# What are poles & zero in Real.......??

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#### koolguy

##### Active Member
O = summing junction with one plus and one minus input

1.) what the use of it??
Gs = G(s) = K/(s*(s+1))
2.) from where??
& this
The z transform of G(s) with unit impulse and sampling period of 1 second is:
G(z)=K*(0.36787944025834*z+0.26424111948332)/(z^2-1.367879441171442*z+0.36787944117144)

If the system is stable the roots of 1+G(z)=0 all lie within the unit circle, so
we compute 1+G(z)=0 for various gains.

With K=1 we have:
1+G(z)=0
where G(z) evaluates to:
(0.36787944025834*z+0.26424111948332)/(z^2-1.367879441171442*z+0.36787944117144)+1=0

#### MrAl

##### Well-Known Member
Hello again,

You ask some very good questions, and i have to ask you a couple too so i know what you want to be able to understand about all this.

1. A summing junction is part of a block diagram that maps out the system we are interested in looking at. Have you used any block diagrams yet?
2. The equation given there is just an example transfer function for the forward path of the system. There are many other possibilities for this equation. I tried to pick one that wasnt too complicated yet at the same time would be complicated enough to show how all this works. Have you worked with any transfer functions yet?
3. To proceed with the analysis, we find the z transform of the system and go from there. Have you done any z transforms yet?

Maybe you are more interested in the pure mathematics of the z transform? It's hard for me to tell what exactly you want unless you describe it very carefully and completely.

#### koolguy

##### Active Member
Hi,
Have you used any block diagrams yet?
No..
I also don't know transfer function!!

#### MrAl

##### Well-Known Member
Hello again,

Ok then what are you using the z transform for?

The block diagram shows us how the system is interconnected. We can then start writing the equations for the system. From that we can see if it is stable or under what conditions it is unstable.

#### koolguy

##### Active Member
Today, i was studding Fourier series in this x(t-t0) = e^-ajw please tell how??

#### birdman0_o

##### Active Member
If you are having this many difficulties perhaps consult your teacher or the course textbook, we don't have enough time to teach you a class over the internet (or the will).

#### MrAl

##### Well-Known Member
Today, i was studding Fourier series in this x(t-t0) = e^-ajw please tell how??

If you mean x is the unit impulse, then we get e^-ajw because the Fourier transform of x(t-t0) produces a sifting integral, where the integral produces a particular value of f(t). The result is then e^-jwt0.
Im not always sure what you are asking because your questions are not as detailed as they should be.

#### koolguy

##### Active Member
If you mean x is the unit impulse, then we get e^-ajw because the Fourier transform of x(t-t0) produces a sifting integral, where the integral produces a particular value of f(t). The result is then e^-jwt0.

How, please give some cool example why e^-axj comes?? this is my question

#### MrAl

##### Well-Known Member
How, please give some cool example why e^-axj comes?? this is my question

Starting with the Fourier Transform integral for the unit impulse:

[LATEX]\int_{\ \ -\infty }^{\ \ \ \ \ \ \ \infty }\delta\left( t-t0\right) \,\ {e}^{-i w t }\ \ dt[/LATEX]

Since the amplitude inside the integral is zero everywhere except when [LATEX]t=t0[/LATEX], we can integrate from [LATEX]t0\ to\ t0+\delta t[/LATEX] as an approximation, but since the amplitude is also infinite at that point we must then also multiply by [LATEX] \frac{1}{\delta t} [/LATEX] so that when [LATEX] \delta t [/LATEX] approaches zero we get an infinitesimally short time period inside the integral and outside we get an infinity so we end up with an equivalent integral. In other words, we approximate the unit impulse as a pulse that is [LATEX] \delta t [/LATEX] wide and [LATEX] \frac{1}{\delta t} [/LATEX] high and then integrate. We must then later take the limit as we let [LATEX] \delta t [/LATEX] approach zero because as [LATEX] \delta t [/LATEX] approaches zero our approximate impulse approaches a true impulse. The integral then becomes:

[LATEX] \left( \frac{1}{\delta t}\right) \int_{t0}^{\ \ \ \ \ \ \ t0+\delta t}\ {e}^{-i w t} \ \ dt\[/LATEX]

We evaluate that and then finally we must take the limit as [LATEX]\delta t[/LATEX] goes to zero, and we get what we expected:

[LATEX]\lim_{\delta t\to 0 }\ \ \left( \frac{i}{w \ \delta t}\right) \left({e}^{-i w \left( t0 + \delta t \right)}\ - \ {e}^{-i w t0 \right)\ \ = \ \ {e}^{-i w t0}[/LATEX]

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#### birdman0_o

##### Active Member
Look into the definition of a Fourier transform and it's derivation, then you will understand.

#### MrAl

##### Well-Known Member
Hello birdman,

See my post just before yours #### koolguy

##### Active Member
Hello Sir,

In practical how shifting in time T0 in time domain produce change in add. of e^-ajw in freq. domain??how??

