Question 1 |
The electrical system shown in the figure converts input source current i_{s}(t) to output voltage v_{o}(t)
Current i i_{L}(t) in the inductor and voltage v_{c}(t) across the capacitor are taken as the state variables, both assumed to be initially equal to zero, i.e., i_{L}(0) =0 and v_{c}(0)=0. The system is
Current i i_{L}(t) in the inductor and voltage v_{c}(t) across the capacitor are taken as the state variables, both assumed to be initially equal to zero, i.e., i_{L}(0) =0 and v_{c}(0)=0. The system is
completely state controllable as well as completely observable | |
completely state controllable but not observable | |
completely observable but not state controllable | |
neither state controllable nor observable |
Question 1 Explanation:
\begin{aligned} i_{s} &=v_{i}+v_{c} \\ \therefore \qquad v_{i} &=-v_{c}+i_{s} \\ v_{L} &=L i_{L} \\ v_{L} &=\left(i_{s}-i_{L}\right) \\ \therefore \qquad L i_{L} &=i_{s}-i_{L} \\ \therefore \qquad i_{L} &=-i_{L}+i_{s} \\ v_{o} &=v_{c} \\ X &=\left[\begin{array}{cc} -1 & 0 \\ 0 & -1 \end{array}\right] X+\left[\begin{array}{ll} 1 \\ 2 \end{array}\right] U \\ Y &=[0 \quad 1] X+[0] \cup \\ A &=-I \\ Q_{C} &=\left[\begin{array}{ll} B & A B \end{array}\right]=\left[\begin{array}{ll} 1 & -1 \\ 1 & -1 \end{array}\right] \Rightarrow\left|Q_{c}\right|=0 \\ Q_{o} &=\left[\begin{array}{ll} C^{T} & A^{\top} C^{\top} \end{array}\right]=\left[\begin{array}{cc} 0 & 0 \\ 1 & -1 \end{array}\right] \Rightarrow\left|Q_{o}\right|=0 \end{aligned}
Question 2 |
For the given circuit, which one of the following is the correct state equation?
A | |
B | |
C | |
D |
Question 2 Explanation:
From source transformation,
KVL in loop 1,
2i_{1}=2i+0.5\frac{\mathrm{d} i}{\mathrm{d} t}+V
\frac{\mathrm{d} i}{\mathrm{d} t}=-2V-4i+4i_{1}
KCL at node (a),
i=0.25\frac{\mathrm{d} V}{\mathrm{d} t}+\frac{V-i_{2}}{1}
\frac{\mathrm{d} v}{\mathrm{d} t}=-4V+4i+4i_{2}
\begin{bmatrix} v\\ i \end{bmatrix}=\begin{bmatrix} -4 & 4\\ -2 & -4 \end{bmatrix}\begin{bmatrix} v\\ i \end{bmatrix}+\begin{bmatrix} 0 &4 \\ 4 &0 \end{bmatrix}\begin{bmatrix} i_{1}\\ i_{2} \end{bmatrix}
KVL in loop 1,
2i_{1}=2i+0.5\frac{\mathrm{d} i}{\mathrm{d} t}+V
\frac{\mathrm{d} i}{\mathrm{d} t}=-2V-4i+4i_{1}
KCL at node (a),
i=0.25\frac{\mathrm{d} V}{\mathrm{d} t}+\frac{V-i_{2}}{1}
\frac{\mathrm{d} v}{\mathrm{d} t}=-4V+4i+4i_{2}
\begin{bmatrix} v\\ i \end{bmatrix}=\begin{bmatrix} -4 & 4\\ -2 & -4 \end{bmatrix}\begin{bmatrix} v\\ i \end{bmatrix}+\begin{bmatrix} 0 &4 \\ 4 &0 \end{bmatrix}\begin{bmatrix} i_{1}\\ i_{2} \end{bmatrix}
Question 3 |
Let the state-space representation of an LTI system be \dot{x}(t)=Ax(t)+Bu(t), \dot{y}(t)=Cx(t)+du(t) where A, B, C are matrices, d is a scalar, u(t) is the input to the system, and y(t) is its output. Let B=[0\;0\;1]^T and d=0. Which one of the following options for A and C will ensure that the transfer function of this LTI system is
H(s)=\frac{1}{s^3+3s^2+2s+1}?
H(s)=\frac{1}{s^3+3s^2+2s+1}?
