It's a \$RLC\$ circuit, maybe.
First, from the right graph, when the real part is \$100\Omega\$ or so, the image part range from \$-500\$ to \$500\Omega\$, so i guess it has a \$R\$ in series with a reactive part. And from the the phase graph, it apparently capacitive at low frequency, inductive at high frequency, so it maybe has a \$C\$ and \$L\$ in series. Now the whole circuit should be a \$RLC\$ in series.
Omit the \$L\$ part at low frequency, and omit the \$C\$ part at high frequency, then
$$
R=100\Omega\\
|Z_{x}|=\sqrt{R^2+\frac{1}{w^2C^2}} = 500 \quad \text{when}\quad w=2\pi \times 1Hz\\
|Z_{x}|=\sqrt{R^2+w^2L^2}=500 \quad \text{when}\quad w=2\pi \times 10^3Hz\\
$$
Solve it, we get
$$
C=2.69 \times 10^{-4}\text{F},L=7.8 \times 10^{-2}\text{H}
$$
Because we omitted \$C\$ at high frequency, and omitted \$L\$ at low frequency, there should some error. So, we adjust the value of \$C\$ and \$L\$, finally get your graphs.
$$
C=3 \times 10^{-4}\text{F},L=8 \times 10^{-2}\text{H},R=100\Omega
$$
Why do we have a large negative drop off in the Bode Plot when we have a large positive rise in the time domain?
Because those plots belong to two different domains. A large peak in time domain does not guarantee a peak in frequency domain. Consider unit impulse function as an example and think.
Is it possible to observe the effect of the pole in the time domain?
Yes. Depending on the position of pole various effects can be observed in the time domain. But it doesn't mean that the effect of a pole at s=5 can be observed at specific time like t=5 or t=1/5. See examples:
If there is a pole at right half plane then signal in time domain will be exponentially increasing.
If there are poles on the imaginary axis, then oscillations can be seen in the time domain.
Finally, how do we understand the phase of this function? I can't see any phase effect in the time domain at all.
The phase plot of a transfer function gives the phase delay caused by that system for different frequencies. To see an effect, feed a sinusoidal signal to a system with impulse response \$e^{−30t}\$ and measure the phase difference between input and output.
To understand better, I suggest you to perform an experiment. The impulse response \$e^{−30t}\$ corresponds to a RC low pass filter with RC = 1/30. So implement a this circuit on a breadboard or simulator. Do a frequency sweep and plot the gain and phase with respect to the frequency. Compare with the bode plot.
Best Answer
Note that current is actually common to both C and R, not really as shown in the schematic.
If you compare the input voltage (Vin) with the output voltage (Vout), the input amplitude is always >= output amplitude. At low frequencies, output is much smaller than input.
And at low frequencies, phase of Vout leads phase of Vin:
At high frequency, well into the passband, Vout amplitude is nearly equal Vin amplitude, and phase of Vout approaches that of Vin: