SOLUTION:
In all of these systems, the feedback block diagram is configured as follows
Since we are looking at steady state, we need to obtain the transfer function between \(E\left ( s\right ) \) and \(R\left ( s\right ) \). Given that \(E\left ( s\right ) =R\left ( s\right ) -Y\left ( s\right ) \) and \(Y\left ( s\right ) =E\left ( s\right ) G\left ( s\right ) H\left ( s\right ) \) then we solve these two equations for \(E\left ( s\right ) \) by eliminating \(Y\left ( s\right ) \) giving\begin{align*} E\left ( s\right ) & =R\left ( s\right ) -E\left ( s\right ) G\left ( s\right ) H\left ( s\right ) \\ E\left ( s\right ) \left ( 1+G\left ( s\right ) H\left ( s\right ) \right ) & =R\left ( s\right ) \\ \frac{E\left ( s\right ) }{R\left ( s\right ) } & =\frac{1}{1+G\left ( s\right ) H\left ( s\right ) } \end{align*}
The above is is the transfer function used for the different \(R\left ( s\right ) \) signals: unit step \(u\left ( t\right ) \), ramp \(t\), and parabolic \(t^{2}\).
The open loop transfer function is \(\frac{1000}{\left ( 1+0.1s\right ) \left ( 1+10s\right ) }\). Since the number of poles at zero is zero, the system type1 is zero.
When the input is a unit step \(u\left ( t\right ) \), then \(R\left ( s\right ) =\frac{1}{s}.\) Using the steady state error transfer function found above gives\begin{align*} E\left ( s\right ) & =\frac{R\left ( s\right ) }{1+G\left ( s\right ) H\left ( s\right ) }\\ & =\frac{1}{s}\frac{1}{1+\frac{1000}{\left ( 1+0.1s\right ) \left ( 1+10s\right ) }}=\frac{1}{s}\frac{\left ( 1+\frac{1}{10}s\right ) \left ( 1+10s\right ) }{\left ( 1+\frac{1}{10}s\right ) \left ( 1+10s\right ) +1000}=\frac{1}{s}\frac{\left ( 1+\frac{1}{10}s\right ) \left ( 1+10s\right ) }{s^{2}+\frac{101}{10}s+1001} \end{align*}
We see that the poles are located at \(s=0,s=-5.05\pm 31.233i\). Therefore this is stable \(E\left ( s\right ) \) as the real parts of the poles are negative. We are allowed one pole at the origin. Applying the final value theorem gives\begin{align*} e_{ss} & =\lim _{t\rightarrow \infty }e\left ( t\right ) =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =\lim _{s\rightarrow 0}\frac{1}{1+\frac{1000}{\left ( 1+0.1s\right ) \left ( 1+10s\right ) }}\\ & =\frac{1}{1+\frac{1000}{\lim _{s\rightarrow 0}\left ( 1+0.1s\right ) \left ( 1+10s\right ) }}\\ & =\frac{1}{1+1000} \end{align*}
Hence \[ e_{ss}=\frac{1}{1001}\] When the input is a ramp, then \(R\left ( s\right ) =\frac{1}{s^{2}},\) therefore\begin{align*} E\left ( s\right ) & =\frac{R\left ( s\right ) }{1+G\left ( s\right ) H\left ( s\right ) }\\ & =\frac{1}{s^{2}}\frac{1}{1+\frac{1000}{\left ( 1+0.1s\right ) \left ( 1+10s\right ) }} \end{align*}
There are two poles at the origin and the other two poles are the same as above at \(s=-5.05\pm 31.233i\). Since there are two poles at the origin, the final value is not defined (taken from now on as infinity in order to be compatible with the text book result and notation).
Finally, when the input is \(t^{2}\), then \(R\left ( s\right ) =\frac{2}{s^{3}}\) and\begin{align*} E\left ( s\right ) & =\frac{2}{s^{3}}\frac{1}{1+G\left ( s\right ) H\left ( s\right ) }\\ & =\frac{2}{s^{3}}\frac{1}{1+\frac{1000}{\left ( 1+0.1s\right ) \left ( 1+10s\right ) }}\\ & =\frac{2\left ( 1+0.1s\right ) \left ( 1+10s\right ) }{s^{3}\left ( 1+0.1s\right ) \left ( 1+10s\right ) +1000s^{3}}\\ & =\frac{2\left ( 1+0.1s\right ) \left ( 1+10s\right ) }{s^{3}\left ( s^{2}+10.1s+1001\right ) } \end{align*}
There are now three poles at the origin \(s=0.\) As above, this means the final value is taken as infinity.
The open loop transfer function is \(\frac{1000}{s\left ( 1+10\right ) \left ( s+100\right ) }\). There is one pole at the origin which means the system type is one.