A | |
B | |
C | |
D |
Question 3 Explanation:
\begin{aligned} X(t) &=A x(t)+B u(t) \\ y(t) &=C x(t) \\ B &=\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right] \\ \frac{\gamma(s)}{U(s)} &=\\ \frac{\gamma(s)}{X_{1}(s)} \times \frac{X_{1}(s)}{U(s)}=& 1 \times \frac{1}{s^{3}+3 s^{2}+2 s+1} \\ x_{1}(s)\left(s^{3}+3 s^{2}+2 s+1\right]&=U(s) \\ x_{2} &=\dot{x}_{1}(t) \\ x_{2}(s) &=s x_{1}(s) \\ x_{3} &=\dot{x}_{2}(t) \\ \Rightarrow \quad x_{3}(s) &=s x_{2}(s)=s^{2} X_{1}(s) \\ \text { So, } \quad s x_{3}(s) &=-x_{1}(s)-2 x_{2}(s)-3 x_{3}(s)+u(s) \\ \dot{x}_{3}(t) &=-x_{1}(t)-2 r_{2}(t)-3 x_{3}(t)+u(t) \\ y(t) &=x_{1}(t) \\ \dot{x}(t) &=\left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 1 \\ -1 & -2 & -3\end{array}\right] x(t)+\left[\begin{array}{l}0 \\ 0 \\ 1\end{array}\right] u(t) \\ y(t) &=[1 \quad 0 \quad 0] x(t) \end{aligned}
Question 4 |
The state equation and the output equation of a control system are given below:
\dot{x}=\begin{bmatrix} -4 & -1.5\\ 4 & 0 \end{bmatrix}x+\begin{bmatrix} 2\\ 0\end{bmatrix}u,
y=\begin{bmatrix} 1.5 & 0.625 \end{bmatrix}x.
The transfer function representation of the system is
\dot{x}=\begin{bmatrix} -4 & -1.5\\ 4 & 0 \end{bmatrix}x+\begin{bmatrix} 2\\ 0\end{bmatrix}u,
y=\begin{bmatrix} 1.5 & 0.625 \end{bmatrix}x.
The transfer function representation of the system is
\frac{3s+5}{s^{2}+4s+6} | |
\frac{3s-1.875}{s^{2}+4s+6} | |
\frac{4s+1.5}{s^{2}+4s+6} | |
\frac{4s+1.5}{s^{2}+4s+6} |
Question 4 Explanation:
Transfer function
\begin{aligned} T(s) &=\frac{Y(s)}{U(s)}=\mathrm{C}(s I-A]^{-1} \mathrm{B} \\ A &=\left[\begin{array}{cc} -4 & -1.5 \\ 4 & 0 \end{array}\right] \\ B &=\left[\begin{array}{c} 2 \\ 0 \end{array}\right] \\ C&=[1.5\quad0.625]\\ {[s I-A]} & =\left[\begin{array}{cc}s+4 & 1.5 \\-4 & s\end{array}\right] \\ {[s I-A]^{-1}} & =\frac{1}{\left(s^{2}+4 s+6\right)}\left[\begin{array}{cc}s & -1.5 \\4 & s+4 \end{array}\right] \\ {[s I-A]^{-1} B} & =\frac{1}{s^{2}+4 s+6}\left[\begin{array}{c}2 s \\8 \end{array}\right] & \\ C[s I-A]^{-1} B & =\frac{1}{s^{2}+4 s+6}[1.5 \quad 0.625]\left[\begin{array}{c}2 s \\ 8\end{array}\right] \\ T(s) & =\frac{3 s+5}{s^{2}+4 s+6} \end{aligned}
\begin{aligned} T(s) &=\frac{Y(s)}{U(s)}=\mathrm{C}(s I-A]^{-1} \mathrm{B} \\ A &=\left[\begin{array}{cc} -4 & -1.5 \\ 4 & 0 \end{array}\right] \\ B &=\left[\begin{array}{c} 2 \\ 0 \end{array}\right] \\ C&=[1.5\quad0.625]\\ {[s I-A]} & =\left[\begin{array}{cc}s+4 & 1.5 \\-4 & s\end{array}\right] \\ {[s I-A]^{-1}} & =\frac{1}{\left(s^{2}+4 s+6\right)}\left[\begin{array}{cc}s & -1.5 \\4 & s+4 \end{array}\right] \\ {[s I-A]^{-1} B} & =\frac{1}{s^{2}+4 s+6}\left[\begin{array}{c}2 s \\8 \end{array}\right] & \\ C[s I-A]^{-1} B & =\frac{1}{s^{2}+4 s+6}[1.5 \quad 0.625]\left[\begin{array}{c}2 s \\ 8\end{array}\right] \\ T(s) & =\frac{3 s+5}{s^{2}+4 s+6} \end{aligned}
Question 5 |
A second - order LTI system is described by the following state equations,
\frac{d}{dt}x_{1}(t)- x_{2}(t)=0
\frac{d}{dt}x_{2}(t)+2x_{1}(t)+3x_{2}(t)=r(t)
Where x_{1}(t) and x_{2}(t) are the two state variables and r(t) denotes the input. The output c(t) =x_{1}(t). The system is.
