How to estimate the bandwidth of a square wave

[ZengLei]

Hi all

when i know the rising time of a square wave,how to get its bandwidth?

i see a formula 0.35/Trising=BW

is that correct?
how that formula comes?

thanks~~~~~~~~~~~

Added after 59 minutes:

The following is a relational expression between risetime of pulse waveform and the highest frequency components contained in spectrum;

Fmax = 0.35/Tr
where,
Tr = risetime of pulse (s),
Fmax = the highest frequency components (Hz).

The expression has been mentioned without any explanation in a large number of documents as if it is a matter of course. I do not think its basis is self-explanatory at all.
I think you can see intuitively that waveforms of speedy risetime have high frequency components. However,

1) the reason why Fmax and Tr are in an inverse proportion, and
2) what is the basis of the constant "0.35,"

cannot be matters of course by any means. What kind of basis on earth, could this expression be derived from?

 

[FvM]

As any estimation, the said rule-of-thumb depends on assumptions and has limited accuracy. For a mathematical exact analysis, you may consider fourier series of a periodical trapeziodal pulse shape. It has obviously infinite bandwith, but you can calculate fmax for a given percentage of energy.

If you assume frequency characteristics as second-order bessel or gaussian filter. You get rise times (10%-90%) of the said magnitude with fmax as cut-off frequency.

 

[Kral]

ZengLei,
The formula BW = .35 / Trise is based on the following assumptions:
.
Rise time is defined as the time required for the signal to go from 10% to 90% of its ultimate change.
.
The load on the circuit is a simple lag. In other words, the output is taken downstream of the internal series resistance, and in parallel with a capacitor.
.
Based on these assumptions, the formula can be derived analytically. Let me know if you want the derivation.
Regards,
Kral

 

[ZengLei]

OK,Kral,would u pls give the detailed derivation of the formula.

thanks~~~

 

[FvM]

(ln(0.9)-ln(0.1))/(2•pi)

 

[IanP]

if you know the rising time of a "square wave" it's not a square wave any more since the harmonics spread out to infinity .. and that it's bandwidth ..

just playing devil's advocate ..

cheers,
ianp

 

[ZengLei]

FvM

:
Vin=Uo
Vout=Uo[1-exp(-t/RC)]

T90%=-RCln0.1
T10%=-RCln0.9

Trise=T90%-T10%=RC(ln0.9-ln0.1)=RCln9

1/(2*pi*RC)=K/(RCln9) ==> K=ln9/(2*pi)=0.35

i'm i right?

We can use this formula to estimate the Bandwidth of a one-pole sysytem.

But can we use this formula to estimate the Bandwidth of a square waveform??


Thanks

 

[FvM]

It has all been said! The bandwith estimation is under the assumption of first order system (as I said, second order is basically the same order of magnitude). A square wave has infinite bandwith, there is nothing left to calculate. An ideal trapezoidal waveform has also infinite bandwith. You can read the fourier series from any mathematical handbook.

 

[Kral]

ZengLei,
I didn’t follow your fifth equation, so here’s my development:
Assume the input and output are initially at zero, V is the instantaneous output voltage, Vf is the final output voltage, T is the RC time constant. Then
1) V = Vf[1-exp(-t/T)
2) V = Vf –Vfexp(-t/T)
3) V10 = .1Vf
4) .1Vf-Vf = -Vfexp(-t10/T)
5) .9 = -exp(-t10/T)
6) ln(.9) = -t10/T
7) t10=-Tln(.9)
8) Similarly, t90 = Tln(.1)
9) trise = t90 - t10 = [-Tln(.1) + Tln(.9)]
10) T =1/w = 1/(2 pi f) for a 1st order lag
11) From 9), 10, trise = [-ln(.1) + ln(.9)] / (2 pi f)
12) trise = .35/f
Regards,
Kral

 

[FvM]

I think, correct application of exponential function wasn't the point in doubt, it was the validity of first order approximation.

 

[ismailbtk]

you can use fourier series and transform to estimate that.

 

[smith123]

it will be nice trise = [-ln(.1) + ln(.9)] / (2 pi f)

 

[springf2000]

in fact y can do a fft calculation and decide your bandwidth, depend on your tolerance,because the bandwidth is infinite in theory