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DFT X(K) frequency spectrum is not what was expected


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ganes



Joined: 17 Sep 2009
Posts: 4
Location: singapore


Post17 Sep 2009 17:38   

DFT


hi,

Can someone tells me how X(K) change f we doubled the sampling freuqency
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Distortion-andi



Joined: 16 Sep 2009
Posts: 5
Helped: 1
Location: Holland


Post18 Sep 2009 8:29   

DFT


Can you state it a bit clearer what your up to?
Is X(k) your time or freq domain, when you double the sampling freq your getting a better resolution in the freq domain fs = 44100 -> 22050 Hz, fs2 = 88200 -> than 44100 Hz..it only helps if you wanna get frequencies over the half sampling rate.....i am not a pro but maybe if you state it a bit clearer, i can help you
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Communications_Engineer



Joined: 10 Oct 2008
Posts: 492
Helped: 39


Post18 Sep 2009 9:18   

Re: DFT


ganes wrote:
hi,

Can someone tells me how X(K) change f we doubled the sampling freuqency



Just think about what you are asking. The answer is in the question, if you really know how DFT works
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ganes



Joined: 17 Sep 2009
Posts: 4
Location: singapore


Post19 Sep 2009 8:05   

DFT


hi,

I have a signal x(t)=25cos(10*pi*t)cos(400*pi*t) . i sampled it at sampling frequency of 410 Hz which is just equal to the Nyquist rate. my matlab code to compute the DFT X(K) is
N=82;
n= 0:81;
Ts = 1/410;
t = n*Ts;
x=25*cos(10*pi*t).*cos(400*pi*t);
X=fft(x);
subplot(3,1,1)
stem(n, x)
axis tight
ylabel('x[n]');
xlabel('n');
subplot(3,1,2);
stem(n*410/N, abs(X));
axis tight
ylabel('DFT X(K),Magnitute');
xlabel('f in Hz');
subplot(3,1,3)
stem(n*410/N, angle(X))
ylabel('DFT X(K),Phase');
xlabel('f in Hz');
axis tight

The result is the DFT X(K) frequency spectrum show 3 frequency present in the signal. but i expect only two frequency present (195Hz and 205Hz) . where the hell this third frequency resulted. am i right to say that is it caused by aliasing.

Now if I sampled the signal at 810Hz and other parameter remain unchanged , my DFT X(K) in the frequency domain have duplicate copies( periodic) but the frequency spectrum is not what i expected to be 195Hz and 205Hz. I thought I have prevented Aliasing by sampling higher than the Nyquist Rate.

struggling to understand on this.
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Communications_Engineer



Joined: 10 Oct 2008
Posts: 492
Helped: 39


Post19 Sep 2009 11:20   

DFT


upload the graph. I'll help you
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ganes



Joined: 17 Sep 2009
Posts: 4
Location: singapore


Post20 Sep 2009 5:04   

Re: DFT


hi,


the plot is as follow;

Normal link:
http://images.elektroda.net/11_1253419265.jpg


Normal link:
http://images.elektroda.net/25_1253419432.jpg

thank you so much for yr help.
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Communications_Engineer



Joined: 10 Oct 2008
Posts: 492
Helped: 39


Post20 Sep 2009 9:40   

Re: DFT


While having a look at the Matlab code, I don't think you should expect to see two frequency bins, since you are not 'adding' but 'multiplying' the two discrete-time signals

If you also look at it, this similar to multiplying a low frequency signal with a higher frequency one, something like AM and I believe the graph is also showing the same thing, both in time-domain and in frequency-domain
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Post20 Sep 2009 9:40   

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ganes



Joined: 17 Sep 2009
Posts: 4
Location: singapore


Post21 Sep 2009 3:13   

DFT


hi,

Thanks , I manage to figure out already.
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awanis



Joined: 15 Sep 2009
Posts: 15
Location: malaysia


Post23 Sep 2009 13:54   

Re: DFT


hai...
ihave a question..
how i can solve the signal from X(n) signal to quantization signal
can give me the example of instruction in matlab how to do it..
where X(n) = 6 sin (3πt)-5cos (6πt)+7cos(8πt +π/6)
N=7 with 4 level quantization using tranqated technique
range= ±10V

DFT X(K) frequency spectrum is not what was expected
sample signal of x(n)

DFT X(K) frequency spectrum is not what was expected
dft signal of x(n)
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Communications_Engineer



Joined: 10 Oct 2008
Posts: 492
Helped: 39


Post24 Sep 2009 6:45   

DFT


kindly re-phrase your question. It is difficult to understand what you are asking
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awanis



Joined: 15 Sep 2009
Posts: 15
Location: malaysia


Post24 Sep 2009 13:42   

Re: DFT


ok..
this the question..
the question give X(n) = 6 sin (3πt)-5cos (6πt)+7cos(8πt +π/6)
its want to get the quantized signal,Xq(n) with 4 quantization level using trancated technique.. and the dynamic range is ±10V..
giving N=7..
from the manual calculating i get the value is..
{-2.5,-2.5,-2.5,-2.5,2.5,7.5,7.5..}
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Communications_Engineer



Joined: 10 Oct 2008
Posts: 492
Helped: 39


Post24 Sep 2009 16:28   

DFT


you have 4 quant levels or 4 bit ADC? Both are different things, I hope you are aware of it

Taking your question as it is, I'd say the answer is wrong. 4 quant level for a dynamic range ±10 would be 20/4 values.

What is N? Why is it equal to 7?
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awanis



Joined: 15 Sep 2009
Posts: 15
Location: malaysia


Post24 Sep 2009 16:44   

Re: DFT


N is equal of the dicsrete signal from 0 until 6..(0<N<6)
i get in what u said 4 quant level for a dynamic range ±10 would be 20/4 values.
its true...
if the answer wrong..
how should i get the true answer from calculating and using matlab..?
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Communications_Engineer



Joined: 10 Oct 2008
Posts: 492
Helped: 39


Post25 Sep 2009 9:44   

DFT


You would have made a function to quantize the input signal in Matlab. So, upload the part where you describe the algorithm into Matlab code. You can send it via email if you dont want to upload it here.

I'll point you out where you are doing the mistake. So that you can reverse engineer the idea

I will however, not make a function for you (I think this is part of your homework)


you may also join our yahoo group "telecom_research"
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awanis



Joined: 15 Sep 2009
Posts: 15
Location: malaysia


Post25 Sep 2009 13:47   

Re: DFT


ok...
thanks...
how can i email?
i will send my instruction in matlab..
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