When the input is a unit step, then \(R\left ( s\right ) =\frac{1}{s}\) and\begin{align*} E\left ( s\right ) & =\frac{1}{s}\frac{R\left ( s\right ) }{1+G\left ( s\right ) H\left ( s\right ) }\\ & =\frac{1}{s}\frac{1}{1+\frac{1000}{s\left ( 1+10\right ) \left ( s+100\right ) }}=\frac{1}{s+\frac{1000}{\left ( 1+10\right ) \left ( s+100\right ) }}=\frac{\left ( 1+10\right ) \left ( s+100\right ) }{s\left ( 1+10\right ) \left ( s+100\right ) +1000}=\frac{\left ( 1+10\right ) \left ( s+100\right ) }{11s^{2}+1100s+1000} \end{align*}
The poles are at \(s=-0.9175,s=-99.08\). This is stable \(E\left ( s\right ) \) and we can now apply the final value theorem\begin{align*} e_{ss} & =\lim _{t\rightarrow \infty }e\left ( t\right ) =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =\lim _{s\rightarrow 0}s\frac{\frac{1}{s}}{1+\frac{1000}{s\left ( 1+10\right ) \left ( s+100\right ) }}\\ & =\frac{1}{1+\lim _{s\rightarrow 0}\frac{1000}{s\left ( 1+10\right ) \left ( s+100\right ) }} \end{align*}
Hence \[ e_{ss}=0 \] When the input is a ramp, then \(R\left ( s\right ) =\frac{1}{s^{2}}\) and\begin{align*} E\left ( s\right ) & =\frac{R\left ( s\right ) }{1+G\left ( s\right ) H\left ( s\right ) }\\ & =\frac{1}{s^{2}}\frac{1}{1+\frac{1000}{s\left ( 1+10\right ) \left ( s+100\right ) }}\\ & =\frac{1}{s^{2}+\frac{1000s}{\left ( 1+10\right ) \left ( s+100\right ) }}\\ & =\frac{\left ( 1+10\right ) \left ( s+100\right ) }{s^{2}\left ( 1+10\right ) \left ( s+100\right ) +1000s}\\ & =\frac{\left ( 1+10\right ) \left ( s+100\right ) }{s\left ( 11s^{2}+1100s+1000\right ) } \end{align*}
There is one pole at the origin \(s=0\) and the other two poles are the same at \(s=-0.9175,s=-99.08\). This is stable. Applying the final value theorem gives\begin{align*} e_{ss} & =\lim _{t\rightarrow \infty }e\left ( t\right ) =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =\lim _{s\rightarrow 0}\frac{1}{s}\frac{1}{1+\frac{1000}{s\left ( 1+10\right ) \left ( s+100\right ) }}\\ & =\lim _{s\rightarrow 0}\frac{1}{s+\frac{1000}{\left ( 1+10\right ) \left ( s+100\right ) }}\\ & =\lim _{s\rightarrow 0}\frac{\left ( 1+10\right ) \left ( s+100\right ) }{s\left ( 1+10\right ) \left ( s+100\right ) +1000}\\ & =\frac{\left ( 1+10\right ) \left ( 100\right ) }{1000} \end{align*}
Hence \[ e_{ss}=1.1 \] When the input is a \(t^{2}\), then \(R\left ( s\right ) =\frac{2}{s^{3}}\) and\begin{align*} E\left ( s\right ) & =\frac{R\left ( s\right ) }{1+G\left ( s\right ) H\left ( s\right ) }\\ & =\frac{2}{s^{3}}\frac{1}{1+\frac{1000}{s\left ( 1+10\right ) \left ( s+100\right ) }}\\ & =\frac{2}{s^{2}+\frac{1000s^{2}}{\left ( 1+10\right ) \left ( s+100\right ) }}\\ & =\frac{\left ( 2\right ) \left ( 11\right ) \left ( s+100\right ) }{s^{2}\left ( 11s+2100\right ) } \end{align*}
There are now two poles at the origin. Therefore final value is taken as infinity.
The open loop transfer function is \(\frac{K\left ( 1+2s\right ) \left ( 1+4s\right ) }{s^{2}\left ( s^{2}+s+1\right ) }\). There are two poles at the origin which means the system type is 2.
When the input is a unit step, then \(R\left ( s\right ) =\frac{1}{s}\) and\begin{align*} E\left ( s\right ) & =\frac{R\left ( s\right ) }{1+G\left ( s\right ) H\left ( s\right ) }\\ & =\frac{1}{s}\frac{1}{1+\frac{K\left ( 1+2s\right ) \left ( 1+4s\right ) }{s^{2}\left ( s^{2}+s+1\right ) }}\\ & =\frac{1}{s+\frac{K\left ( 1+2s\right ) \left ( 1+4s\right ) }{s\left ( s^{2}+s+1\right ) }}\\ & =\frac{s\left ( s^{2}+s+1\right ) }{s^{2}\left ( s^{2}+s+1\right ) +K\left ( 1+2s\right ) \left ( 1+4s\right ) } \end{align*}
We have to now assume that \(E\left ( s\right ) \) is stable to be able to apply the final value theorem as this depends on the value of \(k\) which is not given in the problem. Therefore\begin{align*} e_{ss} & =\lim _{t\rightarrow \infty }e\left ( t\right ) =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =\lim _{s\rightarrow 0}s\left ( \frac{1}{s}\frac{1}{1+\frac{K\left ( 1+2s\right ) \left ( 1+4s\right ) }{s^{2}\left ( s^{2}+s+1\right ) }}\right ) \\ & =\lim _{s\rightarrow 0}\frac{s^{2}\left ( s^{2}+s+1\right ) }{s^{2}\left ( s^{2}+s+1\right ) +K\left ( 1+2s\right ) \left ( 1+4s\right ) }\\ & =\frac{0}{K} \end{align*}
Which means \[ e_{ss}=0 \] When the input is a ramp, then \(R\left ( s\right ) =\frac{1}{s^{2}}.\) Applying the final value theorem gives\begin{align*} e_{ss} & =\lim _{t\rightarrow \infty }e\left ( t\right ) =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =\lim _{s\rightarrow 0}s\left ( \frac{1}{s^{2}}\frac{1}{1+\frac{K\left ( 1+2s\right ) \left ( 1+4s\right ) }{s^{2}\left ( s^{2}+s+1\right ) }}\right ) \\ & =\lim _{s\rightarrow 0}s\left ( \frac{1}{s^{2}+\frac{K\left ( 1+2s\right ) \left ( 1+4s\right ) }{\left ( s^{2}+s+1\right ) }}\right ) \\ & =\lim _{s\rightarrow 0}s\left ( \frac{\left ( s^{2}+s+1\right ) }{s^{2}\left ( s^{2}+s+1\right ) +K\left ( 1+2s\right ) \left ( 1+4s\right ) }\right ) \\ & =\lim _{s\rightarrow 0}s\left ( \frac{1}{K}\right ) \end{align*}
