continuous and discrete time systems

Thank you, MrAl, Steve.

MrAl, thanks for showing it in tabular form.

steveB said:

There are many ways to get this expression, including trial and error. However, the method you show here is perfectly fine since the equation is of the form defined. Here P=1/tau and Q=C/tau. You will get a solution of the correct form, and then you need to choose the constant "c" (little c, not big C) to match the initial condition. There is also a time shift that is needed, but if you find this confusing, then just do the case with n=0, and realize that every time step requires the same solution with each value of nT behaving like t=0 for the initial condition.

Click to expand...

I'm sorry but it looks like I have made an **broken link removed** before I can reach anywhere.

In case you have missed it, please have a look on this post.

Thank you for the help.

Regards
PG

Hi,

Taking a quick look, it looks like you did not divide little c by e^(Integral Px dx). You must do that too. Try again.

Thank you, MrAl, for pointing out the silly error.

Kindly help me with this **broken link removed**. Thanks a lot.

Regards
PG

It's not that x(nT) and x(t) have the same value over the entire range nT<t<(n+1)T. Rather, the assumption is that the function x(t) is going to be sampled using zero order hold. Hence, we are (in a sense) approximating x(t) with a staircase type of function x(nT), and the sampling is done such that these functions are required to be equal only when t=nT, where n is an integer. The two functions may be equal when nT<t<(n+1)T, of they may not be equal in that interval, but they are definitely equal at times nT and (n+1)T.

Now, if the staircase function that results from zero order hold sampling, happens to be equal to the original function, then the original solution I gave is an exact solution. A good and famous example is when x(t) is a unit step function. Also, this solution is appropriate when the function x(t) is slowly varying compared with the time constant tau. In this case, we only need to sample at a rate sufficient to capture the bandwidth of the input signal x(t), and the sampling rate can be comparable to, or even greater than, the time constant tau. This would not be possible with the Euler method, or other approximate methods, and there T must be significantly less than the time constant tau.

steveB said:

It's not that x(nT) and x(t) have the same value over the entire range nT<t<(n+1)T. Rather, the assumption is that the function x(t) is going to be sampled using zero order hold. Hence, we are (in a sense) approximating x(t) with a staircase type of function x(nT), and the sampling is done such that these functions are required to be equal only when t=nT, where n is an integer. The two functions may be equal when nT<t<(n+1)T, of they may not be equal in that interval, but they are definitely equal at times nT and (n+1)T.

Now, if the staircase function that results from zero order hold sampling, happens to be equal to the original function, then the original solution I gave is an exact solution. A good and famous example is when x(t) is a unit step function. Also, this solution is appropriate when the function x(t) is slowly varying compared with the time constant tau. In this case, we only need to sample at a rate sufficient to capture the bandwidth of the input signal x(t), and the sampling rate can be comparable to, or even greater than, the time constant tau. This would not be possible with the Euler method, or other approximate methods, and there T must be significantly less than the time constant tau.

Click to expand...

I think I get it **broken link removed**. Thanks a lot for the help.

Regards
PG

PG1995 said:

I think I get it **broken link removed**. Thanks a lot for the help.

Regards
PG

Click to expand...

Yes, that's looks correct. That's the correct idea of what a staircase function looks like when you do the zero-order hold sampling of a signal. We think of x[n] as a discrete time function that only has values where "n" is an integer. But when that signal is put back in the real world where time is continuous, there needs to be a value between the dots, and zero-order hold sampling holds the values constant like a "staircase".

So, the idea of transforming a continuous time system into a discrete time system is based on the premise that you would ideally like the sampled values of both the input and output signals (x[n] and y[n]) on the discrete time system to equal the x(t) and y(t) of the real system. Of course, the input can be handled just by sampling, but the output is more challenging and can not be done perfectly in the general case. However, if the input signal x(nT) happens to exactly equal x(t), even between the integer values, then the solution I gave is exact and y[n] will perfectly match y(t) at the sampled points. The simplest case of this is when x(t) is a unit step function because u(t) and u(nT) are exactly the same functions.

