# Frequency response of the waveguide

So, I have signal in the waveguide that is transmitted by two modes of radiation, for which the delay is for example $$τ1 = 6ns$$, $$τ2 = 6.5ns$$, respectively. And the energy supplied to the receiver by each mode is the same.

My question is about calculating 3dB 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):

$$d = ABS(6ns - 6.5ns)$$

$$d = 0.5$$

And next I multiply my $$d*2$$ and divide one by my result. So:

$$1/(2*d)$$

Ant this gives me $$1GHz$$, and this is perfectly correct anser! But I don't know why this formula works, why we don't use the lambda formula for our waveguide frequency? And what does it mean that the channel was created in the basic band? I have a problem understanding this. Thank you in advance for your help.

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?