On how to formulate and apply maximum likelihood I have just delved into the basics of maximum likelihood estimation and expectation maximization. The latter is really difficult to follow and I am having a tough time in figuring how I can apply the EM method for parameter estimation.
 A: When you use EM to obtain maximum likelihood estimates, you need a variable that describes your observations $x_{n}$, latent variables $z_{n}$ that are in some way related to your observations (e.g. in coin tossing experiments, $\{H, T\}$ are the latent variables and in gaussian mixtures, the mixing coefficients $\pi_{i}$ take the role of latent variables) and the parameters $\theta$ that you are trying to estimate.
At the risk of not answering your question at all, I think you want a maximum likelihood estimate of $\theta$ using EM based on known observations $x_{n}$ that are given by the following equation:
$$x_{t} = s_{t}(\theta_{0}) + n_{t}$$
If that is correct, a general idea is the following. Since $n_{t}$ is white noise $N(0, \sigma)$, $x_{t}$ can be described by a Gaussian $p(x_{t}|s_{t},\theta) = N(s_{t}(\theta), \sigma)$. In the EM formulation, $x_{t}$'s are known variables, $s_{t}$'s are latent variables and $\theta$ is the parameter. It is customary to group the variables $x_{n}$ in a variable $X$ and likewise, latent variables $s_{n}$ are grouped in a variable $S$.
As you should know, the EM algorithm consists of 2 steps: expectation and maximization. In the expectation step, we use an expression $Q$ as a proxy for the likelihood $L(\theta|X) = p(X|\theta)$, that is, the probability of 
getting the known data $X$ given a parameter $\theta$. This is the same likelihood used to obtain maximum likelihood estimates. However, in EM we use this $Q$ instead:
$$Q(\theta|\theta^{\text{old}}) = E_{S|X, \theta^{\text{old}}} \log p(X,S|\theta)$$
This odd-looking expression is actually a lower bound of the likelihood $L(\theta|X)$. Bishop's book contains a good derivation of $Q$.
In order to start the EM magic, you have to choose a random $\theta^{\text{old}}$ and calculate this expectation. Notice that you need $p(X,S|\theta)$ and $p(S|X,\theta^{\text{old}})$. $p(X,S|\theta)$ is equal to $p(X|S,\theta)p(S|\theta)$ and using Bayes' theorem, $p(S|X,\theta^{\text{old}})$ is 
proportional to $p(X|S,\theta^{\text{old}})p(S|\theta^{\text{old}})$.
At this point, I hope it is clear that $p(X|S,\theta)=\prod_{t} p(x_{t}|s_{t},\theta)$, so that part is not hard to calculate. However, $p(S|\theta)$, that is, $\prod_{t}p(s_{t}|\theta)$ is required. I don't know what distribution could be appropriate since this depends on the specifics of your problem so I will assume you know.
By now, you can calculate $Q(\theta|\theta^{\text{old}})$. 
The maximization step is simply:
$$\theta = \text{arg max}_{\theta} Q(\theta|\theta^{\text{old}})$$
This is the new $\theta$ to be used in the expectation step again until convergence. 
That is a general idea of how EM could work in this case. However, maybe you don't know a distribution for $s_{t}$ or it is difficult to calculate the expectation or the maximization step. 
For the big picture, take a look at this nice explanation.
UPDATE
I think you changed the question quite a bit. Are you asking how to calculate maximum likelihood estimates? Basically, you apply a derivative to the likelihood on the parameter you want to estimate:
$$\frac{\partial}{\partial \theta}L(\theta|X) = 0$$
solve it and that's pretty much it. See more examples here.
