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I am tasked in an assignement to select an appropriate rate 1/2 convolution code for a QAM modulated transmitter with the following setup,
However, from my understanding of codes and modulation they can be viewed as individual blocks so it shouldn't matter which code I would select. So I should just select the best BER performance 1/2 convolution code with this setup instead?
Or is my understanding flawed?
Your understanding is a bit flawed. Not a big deal, though.
You cannot add AWGN to bits. One is just discrete values, and the other is a continuously Gaussian-distributed noise. So, you're always adding noise to some continuous-domain representation of your bits. The bit errors then happen on the decision side ("Rectangular QAM demodulator Baseband" in your diagram).
Now, of course it depends on the modulation what errors you get with a given noise realization! For example, you'll find that a 32-PSK has worse bit error performance than your 32-QAM of the same power. Also, you'll find that it's more likely for a symbol that underwent noise to be falsely classified as one of its immediate neighbors (just because in Gaussian noise, high noise amplitudes are less probably than smaller ones). That also leads to the realization that it does matter how you map the bits to the symbols (you might have heard of Gray coding).
So, no, you cannot just ignore the fact that there's a mapping between bits and constellation points in between.
I'm not sure what simulink does if you forget to convert bits to constellation points. It might just be using 0 and 1 as constellation points (that's OOK, and OOK is almost always a bad idea), thereby giving you 2.5 times as much power per bit¹, and of course that makes your transmission more robust than had you used 32-QAM, but also, 5 times slower.
(each bit becomes 1 symbol. There's two equally likely symbols, $0$ and $1$; they have energy $0^2 = 0$ and $1^2 = 1$, respectively, so you get an energy of $1/2$ on average per symbol, and thus also $E_{\text{sym}}=E_b=1/2$ . I'm taking a wild guess, but I'd expect that the 32-QAM modulator you're using is scaled such that its symbol energy is $E_{\text{sym}}=1$; but each symbol carries 5 bits, so $E_b = E_{\text{sym}}/ 5 = 1/5$.)
