GATE EE 2016 Set 2

To find the simplified logical expression, we group the 1s in the Karnaugh map. The goal is to form the largest possible rectangular groups of 1s, where the number of cells in each group is a power of two.

First, we can see a group of four 1s covering the entire bottom row. In this row, the variable $A$ is always 1, while B and C change. This group simplifies to the term $A$.

Next, we can form another group of four by combining the 1s in the first column ($BC=00$) and the last column ($BC=10$). In this wrapped-around group, the variable $C$ is consistently 0, while A and B vary. This gives us the term $\bar{C}$.

Combining the terms from these two essential groups with an OR operation gives the final simplified expression: $A + \bar{C}$.

This circuit is an inverting amplifier configuration where the gain is the negative ratio of the feedback impedance ($Z_f$) to the input impedance ($Z_{\in}$). Here, the feedback impedance is the parallel combination of resistor $R_2$ and capacitor $C$. The transfer function $H(j\omega) = V_{out}/V_{\in}$ is calculated as:

$H(j\omega) = -\frac{Z_f}{Z_{\in}} = -\frac{R_2 || (1/j\omega C)}{R_1} = \frac{-R_2/R_1}{1 + j\omega R_2 C}$

As frequency $\omega$ approaches zero, the gain approaches its maximum value of $-R_2/R_1$. As frequency increases towards infinity, the gain magnitude approaches zero. This characteristic of passing low-frequency signals while attenuating high-frequency ones defines a low-pass filter.

To find the primary current $I$, we can analyze an equivalent circuit by referring the secondary impedance to the primary side. The turns ratio is $n = N_2/N_1 = 100$.

First, let's calculate the secondary impedance in phasor form: $Z_S = R - jX_C = (80 - j40) \, \text{k}\Omega$.

This impedance, when referred to the primary side, becomes $Z'_S = \frac{Z_S}{n^2} = \frac{80000 - j40000}{100^2} = (8 - j4) \, \Omega$.

The total impedance seen by the source is the sum of the primary inductor's impedance and this referred impedance: $Z_{total} = jX_L + Z'_S = j10 + (8 - j4) = (8 + j6) \, \Omega$.

The magnitude of this impedance is $|Z_{total}| = \sqrt{8^2 + 6^2} = \sqrt{100} = 10 \, \Omega$.

Finally, the rms value of the current $I$ is given by Ohm's law: $I = \frac{V_{rms}}{|Z_{total}|} = \frac{100 \, \text{V}}{10 \, \Omega} = 10 \, \text{A}$.

To find the system's response, we can analyze the problem in the Laplace domain. First, we take the Laplace transform of the differential equation, which gives $sY(s) + \frac{1}{6}Y(s) = 3X(s)$. The input signal $x(t)=3e^{-\frac{t}{3}}u(t)$ has the transform $X(s) = \frac{3}{s+1/3}$.

Substituting $X(s)$ into the system equation and solving for the output $Y(s)$ yields:
$Y(s)\left(s+\frac{1}{6}\right) = 3\left(\frac{3}{s+1/3}\right)$, so $Y(s) = \frac{9}{\left(s+\frac{1}{6}\right)\left(s+\frac{1}{3}\right)}$.

Next, we use partial fraction expansion to simplify $Y(s)$:
$Y(s) = \frac{A}{s+1/6} + \frac{B}{s+1/3} = \frac{54}{s+1/6} - \frac{54}{s+1/3}$.

Finally, taking the inverse Laplace transform of each term gives the time-domain response:
$y(t) = \left(54e^{-\frac{t}{6}} - 54e^{-\frac{t}{3}}\right)u(t)$.

Multiplying the signal $x(t)$ by a cosine function is a classic example of amplitude modulation. This operation shifts the frequency spectrum of the original signal.

First, let's find the frequency of the cosine carrier, $\cos(2000\pi t)$. From the standard form $\cos(2\pi f_c t)$, we see that $2\pi f_c = 2000\pi$, which gives a carrier frequency of $f_c = 1000 \text{ Hz}$, or 1 kHz.

