Transfer Function From State Space Representation MCQ Quiz - Objective Question with Answer for Transfer Function From State Space Representation - Download Free PDF

Concept:

A transfer function (TF) is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output] / L[input]

$TF=\frac{C\left(s\right)}{R\left(s\right)}$

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, transfer function is also known as impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Calculation:

Given differential equation is,

$3{\displaystyle \frac{dy}{dt}}+2y=u$

Laplace transform of the above equation is given by

${\displaystyle \frac{dy}{dt}}=sY(s)$, y(t) = Y(s), u(t) = U(s)

⇒ 3s Y(s) + 2 Y(s) = U(s)

⇒ Y(s) (3s + 2) = U(s)

∴ Transfer function is given by

$\Rightarrow TF=\frac{Y\left(s\right)}{U\left(s\right)}=\frac{1}{3s+2}$

Calculation:-

$A=\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]$

(sI – A)^-1 can be given as:

$=\frac{1}{s^2+4}\left[\begin{array}{cc}s& 1\\ -4& s\end{array}\right]$

ϕ (t) = L^-1 {(sI – A)^-1}

$=\left[\begin{array}{cc}\cos 2t& \frac{1}{2}\sin 2t\\ -2\sin 2t& \cos 2t\end{array}\right]$

x(t) = ϕ(t) x(0)

$=\left[\begin{array}{cc}\cos 2t& \frac{1}{2}\sin 2t\\ -2\sin 2t& \cos 2t\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]$

$=\left[\begin{array}{c}\cos 2t+0.5\sin 2t\\ -2\sin 2t+\cos 2t\end{array}\right]$

y = x₁ – x₂

= cos 2t + 0.5 sin 2t + 2 sin 2t – cos 2t

= 2.5 sin 2t

Peak value = 2.5

$A=\left[\begin{array}{cc}-2& -4\\ 1& 0\end{array}\right]$

$sI-A=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}-2& -4\\ 1& 0\end{array}\right]$

${\left(sI-A\right)}^{-1}=\frac{1}{s(s+2)+4}\left[\begin{array}{cc}s+2& 4\\ -1& s\end{array}\right]$

The characteristics equation is:

s(s + 2) + 4 = 0

s² + 2s + 4 = 0

Comparing with: 

s² + 2ζ ω_n + ω_n² = 0

ω_n = 2 rad/s

ζ = 0.5 

Transfer function = C (sI - A)-1 B + D

In the given question,

$A=\left[\begin{array}{cc}0& -1\\ 1& -2\end{array}\right],B=\left[\begin{array}{c}1\\ 2\end{array}\right]$

$C=\left[\begin{array}{cc}1& 1\end{array}\right],D=0$

$sI-A=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}0& -1\\ 1& -2\end{array}\right]$

$sI-A=\left[\begin{array}{cc}s& 1\\ -1& s+2\end{array}\right]$

${\left(sI-A\right)}^{-1}=\frac{1}{s(s+2)+1}\left[\begin{array}{cc}s+2& -1\\ 1& s\end{array}\right]$

The characteristics equation is:

s² + 2s + 1 = 0

(s + 1)² = 0

The roots are -1, -1

Concept:

Transfer function from the state model:

In general the state model is defined as:

ẋ(t) = A x(t) + B u(t)       ---------   (1)

y(t) = C x(t) + D u(t)       ---------   (2)

Taking Laplace transform of equation (1) with initial condition zero, we get:

sX(s) = A X(s) + B U(s)

X(s) [sI - A] = B U(s)

⇒ X(s) = (sI - A)^{-1 .}B U(s)       --------    (3)

Taking the Laplace transform of equation (2), we get:

Y(s) = C X(s) + D U(s)

Substituting value of X(s) from equation (3), we get:

Y(s) = C (sI - A)^-1 B U(s) + D U(s)

$T.F=\frac{Y\left(s\right)}{U\left(s\right)}=C\left[{\left(sI-A\right)}^{-1}\right]B+D$

Application:

In the question no disturbance matrix is given

So, D = 0.

