GATE ECE 2016 Set 2

A key property of eigenvalues is that a matrix has an eigenvalue of zero if and only if its determinant is zero. Therefore, we can find the value of $x$ by setting $\det(A)=0$.

$\det(A) = \left| \begin{array}{ccc} 3 & 2 & 4 \\ 9 & 7 & 13 \\ -6 & -4 & -9+x \end{array}\right| = 0$

We can calculate the determinant by expanding along the first row:
$3(7(-9+x) - 13(-4)) - 2(9(-9+x) - 13(-6)) + 4(9(-4) - 7(-6)) = 0$

Simplifying the expression within each term gives:
$3(7x - 11) - 2(9x - 3) + 24 = 0$

This equation expands to $21x - 33 - 18x + 6 + 24 = 0$, which simplifies to $3x - 3 = 0$. Solving for $x$ gives the final answer, $x=1$.

The function $f(z)$ is a sum of two terms: $g(z) = 2z^3$ and $h(z) = b|z|^3$.

The first term, $2z^3$, is a polynomial, which is known to be analytic everywhere in the complex plane.

The second term, $|z|^3$, is a real-valued function that depends on both $z$ and its conjugate $\bar{z}$ (since $|z|^3 = (z\bar{z})^{3/2}$). A function that depends on $\bar{z}$ is generally not analytic. In fact, $|z|^3$ is only complex-differentiable at the origin ($z=0$) and is not analytic anywhere.

For the entire function $f(z)$ to be analytic, the non-analytic part must be eliminated. This can only happen if the coefficient of the non-analytic term is zero. Thus, we must have $b=0$, which reduces $f(z)$ to the analytic function $2z^3$.

To understand how the function $f(x)=x^{3}-3x^{2}+1$ behaves, let's analyze its slope by finding its derivative, $f'(x) = 3x^2 - 6x$.

The function's direction can change only where its slope is zero. We find these "critical points" by setting the derivative to zero: $3x^2 - 6x = 0$, which factors into $3x(x-2)=0$. This gives us critical points at $x=0$ and $x=2$, both of which lie within the interval $[-1, 3]$.

To determine if these are maxima or minima, we use the second derivative test. The second derivative is $f''(x) = 6x - 6$. At $x=0$, $f''(0) = -6$ (negative), indicating a local maximum. At $x=2$, $f''(2) = 6$ (positive), indicating a local minimum.

This means the function increases until the maximum at $x=0$, decreases until the minimum at $x=2$, and then increases again as it approaches $x=3$.

To solve this, we can determine how many times the graphs of the functions $y = \sin(x)$ and $y = x/2$ intersect.

An immediate solution is visible at the origin, since $\sin(0) = 0$ and $0/2 = 0$, so $x=0$ is one solution.

Now let's consider the slopes. At $x=0$, the slope of $y=\sin(x)$ is $1$, while the slope of the line $y=x/2$ is $1/2$. Because the sine curve initially rises more steeply than the line, but is then bounded between $-1$ and $1$, the graphs must cross exactly once for $x>0$.

Since both $\sin(x)$ and $x/2$ are odd functions (symmetric about the origin), for every positive solution there must be a corresponding negative solution. Thus, we have one positive solution, one negative solution, and the solution at zero, for a total of three distinct solutions.

Let's begin by finding the magnitude of the vector $I$. To simplify the algebra, we can work with the magnitude squared, $|I|^2$.
$|I|^2 = (15\cos(\omega t))^2 + (5\sin(\omega t))^2 = 225\cos^2(\omega t) + 25\sin^2(\omega t)$.
Using the identity $\sin^2(\omega t) = 1 - \cos^2(\omega t)$, we can express the magnitude solely in terms of cosine:
$|I|^2 = 225\cos^2(\omega t) + 25(1 - \cos^2(\omega t)) = 200\cos^2(\omega t) + 25$.
The magnitude $|I|$ is at its minimum when $|I|^2$ is at its minimum. This occurs when the term $\cos^2(\omega t)$ is minimized, which means $\cos^2(\omega t) = 0$.
This condition is met when $\omega t = 90^\circ$. At this time, the vector components are $I_x = 15\cos(90^\circ) = 0$ and $I_y = 5\sin(90^\circ) = 5$.
The vector becomes $I = 0\hat{x} + 5\hat{y}$, which points purely along the positive y-axis, making an angle of $90^\circ$ with the x-axis.