The original signal $x(t)$ has a maximum frequency of $f_m = 5$ kHz. Modulation creates new frequency components by shifting the original spectrum up and down by $f_c$. The new highest frequency will be the sum of the original maximum frequency and the carrier frequency: $f_{max} = f_m + f_c = 5 \text{ kHz} + 1 \text{ kHz} = 6 \text{ kHz}$.

Let's express the function using the real and imaginary components of $z$. By setting $z = x + iy$, its complex conjugate is $z^* = x - iy$.
The function can then be rewritten as $f(z) = (x + iy) + (x - iy) = 2x$.

The resulting function has a real part $u(x, y) = 2x$ and an imaginary part $v(x, y) = 0$. Since both $u$ and $v$ are simple polynomials, the function $f(z)$ is continuous everywhere.

To be analytic, the function must satisfy the Cauchy-Riemann equations. We calculate the partial derivatives: $\frac{\partial u}{\partial x} = 2$, $\frac{\partial u}{\partial y} = 0$, $\frac{\partial v}{\partial x} = 0$, and $\frac{\partial v}{\partial y} = 0$.
The first Cauchy-Riemann equation, $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$, leads to the statement $2=0$, which is never true. Since the equations are not satisfied anywhere, the function is nowhere analytic.

Let $\lambda$ be an eigenvalue of the matrix $P$. A fundamental property in linear algebra states that if a matrix satisfies a polynomial equation, its eigenvalues must also satisfy that same scalar polynomial equation.

We are given the matrix equation $P^3 = P$. Therefore, the eigenvalue $\lambda$ must satisfy the corresponding scalar equation $\lambda^3 = \lambda$.

To find the possible values of $\lambda$, we solve this equation:
$\lambda^3 - \lambda = 0$
$\lambda(\lambda^2 - 1) = 0$
$\lambda(\lambda - 1)(\lambda + 1) = 0$

This equation yields the possible values for the eigenvalues: $\lambda = 0$, $\lambda = 1$, and $\lambda = -1$. The set of eigenvalues for the 3x3 matrix must consist of values from this set.

To solve this differential equation, we'll use the Laplace Transform method. Applying the transform to each term of the equation gives:
$\$[s^2Y(s) - sy(0) - y'(0)]\$ + 2[sY(s) - y(0)] + Y(s) = 0$
Next, we plug in the initial conditions $y(0) = 0$ and $y'(0) = 1$:
$\$[s^2Y(s) - 0 - 1]\$ + 2[sY(s) - 0] + Y(s) = 0$
Now, we can algebraically solve for $Y(s)$:
$Y(s)(s^2 + 2s + 1) = 1 \implies Y(s) = \frac{1}{(s+1)^2}$
Finally, we find the solution $y(t)$ by taking the inverse Laplace transform of $Y(s)$. From standard transform pairs, we know this corresponds to:
$y(t) = te^{-t}u(t)$

This problem asks for the value of a line integral of a vector field, $\vec{F} = \langle 2xy^2, 2x^2y, 1 \rangle$. A quick check shows this field is conservative (its curl is zero), which means the integral's value is independent of the path taken and depends only on the start and end points.

For a conservative field, we can find a scalar potential function $\phi$ such that $\vec{F} = \nabla\phi$. By integrating the components of $\vec{F}$, we find this potential function is $\phi(x,y,z) = x^2y^2 + z$.

The Fundamental Theorem for Line Integrals allows us to evaluate the integral simply by finding the difference in the potential function at the endpoints.

The value is thus $\phi(\text{end point}) - \phi(\text{start point})$, which is $\phi(1,1,1) - \phi(0,0,0)$.
This calculates to $(1^2 \cdot 1^2 + 1) - (0^2 \cdot 0^2 + 0) = 2$.

The given condition, $f(-x) = -f(x)$, is the mathematical definition of an odd function.

