it is proportional to the square root of it. For a PCB with a dielectric constant of 4 (like FR4 which is in the range of 3 to 5) the propagation delay doubles.

A rising edge with a risetime of 1ns would occupy a trace length of 1ns/(2*85ps) ~ 6in (~ 15cm). At the driving side the signal is already high when at a 6in distance it just starts to rise.

So a 6in (15cm) track clearly is too long, since the potential varies from low to high along the track.

If the length of the track is between 1/6 or 1/4 of the effective length of a feature like an edge a system can be regarded as lumped.

So the upper limit for the example given above is between 6in / 6 (= 1 in, ~2.5cm) and 6in /4 (= 1.5in, ~4cm) for a trace on a PCB with a dielectric constant of 4.