# How is rise time related to bandwidth of the signal?

Say, I want to limit the rise time of my digital signal edges to avoid dealing with transmission line effects.

How do I determine the maximum frequency of harmonics in my signal knowing that my rise time is, say 5ns?

How do I determine corner frequency of my low pass filter knowing that hold time on the receiver chip is, say 10ns?

In wikipedia I've found the formula

$$BW=\frac{0.34}{t_{rise}}$$

does it apply in this case?

Edit

I failed to make myself clear, so I'll try to explain my line of thought.

Say, I have a 30HMz signal and my trace length is well below 1/10 of the wavelength. So I don't have to deal with transmission line effects in respect to that. But my edges are steep - 5ns. This adds some high frequency components into my signal that potentially will suffer from transmission line effects.

My idea is that I slow down edge transitions up to a point where I don't have to deal with transmission line phenomena. The question is twofold:

• how do I calculate fastest rise/fall time that with the given trace length would enable me to traeat my circuit as "lumped"?
• how do I slow down the rise/fall time?

Rise/fall time is time for voltage to change from 10% to 90% of max value. I know how to calculate the approximate speed of signal on FR4 board.

-
I've seen such question before on this site, but I can't find it right now. – AndrejaKo Dec 14 '12 at 12:51
I could not find it too, so that's why I'm asking :) – miceuz Dec 14 '12 at 13:42
edge rates are the principal mechanism for EMI/RFI so there is good reasons to look at this. Series resistors will help you slow it down, and some drivers also have variable current drive and or edge rate control. – placeholder Dec 14 '12 at 19:29

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.

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.

-
Whoever downvoted this, if you disagree or think something is wrong, it would be useful to explain that. – Olin Lathrop Dec 14 '12 at 16:05
Wasn't me that downvoted, but there is a relationship between rise time and bandwidth and it's called Fourier transform. The formula the Op mentions comes from that (it's a simplification of sorts). – Gustavo Litovsky Dec 14 '12 at 16:57
@OlinLathrop I've changed my question to sound like I know what I'm talking about. – miceuz Dec 14 '12 at 17:42
@gl38: You can't describe a relationship between two properties until those properties are defined. Just "rise time" is not really defined, so you can't say how it relates to frequency, and certainly can't start doing Fourier transforms on it. – Olin Lathrop Dec 14 '12 at 17:57
@Olin: You can mathematically define a function that creates an equivalent pulse with the same rise time and then you perform Fourier transform on it. From that you will obtain the frequency spectrum and hence information about the bandwidth. You're right that he needs to define the actual rise time but it's usually the rise 10% to 90% – Gustavo Litovsky Dec 14 '12 at 18:01

That formula is what we typically call the knee frequency. It's based on the 10%-90% rise time of the signal and is usually used as an approximation to tell us what the highest frequency of interest might be in a digital signal we're using. Or said a better way where most of the high frequency energy content of that signal can be found. If your channel can pass that bandwidth then theoretically you will not see any roll of or rise time degradation of the signal. Of course in practice there are other things like reflections that can affect your signal. Here's Tom D at Mentor giving a good explanation of it on SI-LIST.

I'd be more interested to know the length and material used for your channel. Is it long enough that you need to consider transmission line effects (longer than a quarter wavelength, some would say 1/6th wavelength). I don't know what you're doing from your post so just trying to give some general advice. Trying to slow down your rise time in some way if you don't need it, in and of itself is not a bad idea provided your driver can handle the filter load you use without blowing up.

Why not just make sure you use a proper transmission line structure / cable and terminate properly? I'm sure you have your reasons for your project so just a suggestion ;)

-
Thanks for explanation about energy content. I'm not building anything in particular, just trying to wrap my head around issues with high speed digital signal routing and ways to deal with it, I already know about 1/nth of wavelength rule. – miceuz Dec 14 '12 at 17:36
Ah ok well in that case if you don't have them already two good books to have would be Signal and Power Integrity simplified by Eric Bogatin, or High-Speed Digital Design by Howard Johnson. Their websites are pretty good to bethesignal.com/bogatin/index.php and sigcon.com/Pubs/pubsKeyword.htm. I learned a lot when I first started reading through those publications on sigcon. Also you might want to join freelists.org/archive/si-list it's an email list of a lot of SI guys that will answer questions or just a place to learn. – Some Hardware Guy Dec 14 '12 at 18:33

The formulae you cite is used for the BW of signals that will be involved in emission from edges. And there are some assumptions built into it, like for example, Most digital signals in mid swing look like a current source into a capacitor (i.e. a linear ramp) that taper off on the top and bottom. It is also valid to use that for your transmission line concern for reflections etc. and roll off.

But it doesn't speak to harmonics which will be ~ 1/t(rise). i.e. you'll see this 200 MHz spurs in the spectrum.

For the receiver you have to be looking at the eye diagram to ensure your hold times are met. And this is a time domain scenario. So you can have circuit elements in there that help meet your timing and are not visible in the frequency side of things. So your BW can be used to describe things in it's interaction with the hold time, but you can't necessarily derive hold time directly from BW. Modeling or test benches are the way to go here.

-

Not sure I read all of the posts in their entirety, but with regard to the original post (from the guy with the 5ns rise time). You should read books by Dr. Howard Johnson or Lee Ritchey. They explain this in detail.

Don't try to slow down the signal, there's not need for that except under special circumstances.

If you want to get out of the theoretical muck and find a practical solution you can use this: If the length of the trace is longer than 1/5 of the flight time represented by the risetime of the edge, you have a transmission line and need termination. In a practical case, using FR4 or equivalent material, with a dielectric constant of around 4 to 4.6, the travel time is around 5.5 inches per nanosecond. For a 5ns rise time, you have an edge transition that is around 27.5 inches long. If you take 1/5th of that you get 5.5 inches. So if your PWB trace is longer than 5.5 inches, you should use a series terminating resistor to match the impedance (for a point to point connection).

If you have 50 ohm traces, the resistor should be 50 ohms minus the source impedance of the driver (for reflected wave series termination). Start with a 20 ohm resistor. If you get excessive overshoot (more than 5%), make it bigger, if you get a rolled off edge, make it smaller. It does not need to be perfect to get good results. Ideally, use Hyperlynx software to simulate and get close to perfect results every time.

BTW, that equation .34/Trise, is a valid equation for practical applications. And rise time is generally agreed to be the time from 10% to 90% of the signal voltage (any exceptions don't apply to what you are doing). To be more conservative in your designs, use .5/Trise.

-
BTW - this discussion has nothing to do with the wavelength of the energy in the signals, it is all about propagation delay on the PWB. – Charles F Mar 17 at 14:27
Welcome to EE.SE, Charles. No need to comment on your own post. You can edit it. – transistor Mar 17 at 14:31