(New page: Hi, an important fact to keep in mind when studying for the exam that is coming up is to remember that you cannot take the fourier transform of a signal using the table in the book when th...) |
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here we use the shifting property in the table to determine the fourier transform | here we use the shifting property in the table to determine the fourier transform | ||
| − | we use: x[n-no] <math> \to </math> | + | we use: x[n-no] <math> \to e^{-jwno} X(w)</math> |
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| + | = X(w) = <math> \frac{sin[w(N1+1/2)]}{sin(w/2)} </math> | ||
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| + | making no = 4 <math> \to </math> | ||
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| + | x[n-no] <math> \to e^{-jwno} X(w) </math> | ||
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| + | X(w) = <math> \frac{sin[w(4/2+1/2)]}{sin(w/2)} </math> | ||
| + | |||
| + | X(w) = <math> e^{jw4} * \frac{sin[w(5/2*w)]}{sin(w/2)} </math> | ||
Latest revision as of 18:20, 22 October 2008
Hi, an important fact to keep in mind when studying for the exam that is coming up is to remember that you cannot take the fourier transform of a signal using the table in the book when the signal is in CT form.
However, if it is in DT then we can by all means use the set transforms in the book or formula sheet to write out the transform.
Otherwise we would have to justify that it is the correct transform.
I guess i'll do an example that uses the books set transforms to show that a DT signal doesnt need to be justified in this sense.
EX
x[n] = u[n-1] - u[n-5]
here we use the shifting property in the table to determine the fourier transform
we use: x[n-no] $ \to e^{-jwno} X(w) $
= X(w) = $ \frac{sin[w(N1+1/2)]}{sin(w/2)} $
making no = 4 $ \to $
x[n-no] $ \to e^{-jwno} X(w) $
X(w) = $ \frac{sin[w(4/2+1/2)]}{sin(w/2)} $
X(w) = $ e^{jw4} * \frac{sin[w(5/2*w)]}{sin(w/2)} $