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#### birdman0_o

##### Active Member
Mr. Al I feel you are being too kind to this student and he will only learn how to beg for answers. As well you only proved the derivation for the unit step in the Fourier domain, not why the Fourier transform of f(x) includes e^-jwx or why it is even an integral. IMO he should read into this on his own if he wants to understand.

Links like the following Fourier Transform -- from Wolfram MathWorld are great.

#### MrAl

##### Well-Known Member
Hello Sir,

In practical how shifting in time T0 in time domain produce change in add. of e^-ajw in freq. domain??how??

You are going to have to look back at my previous post, post #29 where i explained that. With t0=a, the transform becomes e^-ajw in the limit.
Another view is that multiplication of f(t) by e^-jwT delays the impulse by T units of time.

#### koolguy

##### Active Member
Hello,

The Forum are made for sharing Knowledge clearing Doubt..!!

Mr. Al I feel you are being too kind to this student and he will only learn how to beg for answers. As well you only proved the derivation for the unit step in the Fourier domain, not why the Fourier transform of f(x) includes e^-jwx or why it is even an integral. IMO he should read into this on his own if he wants to understand.

Links like the following Fourier Transform -- from Wolfram MathWorld are great.

#### koolguy

##### Active Member
Yeah, i Know but i want to know it in practically how change in time produce exponential multiplication??

#### MrAl

##### Well-Known Member
Hi again,

You're going to have to be more explicit with your question because i cant figure out what you are asking again.
Perhaps send me a PM in your native language and i'll try to make some sense out of it,

#### MrAl

##### Well-Known Member
Mr. Al I feel you are being too kind to this student and he will only learn how to beg for answers. As well you only proved the derivation for the unit step in the Fourier domain, not why the Fourier transform of f(x) includes e^-jwx or why it is even an integral. IMO he should read into this on his own if he wants to understand.

Links like the following Fourier Transform -- from Wolfram MathWorld are great. Sometimes foreign language speakers appear to be very rude when really they just dont understand the nuances of our own first language. I dont like to make too harsh of a judgment. I also like to refresh on some of these subjects myself • mfayek

#### MrAl

##### Well-Known Member
Yeah, i Know but i want to know it in practically how change in time produce exponential multiplication??

Hi again,

Ok, i took a look at your PM and i think i know what you are asking here now. You probably want to see how the factor e^-ajw really affects the function as in F(w)*e^-ajw.
I'll see if i can come up with a good example of this to illustrate a little better.

The simplest way to understand it is using Laplace instead of Fourier. If we multiply F(s) times e^-sT, in the time domain we get f(t) delayed by time T. This of course means that if we had a time function like sin(wt) and multiplied its transform Laplace(sin(wt)) times e^-sT, we would end up with sin(wt) but delayed by T, which would mean the time function does not start (remains at 0) until we reach the time t=T, and then it starts and appears as it normally would).
For example, if we set the frequency to 1Hz in sin(wt) we would of course get a sine wave that starts out at t=0 and rises up as time progresses and then peaks with amplitude 1 and then later -1 and then back to zero, because it's just a sine wave. But in Laplace(sin(wt))*e^-sT with T=1, we would get a wave that is zero at t=0 and STAYS at zero until we reach the point t=1 and then and only then would the sine wave start rising and proceed in the normal way (it would again reach zero at t=2 because freq=1).
It should be pointed out however that this is not the same as sin(wt+TH), where TH is a phase shift, because that allows the sine to start at t=0 (even though it is now phase shifted) and that is not the same as using a delay of e^-sT.
The interesting thing about this is that we can have a sine wave that starts at just about any time we want just by setting T=time delay. If we say set T=0.5 in the above we would get a wave that starts out at zero and stays at zero until t=0.5 seconds and then and only then the sine wave would start to rise and take on its normal shape. This would look like a sine wave shifted by 180 degrees with the exception that it is zero for the first 0.5 seconds.
The interesting consequence of being able to use a delay like this is that we can generate other kinds of waves too where we simply add delayed (and possibly inverted) waves together.

I'll try to work up a better example of this at some point.

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#### koolguy

##### Active Member
HI Sir,

You are giving good example, but for me it will be better if you have a small images showing this property in real world, it will be far better....

Thanks very much..!!

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