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Bug introduced in 9.0 or earlier and fixed in 11.0.0


Let $h[n]=(\frac{1}{2})^n u[n]-3(\frac{1}{2})^{n-1}u[n-1]$
$g[n]=3^n u[n]-\frac{1}{2}3^{n-1}u[n-1]$

I convolved $h[n]$ and $g[n]$ in Mathematica and obtained the following result:

h = (1/2)^n UnitStep[n] - 3*(1/2)^(n - 1) UnitStep[n - 1]
g = 3^n UnitStep[n] - (1/2)*3^(n - 1) UnitStep[n - 1]
DiscretePlot[DiscreteConvolve[h, g, n, m], {m, -5, 5}]

Here is the plot that Mathematica showed me:

enter image description here However, after doing the convolution by hand I obtained a discrete delta function. (i.e. $h[n]*g[n]=\delta[n]$), which I know is the correct answer.

Why is Mathematica giving me an incorrect answer for the convolution?

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  • $\begingroup$ Did you try on a fresh kernel? I get a different plot than you do, where the convolution is indeed zero for positive m and 1 for m==0. For negative m I see negative values. $\endgroup$ – Marius Ladegård Meyer Apr 18 '16 at 16:46
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This is a bug in DiscreteConvolve[].

The bug is caused by a missing condition (m>=0) in one term of the answer returned by DiscreteConvolve[] for your example.

A workaround for the problem is to apply PiecewiseExpand[] to the first two arguments of DiscreteConvolve[] as shown below.

h = (1/2)^n UnitStep[n] - 3*(1/2)^(n - 1) UnitStep[n - 1];

g = 3^n UnitStep[n] - (1/2)*3^(n - 1) UnitStep[n - 1];

DiscreteConvolve[PiecewiseExpand[h], PiecewiseExpand[g], n, m] // InputForm

(*Piecewise[{{3^m, m == 0}}, 0]*)

Table[%, {m, -5, 5}]

(*{0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0}*)

Sorry for the confusion caused by this problem.

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