To maximize the power absorbed by the load, we can apply the consequences of the maximum power transfer theorem. This theorem implies that for maximum power delivery, the voltage from the source ($V_S$) is split equally between the internal resistance ($R$) and the load.

This means the voltage drop across the resistor $R$ must be half of the source voltage, or $V_S/2$. The current flowing through the entire circuit, including the resistor $R$, is determined by the constant current load, $I_L$.

Using Ohm's law for the resistor $R$, we can write:
$V_R = I_L R$

Setting this voltage drop equal to $V_S/2$ gives the condition for maximum power:
$I_L R = \frac{V_S}{2}$

Solving for $I_L$, we find:
$I_L = \frac{V_S}{2R}$

To analyze this RC circuit, we determine the capacitor's voltage at three key points in time: the initial state, the final state, and the transient behavior described by the time constant.

First, for $t < 0$, the switch is at position 1 and the circuit is in steady state. The capacitor acts as an open circuit, and its initial voltage is determined by the voltage divider on the left: $V_C(0^-) = 10\text{V} \times \frac{2\Omega}{3\Omega + 2\Omega} = 4\text{V}$. Since capacitor voltage cannot change instantly, $V_C(0^+) = 4\text{V}$.

Next, as $t \to \infty$, the switch is at position 2 and the circuit reaches a new steady state. The capacitor is again an open circuit, so no current flows through the 4Ω resistor. The capacitor's final voltage is the voltage across the 2Ω resistor in parallel with the 5A source: $V_C(\infty ) = 5\text{A} \times 2\Omega = 10\text{V}$.

The time constant for $t > 0$ is $\tau = R_{Th}C$. The Thevenin resistance seen by the capacitor is the sum of the 4Ω and 2Ω resistors, as the current source is open-circuited: $R_{Th} = 4\Omega + 2\Omega = 6\Omega$. Thus, $\tau = 6\Omega \times 0.1\text{F} = 0.6\text{s}$.

Finally, we use the general formula for the capacitor voltage transient response: V_C(t) = V_C(\infty ) + \[V_C(0^+) - V_C(∞)]$e^{-t/\tau}$. Substituting the values we found, we get$V_C(t) = 10 + [4 - 10]e^{-t/0.6}$, which simplifies to$V_C(t) = 10 - 6e^{-t/0.6}$ V.

In a parallel RLC circuit operating at resonance, the reactive currents through the inductor and capacitor are equal and opposite, effectively canceling each other out. This means the entire current from the source, $I$, flows through the resistor, so $|I_R| = |I|$.

The magnitude of the current flowing through the inductor at resonance is the quality factor, $Q$, multiplied by the total source current: $|I_L| = Q|I|$.

Therefore, the ratio we need to find is simply the quality factor of the circuit:
$\frac{|I_L|}{|I_R|} = \frac{Q|I|}{|I|} = Q$
For a parallel RLC circuit, the quality factor is calculated as $Q = R\sqrt{C/L}$. Substituting the given component values:
$Q = 10 \sqrt{\frac{10 \times 10^{-6}}{10 \times 10^{-3}}} = 10 \sqrt{10^{-3}} \approx 0.316$

The Z-parameters for the given T-network can be determined by inspection. The off-diagonal parameters are equal to the shunt impedance, $Z_{12} = Z_{21} = Z_c$. The diagonal parameter $Z_{22}$ is the sum of the impedances in the second loop, so $Z_{22} = Z_b + Z_c$.

From the provided matrix, we can directly identify $Z_{12} = j\omega$, which means $Z_c = j\omega$.
We also see that $Z_{22} = 3 + 2j\omega$.
Substituting our known values into the equation for $Z_{22}$ gives:
$3 + 2j\omega = Z_b + j\omega$
Solving for $Z_b$, we find $Z_b = 3 + j\omega$.

The problem states that $Z_b(j\omega) = R_b + j\omega$. By comparing this form to our calculated expression for $Z_b$, we can see that the resistive part is $R_b = 3~\Omega$.

Calculating the energy of this signal by integrating $|x(t)|^2$ is challenging. A much easier approach is to use Parseval's theorem and work in the frequency domain. The total energy $E$ is given by $E = \frac{1}{2\pi} \int_{-\infty }^{\infty } |X(\omega)|^2 d\omega$, where $X(\omega)$ is the Fourier transform of $x(t)$.

