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ZMT zurich med tech

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  4. What is the mismatch efficiency?

What is the mismatch efficiency?

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  • R Offline
    R Offline
    rose
    wrote on last edited by
    #1

    When I moved to the radiation report on Analysis tab, I found mismatch efficiency, but don't know what it means. I would like to know Total available power as well.

    Thank you.

    SylvainS 1 Reply Last reply
    0
    • R rose

      When I moved to the radiation report on Analysis tab, I found mismatch efficiency, but don't know what it means. I would like to know Total available power as well.

      Thank you.

      SylvainS Offline
      SylvainS Offline
      Sylvain
      ZMT
      wrote on last edited by
      #2

      @rose I would recommend reading that section of the Sim4Life Manual:
      2.5 EM FDTD (Finite-Difference Time-Domain) Solver » 2.5.1 EM FDTD Simulation Theory » 2.5.1.4 Power Balance

      In particular, this flow chart taken frem IEEE 145-2013:
      7d5b7455-bb77-4ce9-96a2-0a0ec98e4bf4-image.png

      where c86643f7-54a4-43c6-8728-db27fa55a170-image.png is the mismatch efficiency (ratio of input power Pin to total available power Pa)

      R 1 Reply Last reply
      0
      • SylvainS Sylvain

        @rose I would recommend reading that section of the Sim4Life Manual:
        2.5 EM FDTD (Finite-Difference Time-Domain) Solver » 2.5.1 EM FDTD Simulation Theory » 2.5.1.4 Power Balance

        In particular, this flow chart taken frem IEEE 145-2013:
        7d5b7455-bb77-4ce9-96a2-0a0ec98e4bf4-image.png

        where c86643f7-54a4-43c6-8728-db27fa55a170-image.png is the mismatch efficiency (ratio of input power Pin to total available power Pa)

        R Offline
        R Offline
        rose
        wrote on last edited by
        #3

        @sylvain Thank you! It really helped.

        I'm working on the dipole simulation and I got a proper(I mean reasonable) results. After the simulation, I changed some parameter and simulated.

        Although I changed only the length of the model and frequency from 0.3GHz to 3.0GHz, Total Available Power(TAP) changed from 1.3W to 29.2W and Gain has doubled.

        I would like to know why it happened.

        SylvainS 1 Reply Last reply
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        • R rose

          @sylvain Thank you! It really helped.

          I'm working on the dipole simulation and I got a proper(I mean reasonable) results. After the simulation, I changed some parameter and simulated.

          Although I changed only the length of the model and frequency from 0.3GHz to 3.0GHz, Total Available Power(TAP) changed from 1.3W to 29.2W and Gain has doubled.

          I would like to know why it happened.

          SylvainS Offline
          SylvainS Offline
          Sylvain
          ZMT
          wrote on last edited by
          #4

          @rose your simulations are most likely done with a 1V voltage at the excitation source (that's the default setting). This is perfectly fine from a numerical point of view, but in practice (e.g., if you wanted to compare with measurements) you probably want to normalize your results. In the Far Field Sensor, for example, you can normalize to a Total Available power of 1W (say). This way, you will be comparing apples to apples.

          That will not change the value for the Gain, though (since it is a power ratio, independent of scaling). This must come from either a difference in some aspect ratio of your dipole, or simply that you did not change the dimensions in the same proportion (1/10th) as you changed the frequency.

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