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In honor of Laplace, I'm building this imaginary space that will be a complete screen of symbol to be read and encode by the two end computer that will transfer the images. In the imaginary space of 4K, there is the place for all characters of this planet in 20X20 format pixels, of about 20,736 completed characters. All the screen possibility will be encoded materially into a memory read only device, that will trig the data stored into by the call of the transfer screen number. In the case of Mars, that should be a 7 X 7 low dispersion laser array. This system is now improved, by the use of single pixel (bit) in different screen space (memory available), for the different use.



For a memory size of 2m, where in the case of a cellular phone, “m” should be 32, the bus width addressable memory, so the size of the packets, will be 216 bit. This fact, because the number of packets to be compare with the data for matching them, to achieved a 32 bit number (integer:”m”), will be 216 too, to maximize the bandwidth, that itself is achieved by multiplication of those both number that are related by addition of exponent, about the memory size :)

The number of operation per second is given by the switching speed of the transistorized system, built to perform that screening test. Operation per second = 232 * 25E-11 s = 0,931 Hz

In conclusion, the imaginary bandwidth (bit/s) is equal to the bus width addressable(216) * Operation per second. And the real bandwidth is equal to the memory width(32) * (operation per second). The quotient of both, gives the final ratio, the CATS-RABBITS ratio, of the effective compression rate of the communication...

That ratio is from 1 to 186 361 for huge device :)


LPDDR4: 32 G bytes / 0,25 ns : 1 W N.B.: This 0,25 ns could theoretically be increase by burst data to 0,032 ns by adding a memory complex to compare with the packet within a 4096 Bus width, all that modifiable :)

My approximate evaluation for the power consumption, will include the others parts of the device, that will be estimated by doubling the power consumption of the memory. For a cell phone, with a factor CHATS-LAPINS of 2048X, it should be about 2 W, for a 4 G bytes memory, remembering the reader that the ratio shall be calculated with bits values...

I must remember the reader, that this ratio is applied on the real bandwidth, so, without parallelism of some devices, you won't achieved any increase in effective data rate in the phone, but only reduced the cost of the transmission by 2048 :)

With a 5 M bytes transmission rate device, the cost of decreasing the rate and maintaining the total phone data rate, should be:


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