Thank you, Steve.

I'm sorry but I'm **broken link removed** again. Please help me if possible. Thank you.

Regards
PG

Q1: There is a general definition for "staircase function" that mathematicians use, but I'm using the term a little more specifically. In this case I consider a staircase function to have equal time durations for each "step or stair". Then, I assume that the function is generated by taking x[n] in discrete time and putting it in continuous time by letting x(t)=x[n] for nT<=t<(n+1)T. This clearly defines which endpoint is included in each step.

Q2: Here you ask how y(t) get's replaced with y[n+1]. Essentially, you are finding the exact solution for the differential equation during each step where x(nT) is constant. Then you splice each solution together where the end of one solutions becomes the initial value for the start of the next solution. This means that y(t) is continuous, even though x(nT) is not continuous. However, we are looking for a solution for y[n] in discrete time, not y(t) in continuous time. Hence, we want to do a zero order hold sample on y(t) and turn it into a staircase function. This is why the value at y((n+1)T) is needed. We don't really care what y(t) does between nT and (n+1)T. We have the initial value y(nT) and we want to project forward and get the value y((n+1)T). Then, we simply realize that in discrete time, we only use the integer n to represent time. Hence, y((n+1)T) becomes y[n+1].

You also noted that we are projecting the value of y[n+1] from the values of x[n]. This is correct.

Hi

I don't think this system is stable, or is it? Do you find my working correct?

Please note that the equation I'm using for determining stability has been taken from Reference #1 and it says that the equation is for LTI system. If you look at the solution given at the bottom you can see that it's not an LTI system but still it's a stable system. So, perhaps, I shouldn't have used that equation for stability in first place. What do you say? Please let me know. Thanks.

Regards
PG

Helpful links:
1: BIBI Stability

PG,

This is a good question. I'm answering this quickly (2 mins of reading and responding), but this is a problem that requires careful thought and more time than I put in. So, I could be making a mistake, but lets talk it out and see if what I say makes sense to you.

First, you are correct to question the use of a rule for LTI system, when the system in nonlinear. So, throw the rule away in this case because it tells you nothing.

So, what rule do we use. The BIBO principle is a sound one I believe, so if you can find any bounded input that creates an unbounded output, then you would have proved that the system is unstable. So, try a simple example of a bounded input, for example, the step function u[n]. Will the output be bounded then?

I'm in a hospital waiting room while my wife gets an outpatient procedure. So, I'm bored and thought I'd continue on here. I'm assuming you are busy and didn't get to this, or perhaps you solved it no problem. But, just to finish the line of thinking, I'll continue.

So, above I posed the question of whether the output is bounded if the input is a bounded signal u[n]. There are two possible answers, so without answering it myself, let's explore the thought paths for each answer.

If you answer no, then you have discovered an example of a bounded input with unbounded output, and the system is unstable in the BIBO sense. Case closed - problem complete!

However, if you answer yes, you have not proven that the system is stable. This might be a good indicator that the system is stable, but it is not proof. So, if you are forced down this path, how do you prove it? Well, in general this might not be easy. You could keep searching for bounded input signals that result in unbounded output signals, if you suspect that the system is unstable, but if you suspect the system is stable, then you need a formal proof because failure to find an example bounded input that causes an unbounded output is not proof that it is stable in the BIBO sense.

Formal proofs may be easy for some systems and maybe less easy for other systems. But, we won't go down those paths unless needed.

Thank you, Steve.

I hope your wife is feeling better now. I wasn't busy yesterday. It's just that these days I feel some much sleepy. Like yesterday, I slept around 8:00 PM and woke next morning (or, before noon) around 11:00 AM.

You can say that I have solved it. I didn't think about it much. I followed what you said in the previous post and was able to get across though I understand what you are saying now. It was the same problem with Fermat's last theorem because for some many years no one was able to prove or disprove it. Thank you.