A key property of Fourier series is that they reflect the symmetry of the function they represent. The Fourier series of an odd function simplifies because the coefficients for the constant term ($a_0$) and the cosine terms ($a_k$) all become zero.

This leaves only the sine terms, as sine is also an odd function. The resulting series is therefore a sum of sine functions, which has the general form:
$f(x) = \sum_{k=1}^{\infty } b_k \sin(kx)$

In a series AC circuit, voltages add as phasors. The source voltage $\vec{V_S}$ is the vector sum of the voltage across the resistor $\vec{V_R}$ and the voltage across the coil $\vec{V_C}$. This means the arithmetic sum of their magnitudes must be greater than the source voltage magnitude, so $V_R + V_C > V_S$. This principle immediately eliminates options A ($65+35=100$) and B ($50+50=100$), as equality only holds for purely resistive circuits.

Now, let's compare the remaining options. Option D presents values of $60$ V and $80$ V. These satisfy the Pythagorean theorem, $60^2 + 80^2 = 100^2$, which would be true if the resistor voltage and coil voltage were perpendicular. This only occurs if the coil is a pure inductor. However, the diagram clearly depicts a practical coil with its own internal resistance.

Since the coil is not ideal, option D is incorrect. By eliminating the other three options based on physical principles, option C is the only one that remains. It correctly reflects that the arithmetic sum of voltages ($60+90=150$ V) is greater than the source voltage.

The average power is determined by the interaction of voltage and current components at the same frequency. Since the voltage $v(t)$ contains only the fundamental frequency ($\omega$), we only need to consider the current component at that same frequency, which is $10 \sin(\omega t - 60^{\circ})$. The higher-frequency harmonics in the current ($3\omega$ and $5\omega$) do not contribute to the average power.

We calculate the average power using the RMS values of the fundamental voltage and current.
The formula is $P = V_{rms} I_{rms} \cos(\phi)$, where $\phi$ is the phase difference between the voltage and current.

Here, $V_{rms} = \frac{100}{\sqrt{2}}$ V and $I_{rms} = \frac{10}{\sqrt{2}}$ A for the fundamental components.
The phase angle difference is $\phi = |0^{\circ} - (-60^{\circ})| = 60^{\circ}$.
Therefore, the average power is $P = \left(\frac{100}{\sqrt{2}}\right) \left(\frac{10}{\sqrt{2}}\right) \cos(60^{\circ}) = \frac{1000}{2} \times 0.5 = 250$ W.

Remote backup protection is provided by relays at adjacent buses that operate with a time delay if the primary protection fails. The primary protection for the line segment including relay $R_4$ is provided by $R_3$ and $R_4$. If a fault occurs on this line and $R_4$ fails to operate, the fault current will continue to flow from the middle bus to the faulted line.

This fault current is fed to the middle bus from two directions:

1. From source $S_1$ via the upper parallel line, passing through relay $R_1$.
2. From source $S_2$ via the line connecting the right two buses, passing through relay $R_6$.

The relays at the remote buses that see this persistent fault current are $R_1$ and $R_6$. They are configured to operate after a time delay to serve as remote backup for $R_4$. Relays $R_2$ and $R_5$ are at the same bus as $R_4$ and would be involved in local backup, not remote backup. Thus, the remote backup relays for $R_4$ are $R_1$ and $R_6$.

The size of the Jacobian matrix is determined by the number of unknown variables in the load flow equations. For an $N$-bus system, there are $2N$ potential variables: a voltage magnitude $|V|$ and an angle $\delta$ for each bus.

The number of unknowns is reduced based on the type of each bus:

1. Slack Bus: One generator bus is designated as the slack bus. Both its $|V|$ and $\delta$ are specified, contributing 0 unknowns.
2. PV (Generator) Buses: For the remaining $10 - 1 = 9$ generator buses, $|V|$ is specified but $\delta$ is unknown. This gives 9 unknowns.
3. PQ (Load) Buses: For the remaining $100 - 10 = 90$ load buses, both $|V|$ and $\delta$ are unknown, contributing $90 \times 2 = 180$ unknowns.

