# When Two Sampled Sinusoidal Are Orthogonal?

If two analog sinusoidal are orthogonal with duration T, their minimum frequency difference should be 1/2T in case of no phase offset between them.

And if there is phase offset, the difference is 1/T.

And in N sampled domain the orthogonality depends on sampling frequency.

The condition is $$f_{s}=Nf_{k}/k$$ obtained from $$f_{k}=kf_{s}/N$$.

($$f_{s}$$:sampling frequency, $$f_{k}$$: frequency difference, $$k$$: integer bigger than 0, $$N$$:# of sample)

Ex.)$$N=2048, k=1, f_{k}=15kHz$$ then $$f_{s}=30.72MHz$$

I like to know how $$f_{k}=kf_{s}/N$$ is derived

Thanks

If you have two sinusoidal length $$N$$ sequences

$$x_1[n]=\sin\left(n\omega_1+\phi_1\right),\qquad n\in[0,N-1]\\ x_2[n]=\sin\left(n\omega_2+\phi_2\right),\qquad n\in[0,N-1]\tag{1}$$

then for orthogonality we require

$$\sum_{n=0}^{N-1}x_1[n]x_2[n]=0\tag{2}$$

For the given sequences $$(1)$$ we obtain

\begin{align}\sum_{n=0}^{N-1}x_1[n]x_2[n]=&\sum_{n=0}^{N-1}\sin\left(n\omega_1+\phi_1\right)\sin\left(n\omega_2+\phi_2\right)\\=&\frac12\sum_{n=0}^{N-1}\cos\left[n(\omega_1-\omega_2)+\phi_1-\phi_2\right]-\\&\frac12\sum_{n=0}^{N-1}\cos\left[n(\omega_1+\omega_2)+\phi_1+\phi_2\right]\tag{3}\end{align}

where I've used

$$\sin(\alpha)\sin(\beta)=\frac12\big[\cos(\alpha-\beta)-\cos(\alpha+\beta)\big]\tag{4}$$

The two terms on the right-hand-side of $$(3)$$ have the form

$$\sum_{n=0}^{N-1}\cos(n\omega+\phi)=\cos\left(\frac{N-1}{2}\omega+\phi\right)\frac{\sin(\omega N/2)}{\sin(\omega/2)}\tag{5}$$

For large $$\omega$$ (i.e., $$\omega$$ close to $$\pi$$), the magnitude of the term $$(5)$$ becomes small. Assuming that $$\omega_1+\omega_2$$ is large (but not greater than Nyquist), people tend to neglect the last term on the right-hand side of $$(3)$$.

From $$(5)$$, the first term on the right-hand side of $$(3)$$ is zero if

$$\frac{\sin((\omega_1-\omega_2) N/2)}{\sin((\omega_1-\omega_2)/2)}=0\tag{6}$$

is satisfied, i.e., for

$$\omega_1-\omega_2=\Delta\omega=\frac{2\pi k}{N},\quad k\in\mathbb{Z},\quad k\neq 0\tag{7}$$

With $$\Delta \omega=2\pi\Delta f/f_s$$ we obtain for the frequency difference

$$\Delta f = \frac{kf_s}{N},\quad k\in\mathbb{Z},\quad k\neq 0\tag{8}$$

Note that choosing the frequency difference according to $$(8)$$, the two sinusoids will be approximately orthogonal. If we don't neglect the second term on the right-hand side of $$(3)$$ we can make them exactly orthogonal. However, in that case we cannot anymore freely choose one of the two frequencies, but the frequencies need to satisfy

$$f_1=\frac{(k_2-k_1)f_s}{2N},\quad f_2=\frac{(k_1+k_2)f_s}{2N},\\\quad k_1,k_2\in\mathbb{Z},\quad k_2>k_1\tag{9}$$

• Your reply is very useful to me!! Awesome!! Thanks!! – KHS May 14 '19 at 23:37