0
$\begingroup$

If sampling frequency is fixed, how can I generate a frequency domain signal of a particular bandwidth from a time domain signal of dimension $1\times N$? e.g. if $f_s=100\,\text{kHz}$, how can I generate a signal of $30\,\text{kHz}$ bandwidth? or of $15\,\text{kHz}$?

$\endgroup$
  • $\begingroup$ Your question is unclear. Are you asking "what is a filter"? Are you asking "what is the bandwidth of a signal"? $\endgroup$ – Marcus Müller Feb 24 '17 at 11:11
  • $\begingroup$ I want to generate a signal with bandwidth=30kHz and sampling frequency=100kHz. Is the bandwidth in my control? and can I generate a signal with bandwidth=15kHz and sampling frequency=100kHz? $\endgroup$ – anonymous Feb 24 '17 at 11:14
  • $\begingroup$ I'm taking fft of a [1xN] vector in time domain. can i control the bandwidth of it in frequency domain? $\endgroup$ – anonymous Feb 24 '17 at 11:15
  • $\begingroup$ If N is finite, bandwidth will always be a constant, infinity, unless you specify a stop-band noise floor. $\endgroup$ – hotpaw2 Feb 24 '17 at 13:41
  • $\begingroup$ I am guessing your are asking about "Frequency Resolution"? If so, in order to achieve a frequency resolution of F, the time length T of your signal in your data-set must be greater than 1/F. If you window, the time length will need to be longer depending on the window. So for your case sampling at 100KHz, to achieve 30KHz resolution you must have at least 1/30KHz = 33.33 us of data, or at least 4 samples at 100 KHz, Please clarify if this is what you are indeed looking for and I will offer a more detailed answer below. Otherwise, do you want to generate an actual noise signal with that BW? $\endgroup$ – Dan Boschen Feb 24 '17 at 15:20
3
$\begingroup$

given sample rate $f_\text{s}$, this

$$ x(t) = \sum\limits_{n=-N}^{+N} x[n] \operatorname{sinc}\left(2 B \left(t-\tfrac{n}{f_\text{s}}\right)\right) $$

is bandlimited to $\pm B$, no matter what the numbers $x[n]$ are. $N\ge0$ can be as little or as large as you want.

it might have bandwidth of $B$ if the values of $x[n]$ are sufficiently white or with high-enough content at high frequencies. but the one-sided bandwidth will not exceed $B$.

forgot to be explicit. the sinc function is:

$$ \operatorname{sinc}(u) \triangleq \begin{cases} \frac{\sin(\pi u)}{\pi u} \qquad & u \ne 0 \\ \\ 1 & u = 0 \\ \end{cases} $$

$\endgroup$
  • $\begingroup$ I see great minds think alike :) $\endgroup$ – Robert L. Feb 25 '17 at 23:13
1
$\begingroup$

If your goal is to make a signal with a specified bandwidth, i.e. 30 kHz or 15 kHz, then you need to specify certain things:

What definition of bandwidth are you using? Baseband bandwidth (which is measured from DC to $f_{max}$, or modulation bandwidth? (Which is twice the baseband bandwidth.)

What kind of signal is it? If you aren't picky, you could generate a stream of raised-cosine pulses. It is easy to specify the bandwidth of said pulses by adjusting the symbol period $T$. For example, if $\beta=1$ and you need a baseband bandwidth of 30 kHz, you could set $T=1/30000\approx 33 \mu$s. Because this would not generate an integer number of samples per symbol (samples per symbol = $100000/30000\approx 3.3333$) ), you might want to adjust the symbol period $T$ to make it an integer ratio.

$\endgroup$

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