Summing the unknowns from all buses gives $0 + 9 + 180 = 189$. The Jacobian matrix is a square matrix with dimensions equal to the number of unknowns, resulting in a size of $189 \times 189$.

First, we determine the line's natural loading capability by calculating its characteristic impedance ($Z_c$) and Surge Impedance Loading (SIL).
The characteristic impedance is $Z_c = \sqrt{L/C} = \sqrt{1.6 \times 10^{-3} / 10 \times 10^{-9}} = 400 \, \Omega$.
The SIL is the power level at which the line's reactive power is balanced: $SIL = V_{LL}^2 / Z_c = (400 \, \text{kV})^2 / 400 \, \Omega = 400 \, \text{MW}$.

The line is delivering a 300 MW load, which is less than its SIL of 400 MW. When a transmission line is loaded below its SIL, it generates more reactive power than it consumes, exhibiting a net capacitive effect. To counteract this and absorb the excess reactive power to prevent voltage rise, we must connect a shunt inductor. Therefore, the required compensation is inductive.

The two dielectric materials, A and B, are arranged in a parallel configuration between the capacitor plates. For components in parallel, the potential difference (voltage) $V$ across them is identical.

The distance $d$ between the plates is also the same for both regions. The relationship between the uniform electric field $E$, voltage $V$, and plate separation $d$ is given by $E = V/d$. Since both $V$ and $d$ are the same for regions A and B, the electric field must also be the same in both. Thus, the electric field in region B is equal to the field in region A.

The magnitude of the sinusoidal short-circuit current changes over time, limited by different reactances at different stages.

The initial value of the current, called the sub-transient current ($I''$), is determined by the sub-transient reactance ($X_d''$).
$I'' = \frac{V}{X_d''}$
The final or steady-state value of the current ($I$) is determined by the synchronous reactance ($X_d$).
$I = \frac{V}{X_d}$
The ratio of the initial to the final sinusoidal current is found by dividing these two equations. The voltage term $V$ cancels out.
$\frac{I''}{I} = \frac{V/X_d''}{V/X_d} = \frac{X_d}{X_d''}$
Substituting the given per-unit values, we get:
$\frac{I''}{I} = \frac{1.0}{0.2} = 5.0$

For a linear time-invariant system, the steady-state response to a sinusoidal input is a sinusoid at the same frequency, but with its amplitude and phase altered by the system's frequency response. The input signal $\cos(t)$ has an angular frequency of $\omega = 1$ rad/s.

To find the system's effect, we first determine the frequency response by setting $s = j\omega$ in the transfer function:
$H(j\omega) = \frac{1}{1+j\omega}$

The output amplitude, $A$, is the magnitude of the frequency response evaluated at the input frequency, $\omega = 1$.
$A = |H(j1)| = \left|\frac{1}{1+j1}\right| = \frac{1}{\sqrt{1^{2+}1^2}} = \frac{1}{\sqrt{2}}$

Thus, the value of A is approximately $0.707$.

The highly inductive load ensures the DC output current is a smooth, constant value, which is given as $I_0 = 100$ A.

In a three-phase bridge rectifier, each diode is active for one-third of the total AC cycle. This means each diode conducts the full DC current for a period of $120^\circ$, or $2\pi/3$ radians, and carries no current for the remaining $240^\circ$.

To find the RMS current in a single diode, we average the squared current over a full cycle and take the square root. Since the current is non-zero only during the conduction interval, the calculation is:
$I_{D,rms} = \sqrt{\frac{1}{2\pi} \int_{0}^{2\pi/3} I_0^2 \,d(\omega t)}$
Solving the integral gives:
$I_{D,rms} = \sqrt{\frac{I_0^2}{2\pi} \left[ \omega t \right]_0^{2\pi/3}} = \sqrt{\frac{I_0^2}{2\pi} \cdot \frac{2\pi}{3}} = \frac{I_0}{\sqrt{3}}$
Plugging in the known DC current:
$I_{D,rms} = \frac{100}{\sqrt{3}} \approx 57.74 \text{ A}$

The peak voltage stress on the switch S occurs when it is in the OFF (non-conducting) state. During this phase, the energy stored in the inductor forward-biases the diode D, causing it to conduct. We can assume this ideal diode acts as a short circuit when conducting.

