Checking if u[n] and v[n] are orthogonal

Question

Let $x[n]$ and $y[n]$ be two orthogonal baseband signals, with bandwidth B. Check if the following signals are orthogonal:

$u[n] = x[n]*\cos(2*\pi*f_c*n)$

$v[n] = y[n]*\cos(2*\pi*f_c*n)$

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I'm not sure what the problem asks to be honest. 1) What exactly does he mean by bandwidth?? I only know the definition of a bandwidth of a sinc function, but how can I think of it generally?

2) I assumed the case for a rectangular function and reduced the problem to:

$\sum_{n=-\frac{1}{B}}^{n=\frac{1}{B}}\frac{1}{2}\cdot x[n]\cdot y[n]\cdot \cos(4\cdot \pi\cdot f_c\cdot n)$

But I don't know where to go from there. It's also weird for me to evaluate since $\frac{1}{B}$ is likely not an integer.

EDIT: also, $f_c$ >> B

Answer 1

It's essentially asking you to show that two signals that are orthogonal in the baseband (their original form) remain orthogonal after amplitude modulation process. In other words, amplitude modulation preserves orthogonality.

Think about the question this way. You have two signals that are orthogonal, meaning that

$$ \sum_{m=-\infty}^{\infty}x[m]y[m] = 0 $$

This is a bit ugly, since we have to deal with infinities. However, your problem also says that they both have bandwidth $B$, which means that in frequency domain they will have finite width.

Let's define $\mathscr{B}$ as

$$ \mathscr{B} = \frac{B}{F_s}$$

which means that in your discrete Fourier transform signals $X[k]$ and $Y[k]$, the only non-zero samples exist for $-\frac{\mathscr{B}}{2} \le k \le \frac{\mathscr{B}}{2}$. Since DFT is an orthogonal transform, signals orthogonal in time domain will remain orthogonal in frequency domain, i.e.

$$ \sum_{l=-\infty}^{\infty}X[l]Y[l] =\sum_{l=-\mathscr{B}/2}^{\mathscr{B}/2}X[l]Y[l] = \sum_{m=-\infty}^{\infty}x[m]y[m] = 0 $$

We know that multiplying a signal by a cosine shifts is away from DC in frequency domain, and the fact that the modulating frequency $f_c$ is larger than $B$ means that this won't cause any problems with overlapping modulated bands (think about what would happen if this weren't the case).

I think this gives you enough hits to take it from here.

Answer 2

Signals $x$ and $y$ are orthogonal which means that $\displaystyle \sum_n x[n]y[n] = 0.$ Note that the range of summation is whatever is needed for equality to hold. Signals $u$ and $v$ given by $$u[n] = x[n]\cos(2\pi f_c n), \quad v[n]= y[n]\cos(2\pi f_c n),\ \forall n$$ are orthogonal if $\displaystyle \sum_n u[n]v[n] = \sum_n x[n]y[n] \cos^2(2\pi f_c n) = 0.$ Now, if $f_c$ is a half-integer, that is, $f_c = \frac{m}{2}$ where $m$ is some integer (even or odd), then $2\pi f_c n = \pi (mn)$ is an integer multiple of $\pi$, and so we see that $\cos^2(2\pi f_c n) = 1$ for all $n$. It follows that $\displaystyle \sum_n x[n]y[n] \cos^2(2\pi f_c n) = \sum_n x[n]y[n] = 0$ since $x$ and $y$ have been assumed to be orthogonal. For other values of $f_c$, it is not possible to guarantee that $u$ and $v$ are orthogonal for all choices of orthogonal signals $x$ and $y$. However, for any given specific value of $f_c$, it should be possible to find several pairs of orthogonal $x$ and $y$ for which $u$ and $v$ are orthogonal

Answer 3

Let x[n] and y[n] be two orthogonal baseband signals, with bandwidth B. Check if the following signals are orthogonal:

$u[n]=x[n]∗cos(2∗π∗fc∗n)$ $v[n]=y[n]∗cos(2∗π∗fc∗n)$

Answer: For $u[n]$ to be orthogonal to $v[n]$ Inner Product = $<u[n],v[m]>$

By using Parsevals theorem Compute Integration (from -Inf to Inf) {U(f) * conj(V(f))} df

Here $U(f)$ is frequency domain of $u[n]$ = $(X[f-fc]+X[f+fc])/2$

Here $V(f)$ is frequency domain of $v[n]$ = $(Y[f-fc]+Y[f+fc])/2$

Inner Product will be = Integration (X[f-fc].Y[f-fc] + X[f+fc].Y[f+fc])/4 = 0