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Wavelength 100mm does not correspond to frequency 29.9792 GHz.
Do you mean that the length od the dipole is 100 mm?
Please fix the simulation errors.
The diagram should look smooth. If not, try to plot more points.
There seems to be some mistake in the design of the antenna or during setting up parameters for the simulation, as your simulation got terminated due to insufficient availability of the memory (RAM).
Try simulating it on better machine and also change your parameters of Infinte sphere. Theta: -180 to 180 Step size: 2 and Phi :0 to 90 Step Size: 90.
Later post your results...
Looking at the diagram, I can see you are using HFSS. I'm puzzled why you use such a program, when you have to ask what the radiation pattern of a dipole should look like. You are wasting your time.
As someone else has pointed out, 29.9792 MHz is not a wavelength of 100 mm. I'm not going to work out the wavelength, but it will be very close to 10 m. If the wavelength is 10 m, and your dipole is 100 mm long, then it is very short, and it's radiation pattern would be close to to that of a half-wave dipole, but an even better approximation would be a Hertzian dipole.
I'm interested why you are using HFSS for this. Are you at college, and a lecturer told you to use HFSS? Are you working, and have been asked by your employer to use it? I know you wont admit this, even if it was true, but perhaps you have got a free copy of HFSS. If you have, then I'd suggest you stop using it now - not because you are breaking the law, but simply because you are not using the right tool to analyze antennas, given your knowledge of antennas.
What you are doing is an equivalent of using an Airbus A380 flight simulator to learn how to fly a basic aircraft. If you don't know anything about flying a plane, you will not gain much by using an A380 flight simulator.
Perhaps others will see it differently, but I personally don't think you should be using HFSS.