SOLUTION:
The second condition which says that the closed loops should approximate the open loop response, implies that we should use \(H_{2}\left ( s\right ) =\frac{1}{G_{1}G_{2}}\), i.e. to apply the inversion. This is because \(\frac{Y\left ( s\right ) }{R\left ( s\right ) }=\left . \frac{H_{1}G_{1}G_{2}}{1+H_{1}G_{1}G_{2}H_{2}}\right \vert _{N=0}\) and this becomes \(\frac{Y\left ( s\right ) }{R\left ( s\right ) }\approx G_{1}G_{2}\) when we set \(H_{2}=\frac{1}{G_{1}G_{2}}\) and also by making \(H_{1}=\alpha \) where \(\alpha \) is a large gain. So now we just need to worry about finding \(\alpha \) s.t. \(\left \vert \frac{Y\left ( j\omega \right ) }{N\left ( j\omega \right ) }\right \vert \leq 0.01\) for all \(\omega >0\).
We know that \(\frac{Y\left ( s\right ) }{N\left ( s\right ) }=\frac{G_{2}}{1+G_{1}G_{2}H_{2}H_{1}}\), but since we are using the inversion, this reduces to\[ \frac{Y\left ( s\right ) }{N\left ( s\right ) }=\frac{G_{2}}{1+H_{1}}\] By setting \(H_{1}\left ( s\right ) =\alpha \) and using \(G_{2}=\frac{1}{s+2}\) and moving to the frequency domain, the above becomes\[ \frac{Y\left ( j\omega \right ) }{N\left ( j\omega \right ) }=\frac{\frac{1}{j\omega +2}}{1+\alpha }=\frac{1}{\left ( j\omega +2\right ) \left ( 1+\alpha \right ) }=\frac{1}{\left ( 1+\alpha \right ) j\omega +2\left ( 1+\alpha \right ) }\] Taking the magnitude\[ \left \vert \frac{Y\left ( j\omega \right ) }{N\left ( j\omega \right ) }\right \vert =\frac{1}{\sqrt{\left ( 1+\alpha \right ) ^{2}\omega ^{2}+4\left ( 1+\alpha \right ) ^{2}}}\] We want the above to be smaller than \(0.01\) for all \(\omega \), which implies\begin{align*} \frac{1}{\sqrt{\left ( 1+\alpha \right ) ^{2}\omega ^{2}+4\left ( 1+\alpha \right ) ^{2}}} & \leq 0.01\\ \frac{1}{\left ( 1+\alpha \right ) ^{2}\omega ^{2}+4\left ( 1+\alpha \right ) ^{2}} & \leq 0.01^{2}\\ \left ( 1+\alpha \right ) ^{2}\omega ^{2}+4\left ( 1+\alpha \right ) ^{2} & \geq 10000\\ \omega ^{2} & \geq \frac{10000-4\left ( 1+\alpha \right ) ^{2}}{\left ( 1+\alpha \right ) ^{2}}\\ \omega ^{2} & \geq \frac{10000}{\left ( 1+\alpha \right ) ^{2}}-4\\ \omega & \geq \sqrt{\frac{10000}{\left ( 1+\alpha \right ) ^{2}}-4} \end{align*}
The smallest \(\alpha \) to allow the above is when \(\omega =0\), hence we need to solve for \(\alpha \) from\begin{align*} \sqrt{\frac{10000}{\left ( 1+\alpha \right ) ^{2}}-4} & =0\\ \frac{10000}{\left ( 1+\alpha \right ) ^{2}}-4 & =0\\ \frac{1}{\left ( 1+\alpha \right ) ^{2}} & =\frac{4}{10000}\\ \left ( 1+\alpha \right ) ^{2} & =\frac{10000}{4}=2500\\ 1+\alpha & =50 \end{align*}
Hence\[ \alpha \geq 49 \] Therefore \(H_{1}\left ( s\right ) =\alpha \) where \(\alpha \geq 49\) and \(H_{1}\left ( s\right ) =\frac{1}{G_{1}\left ( s\right ) G_{2}\left ( s\right ) }\). This complete this part.
