Return to Revisions

If the cut off for the low pass filtering (partial sum) is sufficiently high, the number of zeros of $f(M)$ is equal to the number of sign changes of $M$. Although the Gibb's phenomenon comes into play, as long as the cut off frquency is high enough, the Gibb's oscillations do not affect the zero crossings of $f(M)$ as the wave $M$ is binary. Remember that the max over shoot of Gibbs overshoot is only approximately $9%$ percent of the jump amount.