# Signal Sampling

I am new in this field (junior year) and I have a question may a little lazy for the majority of the forum.

I have a speech signal $x(t)=0.4\cos(100πt)+0.3\cos(1800πt)+0.2\cos(980πt)$ and I know that the sampling frequency is $F_s=1000$

If I want to find the signal form after sampling the only thing I have to do is replace $t$ with $κT$

$x(κT)=0.4\cos(100πκT)+0.3\cos(1800πκT)+0.2\cos(980πκT)$ but $T=1/1000$ so $x(k)= 0.4\cos(0.1πκ )+0.3\cos(0.18πκ )+0.2\cos(0.98πκ )$

Is the solution and the notation OK?

• Seems normal to me Jan 28 '15 at 16:40
• Have you heard of the sampling theorem? How do the frequencies of the sinusoids compare to the sampling frequency? Will there be aliasing? Jan 28 '15 at 17:03

First there was a mistake in the simplification - when you divided by $T$. It should be $x(k)= 0.4\cos(0.1πk )+0.3\cos(1.8πk )+0.2\cos(0.98πk )$.
Notice that you sampling frequency is 1000 Hz, but your signal contains frequencies: of 50, 900 and 490 Hz. Therefore the 900 Hz will be aliased. To see what frequency it gets aliased to consider $$\cos (2\pi900nT) = \cos (2\pi900nT+mnT2\pi)$$
if $mnT$ is an integer, but $n$ is always an integer and $T$ is fixed by your problem. Therefore we require $m=\frac{p}{T}$, where $p$ is and integer. In other words $m$ must be a multiple of your sampling frequency.
In this case let $m=-1000$ then we have $$\cos (2\pi900nT-1000nT2\pi)= \cos (2\pi(900-1000)nT)$$ Performing the final simplification we get: $$\cos (2\pi900nT)= \cos (-2\pi100nT) =\cos (2\pi100nT)$$ Note - the last equality just uses the properties of cosine - even function. Therefore your 900Hz frequency looks exactly like a 100 Hz tone when sampled at 1000 Hz.