ABSTRACT
The integrator Bode plot is shown in Figure 15. The line will cross the 0 dB line at co= 1 radjsec since the magnitude of 1/jco = 1 and the log of 1 is 0. Remember that this is the amount added to the total response for each integrating factor in our system. The phase angle contribution was explained earlier using phasors, where we saw that each s (or jco in our case) contributes 90 degrees of phase. That is why two imaginary numbers multiplied by each other equals -1 (a phasor of magnitude one along the negative real axis). When the imaginary number is in the denominator, the angle contribution becomes negative instead of positive (j * j = -1 is still true, it just rotates clockwise -90 degrees for each s instead of counterclockwise). Understanding this concept makes the remaining terms easy to describe.