Regards
PG

Hi

Could you please help me with these queries? It might look like so many queries for a single post but many of the queries are very simple as you will see. Thanks a lot for your help.

Regards
PG

PS: This text might be useful for Q2.

Q1: Yes, correct.

Q2: I think when you first start, real sinusoids are more straightforward and less confusing conceptually. However, once you gain experience, the simplifications of using complex exponential functions is very desirable and makes that approach seem much more natural. But, this is a matter of perception.

Q3: Sounds good.

Q4: Yes, basically that's it. The output is the input modified by a amplitude change and phase shift.

Q5: Actually, I don't know that symbol and never use it myself. If I've seen it before, I guess I didn't pay any attention to it. My best guess is that it is similar to the equal sign with 3 dashes, which means equal to by definition.

Q6: I guess so from that representation. But, you could derive the result without every considering complex exponential functions, if you wanted to.

Q7: Yes, if I understand you, you are correct. Maybe you are starting to see just how simple this stuff can be once you get familiar and comfortable with it. The math can look very complex, but it is really saying very simple and basic things and the simplicity comes mostly from linearity (via superposition principle) and some from time invariance as well.

Thanks a lot, Steve.

Could you please help me with these queries too? Many thanks.

Regards
PG

Hi

I thought I should let you know about the previous post in case you have missed it. There is no hurry if you are busy and please don't mind it. Thanks.

Regards
PG

Hmmm, somehow I did miss it, but I'm not sure why. Strange because the alert feature didn't notify me. How did you know I missed it?

In answering here, I should point out that I'm wondering if the highlighting is blocking some of the equations, or maybe there is a typo there.

Q1: In general, the magnitude of exp(jx) is always one, which results from the well known trig identity sin^2+cos^2=1

Q2: Here is where I'm wondering if there is a typo because I would have expected the h[n] to have an absolute value around it in the right hand side of that equation. In that case the inequality you asked about in the previous post would be the basis to say that is correct. As written, it does not look correct to me.

Q3: That looks to me like a condition for convergence. There are many different definitions for stability, and that one is the BIBO condition, provided that the absolute value is put around the h[k]. So, again I think there is a typo there.

Q4 and Q5: First, let's be careful here as to what w=0 means in each case. In the Q4 case, w=0 implies that x[n]=1 and in the Q5 case, w=0 implied x[n]=0. So, in the first case, delaying a constant value does not change the function at all, so the transform should not change. In the second case, the transform of 0 is 0. The two answers shown are consistent with what we expect. I'm not sure what else to say about these cases.

Now, if you allow w to be other than zero, we can say a lot more about what the transform relations mean.

Thanks a lot.

steveB said:

How did you know I missed it?

Click to expand...

Actually I posted this set of queries before the other queries yesterday. You had helped me with those queries which were posted later so I thought that you might have missed it.

steveB said:

In answering here, I should point out that I'm wondering if the highlighting is blocking some of the equations, or maybe there is a typo there.

Click to expand...

There is some issue with the slides. Some of the entries are missing. If you look closely, you can notice that there is a single vertical bar around h[n] on the right but the bar on the left is missing. Thank you.

Regards
PG

Thank you.

steveB said:

Q4 and Q5: First, let's be careful here as to what w=0 means in each case. In the Q4 case, w=0 implies that x[n]=1 and in the Q5 case, w=0 implied x[n]=0. So, in the first case, delaying a constant value does not change the function at all, so the transform should not change. In the second case, the transform of 0 is 0. The two answers shown are consistent with what we expect. I'm not sure what else to say about these cases.

Click to expand...

Please have a look on these follow-on queries. Thanks.

Regards
PG

First, x[n]=1 is not a unit step function. It is simply a constant value for all time.

Case #1 is x[n]=1, but case 2 is the first difference which is basically a derivative of a constant function. The derivative of a constant is zero.