Hence \[ e_{ss}=0 \] When the input is \(t^{2}\) then \(R\left ( s\right ) =\frac{2}{s^{3}}\) and\begin{align*} e_{ss} & =\lim _{t\rightarrow \infty }e\left ( t\right ) =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =\lim _{s\rightarrow 0}s\left ( \frac{2}{s^{3}}\frac{1}{1+\frac{K\left ( 1+2s\right ) \left ( 1+4s\right ) }{s^{2}\left ( s^{2}+s+1\right ) }}\right ) \\ & =\lim _{s\rightarrow 0}s\left ( \frac{2}{s}\frac{\left ( s^{2}+s+1\right ) }{s^{2}\left ( s^{2}+s+1\right ) +K\left ( 1+2s\right ) \left ( 1+4s\right ) }\right ) \\ & =\lim _{s\rightarrow 0}\left ( 2\frac{1}{k}\right ) \end{align*}
Hence \[ e_{ss}=\frac{2}{K}\]
Summary of results from the above parts is
| \(G\left ( s\right ) H\left ( s\right ) \) | system type (number of poles at origin) | \(e_{ss}\) step | \(e_{ss}\) ramp | \(e_{ss}\) \(t^{2}\) | |
| part(a) | \(\frac{1000}{\left ( 1+0.1s\right ) \left ( 1+10s\right ) }\) | \(0\) | \(\frac{1}{1001}\) | \(\infty \) | \(\infty \) |
| part(b) | \(\frac{1000}{s\left ( 1+10\right ) \left ( s+100\right ) }\) | \(1\) | \(0\) | \(\frac{11}{10}\) | \(\infty \) |
| part(c) | \(\frac{K\left ( 1+2s\right ) \left ( 1+4s\right ) }{s^{2}\left ( s^{2}+s+1\right ) }\) | \(2\) | \(0\) | \(0\) | \(\frac{2}{K}\) |
From the above table, we see that as the system type (number of poles at origin of the open loop \(G\left ( s\right ) H\left ( s\right ) \)) increases, then the system can handle more signal types while still producing zero steady state error (this is good). The input signal that gives constant (non zero) steady state error per system type is summarized below.
| System type (open loop number of poles at origin) | Input that gives constant nonzero \(e_{ss}\) |
| \(0\) | step \(\left ( t^{0}\right ) \) |
| \(1\) | ramp \(\left ( t^{1}\right ) \) |
| \(2\) | parabolic \(\left ( t^{2}\right ) \) |
So the relation between number of poles at origin of open loop and the type of signal that gives constant non zero steady state error can be written as
if the system type is \(m\) then nonzero constant \(e_{ss}\) is generated by signal \(t^{m}\).
SOLUTION:
Let the first input \(R\left ( s\right ) \) be \(U_{1}\left ( s\right ) \) and the second input (the constant \(0.25\)) be \(U_{2}\left ( s\right ) \), then\[ Y\left ( s\right ) =\left [ \left ( 2+\frac{1}{s}\right ) E\left ( s\right ) +U_{2}\left ( s\right ) \right ] G\left ( s\right ) \] And \[ E\left ( s\right ) =U_{1}\left ( s\right ) -Y\left ( s\right ) \] Hence\begin{align} E\left ( s\right ) & =U_{1}\left ( s\right ) -\left [ \left ( 2+\frac{1}{s}\right ) E\left ( s\right ) +U_{2}\left ( s\right ) \right ] G\left ( s\right ) \nonumber \\ & =U_{1}\left ( s\right ) -\left ( 2+\frac{1}{s}\right ) E\left ( s\right ) G\left ( s\right ) -U_{2}\left ( s\right ) G\left ( s\right ) \nonumber \\ E\left ( s\right ) \left ( 1+\left ( 2+\frac{1}{s}\right ) G\left ( s\right ) \right ) & =U_{1}\left ( s\right ) -U_{2}\left ( s\right ) G\left ( s\right ) \nonumber \\ E\left ( s\right ) & =\frac{U_{1}\left ( s\right ) -U_{2}\left ( s\right ) G\left ( s\right ) }{1+\left ( 2+\frac{1}{s}\right ) G\left ( s\right ) }\tag{1} \end{align}
To obtain the error transfer function from \(E\left ( s\right ) \) to \(U_{1}\left ( s\right ) \), the input \(U_{2}\left ( s\right ) \) is set to zero. To obtain the error transfer function from \(E\left ( s\right ) \) to \(U_{2}\left ( s\right ) \), the input \(U_{1}\left ( s\right ) \) is set to zero. Applying these to (1) gives\begin{align*} \left . \frac{E\left ( s\right ) }{U_{1}\left ( s\right ) }\right \vert _{U_{2}=0} & =\frac{1}{1+\left ( 2+\frac{1}{s}\right ) G\left ( s\right ) }\\ \left . \frac{E\left ( s\right ) }{U_{2}\left ( s\right ) }\right \vert _{U_{1}=0} & =\frac{-G\left ( s\right ) }{1+\left ( 2+\frac{1}{s}\right ) G\left ( s\right ) } \end{align*}
