Hello Dear Friends,
How can it be solved that the phase - frequency graph of the bode diagram in Mathcad looks like the phase - frequency graph of the bode diagram in Matlab?
Calculations and graphs were made in Mathcad 15.
Attached is the Mathcad 15 file with the details.
Thank you.
Best regards.
I see in the magnitude - frequency graph of the bode diagram that the figure (straight line) intersects the 0db axis, and therefore there should be a certain frequency value for this.
At least at first glance, I don't see a problem with that.
@Cornel wrote:
I see in the magnitude - frequency graph of the bode diagram that the figure (straight line) intersects the 0db axis, and therefore there should be a certain frequency value for this.
Sure!
I also hope that you see that the frequency is about 1.6*10^-1 Hz!
So why are you looking for it in the range from 10 to 10^6
or in the range from 10^2 to 10^6
or in the range from 10^1 to 10^7
You should fully understand the meaning of the arguments of the functions you were given and also have a good understanding of how they work. Otherwise you may ask here if something is still unclear.
@Werner_E, @LucMeekes, @Fred_Kohlhepp
I don't understand why the following thing happens:
The functions I use for calculating Gain Margin (GM) and Phase Margin (PM) are:
The Plant Transfer Function is:
Case 1: fmax = 10e6 => GM_Tc = -22.627 db & f0_Phase_Plot_Tc = 2.471 kHz.
Case 2: fmax = 10e8 => GM_Tc = 138.668 db & f0_Phase_Plot_Tc = 52710 kHz.
Does anyone have any idea what is the reason for this happening in the case of GM?
Calculations and graphs were made in Mathcad 15.
Attached is the Mathcad 15 file with the details.
Thank you.
Best regards.
The phase of your Tc only dips across the 180° border in a very narrow frequency range, from 1894 Hz to 2795 Hz. The root function that tries to find the crossing needs two endpoints, one on each side of the crossing.
If you stretch your range too far, root will not find any crossing:
It requires a more versatile root finding function to do what you need.
Success!
Luc
@Werner_E, @LucMeekes, @Fred_Kohlhepp
I think there is something wrong with these two functions, how they are built, but I do not know what, at least so "foreshadowed" and I deduce from my more detailed checks by the last resort.
Let's take the following example:
The Plant Transfer Function is:
The axis cutting frequency of 0 decibels on the magnitude-frequency plot gives well, as it results from the function.
The phase-frequency plot does not even cut the axis at -180 degrees so it cannot define a cut-off frequency for this graph, but the function returns a frequency value (this means that the graph cuts the axis at -180 degrees, which is not true).
Reading the value in degrees corresponding to the cutoff frequency read from the magnitude - frequency graph (approximately 13.614kHz) results in a value of -56,259 degrees, which means that the phase margin (PM) is 180 degrees - 56,258 degrees = 123,741 degrees, and the result of the function is -103,666 degrees, and 76,313 degrees, respectively, which is a discrepancy.
Hmm, something strange is happening, because it does not shoot the values read from the graph with the values calculated by functions ...
I was so happy :)) because it seemed to work great 🙂
What do you think is missing or wrong?
Thank you.
Best regards.
I have a mistake regarding the previous comment:
The Phase Margin (PM) is well calculated with the corresponding function used.
In the previous comment, from the graph, I read at the frequency of approximately 1.3 kHz, but it must be at the frequency of approximately 13 kHz (as much as the cutting frequency at 0 decibels is read from the magnitude-frequency characteristic).
But the problem remains in calculating the Gain Margin (GM) with the corresponding function used
This is because the phase-frequency graph does not cut the -180 degree line, which should normally lead to an infinite Gain Margin (GM) (or at least a large value).
As a check, I put the transfer function corresponding to Gc (s) in Matlab and drew the corresponding bode diagrams.
Mathcad:
Matlab:
Therefore, the function that I suspect would not have good results is:
Any suggestions/ideas/opinions?
Thank you.
@Werner_E, @LucMeekes, @Fred_Kohlhepp, @ttokoro, @-MFra-
Ah, yes, the graphics are fine. That was not the idea, at least not in the last posts.
The graphs were discussed at the beginning of this topic, and, in principle, it was solved.
The ideas from the last posts and comments were in the context in which I could evaluate the two characteristics in certain points of them, in the sense of finding through certain functions that solve for the Phase Margin (PM) and the Gain Margin (GM) for different transfer functions in an automatic (or at least semi-automatic) way and not reading fully manually on the graphs.
Which sounds good, that is, okay with what it looks like on the graphs.
A few more things appear, such as:
1) If the phase-frequency graph does not cut the -180 degree axis then GM should return the value of infinity.
2) If the phase-frequency graph cuts the -180 degree axis by more than one point (2.3, etc., as shown below (*)), then a single point should be extracted or the frequency search range should be narrowed only for a single cut of the graph with the axis of -180 degrees, to get a correct answer.
But it's okay because they keep you awake and attentive to what you're doing, and then they make you not forget how to calculate the Phase Margin (PM) and the Gain Margin (GM) from the graphs when you do not have a computer :).
((*)
For example, Matlab takes into account, at least as far as I can see until now, the last point that cuts the -180 degree axis.
For the above completeness in Mathcad:
)
but as @LucMeekes says:
Even so, it works well.
I'm pleased.
I am satisfied with the suggestions and solutions that came from the other colleagues and that were referred to in this topic. Also, as I said above, the results are good, from the latest research.
Thank you all.
Best regards.
