4.13-4 Use the same m¡(t) and fc = 500 Hz from Problem 4.13-1. Now consider another message signal m_{2}(t)=\sin (50 \pi t)[u(t+0.04)-u(t-0.04)] \text { (a) QAM signal } s(t)=m_{1}(t) \cos

2 \pi f_{c} t+m_{2}(t) \sin 2 \pi f_{c} t \text { in time domain and frequency domain. } \text { (b) The signal } e_{1}(t)=2 s(t) \cos 2 \pi f_{c} t \text { in time and frequency response. } \text { (c) The signal } e_{2}(t)=2 s(t) \sin 2 \pi f_{c} t \text { in time and frequency response. } (d) The two recovered message signals in time domain.