GATE EE 2016 Set 1

To find the maximum value of a function on a closed interval, we must evaluate it at the interval's endpoints and at any critical points that lie within the interval.

First, we find the derivative of $f(x) = x^3 - 3x^2 + 2x$, which is $f'(x) = 3x^2 - 6x + 2$.
Setting the derivative to zero, $f'(x) = 0$, gives critical points at $x = 1 \pm \frac{1}{\sqrt{3}}$.
Only the critical point $x = 1 + \frac{1}{\sqrt{3}}$ falls within the interval $[1, 2]$.

Now, we compare the function's value at the endpoints and this critical point:

• $f(1) = 1(1-1)(1-2) = 0$
• $f(2) = 2(2-1)(2-2) = 0$
• $f(1 + \frac{1}{\sqrt{3}}) = -\frac{2}{3\sqrt{3}}$ (a negative value)

Comparing the values $\{0, 0, -\frac{2}{3\sqrt{3}}\}$, the maximum value is clearly 0.

Let the given 3x3 matrix be denoted by $A$. Since all rows of $A$ are identical, the matrix has a rank of 1. A key property of an $n \times n$ matrix is that if its rank is $r$, it will have $n-r$ eigenvalues equal to zero. For our matrix, $n=3$ and $r=1$, so there are $3-1=2$ zero eigenvalues.

The sum of a matrix's eigenvalues is equal to its trace (the sum of its diagonal elements). The trace of $A$ is $1+1+1=3$. Since we already know two eigenvalues are 0, the third eigenvalue, $\lambda_3$, must satisfy $0+0+\lambda_3=3$. Therefore, the only non-zero eigenvalue is 3.

This problem is a direct application of the frequency shifting property of the Laplace transform, which states that $L\{e^{at}f(t)\} = F(s-a)$.

First, let's find the Laplace transform of the base function, $f(t)=\sin(5t)$. Using the standard transform pair, we get $F(s) = L\{\sin(5t)\} = \frac{5}{s^{2+}25}$.

Next, we apply the frequency shift caused by the $e^{2t}$ term. Here, $a=2$, so we replace every $s$ in $F(s)$ with $(s-2)$.
$L\{e^{2t}\sin(5t)\} = \frac{5}{(s-2)^{2+}25}$
To get the final form, we simply expand the denominator:
$\frac{5}{s^2 - 4s + 4 + 25} = \frac{5}{s^2 - 4s + 29}$

To solve this homogeneous linear differential equation, we first find the roots of its characteristic equation: $m^2 + 2m + 1 = 0$. This equation factors into $(m+1)^2=0$, revealing a repeated root of $m=-1$.

The general solution for a repeated root is of the form $y(t) = (c_1 + c_2t)e^{mt}$. For our case, this becomes $y(t) = (c_1 + c_2t)e^{-t}$.

We use the given conditions to find the constants $c_1$ and $c_2$. The condition $y(0)=1$ implies $(c_1 + 0)e^0 = 1$, which gives $c_1=1$. With this, our solution becomes $y(t) = (1 + c_2t)e^{-t}$.

Next, using $y(1)=3e^{-1}$, we get $(1+c_2)e^{-1} = 3e^{-1}$. This simplifies to $1+c_2=3$, so $c_2=2$. Thus, the particular solution is $y(t) = (1+2t)e^{-t}$.

Finally, we evaluate the solution at $t=2$ to find the answer: $y(2) = (1+2(2))e^{-2} = 5e^{-2}$.

To solve this integral, we will apply the Cauchy Residue Theorem. First, let's identify the poles of the integrand, which are the values of $z$ that make the denominator zero. The poles are $z = 1/2$ and the roots of $z^2 - 4z + 5 = 0$, which are $z = 2 \pm i$.

Next, we determine which of these poles lie inside the contour $|z|=1$. The pole $z=1/2$ is inside, as $|1/2| < 1$. The poles $z = 2 \pm i$ are outside, since $|2 \pm i| = \sqrt{2^2 + (\pm 1)^2} = \sqrt{5} > 1$.

