# DTFT 2 Negative Exponents.

Cuthbert Nyack
The DTFT of a negative exponent f(t) = e-at is given by The DTFT has maxima where e-jwT = 1, ie wT = 2np or w = n 2p/T = n ws and minima where e-jwT = -1, ie wT = (2n + 1)p or w = (n + ½) 2p/T = (n + ½) ws.

One characteristic of this signal is that the spectrum decreases as the first power of the frequency because of the discontinuity at the origin. It is difficult to sample this signal without aliasing effects. One way to limit the spectrum is to subtract a second exponent to give
f(t) = e-at - e-bt where b > a. This has the effect of removing the discontinuity at the origin resulting in a spectrum which reduces faster with frequency.
The applet below shows the frequency spectrum of f(t). The magnitude of the spectrum is shown in green. The phase of the spectrum is in red. Real part is in cyan. and imaginary part in yellow. The samples are in magenta and f(t) in brown.

eg parameters:-
(0.3, 0.05, 0.5, 100.0, NA...) This shows that both the FT and the DTFT are in close agreement up to half the sampling frequency and that both are negligible at half the sampling frequency.
(0.3, 0.25, 1.0, 20.0, NA...) This shows the symmetry of the different components in the spectrum.