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So, I have signal in the waveguide that is transmitted by two modes of radiation, for which the delay is for example \$\tau_1=6\,\text{ns},\, \tau_2=6.5\,\text{ns}\$, respectively. And the energy supplied to the receiver by each mode is the same.

My question is about calculating \$3\,\text{dB}\$ width of the bandwidth of the channel (containing the constant component - but I'm not sure what that means). I know I need to substract this two modes one from the other (with an absolute value):

\$ d=|6\,\text{ns}−6.5\,\text{ns}| \$

\$ d=0.5\,\text{ns}\$

And next I multiply my \$d\cdot2\$ and divide one by my result. So:

\$\frac1{2d}\$

Ant this gives me \$1\,\text{GHz}\$, and this is the perfectly correct answer!

But I don't know

  • why this formula works?
  • why we don't use the lambda formula for our waveguide frequency?
  • what does it mean that the channel was created in the basic band?
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  • \$\begingroup\$ what is \$\lambda\$ within your waveguide? \$\endgroup\$
    – Marcus Müller
    Commented Nov 11, 2018 at 15:11
  • \$\begingroup\$ You don't mention "basic band" anywhere in your question, and as far as I can tell, it's not a common term. Can you cite exactly where that comes from? We know base band in the context of channels (but that has not much to do with waveguides) or we know fundamental modes in waveguides. Maybe you meant one of these? \$\endgroup\$
    – Marcus Müller
    Commented Nov 11, 2018 at 15:13
  • \$\begingroup\$ @Marcus Müller using Lambda, we can calculate wavelength of the wave - so the frequency. Here is the formula \$\endgroup\$
    – JimPanse
    Commented Nov 11, 2018 at 15:15
  • \$\begingroup\$ so, how does that formula relate to your \$\tau\$? (it's really all a matter of replacing \$f\$ and other variables with your given values; I'm asking this so you can do it on your own!) \$\endgroup\$
    – Marcus Müller
    Commented Nov 11, 2018 at 15:16
  • \$\begingroup\$ @Marcus Müller yeah, but I don;t know what to do with my c which is 3*10^8, because I think I can't put this in my formula \$\endgroup\$
    – JimPanse
    Commented Nov 11, 2018 at 15:23

1 Answer 1

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Okay, I found the formula, because I did some research, so it is said that:

"It can be shown that almost regardless of the details of the course of the channel characteristic - as long as its frequency response is equal to W - the effect of channel crossing is to widen the pulse width by time Δτ≈1 / W."

So that almost answers my question, the only problem I have is why Δτ is multiplied by two? Otherwise, it wouldn't match to my result. Can anyone know?

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