We then calculate the residue at the simple pole inside the contour, $z=1/2$:
$\text{Res}_{z=1/2} = \lim_{z \to 1/2} (z - \frac{1}{2}) \frac{2z+5}{(z-\frac{1}{2})(z^{2}-4z+5)} = \frac{2(1/2)+5}{(1/2)^{2-}4(1/2)+5} = \frac{6}{13/4} = \frac{24}{13}$

By the Residue Theorem, the value of the contour integral is $2\pi i$ times the sum of the residues of the poles within the contour. Therefore, the integral evaluates to $2\pi i \left(\frac{24}{13}\right) = \frac{48\pi i}{13}$.

To find the steady-state response of the system to a sinusoidal input, we evaluate its frequency response at the input frequency. The input is $r(t) = \cos(2t)$, which has an amplitude of 1 and an angular frequency of $\omega=2$ rad/s.

First, we find the frequency response function $H(j\omega)$ by substituting $s = j\omega$ into the transfer function $H(s)$:
$H(j\omega) = \frac{j\omega}{2+j\omega}$
Now, we evaluate this at the input frequency $\omega=2$:
$H(j2) = \frac{j2}{2+j2}$
The output amplitude $A$ is the magnitude of $H(j2)$, and the phase shift $\varphi$ is the angle of $H(j2)$.
$A = |H(j2)| = \frac{|j2|}{|2+j2|} = \frac{2}{\sqrt{2^2 + 2^2}} = \frac{2}{\sqrt{8}} = \frac{1}{\sqrt{2}}$
$\varphi = \angle H(j2) = \angle(j2) - \angle(2+j2) = 90^\circ - \tan^{-1}\left(\frac{2}{2}\right) = 90^\circ - 45^\circ = 45^\circ$
Thus, the amplitude $A$ is $\frac{1}{\sqrt{2}}$ and the phase shift $\varphi$ is $+45^\circ$.

The phase cross-over frequency, $\omega_{pc}$, is the frequency at which the system's phase angle equals $-180^\circ$. To find it, we analyze the frequency response function, $G(j\omega)$. The calculation that leads to the answer begins with the function $G(s)=\frac{100}{(s+1)^3}$.

The phase of $G(j\omega) = \frac{100}{(1+j\omega)^3}$ is $-180^\circ$ when its denominator, $(1+j\omega)^3$, is a negative real number. This requires the imaginary part of the denominator to be zero.

Let's expand the denominator: $(1+j\omega)^3 = (1-3\omega^2) + j(3\omega-\omega^3)$.
Setting the imaginary part to zero gives the equation $3\omega - \omega^3 = 0$.
Factoring this as $\omega(3-\omega^2)=0$, we find the non-trivial, positive solution is $\omega_{pc} = \sqrt{3}$ rad/s.

Let's first test for linearity. A system is linear if an input $a x_1(t) + b x_2(t)$ produces the output $a y_1(t) + b y_2(t)$. For our system, the output is $[a x_1(t) + b x_2(t)]\cos(t)$, which simplifies to $a[x_1(t)\cos(t)] + b[x_2(t)\cos(t)]$. This is precisely $a y_1(t) + b y_2(t)$, so the system is linear.

Next, we check for time invariance. A system is time-invariant if delaying the input by $t_0$ simply delays the output by the same amount. The output for a delayed input $x(t-t_0)$ is $x(t-t_0)\cos(t)$. However, the original output delayed by $t_0$ is $y(t-t_0) = x(t-t_0)\cos(t-t_0)$. Since $x(t-t_0)\cos(t) \neq x(t-t_0)\cos(t-t_0)$, the system's behavior changes with time, making it time-varying.

To solve this integral, we first address the argument of the Dirac delta function, which is $2t-2$. We use the scaling property of the delta function, which states that $\delta(at-b) = \frac{1}{|a|}\delta(t - \frac{b}{a})$.

