do not understand how the middle expression was derived from above expression, enter image description here

after all summation\multiplication i get:

$$\frac{e^{-jwN} + e^{jwN} - e^{-jw(N+1)} + e^{jw(N+1)}}{2 - e^{-jw} - e^{jw}}$$

  • $\begingroup$ what was the original question? what's the range of pulse in $n$ ? $\endgroup$ – Fat32 Dec 16 '17 at 0:36
  • $\begingroup$ @Fat32 from -N to N inclusive. But I don't think it really matters, I think it's just a matter of arithmetic manipulation only. $\endgroup$ – qqffx Dec 16 '17 at 0:40
  • $\begingroup$ it does not matter for the DTFT magnitude but it matters for the phase of the filter which is dependent on the summing range. So you have this summation: $$ X(e^{j \omega}) = \sum_{n=-N}^{N} e^{-j \omega n} $$ ? $\endgroup$ – Fat32 Dec 16 '17 at 0:44
  • $\begingroup$ @Fat32 yes, I add the whole derivation) I just didn't get how they transform this to Euler function for sinusoid. So I want to know the trick) $\endgroup$ – qqffx Dec 16 '17 at 0:48

First, the following trick will be useful: $$ e^{j a} - e^{j b} = e^{j \frac{a+b}{2}} \left( e^{j \frac{a-b}{2}} - e^{j \frac{b-a}{2}} \right) $$

Then simply expand the sum as: $$ X(e^{j\omega}) = \sum_{n=-N}^{N}e^{-j\omega n} = \frac{ e^{-j\omega (-N)} - e^{-j\omega (N+1)} }{1 - e^{-j\omega}} = \frac{ e^{j\omega N} - e^{-j\omega (N+1)} }{1 - e^{-j\omega}} $$

Apply the trick to the numerator and denominator by recognising $a$ and $b$ carrefully: for the numerator $a=\omega N$ and $b=-\omega (N+1)$ whereas for the denominator $a=0$ and $b=-\omega$ which yields:

$$ X(e^{j\omega}) = \frac{ e^{j\omega N} - e^{-j\omega (N+1)} }{1 - e^{-j\omega}} = \frac{ e^{j \frac{\omega N -\omega (N+1) }{2}} \left( e^{j\frac{\omega N +\omega (N+1) }{2}} - e^{-j\frac{\omega N +\omega (N+1) }{2}} \right) }{ e^{-j\omega/2} \left( e^{j\omega/2} - e^{-j\omega/2} \right) }$$

which simplifies to: $$ X(e^{j\omega}) = \frac{ e^{-j\omega/2} \left( e^{j\frac{\omega N +\omega (N+1) }{2}} - e^{-j\frac{\omega N +\omega (N+1) }{2}} \right) }{ e^{-j\omega/2} \left( e^{j\omega/2} - e^{-j\omega/2} \right) } = \frac{ e^{j \omega \frac{2N+1 }{2}} - e^{-j \omega \frac{2N+1}{2}} }{ e^{j\omega/2} - e^{-j\omega/2} } $$

which is: $$ X(e^{j\omega}) = \frac{ \sin( \omega (N+\frac{1}{2}) )}{ \sin( \omega \frac{1}{2}) } $$

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