Question 1: Assuming that the oscillator (perhaps programmable) is
given, what kind of hardware would you suggest for (2) and (3) ?
For 2: All-pass (phase-shifting) filter, a.k.a. Hilbert transformer
For 3: Balanced modulator
Question 2: Does this setup have significant shortcomings (beside cost
of components?)
Yes. While direct conversion to complex baseband is regularly done in the digital domain, it can be quite finicky to get the same concept working well (and reliably) in the analog domain. It's nearly always easier to do the detection at an intermediate RF frequency.
Question 3: is this observation [about square-wave local oscillators] correct?
Yes, except that you want a bandbass filter that's centered on the carrier frequency, not a lowpass filter. Sometimes this filter has a bandwidth of approximately 2B, in which case it needs to be tuned along with the local oscillator. This sort of setup is called a preselector, and is commonly used in superheterodyne receivers, such as those used for the AM and FM broadcast bands.
But sometimes, a fixed bandpass filter that covers the entire band of interest is used instead. This works as long as none of the unwanted "image" bands ever falls into the band of interest. This is more common in 2-way VHF communications systems, such as those used for air traffic control and public service (police, fire).
What is the physical meaning of "first" and "second order"? ... How do I know if a system is first or second order?
A 1st order system has one energy storage element and requires just one initial condition to specify the unique solution to the governing differential equation. RC and RL circuits are 1st order systems since each has one energy storage element, a capacitor and inductor respectively.
A 2nd order system has two energy storage elements and requires two initial conditions to specify the unique solution. An RLC circuit is a 2nd order system since it contains a capacitor and an inductor
Where do equations (1) and (4) come from?
Consider the homogeneous case for the 1st order equation:
$$\tau \frac{dy}{dt} + y = 0$$
As is well known, the solution is of the form
$$y_c(t) = y_c(0) \cdot e^{-\frac{t}{\tau}}$$
which gives physical significance to the parameter \$\tau\$ - it is the time constant associated with the system. The larger the time constant \$\tau\$, the longer transients take to decay.
For the 2nd order system, the homogeneous equation is
$$\tau^2\frac{d^2y}{dt^2} + 2\tau \zeta \frac{dy}{dt} + y = 0$$
Assuming the solutions are of the form \$e^{st}\$, the associated characteristic equation is thus
$$\tau^2s^2 + 2\tau\zeta s + 1 = 0 $$
which has two solutions
$$s = \frac{-\zeta \pm\sqrt{\zeta^2 -1}}{\tau}$$
which gives physical meaning to the damping constant \$\zeta\$ associated with the system.
The transient solutions are, when \$\zeta > 1\$ (overdamped), of the form
$$y_c(t) = Ae^{\frac{-\zeta +\sqrt{\zeta^2 -1}}{\tau}t} + Be^{\frac{-\zeta -\sqrt{\zeta^2 -1}}{\tau}t} $$
when \$\zeta = 1\$ (critically damped), the solutions are of the form
$$y_c(t) = \left(A + Bt\right)e^{-\frac{\zeta}{\tau}t} $$
and when \$\zeta < 1\$ (underdamped), the solutions are of the form
$$y_c(t) = e^{-\frac{\zeta}{\tau}t}\left(A\cos \left(t\sqrt{1 - \zeta^2}\right) + B\sin \left(t\sqrt{1 - \zeta^2}\right) \right)$$
When given a first order system, why is sometimes equation (2) given,
and sometimes equation (3) as the transfer function for this system?
Different disciplines have different conventions and standard forms. Equation (2) looks to me like control theory standard while equation (3) looks like signal processing standard.
Standard forms evolve to fit the needs of a discipline. Further, if a particularly influential person or group develops and uses a particular convention, that convention often becomes the standard. It might be educational to peruse older textbooks and journals to get a sense of how notation and standard forms evolve.
Best Answer
\$\operatorname{rect}(\cdot)\$ is a function that has value \$1\$ if whatever appears inside those parentheses (it's called the argument of the rect function and I used \$\cdot\$ instead of some algebraic variable as a place holder) has value between \$-\frac 12\$ and \$+\frac 12\$. Otherwise, when the argument is strictly smaller than \$-\frac 12\$ (or strictly larger than \$+\frac 12\$), \$\operatorname{rect}(\cdot)\$ has value \$0\$. In your instance, you need to ask
and a little thought will show, I hope, that \$\omega\$ must be in the interval from \$10-\pi\$ to \$10 + \pi\$. If you have trouble deriving this, try and find the value of \$\omega\$ that makes \$\displaystyle \frac{\omega - 10}{2\pi}\$ equal exactly \$-\frac 12\$ and then, lather, rinse and repeat for exactly \$\frac 12\$.