block diagrams etc.

[steveB]

Thank you.

Yes, you are right that it's simple algebra for Q2 but for some reason my simple algebra isn't working! If possible, please help me with it. Thanks.

Regards
PG

Multiply G through on the right hand side.

Add GHC to both sides

Factor C out of left hand side

Divide both sides by R(1+GH)

 

[PG1995]

Many, many thanks, Steve.

Could you please help me with this query? Thank you.

Regards
PG

Added after steveB's post #43 for own personal reference: Actually I had it wrong mathematically. When we need to find roots (and poles are also roots) we first need to factorize a given polynomial - sometimes factors are going to be over real domain and other times over complex domain. Therefore, you first need to factorize 1/{s^2+s} before saying anything about poles/roots. After factorization it becomes 1/s(s+1) and now you are in a position to say anything about the roots.

 

[steveB]

The s^2+s in the denominator is the same as s(s+1) and the poles are at s=0 and s=-1 in either case.

Two poles at zero would be 1/s^2.

 

[PG1995]

Thank you, Steve.

Could you please help me with this query?

Best wishes
PG

 

[MrAl]

Hi again,


To understand these in general the following might help...

FIG 1     +----A--->---o R*A
          |
    R o---+-->--[Sum]---->---o R+H*X
                  |
                  |
                  +--<----H-----o X



FIG2                   +----A--->---o (R+H*X)*A
                       |
    R o---+-->--[Sum]--+-->---o R+H*X
                  |
                  |
                  +--<----H-----o X



FIG 3                     +----A--->---o R*A+H*X*A
                          |
       R o---+-->--[Sum]--+-->---o R+H*X
                     |
                     |
                     +--<----H-----o X


                                             +--<--(-H*X*A)
                                             |
                                             |
FIG 4                     +----A--->-------[Sum]---o R*A
                          |
       R o---+-->--[Sum]--+-->---o R+H*X
                     |
                     |
                     +--<----H-----o X



                                             +--<--(-H*A)--<--o X
                                             |
                                             |
FIG 5                     +----A--->-------[Sum]---o R*A
                          |
       R o---+-->--[Sum]--+-->---o R+H*X
                     |
                     |
                     +--<----H-----o X

In the above Fig 1, we see the original diagram. We see there are two inputs and two outputs.
One output is R*A and the other output is R+H*X. The two inputs are R and X.

In Fig 2, we moved the pickoff point forward so now it is behind the summing junction. So the output of that section is no longer R*A but is (R+H*X)*A which factors into R*A+H*X*A (Fig 3). So what changed? The upper output changed from:
R*A
to:
R*A+H*X*A

So what do we have to do to R*A+H*X*A to get back the original R*A?

Simple, we have to subtract H*X*A.

In Fig 4, we see we've added another summing junction that sums R*A+H*X*A and just -H*X*A. We needed a negatve H*X*A so that it would subtract from R*A+H*X*A and leave us with R*A, the original output.

In Fig 5, we've just arranged it a little differently to show that the other input came from X, and that second X input can be tied to the first X input lower in the diagram. Note that the outputs are the same as they were when we started, the only difference being we moved a pickoff point and added another summing junction.

So the rule for moving a pickoff point forward past a summing junction (behind the summing junction) is that you must then subtract everything 'else' that came into the summing junction from the new path such that the path output is the same as it was before. In the example above we had all positives to start with, but there will sometimes be negatives like -H instead of H, and if we have a negative into a summing junction that just means subtraction so we would have had to add instead of subtract H*X*A like we did above. So it is also important to note the signs carefully.

This is still a good exercise because in real life often we do not have all the outputs that we would like to have because there is usually a limited number of variables being measured and/or we want to keep the measurements count down because we want as few sensors as possible. So we might end up having to tap off of a point that is not ideal so we'd have to know what we have to do to it in order to get the kind of measurement we want from that non ideal point in the system.

I assume you have a list of the other more simple rules. If you care to post them we can take a look.

 

[steveB]

Thank you, Steve.

Could you please help me with this query?

Best wishes
PG

It's hard to guide you on that one because it's not clear where you got the "f" from. It seems you just made up a variable, called it "f" and stuck it in the answer. If you explain what f is and why you think it belongs there, maybe we can guide you.

