# continuous and discrete time systems

Discussion in 'Mathematics and Physics' started by PG1995, Nov 3, 2012.

1. ### PG1995Active Member

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Thank you very much.

Please have a look here. Thanks.

I was trying to do it myself while you posted it and my answer is somewhat different. Please give it a look and let me know where I'm going wrong. I'm still not able to find some general method to align the exponential terms in proper order. Thanks.

Regards
PG

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2. ### steveBWell-Known MemberMost Helpful Member

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For the angle issue, you are relying too much on a calculator and not enough on your brain. The atan function has been defined to be a true "function" so that it only has one value on the y-axis for every value on the x-axis. However, atan(0)=0, pi, 2pi, 3pi ... or even -pi -2pi etc. This is the whole issue of needing the atan2 function that I believe I've mentioned before. This will give answers in the correct quadrant of the unit circle. So retry your calculations in Matlab with atan2, or just think about it.

Code (text):

>> atan2(0.0001,1)

ans =

1.0000e-004

>> atan2(0.0001,-1)

ans =

3.1415

>> atan2(-0.0001,1)

ans =

-1.0000e-004

>> atan2(-0.0001,-1)

ans =

-3.1415

On the other problem, you just have to be careful. You need to pair up equal magnitudes, and then factor out the correct phase to get exp(jx)+exp(-jx), so that you can make a cos(x) out of it. Every problem might be a little bit different, and you have to think about it. There is no rule that says all problems will simplify. Only special cases will work out nicely. But, often school problems are set up to give easy simplifications.

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3. ### PG1995Active Member

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Thanks a lot.

You do have a point and I can see it now. Yes, we have discussed atan2 in the past. I think I get it now. For a minus real number, we can say that phase is 180 degrees.

Sorry to ask you this again. Though you have told me that you could have made a mistake, it's still better to confirm it. Where am I having it wrong? Thanks.

Regards
PG

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5. ### steveBWell-Known MemberMost Helpful Member

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I guess I was doing a different problem. Sorry, I should have shown the starting point, to make it clear.

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6. ### PG1995Active Member

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Thanks.

Please have a look here. It also contain the problem which I suspect you had in mind. Thank you.

Regards
PG

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7. ### steveBWell-Known MemberMost Helpful Member

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Yes, that's the one, and you found my mistake too. Very good!

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8. ### PG1995Active Member

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Thanks.

But what about Q1? Please don't think I'm being impatient! It's just that I want to wrap up this stuff soon.

Regards
PG

9. ### steveBWell-Known MemberMost Helpful Member

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So, what help do you want with it? If you want to check it, i'd say it looks correct. If you want to put it in a better form, I can't say that I see one off hand. Perhaps you might want to write it as 1+exp(j ...)(...), which would indicate an all pass system in parallel with another system in polar form.

This answer is not as nice at the others that you were able to show in a nice simple polar form. This is what I meant before when I said not all systems will simplify nicely. You can find other systems that look less nice than even this one. In those cases, perhaps rectangular form would look neater.

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10. ### PG1995Active Member

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Thank you, Steve.

Regards
PG

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11. ### steveBWell-Known MemberMost Helpful Member

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I would say you basically have this correct. The omega is a kind of normalized frequency and becomes an angle that spans a range of 2pi. Again, this relates back to the Nyquist sampling theorem which says that discrete time sampling at a particular sample rate, has an implied upper limit on frequency. The maximum frequency is half the sample rate. So the transform of the delta function is unity, which is a function that equals 1 over the domain of omega which spans a range of 2pi. Remember that, for continuous time signals, the Dirac delta function transforms to a value of unity over the domain -infinity to +infinity. So, you can think of discrete time as a kind of mapping. The infinite range of frequencies in continuous time gets transformed to plus and minus Fmax, where Fmax is the maximum cyclic frequency allowed by the Nyquist theorem.

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12. ### PG1995Active Member

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Hi

Could you please help me with these queries, Q1, Q2, Q3, and Q4? For some reason, I wasn't able to upload to the forums. Please also note that the document used for Q2 and Q3 contains many mistakes. Thanks.

Regards
PG

Last edited: Nov 19, 2013
13. ### steveBWell-Known MemberMost Helpful Member

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I'm not sure I understand what the question is. I think I understand your point of view on these interpretations, and I don't see any problem with them.

In the first interpretation, applying an impulse function implies providing all frequencies at once with equal amplitudes and phases, and the frequency domain response shows the response of each and all frequencies. The second interpretation is like applying each sinusoid individually, or one at a time, and then the frequency response at that particular frequency is evident in the transform.

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14. ### PG1995Active Member

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Thanks a lot, Steve.

I have updated the previous post. So, when you have time, please try to help me. Thank you.

Regards
PG

15. ### steveBWell-Known MemberMost Helpful Member

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Q2: Yes

Q3: It looks like they are trying to say that magnitude is an even function and phase is an odd function.

Q4: Yes

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16. ### PG1995Active Member

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Thanks a lot.

Is it really so? Is it true for any discrete-time signal? I think it's true but for some reason I can't think about it clearly. If you don't mind, could you please provide me with some simple example? Thank you.

Regards
PG

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Thank you.

Regards
PG

19. ### steveBWell-Known MemberMost Helpful Member

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"n" is just discrete time, which is the discrete time equivalent to "t" in continuous time systems. The differentiation is in the frequency domain operating on the Fourier Transform. This relation shows how multiplying by n in the time domain can be related to derivatives in the frequency domain.

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Thank you.

Regards
PG

21. ### steveBWell-Known MemberMost Helpful Member

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Yes, this will be unbounded. It's an exponential type of function. Exponential functions blow up with the base is greater than zero and the argument is positive. Just plot it and you'll see.

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