The signal $x(t)$ is a sinc-type function. Its Fourier transform, $X(\omega)$, is a rectangular pulse. For the given signal, this pulse has a constant amplitude of $\frac{1}{4}$ over the angular frequency range from $-4\pi$ to $4\pi$, and is zero elsewhere.

Plugging this into the energy formula, the integral becomes straightforward:
$E = \frac{1}{2\pi} \int_{-4\pi}^{4\pi} \left(\frac{1}{4}\right)^2 d\omega = \frac{1}{2\pi} \cdot \frac{1}{16} \cdot \left[4\pi - (-4\pi)\right]$
$E = \frac{1}{32\pi} (8\pi) = \frac{1}{4} = 0.25$ J.

The Ebers-Moll model is a powerful large-signal model because it is derived from the fundamental physics of the BJT's p-n junctions. It represents the transistor as two interconnected diodes (the base-emitter and base-collector junctions) along with two dependent current sources. The equations of this model are general and hold true regardless of whether these junctions are forward-biased or reverse-biased. Consequently, the model accurately describes the transistor's behavior in all of its primary operating modes: active, saturation, and cut-off.

An NMOS transistor operating in the linear region acts as a voltage-controlled resistor. The value of this resistance, $r_{ds}$, is determined by the channel's properties. The formula for this resistance is $r_{ds} = \frac{1}{\mu_n C_{ox} \frac{W}{L}(V_{GS} - V_T)}$.

From this equation, we can see that the resistance $r_{ds}$ is inversely proportional to the device width $W$ and the overdrive voltage $(V_{GS} - V_T)$. It is directly proportional to the device length $L$.

Therefore, increasing $V_{GS}$ strengthens the conductive channel, which in turn decreases the channel resistance. The statement that the resistance increases with $V_{GS}$ is consequently incorrect.

To solve this, we must first determine if the diode is conducting (ON) or not (OFF). We can do this by making an assumption and checking for a logical contradiction.

Let's assume the diode is ON. In this state, it acts like a voltage source with a $0.7$ V drop. This means the voltage at the node connecting $R_1$ and $R_2$ must be $2 \text{ V} - 0.7 \text{ V} = 1.3 \text{ V}$. Based on this, the current $i_2$ flowing down through $R_2$ is $1.3 \text{ V} / 6 \text{ k}\Omega \approx 0.217 \text{ mA}$. However, the current flowing into that same node through $R_1$ is $(2 \text{ V} - 1.3 \text{ V}) / 2 \text{ k}\Omega = 0.35 \text{ mA}$. Since more current flows into the node than leaves through $R_2$, the diode would have to conduct current backward, which is impossible.

Our initial assumption was wrong, so the diode must be OFF. An OFF diode acts as an open circuit, meaning no current flows through that branch. The circuit simplifies to $R_1$ and $R_2$ in series with the 2 V source. The total series resistance is $2 \text{ k}\Omega + 6 \text{ k}\Omega = 8 \text{ k}\Omega$. The current $i_2$ is then found using Ohm's law:
$i_2 = \frac{2 \text{ V}}{8 \text{ k}\Omega} = 0.25 \text{ mA}$

This circuit uses a current mirror configuration to set the current $I_2$. The key relationship comes from applying Kirchhoff's Voltage Law around the base-emitter loop containing both transistors. This reveals that the voltage across resistor $R_2$ is simply the difference between the two base-emitter voltages: $I_2 R_2 = V_{BE1} - V_{BE2}$.

Since the transistors are matched, we can use the diode equation $V_{BE} = V_T \ln(I_C/I_S)$ to express this voltage difference in terms of the currents, resulting in $I_2 R_2 = V_T \ln(I_1/I_2)$. Rearranging this formula to solve for $R_2$:
$R_2 = \frac{V_T}{I_2} \ln\left(\frac{I_1}{I_2}\right)$
Substituting the given values ($V_T = 26 \text{ mV}$, $I_1 = 1 \text{ mA}$, and $I_2 = 100 \text{ \mu A}$):
$R_2 = \frac{26 \times 10^{-3}}{100 \times 10^{-6}} \ln\left(\frac{1 \times 10^{-3}}{100 \times 10^{-6}}\right) = 260 \ln(10) \approx 598.7 \, \Omega$

The mid-band frequency range of an amplifier is where the voltage gain is maximal and constant. This behavior is defined by the reactances of the circuit's capacitors, given by $X_C = \frac{1}{j2\pi fC}$.

