The quantum Heisenberg uncertainty relations are completely analogous to the Fourier uncertainty relations which are often used in time series analysis and signal processing. Both statements are about the width of wave packets. I think the term uncertainty is a misnomer. As @GuyInchbald stated, it is a historical artifact. The modern understanding is that the Heisenberg uncertainty relations are fundamentally about wavefunctions, not measurements.
You describe an experiment where you start with no wave, and then begin to continuously generate waves that propagate away. If this were a quantum mechanical wavefunction, you would be continuously changing the quantum state. The wavefunction uniquely determines the state. When you add another cycle to the wavefunction, you've made a different quantum state.
... when measuring this wave the wave itself has not changed, only how we are measuring it.
This statement is not true.
What if you make a localized wave packet and send it away? The classical wave equation would say that the wave packet would preserve its shape, so its width would not change. For a wavefunction, the Schrödinger equation would say that the wave would spread out as it propagates away. As time passes the state evolves, its width increases. The $\Delta x$ and $\Delta p$ for a propagating wavepacket would change.
I have gained a lot of insight by thinking about the connection between Fourier and Heisenberg uncertainty and by applying uncertainty reasoning to situations where no measurements occur.
Fourier uncertainty
A sine wave $\psi(t) = \sin(\omega_0 t)$ has a single frequency $\omega_0$, but it is spread out over all time. We can define the width of any signal in time, by its RMS width
$$\Delta t = \sqrt{\left< (t-\left<t\right>)^2 \right>},$$
where the brackets $\left<\right>$ denote the average. For the sine wave this would give $\Delta t \rightarrow \infty$.
We can think of the time representation of the wave as specifying a basis to describe the wave. Each instant of time gets a basis vector, and the amplitude of the wave at that time is one element of the vector.
If we take the Fourier transform of the sine wave we would get a frequency representation:
$$\tilde\psi(\omega) = \delta(\omega-\omega_0),$$
where $\delta$ is the Dirac delta function. In the frequency domain the sine wave is zero everywhere except $\omega_0$. It is infinitely thin.
$$\Delta \omega = \sqrt{\left< (\omega-\left<\omega\right>)^2 \right>} = 0.$$
This frequency representation is a new basis for the wave. There is a basis vector for each frequency value. In this basis it takes just one basis vector to define the wave, the one for $\omega_0$.
If we have a finite wave packet, like a Gaussian enveloped sine wave
$$\psi(t) = e^{-t^2/(2\sigma^2)}\sin(\omega_0 t),$$
the wave will have a finite width in both the time and frequency representation. This wave has some spread over multiple $\omega$s.
Using the RMS width it can be shown that
$$\Delta t\,\Delta\omega \ge \frac{1}{2}.$$
This is a general property of Fourier transforms that tells us about the RMS width of waves in the two representations.
Heisenberg uncertainty
Now what if our wave were a quantum mechanical wavefunction? The same thing would be true. If we multiply the Fourier uncertainty statement by $\hbar$, we get
$$\Delta t\,(\hbar\Delta\omega) = \Delta t\, \Delta E \ge \frac{\hbar}{2}.$$
This says the width of a wavefunction in its energy representation is related to its width in its time representation. An energy eigenstate has $\Delta E=0$, so $\Delta t \rightarrow \infty$. The wavefunction for an energy eigenstate extends infinitely forward through time. This is equivalent to the statement that energy eigenstates are stationary states.
There's no reference to measurement in the previous statement. It is a statement about the wavefunction when written in two different bases: $\psi(t)$ and $\tilde\psi(E)$.
If the wavefunction describes a mixture of energy states, it has finite width in time. It is not stationary.
$$\psi(E) = A\delta(E-E_1) + B\delta(E-E_2) \implies \Delta E > 0 \implies \Delta t\,\,\mathrm{finite}$$
There's an analogous Fourier relation for position $x$ and wavenumber $k = \frac{2\pi}{\lambda}$, which leads to the quantum position momentum uncertainty relation since $p = \hbar k$:
$$\Delta x\, \Delta k \ge \frac{1}{2}\quad\implies\quad \Delta x\, \Delta p \ge \frac{\hbar}{2}.$$
This uncertainty relation tells us about the width of of the wavefunction in two different basis representations: $\psi(x)$ and $\tilde\psi(p)$.
