Hi all, Just looking for a quick confirmation: for a function f1(t)=g(t)u(t), f2(t)=g(t-c) u(t-c), where "u(t)" is the Heaviside function, also called step function. Let G(f) be the Fourier transform of g(t), U(f) be the Fourier Transform of u(t). The two approaches: 1. First find Fourier transform of f1(t)=g(t)u(t), that's to say, G(f) convolve with U(f), then multiply it by exp(-j*2*pi*f*c). 2. First find Fourier transfrom of g(t-c), which is exp(-j*2*pi*f*c)G(f), and then find the Fourier transform of u(t-c), which is exp(-j*2*pi*f*c)U(f), finally convolve exp(-j*2*pi*f*c)G(f) and exp(-j*2*pi*f*c)U(f) to obtain the overall Fourier Transform. Are the above two approaches the same? Thx

# are these two Fourier transform the same thing?

Started by ●August 17, 2006

Reply by ●August 17, 20062006-08-17

Lucy wrote:> Hi all, > > Just looking for a quick confirmation: > > for a function > > f1(t)=g(t)u(t), > f2(t)=g(t-c) u(t-c), > > where "u(t)" is the Heaviside function, also called step function. > > Let G(f) be the Fourier transform of g(t), U(f) be the Fourier > Transform of u(t). > > The two approaches: > > 1. First find Fourier transform of f1(t)=g(t)u(t), that's to say, G(f) > convolve with U(f), then multiply it by exp(-j*2*pi*f*c). > > 2. First find Fourier transfrom of g(t-c), which is > exp(-j*2*pi*f*c)G(f), and then find the Fourier transform of u(t-c), > which is exp(-j*2*pi*f*c)U(f), finally convolve exp(-j*2*pi*f*c)G(f) > and exp(-j*2*pi*f*c)U(f) to obtain the overall Fourier Transform. > > Are the above two approaches the same? > > Thx >Yes it is. I use 'w' instead of '2*pi*f' below. For method 2) you find for the convolution (operator 'X'): F2(w) = [exp(-iwc)G(w)] X [exp(-iwc)U(w)] If you write out the convolution where uppercase 'W' is used as integration variable (like the 'tau' you're used to in the time domain): F2(w) = int exp(-iWc)G(W) exp(-i(w-W)c)U(w-W) dW For the two exponent functions, the uppercase 'W' cancels out, and you only have the 'w' which does not contain the integration variable 'W', so you can take this part out of the integration: F2(w) = exp(-iwc) int G(W)U(w-W) dW = exp(-iwc) [G(w) X U(w)] This exactly the method used in 1), convolution of G(w) and U(w), then multiply by exp(-iwc). Jeroen

Reply by ●August 18, 20062006-08-18

Lucy wrote:> Hi all, > > Just looking for a quick confirmation: > > for a function > > f1(t)=g(t)u(t), > f2(t)=g(t-c) u(t-c), > > where "u(t)" is the Heaviside function, also called step function. > > Let G(f) be the Fourier transform of g(t), U(f) be the Fourier > Transform of u(t). > > The two approaches: > > 1. First find Fourier transform of f1(t)=g(t)u(t), that's to say, G(f) > convolve with U(f), then multiply it by exp(-j*2*pi*f*c). > > 2. First find Fourier transfrom of g(t-c), which is > exp(-j*2*pi*f*c)G(f), and then find the Fourier transform of u(t-c), > which is exp(-j*2*pi*f*c)U(f), finally convolve exp(-j*2*pi*f*c)G(f) > and exp(-j*2*pi*f*c)U(f) to obtain the overall Fourier Transform. > > Are the above two approaches the same?Yes. An even simpler approach comes when you recognize that the step is the integral of an impulse. Jerry -- Engineering is the art of making what you want from things you can get. �����������������������������������������������������������������������

Reply by ●August 18, 20062006-08-18

Lucy wrote:> Hi all, > > Just looking for a quick confirmation: > > for a function > > f1(t)=g(t)u(t), > f2(t)=g(t-c) u(t-c), > > where "u(t)" is the Heaviside function, also called step function. > > Let G(f) be the Fourier transform of g(t), U(f) be the Fourier > Transform of u(t). > > The two approaches: > > 1. First find Fourier transform of f1(t)=g(t)u(t), that's to say, G(f) > convolve with U(f), then multiply it by exp(-j*2*pi*f*c). > > 2. First find Fourier transfrom of g(t-c), which is > exp(-j*2*pi*f*c)G(f), and then find the Fourier transform of u(t-c), > which is exp(-j*2*pi*f*c)U(f), finally convolve exp(-j*2*pi*f*c)G(f) > and exp(-j*2*pi*f*c)U(f) to obtain the overall Fourier Transform. > > Are the above two approaches the same? > > Thx

Reply by ●August 18, 20062006-08-18

Lucy wrote:> Hi all, > > Just looking for a quick confirmation: > > for a function > > f1(t)=g(t)u(t), > f2(t)=g(t-c) u(t-c), > > where "u(t)" is the Heaviside function, also called step function. > > Let G(f) be the Fourier transform of g(t), U(f) be the Fourier > Transform of u(t). > > The two approaches: > > 1. First find Fourier transform of f1(t)=g(t)u(t), that's to say, G(f) > convolve with U(f), then multiply it by exp(-j*2*pi*f*c). > > 2. First find Fourier transfrom of g(t-c), which is > exp(-j*2*pi*f*c)G(f), and then find the Fourier transform of u(t-c), > which is exp(-j*2*pi*f*c)U(f), finally convolve exp(-j*2*pi*f*c)G(f) > and exp(-j*2*pi*f*c)U(f) to obtain the overall Fourier Transform. > > Are the above two approaches the same? > > ThxHello Lucy, For what its worth, the theorem that relates a shift in one domain to the complex modulation in the other domain is known as the Heaviside shifting theorem. Clay S. Turner