Question 1. \(X\) is a uniformly distributed random variable that takes values between 0 and 1.
The value of \(E\{X^3\}\) will be
\(0\)
\(1/8\)
\(1/4\)
\(1/2\)
Solution: \(X\) is uniformly distributed between 0 and 1
So probability density function \[f_X(x) =
\begin{cases}
1, & 0 < x < 1 \\
0, & \text{otherwise}
\end{cases}\] So, \[\begin{aligned}
E\{X^3\} &= \int_{0}^{1} x^3 f_X(x) \,dx \\
&= \int_{0}^{1} x^3 (1) \,dx \\
&= \left[\frac{x^4}{4}\right]_{0}^{1} \\
&= \frac{1^4}{4} - \frac{0^4}{4} \\
&= \frac{1}{4}
\end{aligned}\]
The correct option is C
Question 2. The characteristic equation of a \((3\times 3)\) matrix \(P\) is defined as \[a(\lambda) = |\lambda I - P| = \lambda^3 + \lambda^2 + 2\lambda + 1 = 0\] If \(I\) denotes identity matrix, then the inverse of matrix \(P\) will be
\((P^2 + P + 2I)\)
\((P^2 + P + I)\)
\(-(P^2 + P + I)\)
\(-(P^2 + P + 2I)\)
Solution: According to CAYLEY-HAMILTON Theorem every non-singular square matrix satisfies its own characteristic equation.
Characteristic equation \[a(\lambda) =
|\lambda I - P| = \lambda^3 + \lambda^2 + 2\lambda + 1 = 0\] Matrix \(P\) satisfies above equation
\[\begin{aligned}
P^3 + P^2 + 2P + I &= 0 \\
I &= -(P^3 + P^2 + 2P)
\end{aligned}\] Multiply both sides by \(P^{-1}\) \[\begin{aligned}
P^{-1} I &= P^{-1} [-(P^3 + P^2 + 2P)] \\
P^{-1} &= -(P^2 + P + 2I)
\end{aligned}\] The correct option is D.
Question 3. If the rank of a \((5\times 6)\) matrix \(Q\) is 4, then which one of the following statement is correct ?
\(Q\) will have four linearly independent rows and four linearly independent columns
\(Q\) will have four linearly independent rows and five linearly independent columns
\(QQ^T\) will be invertible
\(Q^T Q\) will be invertible
Solution: Rank of a matrix is no. of linearly independent rows and columns of the matrix.
Here Rank \(\rho(Q)=4\) So \(Q\) will have 4 linearly independent rows and four linearly independent columns.
The correct option is A.
Question 4. A function \(y(t)\) satisfies the following differential equation : \[\frac{dy(t)}{dt} + y(t) = \delta(t)\] where \(\delta(t)\) is the delta function. Assuming zero initial condition, and denoting the unit step function by \(u(t)\), \(y(t)\) can be of the form
\(e^{t}\)
\(e^{-t}\)
\(e^{t} u(t)\)
\(e^{-t} u(t)\)
Solution: Given differential equation for the function \[\frac{dy(t)}{dt} + y(t) = \delta(t)\] Taking Laplace on both the sides we have, \[\begin{aligned} sY(s) + Y(s) &= 1 \\ (s + 1)Y(s) &= 1 \\ Y(s) &= \frac{1}{s + 1} \end{aligned}\] Taking inverse Laplace of \(Y(s)\), \[y(t) = e^{-t} u(t), \quad t > 0\] The correct option is D.
Question 5. Consider function \(f(x) = (x^2 - 4)^2\) where \(x\) is a real number. Then the function has
only one minimum
only tow minima
three minima
three maxima
Solution: Given function \[\begin{aligned}
f(x) &= (x^2 - 4)^2 \\
\text{To obtain minima and maxima (set first derivative to zero):} \\
f'(x) &= 2(x^2 - 4)(2x) = 4x(x^2 - 4) \\
f'(x) &= 0 \implies 4x(x^2 - 4) = 0 \\
x &= 0, \quad x^2 - 4 = 0 \Rightarrow x = \pm 2 \\
\text{So, critical points are $x = 0, +2, -2$}. \\
\text{The second derivative is:}
f''(x) &= \frac{d}{dx} \left[4x^3 - 16x\right] \\
f''(x) &= 4x(2x) + 4(x^2 - 4) \\
&= 12x^2 - 16
\end{aligned}\] Using the second derivative test: \[\begin{aligned}
\text{For } x &= 0, & f''(0) = 12(0)^2 - 16 &=
-16 < 0 & (\text{Maxima}) \\
\text{For } x &= +2, & f''(2) = 12(2)^2 - 16 &=
32 > 0 & (\text{Minima}) \\
\text{For } x &= -2, & f''(-2) = 12(-2)^2 - 16
&= 32 > 0 & (\text{Minima})
\end{aligned}\] So \(f(x)\) has only two minima
The correct option is B.