Let's analyze the potentials across the open switch. One terminal of the switch is connected to the positive side of the input source, placing it at a potential of $+24$ V. The other terminal is connected through the now-conducting diode to the output circuit's top rail.

Because this is a buck-boost converter, the output voltage is inverted. The diagram shows an output magnitude of 36 V, so the potential at this top rail is $-36$ V with respect to ground.

The total voltage across the switch is the difference between these two potentials:
$V_{switch} = (+24 \text{ V}) - (-36 \text{ V}) = 24 + 36 = 60 \text{ V}$.

To find the frequency of maximum phase lag, we first determine the network's transfer function, $H(s) = V_o(s)/V_{\in}(s)$. Using the voltage divider rule with $R_1=9\Omega$, $R_2=1\Omega$, and $C=1F$, we get:
$H(s) = \frac{R_2 + 1/(sC)}{R_1+R_2+1/(sC)} = \frac{1+sR_2C}{1+s(R_1+R_2)C}$
This expression is the standard form of a phase-lag compensator, $\frac{1+Ts}{1+\beta Ts}$.
By comparing terms, we find the time constant $T = R_2C = (1)(1) = 1$ s and the attenuation factor $\beta = \frac{R_1+R_2}{R_2} = \frac{9+1}{1} = 10$.
The frequency for maximum phase lag in a lag compensator is given by the formula $\omega_m = \frac{1}{T\sqrt{\beta}}$.
Substituting our values yields the frequency $\omega_m = \frac{1}{1\sqrt{10}} \approx 0.316$ rad/s.

The direction of rotation in this motor is determined by a rotating magnetic field. This field is produced by the phase difference between the currents in the main winding and the auxiliary winding, a shift created by the series capacitor. To reverse the motor, we must reverse the direction of this rotating field. By swapping the terminals of the auxiliary winding, you reverse its current's direction relative to the main winding's current. This flips the phase sequence, causing the resultant magnetic field and the motor to rotate in the opposite direction.

The ideal current source forces a 5 A current to flow counter-clockwise around the single loop. The voltage $V_{out}$ is the potential difference between the top and bottom wires of the circuit.

We can calculate this voltage by starting at the bottom wire (our 0 V reference) and summing the voltage changes along the left-hand path to the top wire.

First, moving across the voltage source provides a gain of 10 V. Next, moving across the 2 Ω resistor against the direction of current flow results in another voltage gain. This gain is calculated using Ohm's law: $V = I \times R = 5 \text{ A} \times 2 \text{ } \Omega = 10 \text{ V}$.

The total output voltage is the sum of these two gains:
$V_{out} = 10 \text{ V} + 10 \text{ V} = 20 \text{ V}$.

For a network with $n$ nodes, the number of independent Kirchhoff's Current Law (KCL) equations is $n-1$. This is because one node is always treated as a reference, making its equation dependent on the others. With 5 nodes, we can write $5-1 = 4$ independent KCL equations.

The number of independent Kirchhoff's Voltage Law (KVL) equations corresponds to the number of fundamental loops in the graph. This can be calculated as the total number of branches ($b$) minus the number of independent nodes ($n-1$). For this network, we have $b - (n-1) = 7 - (5-1) = 3$ independent KVL equations.

Therefore, we have 4 independent KCL equations and 3 independent KVL equations.

The strength of the electric field at the surface of a conductor is determined by the concentration of charge, known as surface charge density ($\sigma$). Charges tend to accumulate at points on a conductor with the sharpest curvature. For the ellipse, the sharpest points are A and C.