In Matrix form, \[ E\left ( s\right ) =\begin{pmatrix} \frac{1}{1+\left ( 2+\frac{1}{s}\right ) G\left ( s\right ) } & \frac{-G\left ( s\right ) }{1+\left ( 2+\frac{1}{s}\right ) G\left ( s\right ) }\end{pmatrix}\begin{pmatrix} U_{1}\left ( s\right ) \\ U_{2}\left ( s\right ) \end{pmatrix} \] But \(U_{1}\left ( s\right ) =\frac{1.5}{s^{2}}\) and \(U_{2}\left ( s\right ) =\frac{0.25}{s}\), and the above becomes\[ E\left ( s\right ) =\begin{pmatrix} \frac{1}{1+\left ( 2+\frac{1}{s}\right ) \frac{100}{\left ( 0.05s+1\right ) ^{2}}} & \frac{-\frac{100}{\left ( 0.05s+1\right ) ^{2}}}{1+\left ( 2+\frac{1}{s}\right ) \frac{100}{\left ( 0.05s+1\right ) ^{2}}}\end{pmatrix}\begin{pmatrix} \frac{1.5}{s^{2}}\\ \frac{0.25}{s}\end{pmatrix} \] Hence \[ E(s) = \overbrace{ \frac{1.5}{s} \frac{\left ( 0.05s+1\right ) ^{2}}{0.0025s^{3}+0.1s^{2}+201s+100.0}}^{E_{1}\left ( s\right )} -\overbrace{\frac{10000}{s^{3}+40s^{2}+80400s+40000}}^{E_{2}\left ( s\right )} \] The poles of the first term are \(-19.751\pm 282.83i\) ,\(s=-0.497,s=0\), Hence this is stable and have at most one pole at origin. Then using F.V.T. gives\begin{align*} e_{ss_{1}} & =\lim _{s\rightarrow 0}sE_{1}\left ( s\right ) \\ & =\lim _{s\rightarrow 0}1.5\frac{\left ( 0.05s+1\right ) ^{2}}{0.0025s^{3}+0.1s^{2}+201s+100}\\ & =0.015\, \end{align*}
For \(E_{2}\left ( s\right ) \), the poles are at , \(s=-19.75\pm 282.83i,s=-0.498,\) Hence this is stable. Therefore using F.V.T. gives\begin{align*} e_{ss_{2}} & =\lim _{s\rightarrow 0}sE_{2}\left ( s\right ) \\ & =\lim _{s\rightarrow 0}\frac{10000s}{s^{3}+40s^{2}+80400s+40000}\\ & =0 \end{align*}
Hence\begin{align*} e_{ss} & =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =0.015-0\\ & =0.015 \end{align*}
SOLUTION:
Let \(G\left ( s\right ) =\frac{N\left ( s\right ) }{D\left ( s\right ) }\). The closed loop transfer function is \[ \frac{Y\left ( s\right ) }{R\left ( s\right ) }=\frac{G\left ( s\right ) }{1+G\left ( s\right ) }=\frac{\frac{N\left ( s\right ) }{D\left ( s\right ) }}{1+\frac{N\left ( s\right ) }{D\left ( s\right ) }}=\frac{N\left ( s\right ) }{D\left ( s\right ) +N\left ( s\right ) }\] We are given that \(D\left ( s\right ) +N\left ( s\right ) =s^{3}+4s^{2}+6s+4\). The error transfer function is \[ \frac{E\left ( s\right ) }{R\left ( s\right ) }=\frac{1}{1+G\left ( s\right ) }=\frac{1}{1+\frac{N\left ( s\right ) }{D\left ( s\right ) }}=\frac{D\left ( s\right ) }{D\left ( s\right ) +N\left ( s\right ) }\] Substituting for \(D\left ( s\right ) +N\left ( s\right ) \) in the above with the given polynomial results in\begin{equation} \frac{E\left ( s\right ) }{R\left ( s\right ) }=\frac{D\left ( s\right ) }{s^{3}+4s^{2}+6s+4}\tag{1} \end{equation} We are told that \(\lim _{s\rightarrow 0}sE\left ( s\right ) =0\) when \(R\left ( s\right ) =\frac{1}{s}\). Applying this to (1) gives\[ E\left ( s\right ) =\frac{1}{s}\frac{D\left ( s\right ) }{s^{3}+4s^{2}+6s+4}\] \(E\left ( s\right ) \) above is stable since the poles are at \(-2,-1\pm i\) with another pole at zero. Hence F.V.T. can be applied to \(E\left ( s\right ) \) \begin{align*} e_{ss} & =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =\lim _{s\rightarrow 0}s\left ( \frac{1}{s}\frac{D\left ( s\right ) }{s^{3}+4s^{2}+6s+4}\right ) \\ & =\lim _{s\rightarrow 0}\frac{D\left ( s\right ) }{4} \end{align*}
We are also told that the above is zero. Hence\[ 0=\lim _{s\rightarrow 0}\frac{D\left ( s\right ) }{4}\] The above implies that \(D\left ( s\right ) \) must contain only \(s\) terms and no constant terms, since we want \(D\left ( s\right ) =0\) when \(s=0\).
Assuming proper transfer function \(G\left ( s\right ) \) where degree of \(N\left ( s\right ) \leq \) degree of \(N\left ( s\right ) \), then \(D\left ( s\right ) \) can be \(s^{3}\) or \(s^{3}+4s^{2}\), or \(s^{3}+4s^{2}+6s,\) since any of these will give \(\lim _{s\rightarrow 0}\frac{D\left ( s\right ) }{4}=0\). But \(D\left ( s\right ) \) can not be \(s^{2}\) for example, else \(G\left ( s\right ) \) will not proper \(G\left ( s\right ) .\)
There are actually an infinite number of \(D\left ( s\right ) \) polynomials which meets this condition (if we use fractions for the coefficients). Below is an example of two possible \(D\left ( s\right ) \) choices and the corresponding \(G\left ( s\right ) \)
\[ D_{1}\left ( s\right ) =s^{3}+4s^{2}+6s \] Then\[ \fbox{$G\left ( s\right ) =\frac{4}{s^3+4s^2+6s}$}\] For steady state when input is ramp, using the above \(G\left ( s\right ) \) gives\begin{align*} e_{ss} & =\lim _{s\rightarrow 0}s\left ( \frac{1}{s^{2}}\frac{s^{3}+4s^{2}+6s}{s^{3}+4s^{2}+6s+4}\right ) \\ & =\lim _{s\rightarrow 0}\frac{s^{2}+4s+6}{s^{3}+4s^{2}+6s+4} \end{align*}
Hence\[ e_{ss}=1.5 \] Another choice is \(D_{2}\left ( s\right ) =s^{3}+4s^{2}.\) Using this, \(G\left ( s\right ) =\frac{6s+4}{s^{3}+4s^{2}}.\) Using this, and when the input is ramp then \begin{align*} e_{ss} & =\lim _{s\rightarrow 0}s\left ( \frac{1}{s^{2}}\frac{s^{3}+4s^{2}}{s^{3}+4s^{2}+6s+4}\right ) \\ & =\lim _{s\rightarrow 0}\frac{s^{2}+4s}{s^{3}+4s^{2}+6s+4}\\ & =0 \end{align*}
So the steady state error for ramp depends on which \(G\left ( s\right ) \) is used.
