The input to a linear delta modulator having a step-size $\Delta=0.628$ is a sine wave with frequency $f m$ and peak amplitude $E_{m}$. If the sampling frequency $f_{s}=40 \; \mathrm{kH} \mathrm{z}$, the combination of the sine-wave frequency and the peak amplitude, where slope overload will take place is
- $0.3 \mathrm{~V} \; 8 \; \mathrm{kHz}$
- $1.5 \mathrm{~V} \; 4 \; \mathrm{kHz}$
- $1.5 \mathrm{~V} \; 2 \; \mathrm{kHz}$
- $3.0 \mathrm{~V} \; 1 \; \mathrm{kHz}$