1
$\begingroup$

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

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

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


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

$\endgroup$
  • $\begingroup$ Are you sure that there is not a "sin" instead of a "cos" in the second equation? $\endgroup$ – jan Aug 28 '13 at 22:43
  • $\begingroup$ unless it's wrong in the exercise, both of them are cosines here $\endgroup$ – triplebig Aug 28 '13 at 22:52
  • 3
    $\begingroup$ Hint: Use the frequency domain and Parseval's theorem. You'll get to the answer in 2 steps. $\endgroup$ – Sudarsan Aug 28 '13 at 23:04
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
-1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Hey, welcome and thank you for contributing. I tried editing your post with some basic LaTeX maths, but honestly, I'm not completely sure what you wanted to write at certain places. The post is not really well readable like this. Could you please edit your question yourself and improve the formatting? Also, from what I remember about orthogonal signals, your post does actually solve the problem, but it would be nice if you could explicitly write an answer to the question posed. $\endgroup$ – penelope Feb 18 '14 at 11:22
  • $\begingroup$ Hi, Actually the question is if u[n] and v[n] are orthogonal. So prove that we need to take inner product of u[n] and v[n].(which can be also done in frequency domain as per parsevals). They are orthogonal under the condition at fc > 2B $\endgroup$ – Amey Naik Feb 23 '14 at 0:32
  • $\begingroup$ I am not saying that your post is wrong. I am saying that the formatting and the lack of explanation are making it very hard to read and understand. Even after @lennon310 edited it, it is still not clearly written. I am asking if you could please edit your post to make it better quality. $\endgroup$ – penelope Feb 23 '14 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.