GATE EE 2017 Set 2

The source inductance, $L$, is the critical element in this circuit. An inductor opposes any instantaneous change in the current flowing through it. This property causes a gradual transfer of the constant load current, $I_o$, from one diode to the other, a phenomenon known as commutation overlap.

At the beginning of the source's positive half-cycle, $D_1$ turns on. However, due to the inductor $L$, the current through $D_1$ cannot instantly become $I_o$. During this ramp-up time, the freewheeling diode $D_2$ must continue to conduct to supply the remaining current to the load. This extends $D_2$'s conduction period to be greater than $180^\circ$.

Similarly, when the source voltage goes negative, the inductor's tendency to maintain current flow forces $D_1$ to conduct for a short period into the negative half-cycle while $D_2$ is turning on. This overlap extends $D_1$'s conduction angle to be greater than $180^\circ$ as well.

To determine the output $Z$, we first translate the logic circuit into a Boolean expression. The final output is from a 3-input NAND gate, with inputs $(\overline{P\bar{Q}})$, $Q$, and $(\overline{QR})$. This gives the initial expression $Z = \overline{(\overline{P\bar{Q}}) \cdot Q \cdot (\overline{QR})}$.

Applying De Morgan's theorem to the entire expression converts the outer NAND into an OR operation: $Z = \overline{\overline{P\bar{Q}}} + \bar{Q} + \overline{\overline{QR}}$. The double negations cancel out, simplifying the expression to $Z = P\bar{Q} + \bar{Q} + QR$.

We can simplify this by factoring $\bar{Q}$ from the first two terms, yielding $Z = \bar{Q}(P+1) + QR$. Since anything OR-ed with 1 is 1 in Boolean logic, $P+1=1$, which reduces our expression to $Z = \bar{Q} + QR$.

Finally, using the distributive law, $A+BC = (A+B)(A+C)$, we can rewrite this as $Z = (\bar{Q}+Q)(\bar{Q}+R)$. Because $\bar{Q}+Q=1$, the expression simplifies to $Z = \bar{Q}+R$.

To find the probability of the second ball being red, we use the law of total probability, considering the two possible outcomes for the first discarded ball.

The first ball could have been red (with probability $\frac{5}{10}$), leaving 4 red balls in the remaining 9. Or, the first ball could have been black (with probability $\frac{5}{10}$), leaving 5 red balls in the remaining 9.

We sum the probabilities of these two mutually exclusive paths to getting a red ball on the second draw:
$P(\text{2nd is Red}) = P(\text{1st Red, 2nd Red}) + P(\text{1st Black, 2nd Red})$
$= \left(\frac{5}{10} \times \frac{4}{9}\right) + \left(\frac{5}{10} \times \frac{5}{9}\right) = \frac{20+25}{90} = \frac{45}{90} = \frac{1}{2}$

To determine the steady-state error, we first need to find the system's open-loop transfer function, $G(s)$, from the given closed-loop transfer function, $T(s)$. For a unity feedback system, this relationship is $G(s) = \frac{T(s)}{1-T(s)}$, which gives us:
$G(s) = \frac{\frac{Ks+b}{s^{2}+as+b}}{1 - \frac{Ks+b}{s^{2}+as+b}} = \frac{Ks+b}{(s^{2}+as+b) - (Ks+b)} = \frac{Ks+b}{s(s+a-K)}$

The single pole at the origin ($s$ in the denominator) identifies this as a Type 1 system. For a ramp input, the steady-state error of a Type 1 system is given by $e_{ss} = \frac{1}{K_v}$, where $K_v$ is the velocity error constant.