SOLUTION:
For system 1. Using \(K_{0}=0\) we first obtain expression for \(E\left ( s\right ) \) and \(Y\left ( s\right ) \)\begin{align*} E\left ( s\right ) & =R\left ( s\right ) -Y\left ( s\right ) \\ Y\left ( s\right ) & =E\left ( s\right ) \left ( \frac{K_{1}}{s}\right ) \left ( \frac{1}{4s+1}\right ) \end{align*}
Solving for \(E\left ( s\right ) \) from the above two equations gives\begin{align*} E\left ( s\right ) & =R\left ( s\right ) -\left ( E\left ( s\right ) \frac{K_{1}}{s\left ( 4s+1\right ) }\right ) \\ E\left ( s\right ) \left ( 1+\frac{K_{1}}{s\left ( 4s+1\right ) }\right ) & =R\left ( s\right ) \\ E\left ( s\right ) & =R\left ( s\right ) \frac{s\left ( 4s+1\right ) }{s\left ( 4s+1\right ) +K_{1}} \end{align*}
When \(R\left ( s\right ) =\frac{1}{s}\) we want \(e_{ss}=0\), therefore\begin{align*} e_{ss} & =0=\lim _{s\rightarrow 0}sE\left ( s\right ) \\ 0 & =\lim _{s\rightarrow 0}\frac{s\left ( 4s+1\right ) }{s\left ( 4s+1\right ) +K_{1}}\\ & =\lim _{s\rightarrow 0}\frac{s\left ( 4s+1\right ) }{K_{1}}\\ & =\frac{0}{K_{1}} \end{align*}
The above is true for any \(K_{1}\) since the numerator is already zero. Considering now the ramp input. When \(R\left ( s\right ) =\frac{1}{s^{2}}\) we want \(e_{ss}=1\), hence\begin{align*} e_{ss} & =1=\lim _{s\rightarrow 0}sE\left ( s\right ) \\ 1 & =\lim _{s\rightarrow 0}s\frac{1}{s^{2}}\frac{s\left ( 4s+1\right ) }{s\left ( 4s+1\right ) +K_{1}}\\ & =\lim _{s\rightarrow 0}\frac{4s+1}{s\left ( 4s+1\right ) +K_{1}}\\ & =\frac{1}{K_{1}} \end{align*}
Therefore \[ \fbox{$K_1=1$}\] For system 2
\begin{equation} E\left ( s\right ) =R\left ( s\right ) -Y\left ( s\right ) \tag{1} \end{equation} But \begin{equation} Y\left ( s\right ) =U\left ( s\right ) K_{3}\frac{1}{4s+1} \tag{2} \end{equation} And\[ U\left ( s\right ) =R\left ( s\right ) K_{2}-Y\left ( s\right ) \] Hence (2) becomes\begin{align} Y\left ( s\right ) & =\left ( R\left ( s\right ) K_{2}-Y\left ( s\right ) \right ) \frac{K_{3}}{4s+1}\nonumber \\ Y\left ( s\right ) \left ( 1+\frac{K_{3}}{4s+1}\right ) & =R\left ( s\right ) \frac{K_{2}K_{3}}{4s+1}\nonumber \\ Y\left ( s\right ) & =R\left ( s\right ) \frac{\frac{K_{2}K_{3}}{4s+1}}{1+\frac{K_{3}}{4s+1}}\nonumber \\ & =R\left ( s\right ) \frac{K_{2}K_{3}}{4s+1+K_{3}} \tag{3} \end{align}
Substituting (3) into (1) gives\begin{align} E\left ( s\right ) & =R\left ( s\right ) -R\left ( s\right ) \frac{K_{2}K_{3}}{4s+1+K_{3}}\nonumber \\ E\left ( s\right ) & =R\left ( s\right ) \left ( 1-\frac{K_{2}K_{3}}{4s+1+K_{3}}\right ) \tag{4} \end{align}
When \(R\left ( s\right ) =\frac{1}{s}\) we want \(e_{ss}=0\), hence\begin{align*} e_{ss} & =0=\lim _{s\rightarrow 0}sE\left ( s\right ) \\ 0 & =\lim _{s\rightarrow 0}s\frac{1}{s}\left ( 1-\frac{K_{2}K_{3}}{4s+1+K_{3}}\right ) \\ & =1-\frac{K_{2}K_{3}}{1+K_{3}} \end{align*}
For the above to be true, then\begin{equation} \fbox{$\frac{K_2K_3}{1+K_3}=1$} \tag{5} \end{equation} We now obtain a second equation from the ramp condition. When \(R\left ( s\right ) =\frac{1}{s^{2}}\) we want \(e_{ss}=1\), hence\begin{align*} e_{ss} & =1=\lim _{s\rightarrow 0}sE\left ( s\right ) \\ 1 & =\lim _{s\rightarrow 0}s\frac{1}{s^{2}}\left ( 1-\frac{K_{2}K_{3}}{4s+1+K_{3}}\right ) \\ & =\lim _{s\rightarrow 0}\frac{1}{s}\left ( 1-\frac{K_{2}K_{3}}{4s+1+K_{3}}\right ) \\ & =\lim _{s\rightarrow 0}\frac{1}{s}\left ( \frac{4s+1+K_{3}-\left ( K_{2}K_{3}\right ) }{4s+1+K_{3}}\right ) \end{align*}
Replacing \(K_{2}K_{3}\) in the above with \(1+K_{3}\) found in (5) gives\begin{align*} 1 & =\lim _{s\rightarrow 0}\frac{1}{s}\left ( \frac{4s+1+K_{3}-\left ( 1+k_{3}\right ) }{4s+1+K_{3}}\right ) \\ & =\lim _{s\rightarrow 0}\frac{1}{s}\left ( \frac{4s}{4s+1+K_{3}}\right ) \\ & =\frac{4}{\lim _{s\rightarrow 0}\left ( 4s+1+K_{3}\right ) }\\ 1 & =\frac{4}{1+K_{3}} \end{align*}
Hence \(1+K_{3}=4\) or \[ \fbox{$K_3=3$}\] Now that we found \(K_{3}\) we go back to (5) and solve for \(K_{2}\)\begin{align*} \frac{K_{2}K_{3}}{1+K_{3}} & =1\\ K_{2} & =\frac{1+K_{3}}{K_{3}}\\ & =\frac{1+3}{3} \end{align*}
Hence \[ \fbox{$K_2=\frac{4}{3}$}\] Summary
| \(K_{1}\) | \(K_{2}\) | \(K_{3}\) | |
| system 1 | \(1\) | N/A | N/A |
| system 2 | N/A | \(\frac{4}{3}\) | \(3\) |
For system 1.
