In one of my classes I have learned that if we look at transmitting a binary signal with ASK, we can get the bandwidth (i.e. the width of the main lobe in the frequency domain) with BW = 2 * Fb, where Fb is the bit-rate (i.e. 1/T, where T = 'length' of one bit).

I am a little confused as to how this BW <-> Fb relation comes to be. Intuitively I understand that if we assume a carrier frequency of 2GHz and transmit a signal with a bitrate of 5 bits/s, we simply modulate the amplitude of our 2GHz carrier signal, without any frequency changes. Why, then, do we have a lobe in the frequency domain with a width that is greater than 1? As I understand it, there is only one frequency, the 2GHz signal?.. I know my understanding is wrong, but intuitively I do not understand how we have a variance in frequency, independently of what our bitrate is. How is modulating the amplitude of a 2GHz signal 5 times a second different from doing it 20 times a second when viewed in the frequency domain?

  • 2
    $\begingroup$ "we simply modulate the amplitude of our 2GHz carrier signal, without any frequency changes" That's the thing, you can't modulate the amplitude without creating new frequencies. Changing the amplitude of the carrier is a time-variant process. $\endgroup$
    – MBaz
    Commented Dec 17, 2018 at 23:16
  • $\begingroup$ @MBaz So the widening of the main dome/lobe in the frequency domain at higher information/bit rates is an artifact, so to say, of rapidly modulating the carrier amplitude? Would this mean that if the lobe is centered around frequency F, and there was jamming on F-eps and F+eps for any eps, any receiver would still see a clear signal? $\endgroup$
    – Kugelblitz
    Commented Dec 18, 2018 at 6:07
  • 1
    $\begingroup$ Q.1: Yes. Q.2: As Arash explained, the modulated signal will occupy a certain bandwidth. If the jamming intersects the signal's bandwidth, then it iwll distort the received signal. $\endgroup$
    – MBaz
    Commented Dec 18, 2018 at 14:35
  • $\begingroup$ Perfect, thank you! It makes sense to me now! $\endgroup$
    – Kugelblitz
    Commented Dec 18, 2018 at 17:56

1 Answer 1


Assume you have a pulse of width $T$ which is your bit.

$$w(t)=1\quad for\quad-\frac T2\le t\le\frac T2\quad and\; 0\;elsewhere$$ Your bit in frequency domain is:


Look up Sinc function and you can see $Sinc(T.f)$ main lobe is between $\pm \frac 1T$. As your bit is widened in time, it occupies less bandwidth in frequency domain($T$ increases, $\frac 1T$ decreases) and vice versa. This is a very important result to understand and remember for the rest of your life. It's the reasoning behind relationship of frequency $BW$ and pulse duration $T$. Carrier frequency has nothing to do with it, it just shifts your entire signal spectrum up in frequency to make it suitable for transmission in the physical medium(channel) of your choice.

You should review Fourier transform and its properties. Then look up Convolution theorem, which tells if you multiply signals in time, they get convolved in frequency domain. Then learn about Sinc function and its relationship to Rect function.

  • $\begingroup$ With a review on Convolution and your answer it now makes sense to me, thanks! $\endgroup$
    – Kugelblitz
    Commented Dec 18, 2018 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.