Spectrum of sampled signal

In summary, the conversation discusses finding the spectrum of a signal using the Discrete Fourier Transform and the implications of sampling at a certain frequency. It is determined that a 200Hz component will be present after sampling, but a 6000Hz component will be shifted to 4000Hz due to the Nyquist theorem.
  • #1
frenzal_dude
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0

Homework Statement


Hi, I need to find the spectrum of [tex]s(t)=sin(400\pi t) + 0.5cos(12000\pi t)
[/tex] when sampled at 10KHz.

Homework Equations


Using the Discrete Fourier Transform:

[tex]G_{k}=\sum_{n=0}^{N-1}s_{n}e^{\frac{-j2\pi kn}{N}}[/tex]
where N is the amount of samples taken in the signal duration which is [tex]NT_{s}=\frac{N}{10000}[/tex] and [tex]s_{n}=s(nT_{s})[/tex]

The Attempt at a Solution


I decided to take 10 samples, so N=10, therefore the signal duration is 0.001 seconds
Therefore: [tex]G_{k}=\sum_{n=0}^{9}s_{n}e^{\frac{-j2\pi kn}{10}}[/tex]
[tex]G_{k}=\sum_{n=0}^{9}[sin(400\pi nT_{s}e^{\frac{-j2\pi kn}{10}}+0.5sin(12000\pi nT_{s})e^{\frac{-j2\pi kn}{10}}] [/tex]
[tex]G_{k}=\sum_{n=0}^{9}[sin(\frac{4\pi n}{100})e^{\frac{-j2\pi kn}{10}}+0.5sin(\frac{12\pi n}{10})e^{\frac{-j2\pi kn}{10}}][/tex]
[tex]G_{k}=\frac{1}{2}-0.28e^{\frac{-j\pi k}{5}}+0.403e^{\frac{-j\pi k2}{5}}+0.523e^{\frac{-j\pi k3}{5}}+0.077e^{\frac{-j\pi k4}{5}}+1.09e^{\frac{-j\pi k5}{5}}+0.28e^{\frac{-j\pi k6}{5}}+0.93e^{\frac{-j\pi k7}{5}}+0.999e^{\frac{-j\pi k8}{5}}+0.5e^{\frac{-j\pi k9}{5}}[/tex]
I'm not sure what this all actually means, how can we find the frequency components based on all those exponentials? I know that exponentials in the time domian give a delta in the frequency domain, but these exponentials are already in the frequency domain.
Any help would be greatly appreciated!
Frenzal
 
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  • #2
Hey I think I worked it out, you don't need to use the DFT.
The 200Hz component will be present after sampling because 200Hz is less than half the sampling frequency 10000Hz, however the 6000Hz component will not be present after sampling because 6000Hz is not less than half the sampling frequency, ie you need to sample at at least 12000Hz to recover the whole signal (Nyquist theorem). Instead the 6000Hz component will be shifted to 0.5Fs – (6000Hz-0.5Fs )=5000Hz-(1000Hz)=4000Hz.
 

What is a spectrum of a sampled signal?

The spectrum of a sampled signal refers to the representation of the signal's frequency content. It shows the different frequencies present in the signal and their corresponding amplitudes.

How is the spectrum of a sampled signal obtained?

The spectrum of a sampled signal is obtained through a mathematical process known as the Discrete Fourier Transform (DFT). This converts the signal from a time-domain representation to a frequency-domain representation.

What is the significance of the spectrum of a sampled signal?

The spectrum of a sampled signal is significant because it provides crucial information about the signal's characteristics. It can help identify the dominant frequencies and their amplitudes, which can aid in understanding the behavior of the signal.

What is the difference between the spectrum of a continuous signal and a sampled signal?

The spectrum of a continuous signal is a continuous function, while the spectrum of a sampled signal is a discrete function. This is because a sampled signal only contains a finite number of samples, while a continuous signal has an infinite number of samples.

What factors can affect the spectrum of a sampled signal?

The spectrum of a sampled signal can be affected by various factors, including the sampling rate, the length of the signal, and any noise or interference present in the signal. Non-uniform sampling and aliasing can also impact the spectrum of a sampled signal.

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