\begin{align*} E\left ( s\right ) & =R\left ( s\right ) -Y\left ( s\right ) \\ Y\left ( s\right ) & =E\left ( s\right ) \frac{K_{1}}{s}\frac{\left ( 1+\delta \right ) }{4s+1} \end{align*}
Hence\begin{align*} E\left ( s\right ) & =R\left ( s\right ) -\left ( E\left ( s\right ) \frac{K_{1}}{s}\frac{1+\delta }{4s+1}\right ) \\ E\left ( s\right ) \left ( 1+\frac{K_{1}}{s}\frac{1+\delta }{4s+1}\right ) & =R\left ( s\right ) \\ E\left ( s\right ) & =R\left ( s\right ) \frac{1}{1+\frac{K_{1}}{s}\frac{1+\delta }{4s+1}}\\ & =R\left ( s\right ) \frac{s\left ( 4s+1\right ) }{s\left ( 4s+1\right ) +K_{1}\left ( 1+\delta \right ) } \end{align*}
When \(R\left ( s\right ) =\frac{1}{s}\) then\begin{align*} e_{ss} & =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =\lim _{s\rightarrow 0}\frac{s\left ( 4s+1\right ) }{s\left ( 4s+1\right ) +K_{1}\left ( 1+\delta \right ) }\\ & =\frac{\lim _{s\rightarrow 0}s\left ( 4s+1\right ) }{K_{1}\left ( 1+\delta \right ) }\\ & =\frac{0}{K_{1}\left ( 1+\delta \right ) } \end{align*}
The above is zero for any \(K_{1}\) and any perturbation \(\delta \) since the numerator is already zero. This is the same condition we found in part(a). Perturbing \(K_{0}\) has no effect on the result of \(e_{ss}\) for step input.
For system 2
\begin{equation} E\left ( s\right ) =R\left ( s\right ) -Y\left ( s\right ) \tag{1} \end{equation} But \begin{equation} Y\left ( s\right ) =U\left ( s\right ) K_{3}\frac{1+\delta }{4s+1} \tag{2} \end{equation} And\[ U\left ( s\right ) =R\left ( s\right ) K_{2}-Y\left ( s\right ) \] Replacing \(U\left ( s\right ) \) into (2)\begin{align} Y\left ( s\right ) & =\left ( R\left ( s\right ) K_{2}-Y\left ( s\right ) \right ) K_{3}\frac{1+\delta }{4s+1}\nonumber \\ Y\left ( s\right ) \left ( 1+\frac{K_{3}\left ( 1+\delta \right ) }{4s+1}\right ) & =R\left ( s\right ) \frac{K_{2}K_{3}\left ( 1+\delta \right ) }{4s+1}\nonumber \\ Y\left ( s\right ) & =R\left ( s\right ) \frac{\frac{K_{2}K_{3}\left ( 1+\delta \right ) }{4s+1}}{1+\frac{K_{3}\left ( 1+\delta \right ) }{4s+1}}\nonumber \\ & =R\left ( s\right ) \frac{K_{2}K_{3}\left ( 1+\delta \right ) }{4s+1+K_{3}\left ( 1+\delta \right ) } \tag{3} \end{align}
Substituting \(Y\left ( s\right ) \) from (3) into (1) gives\begin{align} E\left ( s\right ) & =R\left ( s\right ) -R\left ( s\right ) \frac{K_{2}K_{3}\left ( 1+\delta \right ) }{4s+1+K_{3}\left ( 1+\delta \right ) }\nonumber \\ & =R\left ( s\right ) \left ( 1-\frac{K_{2}K_{3}\left ( 1+\delta \right ) }{4s+1+K_{3}\left ( 1+\delta \right ) }\right ) \tag{4} \end{align}
When \(R\left ( s\right ) =\frac{1}{s}\)and using the F.V.T. gives \begin{align*} e_{ss} & =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =\lim _{s\rightarrow 0}s\frac{1}{s}\left ( 1-\frac{K_{2}K_{3}\left ( 1+\delta \right ) }{4s+1+K_{3}\left ( 1+\delta \right ) }\right ) \\ & =\lim _{s\rightarrow 0}\left ( 1-\frac{K_{2}K_{3}\left ( 1+\delta \right ) }{4s+1+K_{3}\left ( 1+\delta \right ) }\right ) \\ & =1-\frac{K_{2}K_{3}\left ( 1+\delta \right ) }{1+K_{3}\left ( 1+\delta \right ) } \end{align*}
For the above \(e_{ss}\) to be zero, then the condition is that \[ \frac{K_{2}K_{3}\left ( 1+\delta \right ) }{1+K_{3}\left ( 1+\delta \right ) }=1 \] or\begin{align*} K_{2}K_{3}\left ( 1+\delta \right ) & =1+K_{3}\left ( 1+\delta \right ) \\ K_{2}K_{3}-K_{3} & =\frac{1}{1+\delta } \end{align*}
Using \(K_{2}=\frac{4}{3}\) and \(K_{3}=3\) found in part (a) then the above becomes\begin{align*} \left ( \frac{4}{3}\right ) 3-3 & =\frac{1}{1+\delta }\\ 1 & =\frac{1}{1+\delta } \end{align*}
But this is impossible since the RHS must be either larger than one or smaller than one (depending on the sign of \(\delta \)). This means if \(K_{0}\) is perturbed from unity, then it is no longer possible to obtain zero steady state error for a step input with the same \(k_{2},k_{3}\).