Coupling and bypass capacitances are intentionally large. In the mid-band, the frequency $f$ is high enough to make their reactance $X_C$ negligibly small, so they act as short circuits.

Device parasitic capacitances are very small and internal to the transistor. In the mid-band, the frequency is not yet high enough for their reactance to become small. Thus, their reactance remains very large, and they behave as open circuits. The effects of these parasitic capacitances only become significant at high frequencies, causing the gain to roll off.

The conversion is from a CMOS inverter to a 2-input NOR gate while maintaining similar worst-case rise and fall times.

Charging Time (Pull-Up):
In the NOR gate, the pull-up network consists of two PMOS transistors in series. For the output to charge (pull-up to $V_{DD}$), both PMOS transistors must be ON. The total resistance is the sum of the two individual resistances. To match the charging time of the inverter, which has one PMOS, the total resistance must be the same. Since resistance $R \propto 1/W$, the width of each PMOS in the NOR gate must be doubled to halve its resistance, so that the series combination ($R' + R'$) has the same resistance as the original inverter's PMOS.

Discharging Time (Pull-Down):
The worst-case (slowest) discharging for the NOR gate occurs when only one of the two parallel NMOS transistors is conducting. To match the worst-case discharge time of the inverter, the resistance of this single conducting NMOS must be equal to the resistance of the single NMOS in the inverter. This means the width of the NMOS transistors should remain unchanged.

The average power dissipated in the resistor is determined by the fraction of time the XOR gate's output is HIGH (5V). This fraction is the duty cycle of the output signal.

First, let's find the instantaneous power dissipated when the output is HIGH:
$P_{ON} = \frac{V^2}{R} = \frac{5^2}{10 \times 10^3 \Omega} = 2.5 \text{ mW}$

The average power is this value multiplied by the signal's duty cycle. By tracing the states of the flip-flops, it's found that the XOR gate's output signal has a repeating pattern that is HIGH for 3 clock cycles and LOW for 2, within a 5-cycle period. Thus, the duty cycle is $\frac{3}{5}$.

Finally, we calculate the average power:
$P_{avg} = P_{ON} \times \text{Duty Cycle} = 2.5 \text{ mW} \times \frac{3}{5} = 1.5 \text{ mW}$

To implement a function using a multiplexer, we treat the specified variables, A and B, as the select lines ($S_1=A, S_0=B$). The task is to find what the data inputs $I_0$ through $I_3$ should be, based on the remaining variable, $C_{\in}$.

We start with the standard equation for a full adder's carry-out:
$C_{out} = AB + BC_{\in} + AC_{\in}$

Now, let's analyze this equation for each combination of the select lines A and B:

• If $(A,B) = (0,0)$, the MUX selects $I_0$. The equation becomes $C_{out}=0$. Thus, we set $I_0=0$.
• If $(A,B) = (0,1)$, the MUX selects $I_1$. The equation becomes $C_{out}=C_{\in}$. Thus, we set $I_1=C_{\in}$.
• If $(A,B) = (1,0)$, the MUX selects $I_2$. The equation becomes $C_{out}=C_{\in}$. Thus, we set $I_2=C_{\in}$.
• If $(A,B) = (1,1)$, the MUX selects $I_3$. The equation becomes $C_{out}=1+C_{\in}+C_{\in}=1$. Thus, we set $I_3=1$.

To find the value of the derivative $\frac{dy}{dt}$ at $t=0^+$, we'll use the Initial Value Theorem (IVT). The Laplace transform of the derivative is $L\left\{\frac{dy}{dt}\right\} = sY(s) - y(0)$.

First, let's determine the output $Y(s)$ for a unit step input, where $U(s) = \frac{1}{s}$:
$Y(s) = G(s)U(s) = \frac{s-2}{(s+1)(s+3)} \cdot \frac{1}{s} = \frac{s-2}{s(s+1)(s+3)}$

Next, we find the initial condition $y(0)$ by applying the IVT to $Y(s)$:
$y(0) = \lim_{s \to \infty } sY(s) = \lim_{s \to \infty } \frac{s-2}{(s+1)(s+3)} = 0$

Now, substitute $y(0)=0$ back into the derivative's transform: $L\left\{\frac{dy}{dt}\right\} = sY(s) - 0 = \frac{s-2}{(s+1)(s+3)}$.

