Sampling and Quantization, Calculating Output

Question

I want to understand this concept and I Multimedia Class and I had a question I missed and it seems I'm missing something. I don't need anyone to do my homework for me, rather, help me understand what I'm missing so I can apply it myself. I think my lack of understanding the concept and seeing it solved differently from online and in class sources. However this is the full question.

• Suppose a signal contains tones (harmonics) at 2, 8, and 10 kHz and is sampled at the rate 12 kHz (and then processed with an antialiasing filter limiting output to 6 kHz). What tones will be included in the output?

My instructors class notes are still too technical for me to get a grasp on it and in the notes it looks like it can be solved using "f(alias)=f(sample)-f(true)". I don't know how to apply that, because I would apply that as so.

*2 kHz = 12 kHz - True*

// subtract 12 from each side then flip signs

== 10 kHz  True


*8 kHz = 12 kHz - True*    
==  4 kHz  True


*10 kHz= 12 kHz - True*    
==  2 kHz True

So I'd get 10kHz, 4kHz, 2kHz

And my guess would be if its under the output of 6 kHz its included? So that would mean 2kHz, and 4kHz are the two tones in the output?

However I had one class mate solved as so

2^8=256             256<10,000          included
2^10=1024           1,024 <10,000       included
2^12=4096           4,096<10,000        included

Where does the 10,000 come from?

And this almost identical problem uses tones at 1, 10, and 21 kHz, still sampled at 12 kHz and solves as so

 1 kHz, 12-10=2 kHz, and 2*12-21=3 kHz tones are present

Answer 1

PROBLEM -- There is signal with three components at the following frequencies

F1 equals to 2 kHz,

F2 equals to 8 kHz, and

F3 equals to 10 kHz.

The signal is first sampled at 12 kHz (Fs) and then low-pass filtered at 6 kHz cut-off.

What frequencies are present in the processed signal?

APPROACH -- To figure out frequency components of the sampled-and-filtered signal, you need to mirror-fold original frequencies above 6 kHz (i.e. above maximum frequency you can discern due to sampling) around multiples of Fs.

ANSWER -- Therefore,

F1 at original frequency 2 kHz is within 0 to 6 kHz and will thus appears as 2 kHz,

F2 at original frequency 8 kHz is above 6 kHz and will thus appears as 4 kHz (4 = 1*12 - 8), and

F3 at original frequency 10 kHz is above 6 kHz and will thus appears as 2 kHz (2 = 1*12 - 10).

NOTE -- Just to illustrate, if you had another, fourth component at F4 = 20 kHz, it would appear as 4 kHz (4 = 2*12 - 20).