One problem with the above inversion method for finding \(H_{2}\left ( s\right ) =\frac{1}{G_{1}G_{2}}\) is that \(H_{2}\left ( s\right ) \) becomes improper:\begin{align*} H_{2}\left ( s\right ) & =\frac{1}{G_{1}G_{2}}=\frac{1}{\frac{s+1}{s^{2}+10s+100}\frac{1}{s+2}}\\ & =\frac{\left ( s^{2}+10s+100\right ) \left ( s+2\right ) }{s+1}\\ & =\frac{s^{3}+12s^{2}+120s+200}{s+1} \end{align*}
\(H_{2}\left ( s\right ) \) is improper, since the numerator has a degree larger than the denominator. This introduces differentiator in the feedback loop which is something we do not like to have.
We will now replace \(H_{2}=\frac{1}{G_{1}G_{2}}\) by \(\left ( \frac{1}{G_{1}G_{2}}\right ) H_{LP}\left ( s\right ) \) where \(H_{LP}\left ( s\right ) =\frac{1}{\left ( \varepsilon s+1\right ) ^{k}}\) is a low pass filter where \(k\) is an integer and \(\varepsilon \) is some parameter, both are positive. The goal is to block high frequency noise content and also make \(\left ( \frac{1}{G_{1}G_{2}}\right ) H_{LP}\left ( s\right ) \) become a proper transfer function. We also want to make sure \(\left \vert \frac{Y\left ( j\omega \right ) }{N\left ( j\omega \right ) }\right \vert \) remain less than \(0.01\).
Let \begin{align*} H_{2}\left ( s\right ) & =\frac{1}{G_{1}G_{2}}\frac{1}{\left ( \varepsilon s+1\right ) ^{k}}\\ & =\frac{s^{3}+12s^{2}+120s+200}{\left ( s+1\right ) }\frac{1}{\left ( \varepsilon s+1\right ) ^{k}} \end{align*}
The degree of the numerator is \(3\). So we want \(k\) to be at least \(2\) (it can be more), so that the denominator has at least degree \(3\) as well. If we want strict proper, then we make \(k=3\). Using \(k=2\) we now have\[ H_{2}\left ( s\right ) =\frac{1}{G_{1}G_{2}}\frac{1}{\left ( \varepsilon s+1\right ) ^{2}}\] Therefore, \(\frac{Y\left ( j\omega \right ) }{N\left ( j\omega \right ) }\) now becomes\begin{align} \frac{Y\left ( s\right ) }{N\left ( s\right ) } & =\frac{G_{2}}{1+G_{1}G_{2}H_{1}\left ( \frac{1}{G_{1}G_{2}}\frac{1}{\left ( \varepsilon s+1\right ) ^{2}}\right ) }\nonumber \\ & =\frac{\frac{1}{s+2}}{1+\frac{\alpha }{\left ( \varepsilon s+1\right ) ^{2}}}\tag{1} \end{align}
Where we used \(\alpha \) for \(H_{1}\). We now move to the frequency domain and take the magnitude in order to solve for \(\varepsilon \). We will use the same \(\alpha \) found in part (1), otherwise, there will be two free parameters to adjust at the same time, which would make this a hard problem, and the problem seems to indicate we are to use same \(\alpha \) value found in part (1) although it did not say that explicitly. Therefore (1) becomes (using \(\alpha =49\))\[ \frac{Y\left ( s\right ) }{N\left ( s\right ) }=\frac{\frac{\left ( \varepsilon s+1\right ) ^{2}}{s+2}}{\left ( \varepsilon s+1\right ) ^{2}+49}\] Hence\begin{align*} \left \vert \frac{Y\left ( j\omega \right ) }{N\left ( j\omega \right ) }\right \vert & =\frac{\left \vert \frac{\left ( \varepsilon j\omega +1\right ) ^{2}}{j\omega +2}\right \vert }{\left \vert \left ( \varepsilon j\omega +1\right ) ^{2}+49\right \vert }\\ & =\frac{\frac{\varepsilon ^{2}\omega ^{2}+1}{\sqrt{\omega ^{2}+4}}}{\left \vert -\varepsilon ^{2}\omega ^{2}+1+2\varepsilon j\omega +49\right \vert }\\ & =\frac{\frac{\varepsilon ^{2}\omega ^{2}+1}{\sqrt{\omega ^{2}+4}}}{\sqrt{4\varepsilon ^{2}\omega ^{2}+\left ( 50-\varepsilon ^{2}\omega ^{2}\right ) ^{2}}} \end{align*}