Applying this property with $a=2$ and $b=2$, we get:
$\delta(2t-2) = \frac{1}{2}\delta(t - \frac{2}{2}) = \frac{1}{2}\delta(t-1)$.

Now, we can substitute this simplified form back into the original integral:
$\int_{-\infty }^{\infty } e^{-t} \left(\frac{1}{2}\delta(t-1)\right) dt = \frac{1}{2}\int_{-\infty }^{\infty } e^{-t}\delta(t-1)dt$.

Finally, we use the sifting property of the delta function, $\int_{-\infty }^{\infty } f(t)\delta(t-t_0)dt = f(t_0)$. This means the integral evaluates the function $e^{-t}$ at $t=1$.

The result is $\frac{1}{2} \times e^{-1}$, which simplifies to $\frac{1}{2e}$.

First, we determine the total temperature range to be measured, which is the difference between the maximum and minimum values: $55^{\circ}C - (-40^{\circ}C) = 95^{\circ}C$.

Next, we calculate the number of discrete steps needed to achieve the desired resolution. This is found by dividing the total range by the resolution: $\frac{95^{\circ}C}{0.1^{\circ}C} = 950$ steps.

An ADC uses binary bits to represent these steps. If an ADC has $n$ bits, it can represent $2^n$ unique levels. We need to find the smallest integer $n$ for which $2^n$ is greater than or equal to the required 950 steps.

Evaluating powers of 2, we see that $2^9 = 512$, which is insufficient. However, $2^{10} = 1024$, which is enough to cover all 950 levels. Therefore, a minimum of 10 bits is necessary.

Let's break down the function of this 2-to-1 multiplexer circuit. The input $B$ serves as the select line, $S$. The data inputs, $I_0$ and $I_1$, are determined by input $A$. Due to the inverter, $I_0 = \bar{A}$, while $I_1 = A$.

The general Boolean expression for a 2-to-1 multiplexer's output is $F = \bar{S}I_0 + SI_1$.

By substituting the specific connections from our circuit into this formula, we get:
$F = \bar{B} \cdot \bar{A} + B \cdot A$

This resulting expression, $\bar{A}\bar{B} + AB$, is the standard definition of the XNOR gate. The XNOR function is the complement of the XOR function, so it can be written as $\overline{A \oplus B}$.

The Zener diode is in breakdown, which sets the base voltage to $V_B = 5.3 \text{ V}$. The emitter voltage is one diode drop below the base, so $V_E = V_B - V_{BE} = 5.3 \text{ V} - 0.6 \text{ V} = 4.7 \text{ V}$. We can now find the emitter current using Ohm's law: $I_E = V_E / 470 \, \Omega = 4.7 \text{ V} / 470 \, \Omega = 10 \text{ mA}$.

Next, we find the base current. The current supplied through the $4.7 \text{ k}\Omega$ resistor is $I_{supply} = (10 \text{ V} - V_B) / 4.7 \text{ k}\Omega = (10 - 5.3) \text{ V} / 4.7 \text{ k}\Omega = 1 \text{ mA}$. This current splits between the base and the Zener diode. Since the Zener current is given as $0.5 \text{ mA}$, the base current is $I_B = I_{supply} - I_Z = 1 \text{ mA} - 0.5 \text{ mA} = 0.5 \text{ mA}$.

Finally, the current gain $\beta$ relates the collector and base currents, and we know $I_E = I_C + I_B = \beta I_B + I_B = (\beta+1)I_B$. Therefore, $\beta+1 = I_E / I_B = 10 \text{ mA} / 0.5 \text{ mA} = 20$, which gives $\beta = 19$.

The electric field described is electrostatic, as it originates from a fixed charge distribution (a uniform ring). All electrostatic fields are conservative. This can be seen from the fact that the field $\vec{E}$ is derived from a scalar potential $\psi$, specifically $\vec{E} = -\nabla \psi$.

A fundamental theorem of vector calculus states that the curl of the gradient of any scalar function is always zero. Applying this principle, we get:
$\nabla \times \vec{E} = \nabla \times (-\nabla \psi) = 0$

Therefore, the curl of the electric field is zero, and so is its magnitude.

