Final answer:
While the detailed steps to plot the Bode plots for the given transfer functions are not provided, the general approach involves identifying poles, zeroes, and gain factors, then plotting their effects on magnitude and phase across the frequency spectrum.
Step-by-step explanation:
To sketch the Bode plot for both magnitude and phase of the given functions, we would apply the principles of logarithmic frequency response analysis. However, the information provided seems to be irrelevant to the actual execution of that task. Bode plots represent the frequency response of a system and involve plotting the magnitude (in decibels) and phase (in degrees or radians) against frequency (usually on a logarithmic scale). For the given transfer functions, we would break down each component (zeroes, poles, gain) and plot their respective effects on the magnitude and phase across the frequency spectrum.
For H(s) = -10³(1 + s/10µ) / (1 + s/10´)(1 + s/10³), we must identify the zeroes at s = -10µ, and the poles at s = -10´ and s = -10³. We would then plot the effects of these zeroes and poles on the magnitude and phase plots. The initial gain factor of -10³ would also affect the overall plot.
For H(s) = 100s / (1 + s/1000)(1 + s/10)(1 + s/10¶), there is a zero at the origin, and poles at s = -1000, s = -10, and s = -10¶. Each of these will contribute to the shape of the Bode plot. The process involves calculating the slopes changes in the magnitude plot and the breakpoints in the phase plot induced by each pole and zero, and the overall effect of the gain of 100s.