transfer function,s-domain

[sotomie]

How this is H(w)=1+jw/wz equal to H(s)=1-s/sz where sz=-jwz?? I can't seem to prove that :/

 

[barry]

That's one goofy problem. Doesn't your original equation reduce to :

H(w)=1+j/z ?

and your second equation reduces to:

H(s)=1-1/z.

And from what I can remember, s=jw, not -jw.

Did you copy this problem correctly?

 

[sotomie]

you got the question incorrectly,it isn't w*z its wz which is the zero of a circuit.I've attached the image for your convenience.

 

[albert22]

In Laplace transform s= a + j w where a (sigma), w (omega) = angular freq.
The limit (when a goes to zero) of H(s) = H(jw)
H(jw) is the frequency response. That is H(s) evaluated on the imaginary axis.
In most of the cases given the transfer function H(s) to obtain H(jw) just replace s by jw. Not by -jw.
I dont think that it is possible to go backwards that is to obtain H(s) from H(jw).
1/(-j*wz)= j / wz. You just multiply numerator and denominator by j.

 

[sotomie]

I still don't get it ,i get how we wrote sz in place of wz but then what about the s in numerator,don't we require another j for it ,so we could write s in place of jw ?
Thanks a lot

 

[zorro]

The figure shows two different cases.
In the left-side plot, there is a real zero at s_Z=-ω_Z (with ω_Z real and positive), i.e. in the left half plane.
In the right-side plot, the zero is real as well, but located at s=ω_Z, i.e. in the right half plane.
What is wrong is " s_Z=-jω_Z ".
The two formulas H(s)=1-s/s_Z and H(s)=1+s/s_Z are a bit confusing too. I would prefer to replace s_Z by -ω_Z in them.

Regards

Z

 

[albert22]

The image is not very clear to me. And I dont know if a transfer function which has only zeros makes sense or if it is realizable.
But do not confuse sz = -j * wz which seems to be a zero of H(s) with the laplace variable s= a + j * w. Neither confuse wz which is the cutoff angular frequency (as seen in the graph) with the variable w.
sz, wz are constants while s and w are variables.
If sz is a zero of H(s) the idea is that when w=wz H(jw))=0 so something seems to be wrong in the expresion.
it could be H(jw)=1-w/wz for example.
Do you have more info besides the figure?

I now see the post from zorro I agree.

 

[sotomie]

Thanks a lot !That cleared my confusion a lot.
View attachment freq_response.pdf
That's the lecture slides that my professor provided us with.Slide 21 is where this topic starts.And i still have the confusion about where did the equation of H(jw) came from originally lol but the confusion about the transformation to s-domain is cleared there must be some mistake about sz=-jwz!

 

[albert22]

I see that the slides are explaining the impact that poles and zeros of the transfer function H(s) as a means to plot the bode graphs.
As far as I know you cannot build a system which has only zeros in H(s) so the professor is going step by step separating the responses of the poles and then the zeros.
The problem is that he is talking about zeros in the half right and left plane but when sz= -jw the zero is not on the right or left plane but exactly at the imaginary axis. Because in that case sigma is zero. They would be in the upper or lower plane instead.
H(s) or H(jw) in the case of electric circuits comes from the analysis of the circuit using the impedance abstraction and kirchoff laws. For example ZL= j * w * L. for H(jw) or s * L for H(s). You have an example at the 2nd slide. But again I am pretty sure there is no physical circuit that can have only zeros in the transfer function. So the circuit is not shown.