The magnetic field ($B$) from the long straight conductor depends on the permeability ($\mu$) of the medium it passes through, according to the formula $B = \frac{\mu I}{2 \pi r}$. We need to compare the field at two adjacent points, one just inside the iron toroid and one just outside in the surrounding air/vacuum.

Just outside the toroid, the medium's permeability is $\mu_0$, so the field is $B_{out} = \frac{\mu_0 I}{2 \pi r}$.
Just inside the soft iron, the permeability is $\mu = \mu_r \mu_0$, so the field is significantly stronger: $B_{\in} = \frac{\mu_r \mu_0 I}{2 \pi r}$.

The ratio of the flux densities at these two adjacent points (which are at the same radius $r$) is:
$\frac{B_{\in}}{B_{out}} = \frac{\frac{\mu_r \mu_0 I}{2 \pi r}}{\frac{\mu_0 I}{2 \pi r}} = \mu_r$
Given $\mu_r = 100$, the ratio is 100.

Let's closely compare the structures of the two infinite circuits. The circuit for $R_B$ is formed by a $1\,\Omega$ resistor connected in parallel with the rest of the network. If you look closely, that "rest of the network" is exactly the infinite ladder circuit for $R_A$.

Therefore, the input resistance $R_B$ is the equivalent of the entire circuit A (with resistance $R_A$) connected in parallel with a $1\,\Omega$ resistor.

Using the formula for two resistors in parallel, we can express this relationship as $R_B = R_A || 1\,\Omega$. This evaluates to $R_B = \frac{R_A \cdot 1}{R_A + 1}$, which gives the final result.

Excellent question! Let's break down the relationship step-by-step for a clearer understanding.

The slip at which maximum torque occurs is determined by the ratio of the rotor's resistance to its reactance: $S_{max,T} = R_2 / X_2$.

In a V/f drive, the supply frequency ($f$) changes, which in turn changes the rotor reactance, as reactance is frequency-dependent ($X_2 = 2\pi f L_2$). The rotor resistance ($R_2$) remains constant. This means the slip for maximum torque is inversely proportional to the frequency: $S_{max,T} \propto 1/f$.

We also know that the motor's synchronous speed ($\omega_s$) is directly proportional to the supply frequency ($\omega_s \propto f$).

Combining these two facts, since $S_{max,T}$ is inversely proportional to frequency and synchronous speed is directly proportional to it, we can conclude that the slip at maximum torque has an inverse relationship with the synchronous speed ($S_{max,T} \propto 1/\omega_s$).

The upper branch has a total resistance of $4\,\Omega + 6\,\Omega = 10\,\Omega$. Because the two branches are in parallel, they share the same voltage. Since the upper branch's resistance is double that of the lower ($5\,\Omega$) branch, it will carry half the current.

Let the current in the upper branch be $I$. Then the current in the lower branch must be $2I$. The power (heat per second) in the $5\,\Omega$ resistor is given as $10\,\text{cal/s}$.
Using the power formula $P = I^2R$, we can write:
$P_5 = (2I)^2 \times 5 = 10$
$4I^2 \times 5 = 10 \implies 20I^2 = 10 \implies I^2 = 0.5$

The heat generated by the $4\,\Omega$ resistor is $P_4 = I^2 \times 4$.
Substituting the value we found for $I^2$, we get $P_4 = 0.5 \times 4 = 2\,\text{cal/s}$.

To find the current $I_1$ supplied by the battery, we can apply Kirchhoff's circuit laws. First, let's use the Current Law (KCL) at the central node. The current entering the node, $I_1$, must equal the sum of the currents leaving it. This gives us the relationship $I_1 = I_2 + I_2 = 2I_2$.

Next, we apply the Voltage Law (KVL) around the outer loop. The total voltage provided by the 1V source must equal the sum of the voltage drops across the three resistors in the loop. This gives us $1 = I_1(1 \Omega) + I_2(1 \Omega) + I_2(1 \Omega)$, which simplifies to $1 = I_1 + 2I_2$.

