Fix linear StepModel to match documentation.

Fix definition and adjust documentation

Added Pipenv files to .gitignore

typo in docstring

Added note to whatsnew.rst

remove pipenv files from .gitignore

doc/whatsnew.rst

@@ -22,7 +22,7 @@ Bug fixes/enhancements:
 - always initialize the variables ``success`` and ``covar`` the ``MinimizerResult`` (reported by Marc W. Pound; PR #771)
 - build package following PEP517/PEP518; use ``pyproject.toml`` and ``setup.cfg``; leave ``setup.py`` for now (PR #777)
 - components used to create a ``CompositeModel`` can now have different independent variables (@Julian-Hochhaus; Discussion #787; PR #788))
-
+- fixed function definition for ``StepModel(form='linear')``, was not consistent with the other ones. (@matpompili; PR #794)
 
 Deprecations:
 

lmfit/lineshapes.py

@@ -396,7 +396,7 @@ def step(x, amplitude=1.0, center=0.0, sigma=1.0, form='linear'):
     Starts at 0.0, ends at `amplitude`, with half-max at `center`, and
     rising with `form`:
 
-    - `'linear'` (default) = amplitude * min(1, max(0, arg))
+    - `'linear'` (default) = amplitude * min(1, max(0, arg + 0.5))
     - `'atan'`, `'arctan'` = amplitude * (0.5 + atan(arg)/pi)
     - `'erf'`              = amplitude * (1 + erf(arg))/2.0
     - `'logistic'`         = amplitude * [1 - 1/(1 + exp(arg))]
@@ -413,6 +413,7 @@ def step(x, amplitude=1.0, center=0.0, sigma=1.0, form='linear'):
     elif form in ('atan', 'arctan'):
         out = 0.5 + arctan(out)/pi
     elif form == 'linear':
+        out = out + 0.5
         out[where(out < 0)] = 0.0
         out[where(out > 1)] = 1.0
     else:

lmfit/models.py

@@ -1307,7 +1307,7 @@ class StepModel(Model):
         :nowrap:
 
         \begin{eqnarray*}
-        & f(x; A, \mu, \sigma, {\mathrm{form={}'linear{}'}})  & = A \min{[1, \max{(0, \alpha)}]} \\
+        & f(x; A, \mu, \sigma, {\mathrm{form={}'linear{}'}})  & = A \min{[1, \max{(0, \alpha + 1/2)}]} \\
         & f(x; A, \mu, \sigma, {\mathrm{form={}'arctan{}'}})  & = A [1/2 + \arctan{(\alpha)}/{\pi}] \\
         & f(x; A, \mu, \sigma, {\mathrm{form={}'erf{}'}})     & = A [1 + {\operatorname{erf}}(\alpha)]/2 \\
         & f(x; A, \mu, \sigma, {\mathrm{form={}'logistic{}'}})& = A [1 - \frac{1}{1 + e^{\alpha}} ]