We calculate $K_v$ using its definition: $K_v = \lim_{s \to 0} sG(s)$.
$K_v = \lim_{s \to 0} s \left( \frac{Ks+b}{s(s+a-K)} \right) = \frac{0+b}{0+a-K} = \frac{b}{a-K}$

Therefore, the steady-state error is the reciprocal of $K_v$:
$e_{ss} = \frac{1}{K_v} = \frac{a-K}{b}$

To determine the fundamental component of the phase voltage, we start with the Fourier series expansion of the phase voltage waveform for a three-phase inverter in 180° conduction mode. The general expression for the instantaneous n-th harmonic of the phase voltage ($V_{an}$) is:
$V_{an}(t) = \frac{2V_{dc}}{n\pi} \sin(n\omega t)$

The fundamental component is the first harmonic, which we find by setting $n=1$.
$V_{a1}(t) = \frac{2V_{dc}}{1 \cdot \pi} \sin(\omega t) = \frac{2V_{dc}}{\pi} \sin(\omega t)$

From this expression, we can see that the peak value (or amplitude) of this fundamental sinusoidal voltage is the coefficient of the sine term. Therefore, the peak voltage is $\frac{2V_{dc}}{\pi}$.

Let's analyze these vector fields by considering the physical meaning of divergence and curl. Divergence, $\nabla \cdot \vec{F}$, tells us if a field is "spreading out" from a point (like a source). Curl, $\nabla \times \vec{F}$, tells us if a field is "swirling" or rotating around a point.

1. In figure $\vec{X}$, all vectors point directly away from the center. This is a classic example of a source, so the field has non-zero divergence: $\nabla \cdot \vec{X} \neq 0$.

2. In figure $\vec{Y}$, the vectors form closed loops, indicating pure circulation. If you were to place a tiny paddlewheel in this field, it would spin. This means the field has a non-zero curl: $\nabla \times \vec{Y} \neq 0$.

3. In figure $\vec{Z}$, the vectors are spiraling outwards. This field has both an outward component (divergence) and a rotational component (curl). Because there is rotation, the curl must be non-zero: $\nabla \times \vec{Z} \neq 0$.

Therefore, all three conditions presented in the chosen option are true.

Let's analyze the 5-minute traffic cycle. The light is green for the first 2 minutes and red for the next 3. We can represent this on a time interval $[0, 5)$, where the light is green during $[0, 2)$ and red during $[2, 5)$. Since vehicle arrival time $T$ is uniformly distributed, the probability density function is $f(t) = 1/5$ for $t \in [0, 5)$.

The waiting time $W$ is a function of the arrival time $t$. If a vehicle arrives during the green phase ($0 \le t < 2$), it doesn't wait, so $W(t)=0$. If it arrives during the red phase ($2 \le t < 5$), it must wait until the cycle completes at $t=5$, so the waiting time is $W(t) = 5-t$.

To find the expected waiting time $E[W]$, we integrate the waiting time for each possible arrival, weighted by its probability. We only need to consider the interval where waiting occurs:

$E[W] = \int_{0}^{5} W(t) f(t) \,dt = \int_{2}^{5} (5-t) \cdot \frac{1}{5} \,dt$

Now, we solve the integral:
$E[W] = \frac{1}{5} \left[ 5t - \frac{t^2}{2} \right]_{2}^{5} = \frac{1}{5} \left[ \left(25 - \frac{25}{2}\right) - \left(10 - \frac{4}{2}\right) \right]$

$E[W] = \frac{1}{5} [12.5 - 8] = \frac{4.5}{5} = 0.9$ minutes.

The added electrons are all negatively charged and therefore repel each other. Since the sphere is a conductor, these charges are free to move throughout the material. To reach a stable configuration (electrostatic equilibrium), the electrons will move as far apart from each other as possible to minimize their mutual repulsion. For a spherical object, this maximum separation is achieved when all the excess charge is distributed uniformly over its outer surface.

This compensator's transfer function can be analyzed by examining its poles and zeros. The corner frequencies are defined by the poles at $\omega=1$ and $\omega=10$, and the zeros at $\omega=0.1$ and $\omega=100$.

A phase-lead characteristic is created by a zero-pole pair where the zero occurs at a lower frequency than the pole. In this case, the zero at $\omega=0.1$ and the pole at $\omega=1$ form a phase-lead network.

The maximum phase lead for such a network is always found at a frequency between its zero and pole frequencies. Therefore, the phase lead for the entire compensator will reach its maximum in the frequency range between $0.1$ and $1$ rad/s.

