Thanks, it is an exercise in a test a month ago. There is only the sketch. I remember that we use the approximation:
Something like this (not exact!!!) I can't remember it now.
H(s)=A/(s*(s+a)(s_b)(s+c)(s+d)) ≈ A/(s*(s+p)^4)
Just curious, how do you know that the system has five poles? Do you use the Bode plot as you said above?
Hello again,
Well, that makes the problem a whole heck of a lot simpler now
H(s)=1/(s*(s+p)^4)
and you can play with the scale factor A if you wish to change 'a'.
If we can use the quad pole as above we can rap this up in a jiffy by noting a few points similar to what Steve was mentioning, and it does not require too much work.
First, note that the imaginary part is zero for three different values of w: w(inf), w1, and w0. That means we should try to solve for that first.
Now because we have an 's' as a lone factor in the denominator and no 's' in the numerator, we right away satisfy w(inf). So we have that already.
Next, we solve for H(jw) and find the imaginary part:
H(jw)=H(s) with s=jw, then
IP(w)=ImagPart(H(jw))
Now we set that equal to zero:
IP(w)=0
and solve for all the w. The solutions we want are:
w0=(sqrt(2)-1)*p, and
w1=(sqrt(2)+1)*p
We know w1>w0>0 so they are easy to pick out of all the possible solutions.
Now to find a, we solve for the real part:
RP(w)=RealPart(H(jw))
and take the limit as w goes to zero:
limit [w-->0] RP(w)
and we get:
RP(0)=-4/p^5
so a=4/p^5.
Next we can check the response at w=0 to make sure it is infinite, and at w=inf to make sure it is zero, and it is.
To change 'a' independently of w0 and w1, calculate w0 and w1 as above, then include A and make 'a' such that:
a=4*A/p^5
however A will have to be constrained in order to maintain a graph that looks much like that in the picture.
Note: A is the gain in the numerator which replaces that '1' as shown in your quote.
So all you need to do really is find Hjw and the rest is pretty easy. If you have any problems with this just yell