4. Consider the three transfer functions shown below:
The plots will not be reproduced here. However, we can find the roots and describe which picture corresponds with each transfer function.
The root of the equation is at s = -0.5. This is a decaying exponential solution that will be stable. Plotted in the real-imaginary plane places this point on the real axis at -0.5. The middle picture corresponds to this transfer function.
The two roots of this equation are at s = -2 and s = -4. Both of these solutions are exponentially decaying and stable, leading to the response shown in the top figure. The points would lie at -2 and -4 on the real axis.
The roots of the equation are at s = -0.38 and s = -2.62. This leads to a stable decaying solution and is also the top figure. These points would lie at -0.38 and -2.62 on the real axis.