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



1 Answer 1


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


For the given sequences $(1)$ we obtain


where I've used


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

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

  • $\begingroup$ Your reply is very useful to me!! Awesome!! Thanks!! $\endgroup$
    – KHS
    May 14, 2019 at 23:37

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