Systems and signals – integration with dirac’s function and step function

Small signal is relative term. When non-linear device (for e.g. transistor) is DC biased, it stays fairly linear in small region around the bias point. Small signal is the one that fits in the linear region.

http://en.wikipedia.org/wiki/Small-signal_model

I think you'd better not split the functions. If you want to calculate the fourier transform of \$ 0.5^t u(t)\$, then you might put all that into the integral. Therefore, your integral limits will become \$0\$ to \$+\infty\$ because of \$u(t)\$. And your integral result should not be infinity because your function \$0.5^t\$ goes to zero.

Solving the integral:

\$\large\int_0^\infty e^{(-jwt)} . 0.5^t dt\$

the inside part could be rewrite:

\$\huge\frac{e^{(-jwt)}}{2^t} => (\frac{e^{(-jw)}}{2})^t\$

You could name \$\large\frac{e^{(-jw)}}{2} = u\$ so you get integral of \$\large u^t dt = \frac{u^t}{ln(u)} + C\$

Then you get (replacing u):

\$\huge\frac{(\frac{e^{-jw}}{2})^t}{ ln(e^{-jw}/2)}\$

When t goes to \$+\infty\$ you get zero. When it goes to zero you get:

\$\large\frac{1}{ln(e^{-jw}/2)} \$

which left us:

\$\large\frac{1}{-jw - ln(2)}\$

I'm not sure I did not get lost in calculations but I believe its correct.