3 added 399 characters in body
source | link

Because a certain parameter in the time domain is wanted, the most direct method would be to find the function g(t) for the step repsonse and calculate the first derivative. This is, however, a rather involved procedure.

Hence, it is more convenient not to leave the frequency domain and apply the "initial slope theorem" (which immediately gives the slope g′(0) of the step response at t=0):

Similar to the initial value theorem (find (Hs) for s>>>infinity) the initial slope theorem requires to find the value s*H(s) for infinite frequencies.

This calculation step is identical to the last part of the original question leading to g′(0)=0.

Comment: The initial slope theorem can be extended also to higher derivatives (Ref: Active Network Design, Clause S. Lindquist)


Addendum (Time domain):

For the given system function H(s) the corresponding step response function can be found to be

g(t)=1 - exp(Dwot)*{cos(wnt) + [D/SQRT(1-D²)]sin(wnt)}

with the natural frequency wn=wo*SQRT(1-D²).

and the first derivative at t=0 can be found to be

g′(t=0)=woD-Dwn/SQRT(1-D²)]

Replacing wn by wo we get (as expected):

g′(t=0)=0

Because a certain parameter in the time domain is wanted, the most direct method would be to find the function g(t) for the step repsonse and calculate the first derivative. This is, however, a rather involved procedure.

Hence, it is more convenient not to leave the frequency domain and apply the "initial slope theorem" (which immediately gives the slope g′(0) of the step response at t=0):

Similar to the initial value theorem (find (Hs) for s>>>infinity) the initial slope theorem requires to find the value s*H(s) for infinite frequencies.

This calculation step is identical to the last part of the original question leading to g′(0)=0.

Comment: The initial slope theorem can be extended also to higher derivatives (Ref: Active Network Design, Clause S. Lindquist)

Because a certain parameter in the time domain is wanted, the most direct method would be to find the function g(t) for the step repsonse and calculate the first derivative. This is, however, a rather involved procedure.

Hence, it is more convenient not to leave the frequency domain and apply the "initial slope theorem" (which immediately gives the slope g′(0) of the step response at t=0):

Similar to the initial value theorem (find (Hs) for s>>>infinity) the initial slope theorem requires to find the value s*H(s) for infinite frequencies.

This calculation step is identical to the last part of the original question leading to g′(0)=0.

Comment: The initial slope theorem can be extended also to higher derivatives (Ref: Active Network Design, Clause S. Lindquist)


Addendum (Time domain):

For the given system function H(s) the corresponding step response function can be found to be

g(t)=1 - exp(Dwot)*{cos(wnt) + [D/SQRT(1-D²)]sin(wnt)}

with the natural frequency wn=wo*SQRT(1-D²).

and the first derivative at t=0 can be found to be

g′(t=0)=woD-Dwn/SQRT(1-D²)]

Replacing wn by wo we get (as expected):

g′(t=0)=0

2 added 106 characters in body
source | link

I thinkBecause a certain parameter in the time domain is wanted, the initial value theorem cannotmost direct method would be applied here because this theorem only gives the value ofto find the step responsefunction g(t) at t=0for the step repsonse and calculate the first derivative. You needThis is, however, a rather involved procedure.

Hence, it is more convenient not to leave the frequency domain and apply the "slopeinitial slope theorem" (first derivativewhich immediately gives the slope g′(0) of the step response g(tat t=0).:

Hence, as a first step we are requiredSimilar to find g(t) and calculate the first derivativeinitial value theorem (second stepfind (Hs).

The step reponse for a second-order system is:

s>>>infinity) the initial slope theorem requires to find the value g(t)=Aexp(sigma*t)sins*H(wots) for infinite frequencies.

with sigma=realThis calculation step is identical to the last part of the pole=1/2doriginal question leading to (d=damping factor)g′(0)=0.  

Now, it is not a problemComment: The initial slope theorem can be extended also to find the first derivative for ghigher derivatives (t) and set, finallyRef: Active Network Design, t=0Clause S. Lindquist)

I think, the initial value theorem cannot be applied here because this theorem only gives the value of the step response g(t) at t=0. You need, however, the slope (first derivative) of the step response g(t).

Hence, as a first step we are required to find g(t) and calculate the first derivative (second step).

The step reponse for a second-order system is:

g(t)=Aexp(sigma*t)sin(wot)

with sigma=real part of the pole=1/2d (d=damping factor).  

Now, it is not a problem to find the first derivative for g(t) and set, finally, t=0

Because a certain parameter in the time domain is wanted, the most direct method would be to find the function g(t) for the step repsonse and calculate the first derivative. This is, however, a rather involved procedure.

Hence, it is more convenient not to leave the frequency domain and apply the "initial slope theorem" (which immediately gives the slope g′(0) of the step response at t=0):

Similar to the initial value theorem (find (Hs) for s>>>infinity) the initial slope theorem requires to find the value s*H(s) for infinite frequencies.

This calculation step is identical to the last part of the original question leading to g′(0)=0.

Comment: The initial slope theorem can be extended also to higher derivatives (Ref: Active Network Design, Clause S. Lindquist)

    Post Undeleted by LvW
    Post Deleted by LvW
1
source | link

I think, the initial value theorem cannot be applied here because this theorem only gives the value of the step response g(t) at t=0. You need, however, the slope (first derivative) of the step response g(t).

Hence, as a first step we are required to find g(t) and calculate the first derivative (second step).

The step reponse for a second-order system is:

g(t)=Aexp(sigma*t)sin(wot)

with sigma=real part of the pole=1/2d (d=damping factor).

Now, it is not a problem to find the first derivative for g(t) and set, finally, t=0