# continuous and discrete time systems

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

1. ### PG1995Active Member

Joined:
Apr 18, 2011
Messages:
1,645
Likes:
13
Thank you.

That was dumb! A unit step function would have x[n]=1 where n=>0.

I'm still confused. The system y[n] in case #1 gives a delayed output while the system y[n] in case #2 gives first backward difference which is analogous to a derivative operation. But, in my humble opinion, for both cases the input signal is x[n]=1. Where do I have it wrong, or, what am I missing here? Thanks a lot for the help.

Regards
PG

2. ### steveBWell-Known MemberMost Helpful Member

Joined:
Jan 16, 2009
Messages:
1,297
Likes:
629

You do have that right. Sorry, I accidentally wrote x[n] instead of y[n]. In both cases x[n]=1, when you delay that you still have y[n]=1 and when you take the difference operation you have y[n]=0.

• Like x 1
3. ### PG1995Active Member

Joined:
Apr 18, 2011
Messages:
1,645
Likes:
13
Thank you for the clarification.

If you don't mind, I would like to discuss case #1 a little more. For case #1, the input x[n]=1, when w=0, comes out at the output as it is. Perhaps, one reason for this is that the signal x[n]=0, -inf< n <+inf, is symmetric around n=0, so even if it were shifted or delayed we won't be able to discern it, right? But I suspect if it were a unit step function x[n]=1 where n>=0, then the output should be delayed and the signal won't come out as it is. But as it looks I'm wrong because frequency response even in case of a unit step is "1" when w=0. I believe that I'm very confused here and that's why don't really know how to put forward me query properly. But perhaps you can see where I'm having trouble. Thanks.

Regards
PG

Joined:
Jan 12, 1997
Messages:
-
Likes:
0

5. ### steveBWell-Known MemberMost Helpful Member

Joined:
Jan 16, 2009
Messages:
1,297
Likes:
629

OK, most of what you say is right here. It's not the symmetry that matters though. It is the fact that the function x[n]=1 implies a constant value for all time. When you delay that, it looks the same.

But, when you talk about a unit step function, you are not talking about a simple sinusoid any more. You are talking about an infinite number of sinusoids, one at every frequency in a continuum. So, w=0 makes no sense for a unit step function.

Also, remember the delay property of transforms. That is, delay in the time domain is linear phase shift in the frequency domain. So, each frequency of the unit step function is delayed in time by the same time value, but shifted in phase by a different amount. The phase shift is linearly related to frequency.

• Like x 1

Joined:
Apr 18, 2011
Messages:
1,645
Likes:
13

Regards
PG

7. ### PG1995Active Member

Joined:
Apr 18, 2011
Messages:
1,645
Likes:
13
Hi

Just wanted to clear a minor point here. In reply to Q3, you said the following.
How do you differentiate between convergence and stability? I think when in time domain, we are concerned about stability and when in frequency domain, it's the convergence which interests us. Thanks.

If possible, please also help me with these queries, Q1 and Q2.

Regards
PG

#### Attached Files:

File size:
199.8 KB
Views:
129
• ###### dsp_ideal11.jpg
File size:
146.2 KB
Views:
206
Last edited: Nov 9, 2013
8. ### steveBWell-Known MemberMost Helpful Member

Joined:
Jan 16, 2009
Messages:
1,297
Likes:
629
Convergence just means that when you do an integration or a summation, you get a finite answer. So, my interpretation of what they wrote was that they are looking at the "existence" of the frequency response which is really just making sure that the answer converges. The result is interesting because a condition for convergence is related to the condition for stability.

Q1 I am having trouble reading, and your question (which I can read) is not clear to me also.

Q2: An ideal delay is defined (by that book) to be a perfect delay by an integer number of time steps. Ideal means that is is a perfect delay that does not alter the signal in any way and the delay is an integer number and does not have any fractional delays.

The application of delays is basically a memory storage. Since we are dealing with digital systems, either program based or circuit based, we can store old values of any signal and use them for various purposes, such as making filters (as you mentioned).

Another point about delays, which many people do not appreciate, is that they create additional "states" in a digital system. There are many tricks that one can utilize in control theory when one realizes that delays (which are sometimes detrimental in control feedback) can be treated as states in the system (they make a higher order system).

• Like x 1

Joined:
Apr 18, 2011
Messages:
1,645
Likes:
13
Hi

Regards
PG

File size:
439.6 KB
Views:
239
10. ### steveBWell-Known MemberMost Helpful Member

Joined:
Jan 16, 2009
Messages:
1,297
Likes:
629
Q1: Yes, of course the +/- pi range is because of periodicity of

This can also be related to the Nyquist sampling theorem. If you sample at a particular rate, then the max frequency of the signals is only half that. The math needs to reflect this fact by limiting the maximum meaningful frequency in a Fourier transform.

Q2: I don't think I agree, but I'm not sure I understand it either, so I'm not sure. I'll try to read it more carefully later.

• Like x 1
11. ### PG1995Active Member

Joined:
Apr 18, 2011
Messages:
1,645
Likes:
13
Thank you, Steve.

Okay. Please do try to read it again when you get time. I would also say that please try to go through the Q1 again because it looks like you have missed the point which was confusing me. Thanks.

