Fourier sum

In the previous section we superposed uniformly spaced frequencies. Now we will superpose delayed impulses. The frequency function of a delayed impulse at time delay t₀ is $e^{i\omega t_0}$.Adding some pulses yields the ``Fourier sum'':  

$$B(\omega) \eq \sum_n \ b_n \ e^{i\omega t_n} \eq \sum_n \ b_n \ e^{i\omega n \Delta t}$$ (19)

The Fourier sum transforms the signal b_t to the frequency function $B(\omega)$.Time will often be denoted by t, even though its units are sample units instead of physical units. Thus we often see b_t in equations like (19) instead of b_n, resulting in an implied $\Delta t=1$.