I have to tie everything to something physical.
Those terms in your transfer function will each do something, that is perform a math operation, to an input signal.
If there is as "s" in the numerator it will attempt to differentiate the input.
If there is a number added to the "s" in numerator it will pass the DC component of a signal.
If there is an "s" in denominator it will attempt to integrate the input.
If there is a number added to the "s" in denominator the output will approach final value exponentially.
If there is a "s+n", in a term, then there is a time constant associated with the operation it is going to perform.
The time constant is usually called "tau", greek for t.
I am accustomed to the terms having form :
tau * s + 1, usually written ts+1
but often as in your case they are written s + 1/tau instead - author's choice.
Your transfer function
s+2
-----
s+5
i would have written as
.5s+1
------ X (2/5)
.2s+1
The 1 in numerator tells me that for any nonzero input there will be a nonzero final value, ie it passes the DC component.
and the .5 there says there'll be a transient component related to how much the input moves in .5 second
and the (.2s + 1) denominator tells me the output will approach final value with a .2 second time constant.
For a step input,
the ratio of [Tau numerator/Tau denominator] tells how large the initial transient will be, 2.5 X the step. After getting multiplied by that (2/5) in parentheses, 2.5 X (2/5) = 1 (how convenient.)
i say the output will step to 1 and exponentially approach final value with .2 sec time constant, and final value will be 1X (2/5) = .4
so i say: output will step to 1 and decay to .4 with .2 sec time constant.
That's just how i have learned to use these things. Like a stone ax.
----------
Let's see what he did:
He writes for output eq 4.2 :
2/5 + 3/5e-5t
2/5 is the final value
and e-5t is a transient with 0.2 sec time constant, .2 being tau from denominator
the e-5t term starts out at 1 and decays to zero, so when multiplied by his 3/5
starts at 3/5 and decays to zero
so output is the sum of:
DC gain component of 2/5
plus a transient starting at 3/5 and decaying to zero
so output steps to 5/5 and decays to 2/5 on a .2 sec time constant
same as:::: steps to 1 and decay to .4
_----------------------------
point being not a mathematical tap-dance
but
do not despair
this technique is handy for shortcutting around some of that algebra
and it's real handy for building opamp circuits
a resistor in parallel with a cap makes ts + 1 , and t= r X c
so your transfer function could have been realized with a single op-amp, two resistors and two caps
and it's what is called in the trade a "Lead-Lag" and is used for aligning phase
and soon you'll be evaluating stability from just looking at the graphs
and if math challlenged me can learn to use this, you will doubtless go me one better and actually understand it.
I hope somebody else chimes in, I'm actually embarassed at my lack of knowledge here. .
old jim