Finally, apply the IVT to this result to find the initial value of the derivative:
$\left.\frac{dy}{dt}\right|_{t=0^+} = \lim_{s \to \infty } s \cdot L\left\{\frac{dy}{dt}\right\} = \lim_{s \to \infty } \frac{s(s-2)}{(s+1)(s+3)} = \lim_{s \to \infty } \frac{s^{2-}2s}{s^{2+}4s+3} = 1$

To determine the number of encirclements, we can trace the path of the Nyquist plot by checking its start and end points.

First, let's find the starting point of the plot at $\omega=0$. Substituting $s=j0$ into the transfer function gives $G(j0) = \frac{1-0}{4+0} = 0.25$. The plot begins on the positive real axis.

Next, we find the ending point as $\omega \to \infty$. For very large frequencies, the transfer function behaves like $G(s) \approx \frac{-s}{2s} = -0.5$. The plot terminates on the negative real axis.

The path for $\omega \in [0, \infty )$ is a semicircle in the lower-half complex plane, moving from $0.25$ to $-0.5$. The full plot for $\omega \in (-\infty , \infty )$ is symmetrical about the real axis. Since the entire plot is confined to the real-axis interval $[-0.5, 0.25]$, it never encloses the point $-1+j0$.

The theoretical minimum average codeword length for error-free reconstruction is given by the entropy, $H$, of the source. We calculate the entropy using the formula $H = \sum_{i} p_i \log_2(\frac{1}{p_i})$ for the given symbol probabilities.

For the probabilities $\{\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \frac{1}{8}\}$, the entropy is:
$H = \left(\frac{1}{2}\log_2 2\right) + \left(\frac{1}{4}\log_2 4\right) + \left(\frac{1}{8}\log_2 8\right) + \left(\frac{1}{8}\log_2 8\right)$
$H = \left(\frac{1}{2} \cdot 1\right) + \left(\frac{1}{4} \cdot 2\right) + \left(\frac{1}{8} \cdot 3\right) + \left(\frac{1}{8} \cdot 3\right)$
$H = 0.5 + 0.5 + 0.375 + 0.375 = 1.75$
Therefore, the minimum required average codeword length is 1.75 bits per symbol.

To find the required bandwidth, we first need to calculate the bit rate ($R_b$) of the PCM signal. The bit rate is the product of the sampling frequency ($f_s$) and the number of bits per sample ($n$).
$R_b = f_s \times n = 8 \text{ kHz} \times 8 \text{ bits/sample} = 64 \text{ kbps}$

Next, we use the Nyquist formula for the minimum bandwidth of a multi-level signal, which is $BW_{min} = \frac{R_b}{2 \log_2 M}$.
For 4-level PAM, the number of levels $M=4$.
Plugging in the values, we find the minimum bandwidth:
$BW_{min} = \frac{64 \text{ kbps}}{2 \log_2(4)} = \frac{64}{2 \times 2} = 16 \text{ kHz}$

The particle's trajectory is governed by the Lorentz force law, $\vec{F} = q(\vec{v} \times \vec{B})$. We can analyze the effect of this force by breaking the velocity into two components.

The velocity component parallel to the magnetic field, $v_z\hat{z}$, experiences no force, since $q(v_z\hat{z} \times B\hat{z}) = 0$. This means the particle continues to move along the z-axis with a constant velocity $v_z$.

The velocity component perpendicular to the field, $v_x\hat{x}$, creates a force $\vec{F} = q(v_x\hat{x} \times B\hat{z}) = -qv_x B\hat{y}$. This force is always perpendicular to the velocity in the xy-plane, acting as a centripetal force that causes uniform circular motion.

Combining the constant linear motion along the z-axis with the circular motion in the xy-plane results in a helical path.

Let's break down what we can and cannot determine from the given information.

The electric field vector is purely in the $\hat{x}$ direction, which immediately tells us the wave is linearly polarized. The phase velocity is defined as $v_p = \omega/\beta$, and the group velocity is $v_g = (d\beta/d\omega)^{-1}$. Since we know the relationship between $\beta$ and $\omega$, we can directly calculate both of these velocities.

