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enter image description here Can you please tell how use the scaling property to solve this question?? i am new to dsp subject

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  • $\begingroup$ Welcome to DSP.SE! Where are you having trouble? Do you know how to solve for $X_1[k]$ (This is the formula for the DFT, you must certainly have access to that?) Then if you do, you use that to solve for $X_1[8]$ and $X_1[9]$--- with that do you not know how to solve for the magnitude of the ratio of the two? $\endgroup$ Dec 26 '19 at 17:53
  • $\begingroup$ We avoid giving answers that look like homework problems (please see this: dsp.stackexchange.com/help/how-to-ask ) but help those that have already done some amount of research and study on their own but still having trouble---- it doesn't appear that you have done the first part of this yet? Or maybe you need some math help in understanding the DFT equation? Would like to help you! $\endgroup$ Dec 26 '19 at 17:56
  • $\begingroup$ (And may help you to see the periodic frequency property of the DFT when you insert zeros in time ) $\endgroup$ Dec 26 '19 at 18:08
  • $\begingroup$ I can solve this question by using the basic formula of dft, but i want to use scaling property of dft. Here x1(n)=x(n/3), then what how to calculate X1(k)?? $\endgroup$
    – Aman Sinha
    Dec 26 '19 at 20:34
  • $\begingroup$ And X(k)=Xk+N) so using this property i can find X(8) and X(11) easily, but how is X(8) = X1(8) here?? $\endgroup$
    – Aman Sinha
    Dec 26 '19 at 20:37
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Some hints that should help you:

From the scaling property, if we stretch the time axis, we compress the frequency axis.

It is also important to know the periodic property of the DFT: If you extend the frequency axis beyond N samples, the DFT will repeat. Consider a repeating DFT spectrum that extends out 3N, and consider that you now have 3N time samples. You stretched your time waveform from N to 3N, so what occupied 0 to N-1 in the DFT before will now be in the first third of the spectrum (roughly 0 to (N-1)/3). To understand what occurs in the rest you need to be aware of the periodic property of the DFT.

Process some simple DFT's with zero inserts on your own this will also be more obvious to you.

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  • $\begingroup$ I understood by taking 2 point dft, thanks. One more doubt , here interpolation is done in time domain , what if decimation is done then also can i use the same property?? $\endgroup$
    – Aman Sinha
    Dec 26 '19 at 21:04
  • $\begingroup$ @AmanSinha Very similar—-the other thing to know is that whatever is in the DFT from 0 to N-1 must repeat if you extended the DFT samples beyond that (you can see this by looking at the DFT equation carefully and noting how $e^{jn2\pi/N}$ is cyclical.). So if you compress time by decimating you stretch the frequency samples out—- any frequency samples with non zero values that extend beyond N-1 will therefore roll into the equivalent periodic location in 0 to N-1 (aliasing) $\endgroup$ Dec 26 '19 at 21:19
  • $\begingroup$ To avoid this we must low pass filter prior to decimating. And this is similar to interpolating - to get rid of all those replicated spectrum components we must low pass filter. $\endgroup$ Dec 26 '19 at 21:21
  • $\begingroup$ The process is as follows: for Interpolation: insert zeros and then low pass filter. For decimation: low pass filter then throw away samples. $\endgroup$ Dec 26 '19 at 21:22
  • $\begingroup$ Thanks I got it , that's why we use wrap around for getting periodic convolution from linear convolution of two signals, to avoid aliasing. $\endgroup$
    – Aman Sinha
    Dec 26 '19 at 21:28

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