We have detected your current browser version is not the latest one. Xilinx.com uses the latest web technologies to bring you the best online experience possible. Please upgrade to a Xilinx.com supported browser:Chrome, Firefox, Internet Explorer 11, Safari. Thank you!

cancel
Showing results for
Did you mean:
Highlighted
Visitor
10,203 Views
Registered: ‎10-14-2012

## How to solve cube-root problem using IP core

How to solve cube-root  problem using IP core？

4 Replies
Xilinx Employee
10,193 Views
Registered: ‎02-06-2013

## Re: How to solve cube-root problem using IP core

Hi

Which core are you referring here?

Regards,

Satish

--------------------------------------------------​--------------------------------------------
Kindly note- Please mark the Answer as "Accept as solution" if information provided is helpful.

Give Kudos to a post which you think is helpful.
--------------------------------------------------​-------------------------------------------
Visitor
10,182 Views
Registered: ‎10-14-2012

## Re: How to solve cube-root problem using IP core

Those in "CORE Generator"->"Math Functions"

Xilinx Employee
10,177 Views
Registered: ‎02-06-2013

## Re: How to solve cube-root problem using IP core

Hi

We have cores which can do Square root functions only using cordic or Floating point cores and no cores to do the cube root.

Regards,

Satish

--------------------------------------------------​--------------------------------------------
Kindly note- Please mark the Answer as "Accept as solution" if information provided is helpful.

Give Kudos to a post which you think is helpful.
--------------------------------------------------​-------------------------------------------
Contributor
562 Views
Registered: ‎10-18-2018

## Re: How to solve cube-root problem using IP core

Dear @atreide_lawson

We can solve the cuberoot problem with using the basic ADD, DIVIDE and SQUARE IPs available in the IP Catalog.

Consider for instance, that you want to find a cuberoot of a number C. Then, use the following equation Iteratively for say 20-50 Iterations and a fairly acceptable value can be computed.

X_(n+1) = (1/3) * [2*X_(n) + (C/(X_(n)^2))]                                        ---[1]

Here X_(n) is the initial guess of the cuberoot. For more information see the following paper:

Below is a simple MatLab Code for computing 3^1/3.

```clear all;
c=3;
z=1.01;
for i=1:100
z = (1/3)*((2*z) + (c/(z^2)));
end
z;```
Best Regards,
Urvish