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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: enter image description here

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?

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1 Answer 1

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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.

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  • \$\begingroup\$ 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}\$? \$\endgroup\$
    – xenon
    Commented Sep 20, 2011 at 12:48
  • \$\begingroup\$ @xEnOn - depends on the factor \$d\$. If \$d\$ = 0 then \$BW\$ = signal rate. I don't recall where the \$d\$ comes from :-( \$\endgroup\$
    – stevenvh
    Commented Sep 20, 2011 at 13:00
  • \$\begingroup\$ 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? \$\endgroup\$
    – xenon
    Commented Sep 20, 2011 at 13:33
  • \$\begingroup\$ 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... \$\endgroup\$
    – Yasir Ahmed
    Commented Feb 13, 2018 at 11:12

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