However, the problem states both electrodes are at the same potential, meaning they hold charges of the same sign which repel each other. This mutual repulsion pushes the charge on the left electrode as far away as possible from the right electrode. This effect concentrates the charge at point A. Because it is both a sharp point and farthest from the other like-charged electrode, point A develops the highest charge density and, therefore, the maximum electric field.

Let's simplify the given expression using De Morgan's laws. We start with the outermost negation, which covers a large OR operation. Applying De Morgan's law, \overline{X+Y} becomes \bar{X} \cdot \bar{Y}, which transforms the expression into:
$\overline{(a+\bar{b}+c+\bar{d})} \cdot \overline{(b+\bar{c})}$

Next, we apply De Morgan's law to each of these terms. This "distributes" the negation bar, changing the ORs to ANDs and flipping the variables inside.
$(\bar{a} \cdot \overline{\bar{b}} \cdot \bar{c} \cdot \overline{\bar{d}}) \cdot (\bar{b} \cdot \overline{\bar{c}})$

Using the double negation rule (\overline{\bar{x}} = x), this simplifies to:
$(\bar{a} \cdot b \cdot \bar{c} \cdot d) \cdot (\bar{b} \cdot c)$

We can now rearrange the terms: \bar{a} \cdot (b \cdot \bar{b}) \cdot (c \cdot \bar{c}) \cdot d. The expression contains both b \cdot \bar{b} and c \cdot \bar{c}. According to the complement law, any variable ANDed with its inverse is 0. Since we are ANDing terms with 0, the entire expression evaluates to 0.

This op-amp circuit is a differential summing amplifier. We can determine the output voltage, $V_{out}$, by following a two-step process based on ideal op-amp principles.

First, calculate the voltage at the non-inverting (+) terminal, let's call it $V_A$. The inputs $V_1$ and $V_2$ are connected through a voltage divider network. Using superposition, we find $V_A = \frac{4\,\Omega}{1\,\Omega+4\,\Omega}V_1 + \frac{1\,\Omega}{1\,\Omega+4\,\Omega}V_2 = \frac{4}{5}V_1 + \frac{1}{5}V_2$.

Next, we analyze the inverting side. For an ideal op-amp, $V_- = V_+ = V_A$. Applying KCL at the inverting (-) node, we have $\frac{V_3-V_A}{1\,\Omega} + \frac{V_{out}-V_A}{9\,\Omega} = 0$. Solving for $V_{out}$ gives $V_{out} = 10V_A - 9V_3$.

Finally, substitute the expression for $V_A$ into the output equation:
$V_{out} = 10\left(\frac{4}{5}V_1 + \frac{1}{5}V_2\right) - 9V_3 = 8V_1 + 2V_2 - 9V_3$.

The system's input is the product of two signals, $x_{1}(t) \cdot x_{2}(t)$. In the frequency domain, this time-domain multiplication corresponds to the convolution of their spectra: $X_{1}(\omega) * X_{2}(\omega)$.

The bandwidth of a convolved spectrum is the sum of the individual bandwidths. Since the signals have bandwidths $B_{1}$ and $B_{2}$ respectively, the signal entering the filter has a total bandwidth of $B_{1} + B_{2}$.

This signal is then passed through an LTI filter. An LTI system can only alter existing frequency components; it cannot create new ones. Therefore, the bandwidth of the output signal $y(t)$ remains $B_{1} + B_{2}$.

To reconstruct the signal from its samples, the Nyquist theorem states that the sampling rate must be at least twice the signal's highest frequency. This gives a minimum sampling rate of $2(B_{1} + B_{2})$.

This problem can be elegantly solved by viewing the integral through the lens of Fourier Transforms. The definite integral of a function from $-\infty$ to $\infty$ is equivalent to its Fourier Transform evaluated at a frequency of zero ($\omega=0$).

