As per the question,
"V_m (t) = 0.20 \\sin (2 \\times 10^3 t ) + 0.25 \\sin (2\\times 10^3 t) =0.45\\sin(2\\times 10^3t)"
"Vc (t) = 10 \\sin (2\u03c0 \\times 100\\times 10^6 t) V"
Hence, amplitude modulation,
"V_{AM}=(V_c+V_m\\cos (2\\pi f_m t))\\cos(2\\pi f_ct)"
"V_{AM}=10+0.45=10.45V"
band width of AM = "2(f_1+f_2)=(\\dfrac{2\\times 10^3+2\\pi\\times 100\\times 10^6}{\\pi})"
Comments
Leave a comment