Now, the output transfer function will be:

$T\left(s\right)=\frac{Y\left(s\right)}{U\left(s\right)}=C\left[{\left(sI-A\right)}^{-1}\right]B$      -------   (4)

Given that,

$A=\left[\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right],B=\left[\begin{array}{c}1\\ 0\end{array}\right]$ and C = [1 1]

For the given system, we have:

$\left(sI-A\right)=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}-1& 0\\ 0& -2\end{array}\right]$

$\Rightarrow (sI-A)=\left[\begin{array}{cc}s+1& 0\\ 0& s+2\end{array}\right]$

$\Rightarrow {\left(sI-A\right)}^{-1}=\frac{1}{\left(s+1\right)\left(s+2\right)}\left[\begin{array}{cc}s+2& 0\\ 0& s+1\end{array}\right]$

Substituting it in equation (4), we get:

$\frac{Y\left(s\right)}{U\left(s\right)}=\left[11\right]\left[\begin{array}{cc}\frac{1}{s+1}& 0\\ 0& \frac{1}{s+2}\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]$

$=\left[\begin{array}{cc}\frac{1}{s+1}& \frac{1}{s+2}\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]$

$\Rightarrow \frac{Y\left(s\right)}{U\left(s\right)}=\frac{1}{s+1}$

Concept:

A transfer function (TF) is defined as the ratio of Laplace transform of the output to the Laplace transform of the input by assuming initial conditions are zero.

TF = L[output] / L[input]

$TF=\frac{C\left(s\right)}{R\left(s\right)}$

For unit impulse input i.e. r(t) = δ(t)

⇒ R(s) = δ(s) = 1

Now transfer function = C(s)

Therefore, transfer function is also known as impulse response of the system.

Transfer function = L[IR]

IR = L-1 [TF]

Calculation:

Given differential equation is,

$3{\displaystyle \frac{dy}{dt}}+2y=u$

Laplace transform of the above equation is given by

${\displaystyle \frac{dy}{dt}}=sY(s)$, y(t) = Y(s), u(t) = U(s)

⇒ 3s Y(s) + 2 Y(s) = U(s)

⇒ Y(s) (3s + 2) = U(s)

∴ Transfer function is given by

$\Rightarrow TF=\frac{Y\left(s\right)}{U\left(s\right)}=\frac{1}{3s+2}$

$A=\left[\begin{array}{cc}-2& -4\\ 1& 0\end{array}\right]$

$sI-A=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}-2& -4\\ 1& 0\end{array}\right]$

${\left(sI-A\right)}^{-1}=\frac{1}{s(s+2)+4}\left[\begin{array}{cc}s+2& 4\\ -1& s\end{array}\right]$

The characteristics equation is:

s(s + 2) + 4 = 0

s² + 2s + 4 = 0

Comparing with: 

s² + 2ζ ω_n + ω_n² = 0

ω_n = 2 rad/s

ζ = 0.5 

From the given state model

$\begin{array}{ccc}A=\left[\begin{array}{cc}0& 1\\ -\alpha & -2\beta \end{array}\right]& B=\left[\begin{array}{c}0\\ \alpha \end{array}\right]& C=\left[\begin{array}{cc}1& 0\end{array}\right]\end{array}$

Transfer function = C [sI - A]^-1 B + D

$\left[SI-A\right]=S\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}0& 1\\ -\alpha & -2\beta \end{array}\right]$

$=\left[\begin{array}{cc}s& -1\\ \alpha & s+2\beta \end{array}\right]$

$={\left[sI-A\right]}^{-1}=\frac{1}{s\left(s+2\beta \right)\alpha }\left[\begin{array}{cc}s+2\beta & 1\\ -\alpha & s\end{array}\right]$

$\frac{T}{F}=\frac{\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{cc}s+2\beta & 1\\ -\alpha & s\end{array}\right]\left[\begin{array}{c}0\\ \alpha \end{array}\right]}{\left(s\left(s+2\beta \right)+\alpha \right)}$

$=\frac{1}{\left(s^2+2\beta s+\alpha \right)}\left[\begin{array}{cc}1& 0\end{array}\right]\left[\begin{array}{c}\alpha \\ s\alpha \end{array}\right]$

