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guess_me87 Antenna-Theory.com Newbie
Joined: 26 Mar 2009 Posts: 4 Location: ISLAMABAD
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Posted: Fri May 22, 2009 5:25 pm Post subject: Current distribution... |
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Hello SiR...
Kindly will you explain the current distribution in microstrip patch antenna....
As a i tried my level best to understand how it became the cause of multiband but i was not able to to fully understand this concept...
So plzzzz help me.... |
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Schubert Antenna Wizard
Joined: 08 Apr 2009 Posts: 161
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Posted: Sat May 23, 2009 8:49 pm Post subject: Multiband Operation |
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Make sure you understand how a half-wave patch radiates. The current must be zero at the end of the patch, and the voltage is out of phase with the current. Hence the voltage is at a peak at the end of the patch, and a half-wavelength away at the start of the patch, the voltage has equal magnitude but out of phase. It is this voltage, out of phase, which produces fringing fields that coherently add in phase and produce radiation. Using this argument, you should understand that if this patch radiates at f0, it will also radiate at 3*f0. Can you think why it would not radiate well at 2*f0?
If you understand this, then you understand a single frequency. If you observe that a patch radiates at some frequency f1, then find the corresponding wavelength, and try to think how the current would flow on the patch (for your configuration) such that the voltage produces fringing fields that add up in phase to produce radiation.
The current distribution must be zero at the ends of the patch; current can't flow "off" the patch. The voltage is out of phase with the current. You have to think about how the current might flow to produce radiation; this takes thought and creativity. You need to think how the various "modes" or how the current will flow at each frequency; the current distribution will be different at different frequencies. |
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cui Antenna-Theory.com Newbie
Joined: 30 Apr 2013 Posts: 4
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Posted: Wed May 08, 2013 10:30 am Post subject: Re: Multiband Operation |
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| Schubert wrote: | Make sure you understand how a half-wave patch radiates. The current must be zero at the end of the patch, and the voltage is out of phase with the current. Hence the voltage is at a peak at the end of the patch, and a half-wavelength away at the start of the patch, the voltage has equal magnitude but out of phase. It is this voltage, out of phase, which produces fringing fields that coherently add in phase and produce radiation. Using this argument, you should understand that if this patch radiates at f0, it will also radiate at 3*f0. Can you think why it would not radiate well at 2*f0?
If you understand this, then you understand a single frequency. If you observe that a patch radiates at some frequency f1, then find the corresponding wavelength, and try to think how the current would flow on the patch (for your configuration) such that the voltage produces fringing fields that add up in phase to produce radiation.
The current distribution must be zero at the ends of the patch; current can't flow "off" the patch. The voltage is out of phase with the current. You have to think about how the current might flow to produce radiation; this takes thought and creativity. You need to think how the various "modes" or how the current will flow at each frequency; the current distribution will be different at different frequencies. |
I cannot get that why " if the patch radiates at f0, it will also radiate at 3*f0. and would not radiate well at 2*f0?" ?
Can you explain it to me, thanks. |
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Schubert Antenna Wizard
Joined: 08 Apr 2009 Posts: 161
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Posted: Wed May 08, 2013 11:10 pm Post subject: |
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| If you tried to feed the patch at 2*f0, the current would be zero at the feed point. The impedance then would be Z=V/I=infinite, so no power would be delivered. It's the same story as with the dipole antenna. You can watch the antenna-theory.com dipole antenna video on youtube that will further explain this. |
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cui Antenna-Theory.com Newbie
Joined: 30 Apr 2013 Posts: 4
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Posted: Thu May 09, 2013 3:49 am Post subject: |
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| Very helpful, Thank you very much:-) |
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