Re: Fourier transform of the delta function
- To: mathgroup at smc.vnet.net
- Subject: [mg37497] Re: [mg37468] Fourier transform of the delta function
- From: Vladimir Bondarenko <vvb at mail.strace.net>
- Date: Fri, 1 Nov 2002 01:44:09 -0500 (EST)
- Reply-to: Vladimir Bondarenko <vvb at mail.strace.net>
- Sender: owner-wri-mathgroup at wolfram.com
Matt Flax <flatmax at ieee.org> writes on Thu, 31 Oct 2002 04:42:04 MF> Mathematica simply spits back the same input. That's not quite right, in fact, not all is so dark and dim ;-) The following versions 4.2 for Microsoft Windows (June 5, 2002) 4.1 for Microsoft Windows (November 2, 2000) 4.0 for Microsoft Windows (April 21, 1999) return the correct output. In[1] := FourierTransform[DiracDelta[f - (-1 + 10^(n T/c)/b)], f, t] Out[1] = E^(I*(-1 + 10^((n*T)/c)/b)*t)/Sqrt[2*Pi] In[2] := InverseFourierTransform[%, t, f] Out[2] = DiracDelta[-1 + 10^((n*T)/c)/b - f] which is OK as DiracDelta[-z] == DiracDelta[z] . 4.2 for Microsoft Windows (February 28, 2002) returns the sine/cosine variant (Cos[t - (10^((n*T)/c)*t)/b] - I*Sin[t - (10^((n*T)/c)*t)/b])/Sqrt[2*Pi] The Microsoft Windows 3.0 (April 25, 1997) version requires an add-on to be loaded first, and it uses a different normalization. In[1] := <<Calculus`FourierTransform` In[2] := FourierTransform[DiracDelta[f - (-1 + 10^(n T/c)/b)], f, t] Out[2] = E^(-I*(1 - 10^((n*T)/c)/b)*t) In[3] := InverseFourierTransform[%, t, f] Out[3] = DiracDelta[1 - 10^((n*T)/c)/b + f] Alas, as you pointed out, currently, none Mathematica version can calculate FourierTransform[KroneckerDelta[f-(-1 + 10^(n T/c)/b)], f, t] straightforwardly. Best wishes, Vladimir Bondarenko Mathematical and Production Director Symbolic Testing Group Email: vvb at mail.strace.net Web : http://www.CAS-testing.org/ (under development, 95% ready) http://maple.bug-list.org/ (under development, 20% ready) Voice: (380)-652-447325 Mon-Fri 6 a.m. - 3 p.m. GMT ICQ : 173050619 Mail : 76 Zalesskaya Str, Simferopol, Crimea, Ukraine