\frac{d}{dt}x_{1}(t)- x_{2}(t)=0
\frac{d}{dt}x_{2}(t)+2x_{1}(t)+3x_{2}(t)=r(t)
Where x_{1}(t) and x_{2}(t) are the two state variables and r(t) denotes the input. The output c(t) =x_{1}(t). The system is.
Undamped (oscillatory) | |
Under damped | |
Critically damped | |
Over damped |
Question 5 Explanation:
Given state variable equations are as follows:
\begin{aligned} &\frac{d}{d t} x_{1}(t)-x_{2}(t)=0 &\ldots(i)\\ &\frac{d}{d t} x_{2}(t)+2 x_{2}(t)+3 x_{2}(t)=r(t) &\ldots(ii) \end{aligned}
Also given that, input =r(t) and output =x_{1}(t)
By applying Laplace transform to equation (i), we
get,
s X_{1}(s)=X_{2}(s) \qquad \ldots(iii)
By applying Laplace transform to equation (ii).
we get,
\begin{aligned} s X_{2}(s)+2 X_{1}(s)+3 X_{2}(s)&=R(s) \\ x_{2}(s)&=\frac{R(s)-2 X_{1}(s)}{s+3} &\ldots(iv) \end{aligned}
By substituting equation (iv) in equation (iii), we get,
\begin{aligned} s X_{\mathrm{i}}(s)&=\frac{R(s)-2 X_{1}(s)}{s+3}\\ \left(s^{2}+3 s+2\right) X_{1}(s)&=R(s) \end{aligned}
So, the transfer function of the given system is,
\begin{aligned} \frac{X_{1}(s)}{R(s)} &=\frac{1}{s^{2}+3 s+2} \\ v r^{2} &=2 \\ 2 \zeta_{n} &=3 \\ \xi &=\frac{3}{2 \omega_{n}}=\frac{3}{2 \sqrt{2}}=1.06 \end{aligned}
As \xi \gt 1, the given system is overdamped.
\begin{aligned} &\frac{d}{d t} x_{1}(t)-x_{2}(t)=0 &\ldots(i)\\ &\frac{d}{d t} x_{2}(t)+2 x_{2}(t)+3 x_{2}(t)=r(t) &\ldots(ii) \end{aligned}
Also given that, input =r(t) and output =x_{1}(t)
By applying Laplace transform to equation (i), we
get,
s X_{1}(s)=X_{2}(s) \qquad \ldots(iii)
By applying Laplace transform to equation (ii).
we get,
\begin{aligned} s X_{2}(s)+2 X_{1}(s)+3 X_{2}(s)&=R(s) \\ x_{2}(s)&=\frac{R(s)-2 X_{1}(s)}{s+3} &\ldots(iv) \end{aligned}
By substituting equation (iv) in equation (iii), we get,
\begin{aligned} s X_{\mathrm{i}}(s)&=\frac{R(s)-2 X_{1}(s)}{s+3}\\ \left(s^{2}+3 s+2\right) X_{1}(s)&=R(s) \end{aligned}
So, the transfer function of the given system is,
\begin{aligned} \frac{X_{1}(s)}{R(s)} &=\frac{1}{s^{2}+3 s+2} \\ v r^{2} &=2 \\ 2 \zeta_{n} &=3 \\ \xi &=\frac{3}{2 \omega_{n}}=\frac{3}{2 \sqrt{2}}=1.06 \end{aligned}
As \xi \gt 1, the given system is overdamped.
There are 5 questions to complete.