Hence\[ \left \vert \frac{Y\left ( j\omega \right ) }{N\left ( j\omega \right ) }\right \vert ^{2}=\frac{\left ( \varepsilon ^{2}\omega ^{2}+1\right ) ^{2}}{\left ( \omega ^{2}+4\right ) \left ( 4\varepsilon ^{2}\omega ^{2}+\left ( 50-\varepsilon ^{2}\omega ^{2}\right ) ^{2}\right ) }\] We now find \(\omega \) where \(\left \vert \frac{Y\left ( j\omega \right ) }{N\left ( j\omega \right ) }\right \vert \) is maximum, which is the same as where the above is maximum. The above is maximum when the denominator is minimum. Hence\[ \frac{d}{d\omega }\left ( \omega ^{2}+4\right ) \left ( 4\varepsilon ^{2}\omega ^{2}+\left ( 50-\varepsilon ^{2}\omega ^{2}\right ) ^{2}\right ) =0 \] Solving for \(\omega \) from the above using computer algebra (the algebra is too complicated to do by hand. May be there is a short cut) in terms of \(\varepsilon \), and plugging the solution \(\omega _{\max }\) back to \(\left \vert \frac{Y\left ( j\omega \right ) }{N\left ( j\omega \right ) }\right \vert \) and setting the result to \(0.01\) and solving numerically for \(\varepsilon \) that satisfy the equation gives\[ \varepsilon =0.0197 \] To verify this, a small demo was made to plot \(\left \vert \frac{Y\left ( j\omega \right ) }{N\left ( j\omega \right ) }\right \vert \) for different \(\varepsilon \) values. The following plot shows \(\left \vert \frac{Y\left ( j\omega \right ) }{N\left ( j\omega \right ) }\right \vert \) using \(k=2,\varepsilon =0.0197\) and the maximum magnitude was checked to be just less than \(0.01\)
We will now use\begin{align*} H_{1} & =49\\ H_{2} & =\frac{1}{G_{1}G_{2}}\frac{1}{\left ( 0.0197s+1\right ) ^{2}} \end{align*}
And plot \(\left \vert \frac{Y\left ( s\right ) }{R\left ( s\right ) }\right \vert =\left \vert \frac{H_{1}G_{1}G_{2}}{1+H_{1}G_{1}G_{2}H_{2}}\right \vert \) against the \(\left \vert G_{1}G_{2}\right \vert \) to see how good the choice of \(H_{1}\) and \(H_{2}\) are.\begin{align*} \frac{Y\left ( s\right ) }{R\left ( s\right ) } & =\frac{H_{1}G_{1}G_{2}}{1+H_{1}G_{1}G_{2}H_{2}}\\ & =\frac{49G_{1}G_{2}}{1+49G_{1}G_{2}\frac{1}{G_{1}G_{2}}\frac{1}{\left ( 0.0197s+1\right ) ^{2}}}\\ & =\frac{49\frac{s+1}{s^{2}+10s+100}\frac{1}{s+2}}{1+49\frac{1}{\left ( 0.0197s+1\right ) ^{2}}}\\ & =\frac{49\frac{s+1}{s^{2}+10s+100}\frac{1}{s+2}\left ( 0.0197s+1\right ) ^{2}}{\left ( 0.0197s+1\right ) ^{2}+49} \end{align*}
While \[ G_{1}G_{2}=\frac{s+1}{s^{2}+10s+100}\frac{1}{s+2}\] The following plot shows \(\left \vert \frac{Y\left ( s\right ) }{R\left ( s\right ) }\right \vert \) vs. \(\left \vert G_{1}G_{2}\right \vert \) side by side
The following plot shows both on the same plot
The following plot show difference between the magnitudes
Observations:
From the above difference plot, we see that the maximum difference between \(\left \vert G_{1}G_{2}\right \vert \) and the compensated \(\left \vert \frac{H_{1}G_{1}G_{2}}{1+H_{1}G_{1}G_{2}H_{2}}\right \vert \) occurred at around \(\omega =350\) and had value of about \(0.0014\). This value seems relatively small, and seems to indicate that \(H_{1}\) and \(H_{2}\) used for compensation were a good choice.
;
;
;
;