He all, I was wondering how should be the parameters fo the CORDIC divider block in order to get proper results. In this example I´m trying to get 0.1/0.2 = 0.5 but I don´t get it and I don´t know why? Please, does anyone know how to do it? Thank you!!!

EDIT: To take a look to the parameters of the CORDIC block see this question: System Generator: How to configure the CORDIC divider block. Understanding the block parameters

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  • \$\begingroup\$ Does anyone create the tags 'System Generator' and 'cordic'? \$\endgroup\$
    – Peterstone
    Jan 29, 2011 at 8:17
  • \$\begingroup\$ Perhaps you could describe the options available in the CORDIC divider block. \$\endgroup\$
    – tyblu
    Jan 29, 2011 at 19:10
  • \$\begingroup\$ @tyblu: That´s done. \$\endgroup\$
    – Peterstone
    Feb 2, 2011 at 9:04

1 Answer 1


From the latency on the cordic block, that looks as if you've chosen to use a single processing element - that will produce results of limited accuracy. Very limited it appears :)

I tried using 10 elements for example, with [zeros (1,9) 1] for the "latency per processing element" - to get back to your 21 tick latency. Making sure I ran the simulation long enough (oops), I then got a result (using UFixed_16_11 inputs) of 0.501953125. Which is closer...

Does that help?

  • \$\begingroup\$ Yeeaah!!, I´ve put as 'Number of Proccessing Elements:' 10; and I´ve put a Latency for each Processing Element: [0 0 0 0 0 0 0 0 0 0 1] and I´ve gotten 0.498. \$\endgroup\$
    – Peterstone
    Feb 2, 2011 at 9:50
  • 1
    \$\begingroup\$ Excellent! Just for future readers - "[zeros (1,9) 1]" isn't (just) shorthand, you really can type that in the box. Any legal matlab commands can go in the box and as long as they evaluate to a vector, it'll get used by the underlying scripts. \$\endgroup\$ Feb 2, 2011 at 13:46

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