Nondeterministic PDA for the following language with Kleene star I had a question regarding converting a language with the Kleene star production into a PDA. Here's the particular language I was looking at in my textbook:

$$L = (aaa^*bab)$$

My normal approach to constructing PDA's for languages would be to determine the pattern or rule of the language, and think to myself that it's sort of like counting the number of symbols using a stack, and determining whether the stack is empty at the end of the computation.

However, I'm not sure how to construct a PDA for languages like the one above where there isn't a particular rule regarding the $a^*$ production.

How would one normally handle these types of situations? Thank you.
I figured out the answer and hope this will help anyone with similar confusion in the future.

Since a pushdown automaton is essentially a finite automaton with an extra component called the stack which is used for storage, if we are presented with a regular language like the one given above, we simply need to create a PDA that does not use the stack. This concept confused me at first, but here's what I mean:

$$L = (aaa^*bab)$$

$\space$

$$M = (Q, \Sigma, \Gamma, \delta, q_0, z, F)$$
$$Q = \{q_0,\ q_1,\ q_2,\ q_3,\ q_4,\ q_5\}$$
$$\Sigma = \{a,\ b\}$$
$$\Gamma = \{$\}$$
$$z = \$$$
$$F = \{q_5\}$$

$\space$

$$\delta(q_0, a, $) = \{(q_1,\ $)\}$$
$$\delta(q_1, a, $) = \{(q_2,\ $)\}$$
$$\delta(q_2, a, $) = \{(q_2,\ $)\}$$
$$\delta(q_2, b, $) = \{(q_3,\ $)\}$$
$$\delta(q_3, a, $) = \{(q_4,\ $)\}$$
$$\delta(q_4, b, $) = \{(q_5,\ $)\}$$

$\space$

I'm not completely certain about the stack start symbol, because I've noticed some resources use it regardless, while others assume that at the beginning the stack is empty ($\lambda$) and push the start symbol at the first step.

Anyway, I hope this answer helps anyone else.