I agree. For first system, it gives \(e_{ss}=0\) for a step input regardless of the value of \(K_{1}\) or \(K_{0}\) as was shown in part (b) above. But for system two, \(e_{ss}=0\) for step input only when using specific values of \(K_{i}\). Any small change in \(K_{0}\), the steady state error is no longer zero. In other words, system one is more robust in this regard to changes in \(K_{0}\) and it is therefore the preferred system.
SOLUTION:
We first need to find \(\frac{E\left ( s\right ) }{V_{c}\left ( s\right ) }\). From the block diagram2 . \begin{equation} V_{b}\left ( s\right ) =E\left ( s\right ) K_{a}\left ( \frac{K_{m}}{1+s\tau _{m}}\right ) \tag{1} \end{equation} And\begin{equation} E\left ( s\right ) =V_{c}\left ( s\right ) \left ( \frac{K_{g}}{1+s\tau _{g}}\right ) -K_{t}V_{b}\left ( s\right ) \tag{2} \end{equation} Replacing\(\ V_{b}\left ( s\right ) \) in (2) with (1) gives\begin{align*} E\left ( s\right ) & =V_{c}\left ( s\right ) \left ( \frac{K_{g}}{1+s\tau _{g}}\right ) -K_{t}E\left ( s\right ) K_{a}\left ( \frac{K_{m}}{1+s\tau _{m}}\right ) \\ E\left ( s\right ) \left ( 1+K_{t}K_{a}\left ( \frac{K_{m}}{1+s\tau _{m}}\right ) \right ) & =V_{c}\left ( s\right ) \left ( \frac{K_{g}}{1+s\tau _{g}}\right ) \\ \frac{E\left ( s\right ) }{V_{c}\left ( s\right ) } & =\frac{\left ( \frac{K_{g}}{1+s\tau _{g}}\right ) }{1+K_{t}K_{a}\left ( \frac{K_{m}}{1+s\tau _{m}}\right ) } \end{align*}
Hence\[ \fbox{$\frac{E\left ( s\right ) }{V_c\left ( s\right ) }=\left ( \frac{1+s\tau _m}{1+s\tau _g}\right ) \frac{K_g}{1+s\tau _m+K_tK_aK_m}$}\] When \(V_{c}\left ( s\right ) =\frac{25}{s},K_{g}=K_{t}=1\) then \(E\left ( s\right ) \) from above becomes\[ E\left ( s\right ) =\frac{25}{s}\left ( \frac{1+s\tau _{m}}{1+s\tau _{g}}\right ) \frac{1}{\left ( 1+K_{a}K_{m}\right ) +s\tau _{m}}\] The above \(E\left ( s\right ) \) has one pole at the origin, and has a pole at \(s=\frac{-1}{\tau _{g}}\) and a pole at \(s=-\frac{1+K_{a}K_{m}}{\tau _{m}}\). Hence this is stable (assuming \(K_{a}K_{m}>-1\)). Applying F.V.T. gives\begin{align*} e_{ss} & =\lim _{s\rightarrow 0}sE\left ( s\right ) \\ & =\lim _{s\rightarrow 0}25\left ( \frac{1+s\tau _{m}}{1+s\tau _{g}}\right ) \frac{1}{\left ( 1+K_{a}K_{m}\right ) +s\tau _{m}} \end{align*}
Hence\[ e_{ss}=\frac{25}{1+K_{a}K_{m}}\]
When \(e_{ss}\) is one degree per second, then from the above\[ 1=\frac{25}{1+K_{a}K_{m}}\] Or\[ \fbox{$K_aK_m=24$}\]
To find the step response, we find the closed loop \(\frac{V_{b}\left ( s\right ) }{V_{c}\left ( s\right ) }\) transfer function first. Substituting (2) into (1) found in part (a) above to obtain an expression for \(V_{b}\left ( s\right ) \)\begin{align*} V_{b}\left ( s\right ) & =\left ( V_{c}\left ( s\right ) \left ( \frac{K_{g}}{1+s\tau _{g}}\right ) -K_{t}V_{b}\left ( s\right ) \right ) K_{a}\left ( \frac{K_{m}}{1+s\tau _{m}}\right ) \\ V_{b}\left ( s\right ) \left ( 1+\frac{K_{m}K_{t}K_{a}}{1+s\tau _{m}}\right ) & =V_{c}\left ( s\right ) \frac{K_{g}K_{a}K_{m}}{\left ( 1+s\tau _{g}\right ) \left ( 1+s\tau _{m}\right ) } \end{align*}
Hence the closed loop transfer function is\begin{align*} \frac{V_{b}\left ( s\right ) }{V_{c}\left ( s\right ) } & =\frac{\frac{K_{g}K_{a}K_{m}}{\left ( 1+s\tau _{g}\right ) \left ( 1+s\tau _{m}\right ) }}{1+\frac{K_{m}K_{t}K_{a}}{1+s\tau _{m}}}\\ & =\frac{1}{\left ( 1+s\tau _{g}\right ) }\frac{K_{g}K_{a}K_{m}}{\left ( 1+s\tau _{m}\right ) +K_{m}K_{t}K_{a}} \end{align*}
