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I'm given the following transfer function that represents a physical apparatus

[itex]G(s) = \frac{X(s)}{F(s)} = \frac{3.333*10^{4}}{s^{2}}[/itex]

I'm asked to design a lead compensator to achieve closed-loop stability. The requirements are that it must have a transient response of [itex]16%[/itex] overshoot, a settling time of [itex]2 mS[/itex] for a step input.

[itex]ζ = -\frac{ln(\frac{OS}{100})}{\sqrt{∏^{2} + ln^{2}(\frac{OS}{100})}}[/itex]

[itex]ω_{BW} = \frac{4}{T_{s}ζ}\sqrt{(1-2ζ) + \sqrt{4ζ^{4} - 4ζ^{2} + 2}}[/itex]

[itex]Θ_{m} = arctan(\frac{2ζ}{\sqrt{-2ζ^{2} + \sqrt{1 + 4ζ^{4}}}})[/itex]

I've spent some time trying to solve this problem. I've looked through my notes, my textbook, the internet and still am lost. I've looked at the solutions manual and it doesn't seem to help either. So I was hoping that someone could point me in the right direction.

I start off by finding the damping ratio since

[itex]ζ = -\frac{ln(\frac{OS}{100})}{\sqrt{∏^{2} + ln^{2}(\frac{OS}{100})}} = -\frac{ln(\frac{16}{100})}{\sqrt{∏^{2}+ln^{2}(\frac{16}{100})}} ≈ 0.503[/itex]

I know find the bandwidth frequency [itex]ω_{BW}[/itex]

[itex]ω_{BW} = \frac{4}{T_{s}ζ}\sqrt{(1-2ζ) + \sqrt{4ζ^{4} - 4ζ^{2} + 2}} = \frac{4(10^{3} mS)}{(2 mS)(0.503) S}\sqrt{(1-2(0.503)^{2}) + \sqrt{4(0.503)^{4} - 4(0.503)^{2} +2}} ≈ 5,044.122 \frac{1}{S}[/itex]

Next I find the gain evaluated at [itex]jω_{BW}[/itex] [itex]G(jω_{BW})[/itex]

[itex]G(jω_{BW}) = \frac{3.333*10^{4}}{(j5,044.122 \frac{1}{S}} ≈ -0.131*10^{-2}[/itex]

Next I find the phase margin [itex]Θ_{m}[/itex]

[itex]Θ_{m} = arctan(\frac{2ζ}{\sqrt{-2ζ^{2} + \sqrt{1 + 4ζ^{4}}}}) = arctan(\frac{2(0.503)}{\sqrt{-2(0.503)^{2} + \sqrt{1 + 4(0.503)^{4}}}}) ≈ 52.068°[/itex]

I'm not really sure were to go from here but all of the above work seems right and are all formulas that I can find in my book. The solutions manual does this next

[itex]0.013K = \frac{1}{\sqrt{2}}[/itex]

and solves for [itex]K[/itex]. I'm not sure where this formula comes from. The solutions manual says it's the compensators gain. But that doesn't really help me. The equation is clearly

[itex]G(jω_{BW})K = ?[/itex]

I'm not really sure what the right side of the equation is.

Thanks for any help.

## Homework Statement

I'm given the following transfer function that represents a physical apparatus

[itex]G(s) = \frac{X(s)}{F(s)} = \frac{3.333*10^{4}}{s^{2}}[/itex]

I'm asked to design a lead compensator to achieve closed-loop stability. The requirements are that it must have a transient response of [itex]16%[/itex] overshoot, a settling time of [itex]2 mS[/itex] for a step input.

## Homework Equations

[itex]ζ = -\frac{ln(\frac{OS}{100})}{\sqrt{∏^{2} + ln^{2}(\frac{OS}{100})}}[/itex]

[itex]ω_{BW} = \frac{4}{T_{s}ζ}\sqrt{(1-2ζ) + \sqrt{4ζ^{4} - 4ζ^{2} + 2}}[/itex]

[itex]Θ_{m} = arctan(\frac{2ζ}{\sqrt{-2ζ^{2} + \sqrt{1 + 4ζ^{4}}}})[/itex]

## The Attempt at a Solution

I've spent some time trying to solve this problem. I've looked through my notes, my textbook, the internet and still am lost. I've looked at the solutions manual and it doesn't seem to help either. So I was hoping that someone could point me in the right direction.

I start off by finding the damping ratio since

[itex]ζ = -\frac{ln(\frac{OS}{100})}{\sqrt{∏^{2} + ln^{2}(\frac{OS}{100})}} = -\frac{ln(\frac{16}{100})}{\sqrt{∏^{2}+ln^{2}(\frac{16}{100})}} ≈ 0.503[/itex]

I know find the bandwidth frequency [itex]ω_{BW}[/itex]

[itex]ω_{BW} = \frac{4}{T_{s}ζ}\sqrt{(1-2ζ) + \sqrt{4ζ^{4} - 4ζ^{2} + 2}} = \frac{4(10^{3} mS)}{(2 mS)(0.503) S}\sqrt{(1-2(0.503)^{2}) + \sqrt{4(0.503)^{4} - 4(0.503)^{2} +2}} ≈ 5,044.122 \frac{1}{S}[/itex]

Next I find the gain evaluated at [itex]jω_{BW}[/itex] [itex]G(jω_{BW})[/itex]

[itex]G(jω_{BW}) = \frac{3.333*10^{4}}{(j5,044.122 \frac{1}{S}} ≈ -0.131*10^{-2}[/itex]

Next I find the phase margin [itex]Θ_{m}[/itex]

[itex]Θ_{m} = arctan(\frac{2ζ}{\sqrt{-2ζ^{2} + \sqrt{1 + 4ζ^{4}}}}) = arctan(\frac{2(0.503)}{\sqrt{-2(0.503)^{2} + \sqrt{1 + 4(0.503)^{4}}}}) ≈ 52.068°[/itex]

I'm not really sure were to go from here but all of the above work seems right and are all formulas that I can find in my book. The solutions manual does this next

[itex]0.013K = \frac{1}{\sqrt{2}}[/itex]

and solves for [itex]K[/itex]. I'm not sure where this formula comes from. The solutions manual says it's the compensators gain. But that doesn't really help me. The equation is clearly

[itex]G(jω_{BW})K = ?[/itex]

I'm not really sure what the right side of the equation is.

Thanks for any help.

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