$=\frac{\alpha }{\left(s^2+2\beta s+\alpha \right)}$

By comparing the above transfer function with the standard second order transfer function $\frac{{\omega }_n^2}{s^2+2\xi {\omega }_{n}s}$

${\omega }_n=\sqrt{\alpha }$

And 2ξω_n = 2

$\Rightarrow \xi =\frac{\beta }{\sqrt{\alpha }}$

X^’ = Ax + Bu → (1)

Y = Cx + Du → (2)

Apply Laplace transform, to equation (1)

S x(s) = A x(s) B u(s)

⇒ B u(s) = (SI - A) X(s)

⇒ X(s) = (SI - A)^-1 B u(s)

Now apply Laplace transform to equation (2)

Y(s) = C x(s) + D u(s)

⇒ Y(s) = C[(SI - A)^-1] B u(s) + D u(s)

$\frac{d{x}_1}{dt}=2x_1+x_2+1,\frac{d{x}_2}{dt}=-2x_1+1$

$y=3x_1$

Considering the standard equation

$\begin{array}{l}{x}_i=Ax+BU,y=Cx+DU\\ \left[\begin{array}{c}{\dot{x}}_1\\ {\dot{x}}_2\end{array}\right]=\left[\begin{array}{rr}2& 1\\ -2& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]+\left[\begin{array}{c}1\\ 1\end{array}\right]\left[1\right]\\ y=\left[\begin{array}{cc}3& 0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\end{array}\right]\end{array}$

Transform function $C{\left(SI-A\right)}^{-1}B$

$G\left(s\right)=\left[\begin{array}{cc}3& 0\end{array}\right]{\left[\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}2& 1\\ -2& 0\end{array}\right]\right]}^{-1}\left[\begin{array}{c}1\\ 1\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}3& 0\end{array}\right]{\left[\begin{array}{cc}s-2& -1\\ 2& s\end{array}\right]}^{-1}$

$\Rightarrow \frac{1}{s^2-s+2}\left[\begin{array}{cc}3& 0\end{array}\right]\left[\begin{array}{c}s+1\\ s-4\end{array}\right]$

$\Rightarrow \frac{3\left(s+1\right)}{s^2-2s+2}$

From the given state space equations,

$A=\left[\begin{array}{cc}1& 2\\ 2& 0\end{array}\right]$

$B=\left[\begin{array}{c}1\\ 2\end{array}\right]$

$C=\left[\begin{array}{cc}1& 0\end{array}\right]$

Transfer function = C[sI - A]^-1 B + D

$sI-A=\left[\begin{array}{cc}s& 0\\ 0& s\end{array}\right]-\left[\begin{array}{cc}1& 2\\ 2& 0\end{array}\right]=\left[\begin{array}{cc}s-1& -2\\ -2& s\end{array}\right]$

${\left(sI-A\right)}^{-1}=\frac{1}{s^2-s-4}\left[\begin{array}{cc}s& 2\\ 2& s-1\end{array}\right]$

Transfer function $=\left[\begin{array}{cc}1& 0\end{array}\right]\frac{1}{\left(s^2-s-4\right)}\left[\begin{array}{cc}s& 2\\ 2& s-1\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]$

$=\frac{1}{\left(s^2-s-4\right)}\left[\begin{array}{cc}s& 2\end{array}\right]\left[\begin{array}{c}1\\ 2\end{array}\right]$

ẋ = Ax + Bu

y = Cx + Du

Given that, D ≠ 0

y = Cx + Du

The transfer function is calculated as:

$\mathrm{T}\left(\mathrm{s}\right)=\frac{\mathrm{C}\left(\mathrm{s}\right)}{\mathrm{R}\left(\mathrm{s}\right)}=\mathrm{C}{\left[\mathrm{s}\mathrm{I}-\mathrm{A}\right]}^{-1}\mathrm{B}$

$\left[\mathrm{s}\mathrm{I}-\mathrm{A}\right]=\left[\begin{array}{cc}\mathrm{s}+4& 1.5\\ -4& \mathrm{S}\end{array}\right]$