I can just reiterate what the book says. Generally, poles are found by setting the denominator to zero. In this case we do that and then solve for values of "s" that make the denominator zero. This then implies that exp(-s)=1. If we were talking about functions over a real domain s, then obviously we would obtain s=0 because we know that exp(0)=1. However, since we are dealing with functions over a complex domain s, we need to be more careful and look for complex values of s also.

The books clearly shows why s can be j2∏n, where n is any integer. Hence, s=0 is only one of an infinite number of solutions and there are an infinite number of poles, all evenly spaced out on the imaginary axis, and separated by a distance of 2∏.

But, let's be more explicit than the book. The book points out that exp(-σ) ( cos(ω) - j sin(ω)) =1. Since the right hand side is real, the imaginary part of the left hand must be zero. Hence. sin(ω)=0, which implies that any solution we find is limited to values of ω that obey ω=m∏, where "m" is any integer. However, not necessarily are all of these values valid solutions because the real part of the right hand side is 1, and the real part of the left hand side must also be one. Hence exp(σ) cos(m∏)=1. Here we can see that the cosine will be ±1 (1 for even "m" and -1 for odd "m"), and exp(σ) is always positive for any real value of σ. Hence, solutions require σ=0 and m must be even. Well, this converts to s=jn2∏ for any integer n. Notice that we use 2n in place of m, so that we can say n is any integer, rather than having to say that m is any even integer including 0.

 

[PG1995]

Thanks a lot, MrAl, Steve.

If you explain what f is and why you think it belongs there, maybe we can guide you.

Because ω=2∏f, isn't it? But the book doesn't show 'f' anywhere.

Regards
PG

 

[steveB]

Thanks a lot, MrAl, Steve.



Because ω=2∏f, isn't it? But the book doesn't show 'f' anywhere.

Regards
PG

OK, that is true, but all that means is that once we solve for ω, we have also solved for f since f=ω/(2∏). Hence, we can also say the solution is σ=0 and f=n, where n is any integer. We know that s=σ+j2∏f, hence this is the same as saying s=j2∏n.

 

[PG1995]

Thank you.

So, in a way, in the expression ω=2∏n, n=f, and poles will only occur when 'f' is an integer number; and when it's not integer then poles do no occur.

Regards
PG

 

[steveB]

Thank you.

So, in a way, in the expression ω=2∏n, n=f, and poles will only occur when 'f' is an integer number; and when it's not integer then poles do no occur.

Regards
PG

I'm not sure if I understand the phrasing here. Usually people say, "the poles occur at/where f is an integer". If you say, "the poles occur when f is an integer" and if you mean the same thing, then that's OK. However, when you say, "when f is not an integer the poles do not occur", I'm not sure of the meaning. It is true that the poles are not located at non integer frequencies on the imaginary axis, but a system described by the transfer function given, always has poles, even when that system is excited or driven by pure sinusoidal signals with non-integer frequency.

I just want to make sure you are understanding this correctly.

 

[PG1995]

Usually people say, "the poles occur at/where f is an integer". If you say, "the poles occur when f is an integer" and if you mean the same thing, then that's OK.

Yes, I meant this. And I was trying to say that when 'f' is not an integer then we can't have such a compact expression for the location of poles as this one ω=±2n∏. Thank you.

Regards
PG

 

[PG1995]

I assume you have a list of the other more simple rules. If you care to post them we can take a look.

Hi

Sorry for the late reply. Just got free yesterday. I have summarized all the rules for block diagrams I needed so far. Please give it a look and let me know if there is something wrong, especially Rule #7. Thank you.

I'm not sure if I understand the phrasing here. Usually people say, "the poles occur at/where f is an integer". If you say, "the poles occur when f is an integer" and if you mean the same thing, then that's OK. However, when you say, "when f is not an integer the poles do not occur", I'm not sure of the meaning. It is true that the poles are not located at non integer frequencies on the imaginary axis, but a system described by the transfer function given, always has poles, even when that system is excited or driven by pure sinusoidal signals with non-integer frequency.

That day I didn't have time to ask this. But I didn't understand the text in blue. Could you please elaborate on it a bit? Thanks.

Regards
PG