However, the power flux is given by the magnitude of the time-averaged Poynting vector, $S = \frac{1}{2} \frac{|E_x|^2}{\eta}$. This calculation depends on the intrinsic impedance of the medium, $\eta = \sqrt{\mu/\epsilon}$. The problem provides no information about the medium's permeability ($\mu$) or permittivity ($\epsilon$), so we cannot find $\eta$. Without the impedance, the power flux cannot be determined.

For light to be guided through the fiber, the angle of incidence $\theta_i$ must not exceed the fiber's maximum acceptance angle. The sine of this maximum angle is determined by the numerical aperture (NA) of the fiber.

The relationship is given by the formula:
$\sin \theta_{i, \text{max}} = \text{NA} = \sqrt{n_1^2 - n_2^2}$

Substituting the given refractive indices for the core ($n_1 = 1.5$) and cladding ($n_2 = 1.4$):
$\sin \theta_{i, \text{max}} = \sqrt{1.5^2 - 1.4^2} = \sqrt{2.25 - 1.96} = \sqrt{0.29} \approx 0.5385$

To find the angle itself, we take the inverse sine:
$\theta_{i, \text{max}} = \sin^{-1}(0.5385) \approx 32.58^\circ$

To ensure a stable numerical solution using the forward Euler method, we must analyze the stability condition. For a linear differential equation of the form $\frac{dx}{dt} = \lambda x + c$, the method is stable if the time step $h$ satisfies $|1 + h\lambda| \le 1$.

In our specific problem, $\frac{dx}{dt} = -3x + 2$, we can identify the coefficient $\lambda = -3$.

Plugging this into the stability condition gives us $|1 + h(-3)| \le 1$, which simplifies to $|1 - 3h| \le 1$.

Solving this absolute value inequality, we get $-1 \le 1 - 3h \le 1$. Subtracting 1 from all parts gives $-2 \le -3h \le 0$. Finally, dividing by $-3$ and reversing the inequality signs yields $0 \le h \le \frac{2}{3}$.

Therefore, the largest time step $h_{max}$ that can be used without the solution becoming unstable is $\frac{2}{3}$, which is approximately $0.667$.

This line integral over a closed curve is a perfect application of Green's Theorem. The theorem allows us to convert the line integral $\oint_C (P\,dx + Q\,dy)$ into a double integral over the region $R$ enclosed by the curve, using the formula $\iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA$.

For this problem, we have the components $P = xy^2$ and $Q = x^2y$. Let's calculate the necessary partial derivatives.
The partial derivative of $Q$ with respect to $x$ is $\frac{\partial}{\partial x}(x^2y) = 2xy$.
The partial derivative of $P$ with respect to $y$ is $\frac{\partial}{\partial y}(xy^2) = 2xy$.

Substituting these into the formula for Green's Theorem gives:
$\iint_R (2xy - 2xy) \,dA = \iint_R 0 \,dA$
The integral of zero over any region is 0.

To find the probability $P(X+Y \leq 1)$, we must integrate the joint density function over the region where this inequality is true, given the constraints $0 \leq x \leq 1$ and $0 \leq y \leq 1$. This defines a triangular region of integration.

We set up the double integral as follows:
$P(X+Y \leq 1) = \int_{0}^{1} \int_{0}^{1-x} (x+y) \,dy\,dx$
First, we solve the inner integral with respect to $y$:
$\int_{0}^{1} \left[ xy + \frac{y^2}{2} \right]_{y=0}^{y=1-x} \,dx = \int_{0}^{1} \left( x(1-x) + \frac{(1-x)^2}{2} \right) \,dx$
Simplifying the expression inside the integral gives $\int_{0}^{1} \left( \frac{1-x^2}{2} \right) \,dx$.
Finally, integrating with respect to $x$ yields the result:
$\left[ \frac{x}{2} - \frac{x^3}{6} \right]_{0}^{1} = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}$

The trace of a matrix is the sum of its main diagonal elements. We are given that the trace is 14, so we have the equation $a + 5 + 2 + b = 14$, which simplifies to $a + b = 7$.

The given matrix is upper triangular, so its determinant is the product of its main diagonal elements. We are given that the determinant is 100, which gives us the equation $(a)(5)(2)(b) = 100$, simplifying to $ab = 10$.

