Electronic – Frequency response of the waveguide

bandwidthfrequencysignalwaveguide

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?

Best Answer

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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