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The answer is:

$\cos(1250\pi t)$ and $\cos(21250\pi t)$

I’m not able to verify.

My understanding is that when you periodically increase $t$ by $1/10$ ($10$ here is the frequency, so $1/10$ is time period) from $0$ to the signal $\cos(1250\pi t)$, you should get $x(n)=\cos(n\pi /8)$.

However, $\cos(1250\pi (1/10))$ is not same as $\cos((1)\pi /8)$.

So I would like the verification and what's wrong with my reasoning? Any help is appreciated.

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  • $\begingroup$ These are two of infinitely many signals that produce the same sequence! The effect you're looking for is aliasing. $\endgroup$ – Marcus Müller May 9 '17 at 8:45
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Considering the aliasing, here's the simple derivation: Given a cosine-signal $x(t)$ by

$$x(t)=\cos(2\pi f t)$$

which we sample with sampling frequency $F$, i.e. $t=n/F$, we get

$$x[n]=\cos(2\pi f/F n)$$

and if we now factor $f=kF+f_0$ with $k\in\mathbb{N}$ you'll see that

$$x[n]=\cos(2\pi (kF+f_0)/F n)=\cos(2\pi f_0/F n)$$

For an illustration of aliasing, you can have a look at one of my articles on aliasing.

So, cosines that are an integer multiple apart in frequency, yield the same sampled signal.

Now, to have

$$cos(2\pi f_0/F n)=\cos(\pi n/8)$$

You need $2\pi f_0/F = \pi/8$ and hence $f_0 = F/16$. Hence, your proposed solutions are not correct in my opinion, because $1250\pi t=2\pi 625=2\pi(62*F+5)$.

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