Regards
PG

12. ### steveBWell-Known MemberMost Helpful Member

Joined:
Jan 16, 2009
Messages:
1,297
Likes:
629
You'll have to clarify what the other point is because I still don't see it.

13. ### PG1995Active Member

Joined:
Apr 18, 2011
Messages:
1,645
Likes:
13
Okay. I think I get it now. Let me know if I have it correct. And please do give that Q2 another look when you get time because that concept is important for me to understand. Thank you.

Regards
PG

File size:
98.7 KB
Views:
177
14. ### steveBWell-Known MemberMost Helpful Member

Joined:
Jan 16, 2009
Messages:
1,297
Likes:
629
Yes, looks correct to me. You see now why periodicity is not compatible with a function that is zero for omega> pi.

• Like x 1

Joined:
Apr 18, 2011
Messages:
1,645
Likes:
13
Hi again,

Regards
PG

File size:
362.7 KB
Views:
197
16. ### steveBWell-Known MemberMost Helpful Member

Joined:
Jan 16, 2009
Messages:
1,297
Likes:
629
For the original Q2, I'm confused why you say a real number must have a phase of zero. Why can't the phase be 180 degrees (pi)? Or, -180 degrees (-pi) ... etc. ?

For the new Q1, I dont' know why you are having trouble finding the phase of a complex number. I know you learned how to do this years ago.

For the new Q2, wouldn't the Euler relation exp(jx)=cos(x) + j sin(x) allow you to get trigonometric functions?

• Like x 1
17. ### PG1995Active Member

Joined:
Apr 18, 2011
Messages:
1,645
Likes:
13
Thank you, Steve.

Original Q2: But my point was that if a signal is symmetric around n=0 axis or if a signal is even then according to DTFT it's phase will be zero. Do you agree with this? I'm sorry if I'm missing your point.

I'm still confused about Q1 and Q2.

Q1: Yes, I did learn how to find phase of a complex number more than a year ago. But now my knowledge is very much rusty. I understand that the phase is arctan(b/a) for a+ib. Please see my attempt here.

Q2: Yes, one needs to use Euler formula here but the problem is to first get the exponential terms in proper arrangement and this is where I'm having trouble. EDIT: I have tried to solve another similar problem (at the bottom) but I'm stuck again. I'm sure there should be some general way of arranging the exponential terms in a sequence or order which makes the application of Euler formula straightforward. I hope you can see my trouble. Thanks.

Thanks a lot.

Regards
PG

#### Attached Files:

File size:
101.9 KB
Views:
191
• ###### dsp_stem.jpg
File size:
309.4 KB
Views:
193
Last edited: Nov 15, 2013
18. ### steveBWell-Known MemberMost Helpful Member

Joined:
Jan 16, 2009
Messages:
1,297
Likes:
629
Well, let's check it out. What is the phase if we change h[n] to the following?

In other words, make it negative to what it was before.

OK, rusty or not, you do know how to do it. It looks good to me. I didn't check every detail, but the approach is correct, which is most important.

I'm really confused about why you are having trouble. You are trying to find a way to make the application of the Euler formula straightforward. But, what can be more straightforward than taking every A exp(jb) and converting it to A cos(b) + j sin(b). You do realize that the answer is allowed to be complex, ... right? Also, there is no requirement to use sine and cosine forms, and the answer you provided already is correct.

Is there some particular form that you expect to achieve that I'm missing?

• Like x 1
19. ### PG1995Active Member

Joined:
Apr 18, 2011
Messages:
1,645
Likes:
13
Okay. Please have a look here.

I think that you are still missing the point, or perhaps it's just me. Please have a look here and see if you can see where I'm having trouble. Thanks.

Regards
PG

File size:
89.7 KB
Views:
186
File size:
214.9 KB
Views:
170
20. ### steveBWell-Known MemberMost Helpful Member

Joined:
Jan 16, 2009
Messages:
1,297
Likes:
629
OK, so for the first issue, you have shown two problems which differ by a negative sign. Isn't a negative sign the equivalent of a 180 degree phase shift? So, if you say the first one is a phase of zero, then the second one should be a phase of 180 degrees. This was the very first point I was trying to make. Just because something is real, does not mean the phase is zero. It can be zero for positive numbers and 180 degrees (or -180 degrees) for negative numbers.

OK, for the second issue, it seems that you are trying to get a "preferred form" for the final answer. It's always good to simplify expressions when you can. However, is there any rule that says that all cases will lead to nice simplifications? I would tend to say no. But, I'll go back and look at your original problem to see if I can find any simplifications there.

• Like x 1
21. ### steveBWell-Known MemberMost Helpful Member

Joined:
Jan 16, 2009
Messages:
1,297
Likes:
629
To simplify the answer to trig functions, I would recommend the following on that problem. First, you will want to pair up terms with equal amplitudes, so that they can be combined. Then with each pair you need to factor something out, e.g. factor exp(-j 3.5 Omega).

Just doing this quickly in my head, I get something like the following ...

I could have made a mistake, but you will get the point, I think.

Last edited: Nov 15, 2013
• Like x 1