Question 6 \(A\) is \(m \times n\) full rank matrix with \(m > n\) and \(I\) is identity matrix. Let matrix \[A^\dagger = (A^T A)^{-1} A^T\] Then, which one of the following statement is FALSE ?
\(A A^\dagger A = A\)
\((A A^\dagger)^2 = A A^\dagger\)
\(A^\dagger A = I\)
\(A A^\dagger A = A^\dagger\)
Solution: \[\begin{aligned} A' &= (A^T A)^{-1} A^T \\ &= A^{-1} (A^T)^{-1} A^T \quad \text{(This step seems to assume } A^T A = I \text{ or similar simplification)} \\ &= A^{-1} I \\ &= A^{-1} \end{aligned}\]
Put \(A' = A^{-1}\) in all option.
option (A) \[\begin{aligned} A A' A &= A \\ A A^{-1} A &= A \\ I A &= A \\ A &= A \quad \text{(true)} \end{aligned}\]
option (B) \[\begin{aligned} (A A')^2 &= I \\ (A A^{-1})^2 &= I \\ (I)^2 &= I \quad \text{(true)} \end{aligned}\]
option (C) \[\begin{aligned} A' A &= I \\ A^{-1} A &= I \\ I &= I \quad \text{(true)} \end{aligned}\]
option (D) \[\begin{aligned} A A' A &= A' \\ A A^{-1} A &= A' \\ I A &= A' \\ A &= A' \quad \text{(false, since the question likely seeks the one that is NOT true)} \end{aligned}\] The correct option is D.
Question 7.A differential equation \(\mathrm{d}x/\mathrm{d}t = e^{-2t} u(t)\), has to be solved using trapezoidal rule of integration with a step size \(h = 0.01\) s. Function \(u(t)\) indicates a unit step function. If \(x(0^-) = 0\), then value of \(x\) at \(t = 0.01\) s will be given by
\(0.00099\)
\(0.00495\)
\(0.0099\)
\(0.0198\)
Solution:\[\begin{aligned} \frac{\mathrm{d}x}{\mathrm{d}t} &= e^{-2t} u(t) \\ x &= \int e^{-2t} u(t)\, \mathrm{d}t \\ x &= \int_{0}^{t} e^{-2t}\, \mathrm{d}t \quad \text{($u(t)=1$ for $t \geq 0$)} \\ x &= \int_{0}^{t} f(t)\, \mathrm{d}t, \\ t &= 0.01 \text{ s} \end{aligned}\]
From trapezoid rule \[\begin{aligned} \int_{t_0}^{t_0+nh} f(t)\, \mathrm{d}t &= \frac{h}{2} [f(0) + f(0.01)] \\ \int_{0}^{0.01} f(t)\, \mathrm{d}t &= \frac{0.01}{2} [e^0 + e^{-0.02}], \quad h = 0.01 \\ &= 0.0099 \end{aligned}\] The correct option is C.
Question 8.Let \(P\) be a \(2 \times 2\) real orthogonal matrix and \(\vec{x}\) is a real vector \([x_1, x_2]^T\) with length \[\|\vec{x}\| = (x_1^2 + x_2^2)^{1/2}.\] Then, which one of the following statements is correct?
