Solve Periodic Digital Signal: x[n]=1+cos(pi*n/3)+3sin(pi*n/2)

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The periodic digital signal x[n]=1+cos(pi*n/3)+3sin(pi*n/2) has a period of 12, not 8, as the cosine term completes a cycle in 6 samples and the sine term in 4 samples. The least common multiple of these periods is 12, which means the signal repeats every 12 samples. To tabulate the values of x[n] over one complete period, calculate x[0] through x[12]. The calculations for the sine and cosine functions show that they complete their cycles as their arguments range from 0 to 2π. Understanding these periodicities is crucial for accurately analyzing the signal.
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Hey guys, I'm trying to revise for a DSP exam I have this week and I'm looking at past papers I have. I've got the question, and the out and out answer but it doesn't show the workings so I'm struggling to find out how to get there;For the following periodic digital signal;
x[n]=1+cos(pi*n/3)+3 sin(pi*n/2)

i)How many samples are there in one period?
ii)Tabulate the values of x[n] over one complete period

The answer to i) is 8, and I've got the plot for ii) but getting there is just baffling me, so if anyone can shed some light on it, or point me to some reading material, it would be much appreciated.

Cheers.
 
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BarryThomas89 said:
Hey guys, I'm trying to revise for a DSP exam I have this week and I'm looking at past papers I have. I've got the question, and the out and out answer but it doesn't show the workings so I'm struggling to find out how to get there;


For the following periodic digital signal;
x[n]=1+cos(pi*n/3)+3 sin(pi*n/2)

i)How many samples are there in one period?
ii)Tabulate the values of x[n] over one complete period




The answer to i) is 8, and I've got the plot for ii) but getting there is just baffling me, so if anyone can shed some light on it, or point me to some reading material, it would be much appreciated.

Cheers.
Assuming you have writting the problem correctly, I don't get 8 for the period of x[n]. The period of the cosine term is 6 and the period of the sine term is 4. The smallest integer that is divisible evenly by 6 and 4 is 12, not 8.
 
You may be on to something here. Because another thing that confused me about the answers given is that 8 was given for the first part, and the plot had 12 values...
Is it easy to explain how you got there, or do you know anywhere I can read up on it?
 
It's pretty easy to explain, and I pretty much did so in my first reply. I'll see if I can elaborate a bit.

cos(n pi/3) goes through a complete cycle in 6 units. sin(n pi/2) goes through a complete cycle in 4 units. The sum of these functions goes through a complete cycle in the smallest integer that is evenly divisible by 6 and 4, and that's 12.
 
Do they do complete cycles in 6 and 4 because there's 2 pi in one period? Or is that coincidence?
Also, any tips on the second part?
 
Both the sine and cosine function complete a cycle as the argument goes from 0 to 2 pi. For cos(n pi/3), n pi/3 goes from 0 to 2pi as n goes from 0 to 6. For sin(n pi/2), n pi/2 goes from 0 to 2pi as n goes from 0 to 4.

The second part is pretty simple; just make a table of values, calculating x[0], x[1], x[2], ..., x[12].
 
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