Let's consider the function inside the integral, $f(t) = \frac{\sin(2\pi t)}{\pi t}$. This is a form of the sinc function. Its Fourier Transform, $F(\omega)$, is a rectangular pulse. Specifically, the transform pair is:
$\frac{\sin(2\pi t)}{\pi t} \longleftrightarrow \text{rect}\left(\frac{\omega}{4\pi}\right)$
To find the value of $\int_{-\infty }^{\infty } \frac{\sin(2\pi t)}{\pi t} dt$, we simply evaluate the Fourier Transform at $\omega=0$:
$\int_{-\infty }^{\infty } \frac{\sin(2\pi t)}{\pi t} dt = \text{rect}(0) = 1$
The question asks for twice this value, so we have $2 \times 1 = 2$.

To solve this differential equation, we first find the roots of its characteristic (or auxiliary) equation: $m^2 - 4m + 4 = 0$. This equation factors into $(m-2)^2=0$, which gives a repeated real root of $m=2$.

For a repeated root, the general solution has the form $y(x) = (C_1 + C_2x)e^{2x}$.

Now, we apply the initial conditions. The condition $y(0)=0$ gives $0 = (C_1 + C_2 \cdot 0)e^0$, which means $C_1=0$. The solution is now $y(x) = C_2xe^{2x}$.

To use the second condition, we differentiate the solution: $\frac{dy}{dx} = C_2e^{2x} + 2C_2xe^{2x}$. Applying $\frac{dy}{dx}|_{x=0}=1$ yields $1 = C_2e^0 + 0$, so $C_2=1$.

The specific solution is $y(x) = xe^{2x}$. Finally, we evaluate this at $x=1$:
$y(1) = 1 \cdot e^{2(1)} = e^2 \approx 7.389$.

To compute this line integral, we must express everything in terms of the given parameter, $t$. The path is defined by $x=t$, $y=t^2$, and $z=t$, with $t$ running from 0 to 1. From this, we find the differentials: $dx=dt$, $dy=2t\,dt$, and $dz=dt$.

Now, substitute these expressions into the line integral $\int_C 5xz\,dx + (3x^{2+}2y)\,dy + x^2z\,dz$:
$\int_0^1 \left[ 5(t)(t) \cdot (dt) + (3t^{2+}2t^2) \cdot (2t\,dt) + t^2(t) \cdot (dt) \right]\$$
We can simplify the integrand by combining terms and factoring out $dt$:
$\int_0^1 (5t^2 + 10t^3 + t^3) \, dt = \int_0^1 (5t^2 + 11t^3) \, dt$
Finally, we evaluate this standard definite integral:
$\left[ \frac{5t^3}{3} + \frac{11t^4}{4} \right]_0^1 = \frac{5}{3} + \frac{11}{4} = \frac{20+33}{12} = \frac{53}{12} \approx 4.417$

We start with the vector relationship $\binom{a}{b} = P\binom{x}{y}$, which translates into the linear equations $a=3x+y$ and $b=x+3y$.

The problem states that $a^{2+}b^2=1$. By substituting our expressions for $a$ and $b$, we get an equation for the set S in terms of $x$ and $y$: $(3x+y)^2 + (x+3y)^2 = 1$.

Expanding this expression simplifies to $10x^2 + 12xy + 10y^2 = 1$. The $xy$ term indicates this is the equation of a rotated ellipse.

The orientation of the ellipse's axes corresponds to the eigenvectors of the matrix for this quadratic form, which are $\binom{1}{1}$ and $\binom{1}{-1}$.

For an ellipse equation of the form $v^T A v = 1$, the semi-axis length along an eigenvector is inversely proportional to the square root of the corresponding eigenvalue. The eigenvector $\binom{1}{1}$ is associated with the larger eigenvalue ($\lambda=16$), which corresponds to the shorter semi-axis, hence the minor axis.

A valid probability density function (PDF) must integrate to 1 over its entire domain. We can find the value of 'a' by setting up this condition, splitting the integral based on the piecewise definition of $f_X(x)$.

The total probability is $\int_{-\infty }^{\infty } f_X(x) dx = \int_{-\infty }^{0} ae^{4x} dx + \int_{0}^{\infty } \frac{3}{2}e^{-3x} dx = 1$.