Now we must solve the system of equations: $a+b=7$ and $ab=10$. By inspection, the two numbers that add up to 7 and multiply to 10 are 2 and 5. So, the values for $a$ and $b$ are 2 and 5, in some order.

Finally, we calculate the required value, $|a-b|$. This will be $|5-2|$ or $|2-5|$, both of which equal 3.

This complex resistor network can be simplified using techniques like star-delta transformations. The result of these transformations is an equivalent circuit with two main pathways between terminals a and b.

The first pathway consists of two effective resistors in series, each with a value of $(4 \Omega \, || \, 1 \Omega) = \frac{4}{5} \Omega$. The total resistance of this path is $\frac{4}{5} + \frac{4}{5} = \frac{8}{5} \Omega$.

The second pathway is a single effective resistor with the same value of $\frac{4}{5} \Omega$.

The total equivalent resistance is these two pathways combined in parallel:
$R_{eq} = \left(\frac{8}{5}\right) || \left(\frac{4}{5}\right) = \frac{\frac{8}{5} \times \frac{4}{5}}{\frac{8}{5} + \frac{4}{5}} = \frac{32/25}{12/5} = \frac{8}{15} \Omega$.

To find the current flowing through resistor $R_2$, we first need to determine the voltage $V_x$ across $R_3$. We can achieve this by applying Kirchhoff's Voltage Law (KVL) to the entire outer loop of the circuit.

The KVL equation for the outer loop is $60 - V_{R1} - V_{R2} - V_x = 0$. The voltage drops across the resistors, when expressed in terms of $V_x$, lead to the equation $60 - 0.8V_x - 0.6V_x - V_x = 0$.

Combining the terms involving $V_x$ gives $60 - 2.4V_x = 0$. Solving for $V_x$, we find that $V_x = \frac{60}{2.4} = 25$ V.

Finally, since $R_2$ and $R_3$ are in series, the same current flows through them. Using Ohm's Law on $R_3$, the current through $R_2$ is:
$I_{R2} = \frac{V_x}{R_3} = \frac{25 \text{ V}}{5 \, \Omega} = 5 \text{ A}$

The goal is to make the discrete-time filter's impulse response, $g[n]$, identical to the sampled version of the continuous-time filter's impulse response, $h(t)$. First, we find $h(t)$ from $H(s)$ using partial fraction expansion: $H(s) = \frac{1}{s+2} + \frac{1}{s+4}$, so the inverse Laplace transform is $h(t) = (e^{-2t} + e^{-4t})u(t)$.

Next, we sample $h(t)$ at a frequency of $f_s=2$ Hz, which means the sampling period is $T_s = 1/2$ s. The resulting discrete-time sequence is $h[n] = h(nT_s) = h(n/2) = (e^{-n} + e^{-2n})u[n]$.

The Z-transform of this sequence gives the desired discrete-time transfer function:
$G(z) = \mathcal{Z}\{h[n]\} = \frac{z}{z-e^{-1}} + \frac{z}{z-e^{-2}}$

Combining the fractions gives:
$G(z) = \frac{z(z-e^{-2}) + z(z-e^{-1})}{(z-e^{-1})(z-e^{-2})} = \frac{2z^2 - z(e^{-1}+e^{-2})}{z^2 - z(e^{-1}+e^{-2}) + e^{-3}}$

Substituting numerical values yields $\frac{2z^2 - 0.5032z}{z^2 - 0.5032z + 0.04979}$. By comparing the denominator of this result to the given form for $G(z)$, we can identify the constant term $k$ as $e^{-3} \approx 0.0498$.

The new sequence $x_1[n]$ is an "up-sampled" version of $x[n]$, created by inserting two zeros between each of its original samples. This process has a direct and simple effect in the frequency domain.

A key property of the DFT states that up-sampling a sequence by a factor of $L$ (here, $L=3$) causes its DFT spectrum to be repeated $L$ times.

Therefore, the 12-point DFT $X_1[k]$ consists of three back-to-back copies of the original 4-point DFT $X[k]$.
$X_1[k] = \{ \underbrace{12, 2j, 0, -2j}_{\text{1st copy}}, \underbrace{12, 2j, 0, -2j}_{\text{2nd copy}}, \underbrace{12, 2j, 0, -2j}_{\text{3rd copy}} \}$