Now we can solve this system of two equations. Substituting the first equation ($I_1 = 2I_2$) into the second gives $1 = (2I_2) + 2I_2 = 4I_2$. Solving for $I_2$ yields $I_2 = \frac{1}{4}$ A. Finally, we find the battery current $I_1$ using our KCL relationship: $I_1 = 2I_2 = 2 \times \frac{1}{4} = 0.5$ A.

In a load flow study, the voltage angle, $\delta$, is an unknown variable for every bus except the slack bus. This is true for both PV and PQ bus types. Therefore, converting a bus from PV to PQ does not change the number of unknown angles.

For a PV bus, the voltage magnitude, $|V|$, is a specified, known value. For a PQ bus, $|V|$ is an unknown variable that must be calculated.

When two PV buses are converted to PQ buses (typically due to reactive power limit violations), their previously specified voltage magnitudes now become unknown variables. Thus, the number of unknown voltage magnitudes in the system increases by two.

The voltage drop at a healthy bus during a remote fault is the product of the fault current and the transfer impedance between the two buses. We can first calculate this transfer impedance, $Z_{AB}$, using the data from the fault at bus A.

The voltage at bus B during a fault at A is given by:
$V_B = V_{pre-fault} - Z_{AB} \times I_{fault, A}$
$0.8 = 1.0 - Z_{AB} \times 10 \implies Z_{AB} = 0.02$ pu.

By reciprocity, the transfer impedance from B to A is the same ($Z_{BA} = Z_{AB}$). Now, we can find the voltage at bus A during a fault at bus B:
$V_A = V_{pre-fault} - Z_{BA} \times I_{fault, B}$
$V_A = 1.0 - (0.02 \times 8) = 1.0 - 0.16 = 0.84$ pu.

A synchronous generator operating at a lagging power factor supplies both active and reactive power to the grid, making it a source of reactive power. An overexcited synchronous motor runs at a leading power factor, meaning it consumes active power but supplies reactive power to the system, acting as a dynamic capacitor. Therefore, it is also a source. An induction generator, however, lacks a field winding for excitation and must absorb reactive power from the grid to establish its magnetic field. Consequently, it always acts as a sink of reactive power.

For a converter operating in steady state, the inductor volt-second balance principle states that the average voltage across the inductor over a full switching period must be zero. This means the positive area under the $V_L$ curve must equal the magnitude of the negative area.

From the graph, the voltage during the ON-time is $30$ V and during the OFF-time is $-20$ V. Let the duty cycle be $D$. The ON-time is $T_{ON} = D \cdot T_S$ and the OFF-time is $T_{OFF} = (1-D) \cdot T_S$.

Equating the volt-second areas gives:
$30 \cdot T_{ON} = 20 \cdot T_{OFF}$
$30 \cdot (D \cdot T_S) = 20 \cdot ((1-D) \cdot T_S)$

The switching period $T_S$ cancels out, leaving:
$30D = 20(1-D) \implies 30D = 20 - 20D \implies 50D = 20$
Therefore, the duty cycle is $D = \frac{20}{50} = 0.4$.

The direction of the 100 A current is upward, which forward-biases the diode (D) but reverse-biases the IGBT (S). Consequently, the entire current must flow through the diode.

From the diode's V-I characteristic in Figure (c), we can model its on-state behavior as a voltage source in series with a resistor. The threshold voltage is $V_o = 0.7$ V, and the on-state resistance is $R_D = dV/dI = 0.01 \Omega$.

The total voltage drop across the diode for a current of $I_D = 100$ A is:
$V_D = V_o + I_D R_D = 0.7 \text{ V} + (100 \text{ A} \times 0.01 \Omega) = 1.7 \text{ V}$

The conduction power loss in the module is the power dissipated by the diode:
$P = V_D \times I_D = 1.7 \text{ V} \times 100 \text{ A} = 170 \text{ W}$

In a lap-connected motor, the number of parallel paths in the armature winding equals the number of poles. With 4 poles, the total 40 A armature current divides into 4 paths, so the current in each conductor is $40 \text{ A} / 4 = 10 \text{ A}$.

