DFT product of sinusoids

(From Shaums DSP outline, 2nd edition, page 248, problem 6.21)

Book says, evaluate the Sum:

$$S = \sum^{N-1}_{n=0} \Bigg( x_1[n] \ x^{*}_2[n] \Bigg)$$

when:

\begin{aligned} x_1[n] = \cos\left( \frac{2\pi n k_1}{N} \right) \\ \\ x_2[n] = \cos\left( \frac{2\pi n k_2}{N} \right) \end{aligned}

using Property:

$$\sum^{N-1}_{n=0} \Bigg( x_1[n]\ x^{*}_2[n] \Bigg) = \frac{1}{N} \sum^{N-1}_{k=0} \Bigg( X_1[k]\ X_2^{*}[k]\Bigg)$$

I start by converting cosines into complex exponentials:

$$x_1[n] = \cos\left( \frac{2\pi n k_1}{N} \right)$$

$$x_1[n] = 0.5e^{j2\pi n k_1 / N} + 0.5e^{-j 2\pi n k_1 / N}$$

Applying Definition of DFT to find: $$X_1[k] = \sum_{n=0}^{N-1} \Bigg( 0.5e^{j2\pi n k_1 / N} + 0.5e^{-j 2\pi n k_1 / N} \Bigg) e^{-j2\pi nk/N}$$

$$X_1[k] = \sum_{n=0}^{N-1} \Bigg( 0.5e^{-j(2\pi n/N)(k- k_1)} + 0.5e^{-j(2\pi n/N)(k+ k_1)} \Bigg)$$

From this I determine that sum over one period of a complex exponential is zero, except when k is selected to cancel out exponential function input to zero, such as when $$k=k_1$$ and $$k=-k_1$$. Thus:

$$X_1[k] = \begin{cases} \frac{N}{2}&(k=k_1)\ or\ (k=-k_1) \\ 0 & else \end{cases}$$

By similar means: $$X_2[k] = \begin{cases} \frac{N}{2}&(k=k_2)\ or\ (k=-k_2) \\ 0 & else \end{cases}$$

Here's where I get lost:

Books says, therefore:

$$\sum^{N-1}_{n=0} x_{1}[n] x^{*}_{2}[n] = \frac{1}{N}\Bigg[ \frac{N^2}{4} + \frac{N^2}{4} \Bigg] = \frac{N}{2}$$

Where does: $$\frac{N^2}{4} + \frac{N^2}{4}$$ come from?

Actually you have quite done it, but lets indicate.

When the frequencies of both sinusoidal signals $$x_1[n]$$ and $$x_2[n]$$ are the same; i.e., $$k_1 = k_2 = m$$, then the impulses in the corresponding N-point DFT sequences $$X_1[k]$$ and $$X_2[k]$$ will occur at the same index $$k=m$$ and $$k=-m$$ (or $$N-m$$) with a weight of $$N/2$$.

Hence multiplication of $$X_1[k]$$ with $$X_2[k]$$ will be $$N/2 * N/2 + N/2 * N/2 = N^2/2$$, as there are two impulses in the DFT of a cosine signal...

Finally dividing by $$N$$ yields the result as $$N/2$$ when the frequencies are the same and $$0$$ when not.

• would it be: $(1/N)\sum^{N-1}_{k=0}X_{1}[k]X^{*}_{2}[k] = (1/N)(X_{1}[k_1{]}X^{*}_{2}[k_{1}] + X_1[k_{2}]X^{*}_{2}[k_{2}])$ – MrCasuality Jan 27 at 22:08
• $X_1[k_1] = N/2$, and $X_2[k_2] = N/2$, when I plug in... I get a mysterious $X_{1}^{*}[k_2]$ and $X_{2}^{*}[k_1]$ that has a value that is not nice and clean.. – MrCasuality Jan 27 at 22:14
• actually, $X^*_1[k_2]$ should equal zero for $k_2$? no idea.. its a mystery – MrCasuality Jan 27 at 22:20
• assuming $k_1 \ne k_2$ … or maybe its required that $k_1 = k_2$ – MrCasuality Jan 27 at 22:26
• ahh...ok...that's the answer... the product is zero unless $k_1 = k_2$... some type of frequency correlation problem... – MrCasuality Jan 27 at 22:27