Using same assumptions as part (a), and now using that \(\tau _{g}\) is negligible so that \(\frac{1}{\left ( 1+s\tau _{g}\right ) }\approx 1\) in the above, and using \(V_{c}\left ( s\right ) =\frac{1}{s}\) since we are told in this part it is a step input (should we have used \(\frac{25}{s}\) again here? It is not clear, but it says step input so I think \(\frac{1}{s}\) should be used in this part), then the above simplifies to\begin{align*} V_{b}\left ( s\right ) & =\left ( \frac{1}{s}\right ) \frac{K_{a}K_{m}}{\left ( 1+s\tau _{m}\right ) +K_{m}K_{a}}\\ & =\frac{1}{s}\frac{K_{a}K_{m}}{\left ( 1+K_{m}K_{a}\right ) +s\tau _{m}}\\ & =\frac{K_{a}K_{m}}{\tau _{m}}\frac{1}{s}\frac{1}{s+\left ( \frac{1+K_{m}K_{a}}{\tau _{m}}\right ) } \end{align*}
We now need to find the inverse Laplace transform. Using partial fractions\begin{equation} \left ( \frac{K_{a}K_{m}}{\tau _{m}}\right ) \frac{1}{s}\frac{1}{s+\left ( \frac{1+K_{m}K_{a}}{\tau _{m}}\right ) }=\frac{A}{s}+\frac{B}{s+\left ( \frac{1+K_{m}K_{a}}{\tau _{m}}\right ) } \tag{3} \end{equation} Hence\[ A=\lim _{s\rightarrow 0}\left ( \frac{K_{a}K_{m}}{\tau _{m}}\right ) \frac{1}{\frac{1+K_{m}K_{a}}{\tau _{m}}+s}=\frac{K_{a}K_{m}}{1+K_{m}K_{a}}\] And\[ B=\lim _{s\rightarrow \frac{-\left ( 1+K_{m}K_{a}\right ) }{\tau _{m}}}\frac{K_{a}K_{m}}{\tau _{m}}\frac{1}{s}=\frac{K_{a}K_{m}}{\tau _{m}}\frac{1}{\frac{-\left ( 1+K_{m}K_{a}\right ) }{\tau _{m}}}=-\frac{K_{a}K_{m}}{1+K_{m}K_{a}}\] Now that we found \(A,B\) using partial fractions, we replace these values in (3) to obtain \(V_{b}\left ( s\right ) \)\begin{equation} V_{b}\left ( s\right ) =\frac{K_{a}K_{m}}{1+K_{m}K_{a}}\frac{1}{s}-\frac{K_{a}K_{m}}{1+K_{m}K_{a}}\frac{1}{\frac{1+K_{m}K_{a}}{\tau _{m}}+s} \tag{4} \end{equation} Now we can apply inverse Laplace transform. Hence\begin{align*} v_{b}\left ( t\right ) & =\left ( \frac{K_{a}K_{m}}{1+K_{m}K_{a}}-\frac{K_{a}K_{m}}{1+K_{m}K_{a}}e^{-\frac{1+K_{m}K_{a}}{\tau _{m}}t}\right ) u\left ( t\right ) \\ & =\frac{K_{a}K_{m}}{1+K_{m}K_{a}}\left ( 1-e^{-\frac{1+K_{m}K_{a}}{\tau _{m}}t}\right ) u\left ( t\right ) \end{align*}
Let \[ \fbox{$q=\frac{1+K_mK_a}{\tau _m}$}\] and \[ \fbox{$k=\frac{K_aK_m}{\tau _m}$}\] Then \(v_{b}\left ( t\right ) \) can be written as required\[ v_{b}\left ( t\right ) =\frac{k}{q}\left ( 1-e^{-qt}\right ) u\left ( t\right ) \]
We first need to find the steady state \(v_{b}\left ( t\right ) \). From (4) found above in part (c)\[ V_{b}\left ( s\right ) =\frac{K_{a}K_{m}}{1+K_{m}K_{a}}\frac{1}{s}-\frac{K_{a}K_{m}}{1+K_{m}K_{a}}\frac{1}{\frac{1+K_{m}K_{a}}{\tau _{m}}+s}\] Then applying F.V.T. assuming stability \begin{align*} V_{b}\left ( \infty \right ) & =\lim _{s\rightarrow 0}sV_{b}\left ( s\right ) \\ & = \lim _{s\rightarrow 0}s\left ( \frac{K_{a}K_{m}}{1+K_{m}K_{a}}\frac{1}{s}-\frac{K_{a}K_{m}}{1+K_{m}K_{a}}\frac{1}{\frac{1+K_{m}K_{a}}{\tau _{m}}+s}\right ) \\ & =\frac{K_{a}K_{m}}{1+K_{m}K_{a}}\\ & =\frac{k}{q} \end{align*}
Let the settling time be \(t_{s}\), then we want to solve for \(t_{s}\) from\begin{align*} v_{b}\left ( t_{s}\right ) & =0.98V_{b}\left ( \infty \right ) \\ \frac{k}{q}\left ( 1-e^{-qt_{s}}\right ) & =0.98\frac{k}{q}\\ 1-e^{-qt_{s}} & =0.98\\ e^{-qt_{s}} & =0.02 \end{align*}
Taking natural logs on both sides gives\begin{align*} -qt_{s} & =\ln \left ( 0.02\right ) \\ qt_{s} & =3.912 \end{align*}
Hence\[ \fbox{$t_s=\frac{3.912}{\left ( \frac{1+K_mK_a}{t_m}\right ) }$}\] Using \(t_{m}=0.4\) seconds in the above gives\[ t_{s}=\frac{1.5648}{1+K_{m}K_{a}}\] For \(t_{s}\leq 0.04\) then\begin{align*} \frac{1.5648}{\left ( 1+K_{m}K_{a}\right ) } & \leq 0.04\\ 1+K_{m}K_{a} & \geq \frac{1.5648}{0.04}\\ 1+K_{m}K_{a} & \geq 39.12\, \end{align*}
Hence\[ \fbox{$K_mK_a\geq 38.12$}\, \]
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