As a conductor rotates, it moves under alternating North and South poles, which causes the current within it to reverse direction periodically. The frequency of this reversal is related to the motor's speed ($N$) and poles ($P$) by $f = \frac{PN}{120}$. Plugging in the values, we get $f = \frac{4 \times 600}{120} = 20 \text{ Hz}$.

The time period for one full electrical cycle is $T = 1/f = 1/20 \text{ s} = 50 \text{ ms}$. This means the current flows in one direction for half the period ($25 \text{ ms}$) and in the opposite direction for the other half. The resulting waveform is approximately trapezoidal, with a peak of 10 A.

An ideal transformer scales impedance from one port to another. The impedance of the load on the secondary side ($Z_2$) appears as a different impedance when viewed from the primary side ($Z_1$). This relationship is defined by the square of the turns ratio ($n$):
$Z_1 = n^2 Z_2$

In this problem, the load is an inductor with impedance $Z_2 = j\omega L$. Substituting this into the impedance reflection formula gives:
$Z_1 = n^2 (j\omega L)$

By rearranging the terms, we get $Z_1 = j\omega(n^2L)$. This expression has the standard form for an inductor's impedance, $j\omega L_{eq}$. By comparing the two, we find that the equivalent inductance at port 1 is $L_{eq} = n^2L$.

This problem involves calculating the number of ways a candidate can choose their 3 pens. We treat this as a problem of combinations with repetition, where the order of pens doesn't matter.

The total number of unique combinations of 3 pens from 4 available colours ($n=4, r=3$) is given by the formula $\binom{n+r-1}{r}$.
$\text{Total Outcomes} = \binom{4+3-1}{3} = \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$
The favorable outcomes are when all 3 pens have the same colour. This can happen in 4 ways: all black, all blue, all green, or all red.

Therefore, the probability is the ratio of favorable outcomes to total outcomes.
$P(\text{all same colour}) = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{4}{20} = \frac{1}{5} = 0.2$

The sum $S = \sum_{n=0}^{\infty } n\alpha^n$ is a classic arithmetic-geometric series whose closed-form can be found using properties of the Z-transform.

The Z-transform of the sequence $x[n] = n\alpha^n u(n)$ is given by the standard pair $X(z) = \frac{\alpha z^{-1}}{(1-\alpha z^{-1})^2}$. By definition, the Z-transform is also $X(z) = \sum_{n=0}^{\infty } (n\alpha^n) z^{-n}$. To find our specific sum $S$, we can simply evaluate the closed-form expression of the Z-transform at $z=1$.

This gives $S = X(1) = \frac{\alpha}{(1-\alpha)^2}$.

We are given the condition $S=2\alpha$. Setting the two expressions for $S$ equal gives:
$\frac{\alpha}{(1-\alpha)^2} = 2\alpha$

Since we know $0 < \alpha < 1$, $\alpha$ is not zero, so we can divide both sides by $\alpha$:
$\frac{1}{(1-\alpha)^2} = 2$

Taking the square root gives $1-\alpha = \frac{1}{\sqrt{2}}$, where we choose the positive root because $\alpha < 1$. Finally, solving for $\alpha$ yields $\alpha = 1 - \frac{1}{\sqrt{2}} \approx 0.293$.

Here is a clearer, more pedagogically sharp explanation:

A key property of eigenvalues is that if a matrix $A$ has an eigenvalue $\lambda$ with eigenvector $x$, then any polynomial function of the matrix, let's call it $p(A)$, will have the same eigenvector $x$ with a new eigenvalue of $p(\lambda)$.

In this case, the matrix polynomial is $p(A) = A^2 - 3A + 4I$. We can find the new eigenvalues by simply plugging the original eigenvalues of $A$ into this polynomial.

