- #1
EmilyRuck
- 136
- 6
The Transverse resonance method is used to determine the propagation constant of a wave in several waveguides, like the rectangular waveguide, or also dielectric waveguides.
It takes advantage of the fact that a standing wave is present along a certain direction (transverse with respect to the main propagation direction), due to purely reactive loads at both the ends of the transmission line (which represents the wave propagation). http://www0.egr.uh.edu/courses/ece/ECE6351-5317/SectionJackson/Class%20Notes/Notes%2013%20-%20Transverse%20resonance%20method.pptx provides and example in the slides 3-4-5.
In such conditions, the transmission line (ended with the reactive loads) is said to be a resonant structure. It means that [itex]Z_{in}^r (x)[/itex] looking forward must be equal to minus the [itex]Z_{in}^l (x)[/itex] looking backward, so that [itex]Z_{in}^r (x) + Z_{in}^l (x) = 0[/itex] for every position [itex]x[/itex] along the line. Why is this the resonance condition? Which is the relation between this condition and the existence of a standing wave?
I tried to consider a transmission line with total length [itex]2 \ell[/itex] centered in [itex]x = 0[/itex] and ended at both [itex]x = \ell[/itex] and [itex]x = -\ell[/itex] with a short-circuit [itex]Z_L = 0[/itex]. The input impedance in a generic position [itex]x[/itex] is actually
[itex]Z_{in}^l (x) = j Z_0 \tan [ \beta (\ell - |x|)][/itex]
[itex]Z_{in}^r (x) = j Z_0 \tan [ \beta (-\ell + |x|)] = -j Z_0 \tan [ \beta (\ell - |x|)] [/itex]
so the assumption [itex]Z_{in}^r (x) + Z_{in}^l (x) = 0[/itex] seems to be verified. But what is the physical reason for this?
In the linked slides this is derived from the continuity of voltage and current in the position [itex]x[/itex]: [itex]V^r = V^l[/itex] and [itex]I^r = -I^l[/itex], but this can be applied to every transmission line and for me has apparently nothing to do with resonance.
It takes advantage of the fact that a standing wave is present along a certain direction (transverse with respect to the main propagation direction), due to purely reactive loads at both the ends of the transmission line (which represents the wave propagation). http://www0.egr.uh.edu/courses/ece/ECE6351-5317/SectionJackson/Class%20Notes/Notes%2013%20-%20Transverse%20resonance%20method.pptx provides and example in the slides 3-4-5.
In such conditions, the transmission line (ended with the reactive loads) is said to be a resonant structure. It means that [itex]Z_{in}^r (x)[/itex] looking forward must be equal to minus the [itex]Z_{in}^l (x)[/itex] looking backward, so that [itex]Z_{in}^r (x) + Z_{in}^l (x) = 0[/itex] for every position [itex]x[/itex] along the line. Why is this the resonance condition? Which is the relation between this condition and the existence of a standing wave?
I tried to consider a transmission line with total length [itex]2 \ell[/itex] centered in [itex]x = 0[/itex] and ended at both [itex]x = \ell[/itex] and [itex]x = -\ell[/itex] with a short-circuit [itex]Z_L = 0[/itex]. The input impedance in a generic position [itex]x[/itex] is actually
[itex]Z_{in}^l (x) = j Z_0 \tan [ \beta (\ell - |x|)][/itex]
[itex]Z_{in}^r (x) = j Z_0 \tan [ \beta (-\ell + |x|)] = -j Z_0 \tan [ \beta (\ell - |x|)] [/itex]
so the assumption [itex]Z_{in}^r (x) + Z_{in}^l (x) = 0[/itex] seems to be verified. But what is the physical reason for this?
In the linked slides this is derived from the continuity of voltage and current in the position [itex]x[/itex]: [itex]V^r = V^l[/itex] and [itex]I^r = -I^l[/itex], but this can be applied to every transmission line and for me has apparently nothing to do with resonance.
Last edited by a moderator: