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?

The bandwidth doesn't start at DC (0 Hz), but is centered around your carrier frequency, so

$f_{MIN} = f_C - BW/2 = 15Hz - 5Hz = 10Hz$, and
$f_{MAX} = f_C + BW/2 = 15Hz + 5Hz = 20Hz$

or

$BW = f_{MAX} - f_{MIN} = 20Hz - 10Hz = 10Hz$

The minimum bandwidth for ASK is equal to the Baud rate (which in this case = bit rate).

Also, IIRC, d can be a value in [0..1], so isn't restricted to 0 or 1.

• Since the $BW$ is unknown before I have the $f_{max}$ and $f_{min}$, how could I know that for $f_{MIN} = f_C - BW/2$, the $BW/2$ is equals to $5Hz$ so that I could subsequently find the $f_{max}$ and $f_{min}$? – xenon Sep 20 '11 at 12:48
• @xEnOn - depends on the factor $d$. If $d$ = 0 then $BW$ = signal rate. I don't recall where the $d$ comes from :-( – stevenvh Sep 20 '11 at 13:00
• I read that $d$ comes from the implementation ratio. According to the formula, indeed, the bandwidth = signal rate when $d=0$. But why and how is the bandwidth equals to the signal rate? – xenon Sep 20 '11 at 13:33
• ASK is a sine wave modulated by a rectangular pulse train...so in the frequency domain the spectrum is centered around the carrier frequency fc=15Hz and the shape of the spectrum is a sync if the pulse train is of limited duration. If the pulse train is infinite we will get harmonics at the fundamental and its multiples... – Yasir Ahmed Feb 13 '18 at 11:12