For the first eigenvalue $\lambda_1 = 1$, the new eigenvalue is:
$p(1) = (1)^2 - 3(1) + 4 = 1 - 3 + 4 = 2$.

For the second eigenvalue $\lambda_2 = -2$, the new eigenvalue is:
$p(-2) = (-2)^2 - 3(-2) + 4 = 4 + 6 + 4 = 14$.

Crucially, the eigenvectors $x_1$ and $x_2$ do not change. Thus, the new eigenvalues are 2 and 14, corresponding to the original eigenvectors $x_1$ and $x_2$.

A fundamental property of real matrices states that the rank of a matrix $A$ is always equal to the rank of the product $A^T A$. This is expressed by the identity $Rank(A) = Rank(A^T A)$.

In this problem, we are given that the matrix $A$ has a rank of 2. Applying the aforementioned property directly, we can determine the rank of $A^T A$. Since $Rank(A) = 2$, it follows that the rank of $A^T A$ must also be 2. The dimensions of matrix $A$ do not alter this core relationship.

Let's break down the Bode plot to identify the characteristics of the transfer function.

The plot begins with a slope of $+20$ dB/decade, which tells us there is a single zero at the origin (an $s$ term in the numerator). The slope then changes at two corner frequencies, indicating the presence of poles. The total change in slope from the initial $+20$ dB/dec to the final $-40$ dB/dec is $-60$ dB/dec. Since each pole contributes $-20$ dB/dec to the slope, this net change implies there are three poles in the system. The final segment's slope of $-40$ dB/dec specifically suggests that two of these poles are a double pole. Therefore, the system must have one zero at the origin, one single pole, and one double pole. Among the given choices, only option (A) exhibits this structure.

The solution to the unforced state equation $\dot{x}(t)=Ax(t)$ is given by $x(t)=e^{At}x(0)$. Since the system matrix

is diagonal, the state transition matrix $e^{At}$ is simply

. The state vector is then

. Consequently, the output is

. For $t= \log_{e}2$, we evaluate $y(\log_{e}2)=e^{\log_{e}2}+e^{2\log_{e}2}=e^{\log_{e}2}+e^{\log_{e}2^2}$. Using the property $e^{\log_{e}a}=a$, the result is $y(\log_{e}2) = 2 + 2^2 = 2+4=6$.

We can determine the number of encirclements using the Nyquist stability criterion, $N = P - Z$.

First, find $P$, the number of open-loop poles in the right-half plane. The given transfer function, $G(s)H(s)=\frac{s+3}{s^{2}(s-3)}$, has a pole at $s=3$. Thus, $P=1$.

Next, we find $Z$, the number of unstable closed-loop poles, by analyzing the characteristic equation $1+G(s)H(s)=0$, which simplifies to $s^{3-}3s^{2+}s+3=0$. Constructing a Routh-Hurwitz array for this polynomial reveals two sign changes in the first column, which means there are two roots in the right-half plane. Therefore, $Z=2$.

Finally, we calculate the number of encirclements, $N = P - Z = 1 - 2 = -1$. A value of $N=-1$ indicates one encirclement of the critical point $(-1, j0)$ in the clockwise direction.

To find the number of roots with real parts less than -1, we can shift the origin of the s-plane to the point $s=-1$. We do this by defining a new variable, $z=s+1$, which means we substitute $s = z-1$ into the polynomial. This transformation maps the region $\text{Re}(s) < -1$ to the left-half of the new $z$-plane, $\text{Re}(z) < 0$.

The transformed polynomial becomes:
$(z-1)^3 + 5.5(z-1)^2 + 8.5(z-1) + 3 = 0$

Expanding and simplifying this expression, we get:
$z^3 + 2.5z^2 + 0.5z - 1 = 0$

We now use the Routh-Hurwitz criterion to find the number of roots of this new polynomial in the right-half plane. The coefficients $[1, 2.5, 0.5, -1]$ have one sign change. This indicates there is one root in the right-half plane. Since the total number of roots is 3, the other $3-1=2$ roots must be in the left-half plane, corresponding to the original roots with real parts less than -1.