Evaluating these integrals, we get [\frac{a}{4}e^{4x}]_{-\infty }^0 + \[-\frac{1}{2}e^{-3x}]_0^∞ = 1$.

This equation simplifies to $(\frac{a}{4} - 0) + (0 - (-\frac{1}{2})) = 1$, which gives $\frac{a}{4} + \frac{1}{2} = 1$, so $a=2$.

Now, to find $Prob\{X \le 0\}$, we integrate the PDF over the relevant range $(-\infty , 0]$ using the value of 'a' we just found.

Prob\{X \le 0\} = \int_{-\infty }^{0} 2e^{4x} dx = \[\frac{2e^{4x}}{4}]{-∞}^0 = $[\frac{e^{4x}}{2}]{-\infty }^0 = \frac{1}{2} - 0 = \frac{1}{2}$.

To find the driving point input impedance, we need to determine the ratio $Z_{\in} = V_s/I_s$. We can achieve this by applying nodal analysis. Let's label the voltage at the node to the right of the $2\,\Omega$ input resistor as $V_A$.

By applying Kirchhoff's Current Law (KCL) at node $V_A$, we equate the currents entering the node to the currents leaving it:
$I_s + 4V_1 = \frac{V_A}{3} + \frac{V_A}{2+4}$

Next, we write the controlling voltage $V_1$ and the node voltage $V_A$ in terms of the source variables. From the input loop, Ohm's law gives $V_1 = 2I_s$, and Kirchhoff's Voltage Law gives $V_A = V_s - V_1 = V_s - 2I_s$.

Substituting these expressions back into the KCL equation:
$I_s + 4(2I_s) = \frac{V_s - 2I_s}{3} + \frac{V_s - 2I_s}{6}$

Simplifying the equation, we get $9I_s = (V_s - 2I_s)(\frac{1}{3} + \frac{1}{6}) = \frac{1}{2}(V_s - 2I_s)$.
This further simplifies to $18I_s = V_s - 2I_s$, which means $20I_s = V_s$.
Therefore, the input impedance is $Z_{\in} = \frac{V_s}{I_s} = 20 \, \Omega$.

To find the power delivered to the load, we must first determine the current $I_2$ flowing through it. We start with the standard z-parameter equations for the two-port network:
$V_1 = 40I_1 + 60I_2$
$V_2 = 80I_1 + 100I_2$

From the output side of the circuit, Ohm's law gives $V_2 = -I_2 R_L = -20I_2$. By substituting this into the second z-parameter equation, we get $-20I_2 = 80I_1 + 100I_2$, which simplifies to the relationship $I_2 = -\frac{2}{3}I_1$.

Next, we use this relationship to find the input voltage $V_1$:
$V_1 = 40I_1 + 60(-\frac{2}{3}I_1) = 40I_1 - 40I_1 = 0 \text{ V}$.

Now, apply KVL to the input loop: $20 = 10I_1 + V_1$. Since $V_1=0$, we find that $I_1 = 2 \text{ A}$. This allows us to calculate $I_2 = -\frac{2}{3}(2) = -\frac{4}{3} \text{ A}$.

Finally, the average power delivered to the load resistor is calculated as:
$P_L = |I_2|^2 R_L = \left(\frac{4}{3}\right)^2 \times 20 = \frac{16}{9} \times 20 \approx 35.55 \text{ W}$.

For the line currents to be zero, the equivalent impedance of the delta-connected load must be infinite. This can be achieved if each branch of the delta load acts as a parallel resonant circuit. First, we convert the inner star-connected capacitors to an equivalent delta configuration. For a balanced system, each equivalent delta capacitor is $C_\Delta = C/3$.

Now, each branch of the overall delta load has an inductor $L$ in parallel with this capacitance $C_\Delta$. For parallel resonance, the angular frequency $\omega$ must satisfy the condition $\omega = 1/\sqrt{L C_\Delta}$.