Can the DTFT of a Unit Step Function be Simplified for u(n) - u(n-L)?

[JonathanT]

So I'm trying to find the DTFT of the following; where u(n) is the unit step function.

u \left( n \right) =\cases{0&$n<0$\cr 1&$0\leq n$\cr}

I want to find the DTFT of

u \left( n \right) -2\,u \left( n-8 \right) +u \left( n-16 \right)

Which ends up being a piecewise defined function looking like

u \left( n \right) -2\,u \left( n-8 \right) +u \left( n-16 \right) = \cases{1&$0\leq n$\ and \ $n\leq 7$\cr -1&$8\leq n$\ and \ $n\leq 15$\cr}

With the function zero elsewhere.

I plug this into the formula for a DTFT and get the following:

1+{{\rm e}^{-i\omega}}+{{\rm e}^{-2\,i\omega}}+{{\rm e}^{-3\,i\omega}}<br /> +{{\rm e}^{-4\,i\omega}}+{{\rm e}^{-5\,i\omega}}+{{\rm e}^{-6\,i\omega<br /> }}+{{\rm e}^{-7\,i\omega}}-{{\rm e}^{-8\,i\omega}}-{{\rm e}^{-9\,i<br /> \omega}}-{{\rm e}^{-10\,i\omega}}-{{\rm e}^{-11\,i\omega}}-{{\rm e}^{-<br /> 12\,i\omega}}-{{\rm e}^{-13\,i\omega}}-{{\rm e}^{-14\,i\omega}}-{<br /> {\rm e}^{-15\,i\omega}}

This should be correct, however, it is very ugly and I'm looking for a better form for my answer. I cannot reduce the summation using a harmonic series because the coefficient |a| = 1.

I can keep it in summation form but I feel like I'm missing an easy step that can simplify this.

Thanks for any help you can offer.

 

[jbunniii]

Try using this fact to simplify your answer:
$$\sum_{n = 0}^{N-1} x^n = \begin{cases}
\frac{1-x^{N}}{1-x} & \text{ if }x \neq 1 \\
\\
N & \textrm{ if }x = 1
\end{cases}$$

 

[JonathanT]

Well I feel stupid. I saw this formula but like I said in my original post I thought I couldn't use it since series wasn't approaching zero. I guess I was being stupid and thinking about infinite series. Since this is finite of course I can use that.

If my math is right I get

{\frac { \left( 1-{{\rm e}^{-8\,i\omega}} \right) ^{2}}{1-{{\rm e}^{-i\omega}}}}

 

[jbunniii]

Well I feel stupid. I saw this formula but like I said in my original post I thought I couldn't use it since series wasn't approaching zero. I guess I was being stupid and thinking about infinite series. Since this is finite of course I can use that.

If my math is right I get

{\frac { \left( 1-{{\rm e}^{-8\,i\omega}} \right) ^{2}}{1-{{\rm e}^{-i\omega}}}}

I haven't checked your math, but I will mention that you can always simplify expressions involving $1 - e^{-ix}$ by factoring out $e^{-ix/2}$ to obtain $e^{-ix/2}(e^{ix/2} - e^{-ix/2}) = 2i e^{-ix/2} \sin(x/2)$.

 

[JonathanT]

Thanks. I've got it in this form now:

{\frac {2\,i \left( \sin \left( 4\,\omega \right) \right) ^{2}{{\rm e}^{-15/2\,i\omega}}}{\sin \left( 1/2\,\omega \right) }}

After all this I found a formula for any equation in the form u(n) - u(n-L). Thanks for the help.