\(\|P\vec{x}\| \le \|\vec{x}\|\) where at least one vector satisfies \(\|P\vec{x}\| < \|\vec{x}\|\)
\(\|P\vec{x}\| \le \|\vec{x}\|\) for all vector \(\vec{x}\)
\(\|P\vec{x}\| \ge \|\vec{x}\|\) where at least one vector satisfies \(\|P\vec{x}\| > \|\vec{x}\|\)
No relationship can be established between \(\|\vec{x}\|\) and \(\|P\vec{x}\|\)
Solution:Let assume \(P\) is an orthogonal matrix So \(P P^{\mathrm{T}} = I\) \[P = \begin{bmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{bmatrix}\]
\[P \vec{X} = \begin{bmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\]
\[= \begin{bmatrix} x_1 \cos{\theta} - x_2 \sin{\theta} \\ x_1 \sin{\theta} + x_2 \cos{\theta} \end{bmatrix}\]
\[\left\| P\vec{X} \right\| = \sqrt{\left( x_1 \cos{\theta} - x_2 \sin{\theta} \right)^2 + \left( x_1 \sin{\theta} + x_2 \cos{\theta} \right)^2}\]
\[= \sqrt{x_1^2 + x_2^2}\]
\[\left\| P\vec{X} \right\| = \left\| \vec{X} \right\|\]
The correct option is B.
Question 9. The state space equation of a system is described by \[\dot{\mathbf{X}} = A\mathbf{X} + B u, \quad Y = C\mathbf{X}\] where \(\mathbf{X}\) is state vector, \(u\) is input, \(Y\) is output and \[A = \begin{bmatrix} 0 & 1 \\ 0 & -2 \end{bmatrix}, \quad B = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad C = \begin{bmatrix} 1 & 0 \end{bmatrix}\] The transfer function \(G(s)\) of this system will be
\(\displaystyle \frac{s}{s+2}\)
\(\displaystyle \frac{s+1}{s(s-2)}\)
\(\displaystyle \frac{s}{s-2}\)
\(\displaystyle \frac{1}{s(s+2)}\)
Solution: State space equation of the system is given by, \[\dot{\mathbf{X}} = A\mathbf{X} + B u\] \[Y = C\mathbf{X}\]
Taking Laplace transform on both sides of the equations: \[s \mathbf{X}(s) = A\mathbf{X}(s) + B U(s)\] \[(sI - A)\mathbf{X}(s) = B U(s)\] \[\mathbf{X}(s) = (sI - A)^{-1} B U(s)\] \[\therefore \quad Y(s) = C \mathbf{X}(s)\] So \[Y(s) = C(sI - A)^{-1} B U(s)\]
\[\text{T.F} = \frac{Y(s)}{U(s)} = C(sI - A)^{-1}B\]
\[(sI - A) = \begin{bmatrix}s & 0 \\ 0 & s\end{bmatrix} - \begin{bmatrix}0 & 1 \\ 0 & -2\end{bmatrix} = \begin{bmatrix}s & -1 \\ 0 & s+2\end{bmatrix}\]
\[(sI - A)^{-1} = \frac{1}{s(s+2)} \begin{bmatrix}s+2 & 1 \\ 0 & s \end{bmatrix} = \begin{bmatrix} \frac{1}{s} & \frac{1}{s(s+2)} \\ 0 & \frac{1}{s+2} \end{bmatrix}\]
Transfer function
\[G(s) = C [sI - A]^{-1} B = \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{s} & \frac{1}{s(s+2)} \\ 0 & \frac{1}{s+2} \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix}\]
\[= \begin{bmatrix} 1 & 0 \end{bmatrix} \begin{bmatrix} \frac{1}{s(s+2)} \\ \frac{1}{s+2} \end{bmatrix} = \frac{1}{s(s+2)}\] The correct option is D.
Question 10. A unity feedback is provided to the above system \(G(s)\) to make it a closed loop system as shown in figure.
For a unit step input \(r(t)\), the steady state error in the input will be
\(0\)
\(1\)
\(2\)
\(\infty\)
Solution: Steady state error is given by, \[e_{ss} = \lim_{s \to 0}\left[ \frac{sR(s)}{1 + G(s)H(s)} \right]\]
Here, \[R(s) = \mathcal{L}[r(t)] = \frac{1}{s} \quad \text{(Unit step input)}\] \[G(s) = \frac{1}{s(s+2)}\] \[H(s) = 1 \quad \text{(Unity feedback)}\]
So, \[e_{ss} = \lim_{s \to 0} \left[ \frac{s \left(\frac{1}{s}\right)}{1 + \frac{1}{s(s+2)}} \right]\] \[= \lim_{s \to 0} \left[ \frac{s(s+2)}{s(s+2) + 1} \right]\] \[e_{ss} = \lim_{s \to 0} \left[ \frac{s(s+2)}{s(s+2)+1} \right] = 0\]
The correct option is A.