# [Numpy-discussion] phase unwrapping (1d)

Pierre Haessig pierre.haessig@crans....
Thu Jan 17 09:48:36 CST 2013

```Hi Neal,

Le 14/01/2013 15:39, Neal Becker a écrit :
> This code should explain all:
> --------------------------------
> import numpy as np
> arg = np.angle
>
> def nint (x):
>     return int (x + 0.5) if x >= 0 else int (x - 0.5)
>
> def unwrap (inp, y=np.pi, init=0, cnt=0):
>     o = np.empty_like (inp)
>     prev_o = init
>     for i in range (len (inp)):
>         o[i] = cnt * 2 * y + inp[i]
>         delta = o[i] - prev_o
>
>         if delta / y > 1 or delta / y < -1:
>             n = nint (delta / (2*y))
>             o[i] -= 2*y*n
>             cnt -= n
>
>         prev_o = o[i]
>
>     return o
>
>
> u = np.linspace (0, 400, 100) * np.pi/100
> v = np.cos (u) + 1j * np.sin (u)
> plot (arg(v))
> plot (arg(v) + arg (v))
> plot (unwrap (arg (v)))
> plot (unwrap (arg (v) + arg (v)))
I think your code does the job.

I tried the following simplification, without the use of nint (which by
the way could be replaced by int(floor(x)) I think) :

def unwrap (inp, y=np.pi, init=0, cnt=0):
o = np.empty_like (inp)
prev_o = init
for i in range (len (inp)):
o[i] = cnt * 2 * y + inp[i]
delta = o[i] - prev_o

if delta / y > 1:
o[i] -= 2*y
cnt -= 1
elif delta / y < -1:
o[i] += 2*y
cnt += 1

prev_o = o[i]

return o

And now I understand the issue you described of "phase changes of more
than 2pi" because the above indeed fail to unwrap (arg (v) + arg (v)).
On the other hand np.unwrap handles it correctly.

(I still don't know for the speed issue).

Best,
Pierre

-------------- next part --------------
A non-text attachment was scrubbed...
Name: signature.asc
Type: application/pgp-signature
Size: 900 bytes
Desc: OpenPGP digital signature
Url : http://mail.scipy.org/pipermail/numpy-discussion/attachments/20130117/c0836328/attachment.bin
```