In Amplitude Shift Keying, the bandwidth required is given by \$B=(1+d)S\$ where B is bandwidth, S is the signal rate, and d is a value of either \$0\$ or \$1\$.
But how could the required bandwidth be just some multiples of the signal rate when one signal element itself could have higher frequencies?
Say in this picture:
The baud rate is 5, which implies the signal rate is also 5 signals per second. One signal itself would need 3 cycles. Then the worst case scenario is where all the bits are 1 and the thus the highest frequency is \$ 3 \times 5 =15Hz\$ and so the minimum bandwidth required of a channel to allow this modulated signal to pass through is still 15Hz. Is this right?
However, if I were to follow the formula \$B=(1+d)S\$, I would get...
If let \$d=0\$, \$ B=(1+0) \times 5 = 5Hz \$
If let \$d=1\$, \$ B=(1+1) \times 5 = 10Hz \$
But is 5Hz or 10Hz be sufficient for the signal in the picture? Why does the formula say that the bandwidth is either equivalent or maximum 2 times the signal rate? How is it so?