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I have a sample test with an answer but don't understand how they got to the answer:

$x(t)$ has info only between $2 < |\omega| < 4$
$X^F(\omega) = 0$ for other frequencies.

All noise was cleaned from the signal and its CTFT is defined:
$X^F(\omega) = \begin{cases} |\omega|-2 & \text{for 2 < |$\omega$| < 4}\\ 0 & \text{otherwise} \end{cases}$

The clean signal was sampled at $\frac 1 T = \frac {12}\pi$
Draw the DTFT of $X^f(\theta) \text{ for} -3\pi < \theta < 3\pi$

Use a clear coordinate system.

The answer has x-axis span of -9 to 9 and y-axis of 0 to 4.

answer

How was this answer calculated?

I thought I would get a simple repeating copy of the original CTFT, but the original CTFT with $2 < |w| < 4$ is a line beginning at w=2 (on the x-axis) with the value zero, going up until the point $w=4$ where it has the value of 2...

Please help!! Thanks

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closed as unclear what you're asking by Jason R, MBaz, SleuthEye, jojek, Peter K. Feb 1 '16 at 15:52

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    $\begingroup$ It would be much better if you put an image of the plot instead of just describing it. Regarding the question, when you are asked to draw the DTFT of $X_{f}(\theta)$ for $-3\pi < \theta < 3\pi$, it would be enough to draw it between $\pi$ and $-\pi$, as DTFTs are periodic with period $2\pi$. Apart from that, I don't understand what you are trying to say when you write things like "with -9 to -8: zero"... Do you mean that $X_{f}(\theta)=0$ when $-9< \theta <-8$? Because that seems pretty weird to me, as no multiples of $\pi$ appear in that description. $\endgroup$ – Tendero Jan 31 '16 at 0:28
  • $\begingroup$ Thank you!! yes that is what I'm saying. I edited the question and replaced the description with an image of the plot. I think the notation of $\omega$ is what implies the $\pi$. $\endgroup$ – pashute Jan 31 '16 at 6:45
  • $\begingroup$ anybody pease?! $\endgroup$ – pashute Jan 31 '16 at 22:17
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When a continuous time signal is sampled in discrete intervals, the effect in the frequency domain is that the CTFT becomes periodic in frequency - that is, you end up with "copies" of the original CTFT, with each copy centered at frequencies equal to k*1/T (k = any integer).

So, when you sample at 12/pi you will get a copy of the original CTFT centered at w=0, w=12/pi, w=-12/pi, w=24/pi, w=-24/pi, and so on.

Since you have frequency content between 2<|w|<4 in the w=0 case, the w=k*12/pi cases will have frequency content between 2<|w-k*12/pi|<4.

Be careful - if frequency content from multiple copies have frequency content in the same places on the frequency axis, the effect is that they add together. This is called "aliasing" and is usually an unwanted effect of sampling.

Here is a nice graph that shows the effect of sampling a continuous time signal on the FT.

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  • $\begingroup$ Thank you. But you did not answer why I get the linear lines from zero to 4 (and back) when the CTFT was defined as $\omega - 2$... $\endgroup$ – pashute Feb 1 '16 at 9:53
  • $\begingroup$ Here is what's not clear from your answer: a. What is the content of the DTFT in the frequencies where there is content? b. Let me calculate the frequencies where the DTFT does have content, from the formula that you wrote: $2<|w-k*12/pi|<4$ Here goes: for k=0: $2<|w|<4$ - This is not what we see in the graph for k=1: the limits are $2<|w-12/pi||<4$ 12/pi is very close to 4. So, obviously, the two graphs are parallel with the -k*12/pi graph being below. What happened that caused the graph in the answer to "shrink". I answer that (I think) in my answer below. $\endgroup$ – pashute Feb 21 '16 at 21:05
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Thanks @rothloup !
I'm adding a layman's explanation to your answer and explaining the result:

There are four types of Fourier transforms: (that's easy to remember!)

All Fourier transforms work with infinite values so either we add zeroes around the given signal or we repeat the signal to infinity, with the length of the given signal as its periodicity.

The four types of transforms are:

  1. CTFT or simply the FT (Fourier Transform): Continuous, Aperiodic
    Used for a continuous signal with "one time" information in a small region of time. Zeroes added before and after to infinity, for using this transformation.

  2. FS or Fourier Series: Continuous, Periodic
    Used for a continuous periodic signal. We repeat the original (limited time region) signal before and after to infinity, for using this transformation.

  3. DTFT: Discrete, Aperiodic
    Used for a discrete signal with "one time" information in a small region of time. Zeroes added before and after to infinity, for using this transformation.

  4. DFT: Discrete, Periodic
    Used for a periodic discrete signal. We repeat the original (limited time region) signal before and after to infinity, for using this transformation.

The input of the DTFT - is discrete (sampled) input from an aperiodic signal (actually: information in a specified finite region of time, the rest set to zero...).

Because of the infinite zeroes being added, the period becomes extremely large, actually infinitely large, so there is only one period - or in other words the input is actually aperiodic!

Because of the zeroes added, the spacing in the frequency domain of the transform's output become extremely (actually: infinitely) close to each other.

So the output of the DTFT is Continuous and Periodic and also Symetric around the Nyquist number (in our case zero).

(That is why in the computer, where you deal only with sampled discrete data, we use the DFT and not the DTFT)

So if we have the CTFT $X[(\omega]$, we get the DTFT $X(\Omega)$ by setting:
$$\Omega = \omega T$$

In our case: since the samples were $12/\pi$ (in words: 12 times each cycle) and since the DTFT is symetric the period is half of the cycle (the other half is simply its mirror image) so the period is 6 and we multiply the $\Omega$ by that. While the areas with valid info are reduced: Formerly from 2-4, a quarter of the -4 to 4 period span, And become a sixth of the now -3 to 3 span:

$$ X(\Omega) = \begin{cases} 6|\Omega|-2 & \text{for 1 < |$\Omega$| < 2}\\ 0 & \text{otherwise} \end{cases}$$

  • a line from 4 to 0 between $\Omega$ 1 and 2
  • a line from 0 to 4 between $\omega$ -2 to -1. (actually down from 4 to 0 going backwards, from $\Omega$ 2 to 1)
  • and zero: between 0 and 1, between 2 and 3, and between 0 to -1 and -2 to -3.

As seen on the plot between -3 and +3

The pattern is repeated, starting with -3, resulting in:

$$ X(\Omega) = \begin{cases} \frac{6|\Omega|}{k}-2 & \text{for: $6k+1$ < |$\Omega$| < $6k+2$ ($k$ being any positive integer)} \\ 0 & \text{otherwise} \end{cases}$$

As plotted in the answer

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  • $\begingroup$ Excuse me if I'm wrong, but that just doesn't seem right. I don't think you can affirm that $X(\Omega)$ equals what you wrote there. DTFTs are periodic with period $2\pi$ and you don't seem to have had that in mind. Also, the final expression you wrote affirms that $X(\Omega)$ is null for $1<|\Omega |<2$, which isn't true according to the image of your first post. $\endgroup$ – Tendero Feb 2 '16 at 1:33
  • $\begingroup$ OK so then how do you solve the question in the test? (Old test, for learning purposes...) $\endgroup$ – pashute Feb 21 '16 at 20:17

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