Convolution between A(t) and S (t) is defined as integral (-infinity to +infinity) with inside integral we have A(T)S(t-T)dT, now suppose if we have A(T)S(vt-vT) inside that intergal, where 'v' is a constant. Can we still call it as a convolution? if yes how can we take care of 'v' and if its not a convolution how can we manipulate it to make it convolution? This message was sent using the Comp.DSP web interface on www.DSPRelated.com

# Constant in convolution intergal

Started by ●June 2, 2005

Reply by ●June 2, 20052005-06-02

"aries44" <omar_shahid2@hotmail.com> writes:> Convolution between A(t) and S (t) is defined as > > integral (-infinity to +infinity) with inside integral we have > A(T)S(t-T)dT, now suppose if we have A(T)S(vt-vT) inside that intergal, > where 'v' is a constant. Can we still call it as a convolution?No.> if yes how > can we take care of 'v' and if its not a convolution how can we manipulate > it to make it convolution?Let S'(t) = S(vt). Then what you have expressed is the convolution of A(t) with S'(t). -- Randy Yates Sony Ericsson Mobile Communications Research Triangle Park, NC, USA randy.yates@sonyericsson.com, 919-472-1124

Reply by ●June 3, 20052005-06-03

>"aries44" <omar_shahid2@hotmail.com> writes: > >> Convolution between A(t) and S (t) is defined as >> >> integral (-infinity to +infinity) with inside integral we have >> A(T)S(t-T)dT, now suppose if we have A(T)S(vt-vT) inside thatintergal,>> where 'v' is a constant. Can we still call it as a convolution? > >No. > >> if yes how >> can we take care of 'v' and if its not a convolution how can wemanipulate>> it to make it convolution? > >Let S'(t) = S(vt). Then what you have expressed is the convolution of >A(t) with S'(t).shouldnt it be S'(t) = S(vT) instead of S(vt)??????>Randy Yates >Sony Ericsson Mobile Communications >Research Triangle Park, NC, USA >randy.yates@sonyericsson.com, 919-472-1124 >This message was sent using the Comp.DSP web interface on www.DSPRelated.com

Reply by ●June 3, 20052005-06-03