Block Diagram Reduction (Solved Problem 4)

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Control Systems: Block Diagram Reduction (Solved Problem 4)
Topics discussed:
1. Solved Problem based on Elimination of Disturbance (given into the system) in the final output.
2. Overview of Multiple input Single output (MISO) systems.
3. Use of Superposition theorem in order to analyze a MISO system.

Music:
Axol x Alex Skrindo - You [NCS Release]

#ControlSystemByNeso #ControlSystems #BlockDiagrams

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Another great video as always! In case anyone is interested, I thought of a slightly different way to solve it:

As he did in the video, we can combine the two cascaded blocks to get 10/(s(s+10)) for an equivalent block in the middle. Then we can move the signal D*Gd from where it is now (the first summation point) to a point after the cascaded blocks. Doing so would make the signal D*Gd*10/(s(s+10)). But then notice that this signal is being subtracted while the initial signal D is being added. So if we want the effect of the disturbance to be 0, then the sum of these two signals should be zero. In other words, D - D*Gd*10/(s(s+10)) = 0, leading to 1 = 10*Gd/(s(s+10)) and Gd = s(s+10)/10.

PunmasterSTP

I watch your electrical videos . very good content thanks for it.
Please make a q&a video .
Who is created this channel?
Who create videos?
Who are you?
Like this

lordputinrasiaWale

There is another method which is less time consuming, firstly find the following equation:

After that differentiate the equation implicitly to find dC and dD,
dC/dD = [S(S+10)-10Gd]/[S(S+10)+10] = 0,
which means when D(S) changes it will cause no change in C(S), i.e. rate of change is zero, it happens only when the numerator equals zero which yields,
S(S+10)-10Gd = 0, solving for Gd, will give, Gd = S(S+10)/10.

mohamedaitelcaid

its a humble request to complete this subject soon as i have been waiting from last year to get this subject compeleted.

lecturesonelectricalengine

Sir can u solve this question in 5 seconds u dont then how can ee solve this in 5 seconds

toxicgaming

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