${\left[\mathrm{s}\mathrm{I}-\mathrm{A}\right]}^{-1}=\left[\begin{array}{cc}\mathrm{s}& -1.5\\ 4& \mathrm{s}+4\end{array}\right]\frac{1}{{\mathrm{s}}^2+4\mathrm{s}+6}$

${\left[\mathrm{s}\mathrm{I}-\mathrm{A}\right]}^{-1}\mathrm{B}=\left[\begin{array}{c}2\mathrm{s}\\ 8\end{array}\right]\frac{1}{{\mathrm{s}}^2+4\mathrm{s}+6}$

$\mathrm{C}{\left[\mathrm{s}\mathrm{I}-\mathrm{A}\right]}^{-1}\mathrm{B}=\frac{3\mathrm{s}+5}{{\mathrm{s}}^2+4\mathrm{s}+6}$

Concept:

General representation of state model is given below:

(ẋ = A_x + B_y , & y = Cx + Dy)

ẋ → Velocity vector

x → State vector

u → Input vector

y → Output vector

It’s Transfer Function TF

TF = C[sI - A]^-1 B + D

Calculation:

Given state model is:

ẋ₁ = 2x₁ – x₂ + 34

ẋ₂ = -4x₂ – 4 & y = 3x₁ – 2x₂

$\Rightarrow \begin{array}{cccc}A=\left[\begin{array}{cc}2& -1\\ 0& -4\end{array}\right]& ,B=\left[\begin{array}{c}3\\ -1\end{array}\right]& \mathrm{\&}& D=Nullvector\end{array}$

$\left[sI-A\right]=\left[\begin{array}{cc}3-2& 1\\ 0& s+4\end{array}\right]\Rightarrow {\left(sI-A\right)}^{-1}=\frac{1}{\left(s-2\right)\left(s+4\right)}\left[\begin{array}{cc}s+4& -1\\ 0& s-2\end{array}\right]$  

$T.F=\left[3-2\right]\left[\begin{array}{cc}\frac{1}{s-2}& -\frac{1}{\left(s-2\right)\left(s+4\right)}\\ 0& \frac{1}{s+4}\end{array}\right]\left[\begin{array}{c}3\\ -1\end{array}\right]$  

$T.F=\left[\begin{array}{cc}\frac{3}{s-2}& \frac{-2s+1}{\left(s-2\right)\left(s+4\right)}\end{array}\right]\left[\begin{array}{r}3\\ -1\end{array}\right]=\frac{115+35}{\left(s-2\right)\left(s+4\right)}$  

Concept:

X’ = Ax + Bu → (1)

Y = Cx + Du → (2)

Apply Laplace transform, to equation (1)

S x(s) = A x(s) B u(s)

⇒ B u(s) = (SI - A) X(s)

⇒ X(s) = (SI - A)-1 B u(s)

Now apply Laplace transform to equation (2)

Y(s) = C x(s) + D u(s)

⇒ Y(s) = C[(SI - A)-1] B u(s) + D u(s)

Calculation:

From the given state-space representation,

$A=\left[\begin{array}{cc}-3& 1\\ 0& -2\end{array}\right],B=\left[\begin{array}{c}2\\ 1\end{array}\right],C=\left[\begin{array}{cc}1& 0\end{array}\right],D=0$

$\left(sI-A\right)=\left[\begin{array}{cc}s+3& -1\\ 0& s+2\end{array}\right]$

${\left(sI-A\right)}^{-1}=\frac{1}{\left(\mathrm{s}+3\right)\left(\mathrm{s}+2\right)}\left[\begin{array}{cc}s+2& 1\\ 0& s+3\end{array}\right]$

$=\frac{1}{s^2+5s+6}\left[\begin{array}{cc}s+2& 1\\ 0& s+3\end{array}\right]$

$TF=\left[\begin{array}{cc}1& 0\end{array}\right]\frac{1}{s^2+5s+6}\left[\begin{array}{cc}s+2& 1\\ 0& s+3\end{array}\right]\left[\begin{array}{c}2\\ 1\end{array}\right]+0$

$=\frac{2s+5}{s^2+5s+6}$