Quantization problem
without loss of generality, it doesn't matter what \$\varphi\$ is and it can be assumed \$ A_c \ge 0 \$.

since this is a 1-bit quantizer, the sole quantizing threshold is at 0.  then the error will depend on the ratio of the output level of the quantizer and \$ A_c \$.  let the 1-bit quantizer be defined as:

$$ \hat{x}(t) \triangleq \frac{\Delta}{2}\operatorname{sgn}\{ x(t) \}  $$

where \$ x(t) = A_c \sin(2 \pi f_c t + \varphi) \$ and the stepsize \$ \Delta &gt; 0 \$.

and the quantization error is

$$ \epsilon_x(t) \triangleq \hat{x}(t) - x(t) $$

if the output level of the 1-bit quantizer is \$\pm 1\$ (which means \$\Delta=2\$) and \$ A_c \$ is a million, you can expect an awful lotta quantization error.  but this 1-bit quantizer will behave exactly the same as if \$ A_c = 2 \$ in which the quantization error would be far less.  somehow, a 1-bit quantizer has an inherent gain factor in it and that has to be modeled from some knowledge of the nature of the amplitude of \$x(t)\$.