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Question about time domain form and sinc function

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rednewguy

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the transfrom of a pulse is sinc. my doubt is if the centre of main lobe of sinc is not at zero and is at some other frequency say 'Fc'.
then what is its time domain form.

thanks
 

Re: sinc function

What you are doing is translating the function the frequnecy domain, or multiplying it by a sinusoid.That is modulation essentially.The time domain would be a sin wave of the frequency at which the sinc peaks,multiplied by the pulse in time.
So it would be zero at all instants when the pulse doesnt exist, and will have the corresponding sin value at the instant when the pulse exists.
 

    rednewguy

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sinc function

Let's see... In frequency domain, the sinc function with the main lobe at zero is Sinc(W), The sinc function with the main lobe somewhere else , let's say Wo, is Sinc(W-Wo). Using the frequency shift property of the Fourier transform we have:

x(t) ---F--> F(w)

Exp(Jwot)*x(t) ----F--->F(w - wo)

So

G(t) ---F---> Sinc(W)

Exp(Jwot)*G(t) ---F---> Sinc( W -wo)

Where ---F--> represents Fourier transform and G(t) is the function representing a pulse. That's all.
 

    rednewguy

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Re: sinc function

hi!!

if u shift a signal in frequency domain then it is equivalent to multiplying it by an exponential having frequency equal to amount by which it is shifted!!!

so in time domain the signal will be complex not real n hence cannot be represented on time axis alone!!!!
 

Re: sinc function

Let \[x(t)---> X(j \omega)\], then \[e^{j\omega_0 t}x(t)--->X(j(\omega-\omega_0))\]. So, from this you can get the answer you want.

thnx

purna!

Added after 3 minutes:

Let \[x(t)---> X(j \omega)\], then \[e^{j\omega_0 t}x(t)--->X(j(\omega-\omega_0))\]. So, from this you can get the answer you want.

thnx

purna!
 
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Re: sinc function

rednewguy said:
the transfrom of a pulse is sinc. my doubt is if the centre of main lobe of sinc is not at zero and is at some other frequency say 'Fc'.
then what is its time domain form.

thanks

rectangular function....i mean a bandlimted signal for your unlimited signal
 

sinc function

hi
u have shifted ur shape in the frequency so u can multiply it in the time domain by an exponential func. and reversely in the freq.
look at the features of the fourier transform in any signal processing book.
bye
 

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