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I think the Fourier transform for \[f(t)=\frac{1}{t}\] is \[F(\omega) = -j \pi sgn(\omega)\], where \[sgn\] is the sign function. See top of page 2 in **broken link removed** and note that you can multiply both sides of the transform integral by constants. I think this is consistent with what steve10 reported above.
Is this signal valid for all values of time (positive and negative t)?
If the signal does exist for positive values of time and h(t) is indeed an impulse response (with the impulse applied at t=0) then this would be a non-causal system. This just means that there is a response before the signal is even applied!! If this is a causal system then you would have to make the negative portion of the respose equal zero or just h(t) = (1/t) u(t), where u(t) is the unit step function. This would make the evaluation of the integral much easier as you would only have to integrate from zero to \[\infty\].
The fourier transform of 1/Πt is -jsgn(f)
So FT of 1/t will bo -jΠsg(f)
sgn(f) has the property that
sgn(f)=1 f>0
= -1 f<0
The system is a hilbert transform system(i.e) it produces a phase shift of + or - 90 degrees based on wetherthe frequency is - or +.
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