Suppose you have a source of light that emits light with a wavelength of 2 meters, and you set the device to be turned on and switched off alternately. You also set it so that each interval the device is turned on is only long enough for 1 meter to be emitted (1/2 a wavelength). Do you ever observe any photons?
Do you ever observe any photons?
Yes, though probably not for the reason you think.
This is what the wave you're generating looks like. I've included a cosine wave for reference:
So you are still generating a wave, but it has a different shape to a cosine wave. But any wave can be expressed as a sum of sines and cosines by expanding it as a Fourier series. So if we call your signal $S$ the expansion would be:
$$ S(\omega t) = a_0 + a_1\cos(\omega t) + a_2\cos(2\omega t) + a_3\cos(3\omega t) + ... $$
where the numbers $a_n$ are constants (actually the general expression is more complex that this but I've simplified it by assuming your signal is symmetric about $t = 0$). So you are actually transmitting a sum of waves with frequencies that are integer multiples of your base frequency $\omega$. In other words you are sending out a stream of photons with different energies.
I suspect your going in point is that a photon is a particle like a tiny billiard ball, and if you only transmit half a wave how can you be transmitted half a photon? The key thing to remember is that a photon is a quantum object so it is delocalised and doesn't have a well defined position. In effect the photons are spread out over the whole length of the beam of waves you're sending out, and there is no correlation between the position of the photon and the wavelength of the signal. In this case, where you're sending out a series of half waves, the photons are effectively composed of a sum of all the half waves.