The key component in this circuit is the phase-shifting transformer. The complex turns ratio, $\alpha$, causes this device to be non-reciprocal. A network's bus admittance matrix, $Y_{bus}$, is symmetric only if the network is reciprocal, which requires that the transfer admittance $Y_{12}$ equals $Y_{21}$. The phase shifter violates this condition, making $Y_{bus}$ unsymmetric. The bus impedance matrix, $Z_{bus}$, is the inverse of the $Y_{bus}$ matrix. Since the inverse of an unsymmetric matrix is also unsymmetric, $Z_{bus}$ must be unsymmetric too.

A single-phase full-bridge converter rectifies the AC input by utilizing both the positive and negative half-cycles of the supply voltage. This process creates an output DC voltage that has two main pulses for every one cycle of the AC input.

Because the output waveform's pattern repeats twice as often as the input waveform, its fundamental frequency is double the supply frequency. Given a supply frequency of $f_s = 50$ Hz, the fundamental ripple frequency on the DC side is:

$f_{ripple} = 2 \times f_s = 2 \times 50 \text{ Hz} = 100 \text{ Hz}$

We can solve this system of equations using the substitution method. First, isolate $x$ in the linear equation, which gives us $x = 11 - 2y$.

Next, substitute this expression for $x$ into the quadratic equation:
$2(11-2y)^2 + y^2 = 34$

Expanding and simplifying this equation gives a quadratic in terms of $y$:
$2(121 - 88y + 4y^2) + y^2 = 34$
$242 - 176y + 8y^2 + y^2 = 34$
$9y^2 - 88y + 208 = 0$

Solving this quadratic yields two solutions: $y=4$ and $y = \frac{52}{9}$. Since the problem states that $x$ and $y$ must be integers, we select $y=4$.

Using this value, we find $x$:
$x = 11 - 2(4) = 3$

The integer solution is $(x, y) = (3, 4)$. Therefore, the value of $(x+y)$ is $3+4=7$.

To find the partial derivative with respect to $x$, we must differentiate the function while treating the variables $y$ and $z$ as constants.

Our function is $f(x,y,z) = (x^{2}+y^{2}-2z^{2})(y^{2}+z^{2})$. We will use the product rule for differentiation.

The derivative of the first part, $(x^{2+}y^{2-}2z^2)$, with respect to $x$ is $2x$. The derivative of the second part, $(y^{2+}z^2)$, with respect to $x$ is $0$, since it contains no $x$ terms.

Applying the product rule, we get $\frac{\partial f}{\partial x} = (2x)(y^{2+}z^2) + (x^{2+}y^{2-}2z^2)(0)$.
This simplifies the partial derivative to $\frac{\partial f}{\partial x} = 2x(y^{2+}z^2)$.

Now, we evaluate this derivative at the point $(x=2, y=1, z=3)$:
$2(2)(1^2 + 3^2) = 4(1+9) = 4(10) = 40$.

The transfer impedance $Z_{21}$ is defined as the ratio of the output voltage to the input current, $V_2/I_1$, while the output port is open ($I_2=0$). We can simplify this bridged-T circuit by converting the upper Pi-network (composed of the 4Ω, top 2Ω, and right 2Ω resistors) into its equivalent T-network.

The new resistors of the T-network are calculated as follows:
$R_1 = \frac{4 \times 2}{4+2+2} = 1\Omega$ (new center leg)
$R_2 = \frac{4 \times 2}{4+2+2} = 1\Omega$ (new input series resistor)
$R_3 = \frac{2 \times 2}{4+2+2} = 0.5\Omega$ (new output series resistor)

The new $1\Omega$ center leg is now in series with the original vertical 2Ω resistor. Since $I_2=0$, the entire input current $I_1$ flows through this combined $1\Omega + 2\Omega = 3\Omega$ path to ground. As no current flows through the $0.5\Omega$ output resistor, $V_2$ is simply the voltage across this $3\Omega$ path. Therefore, $V_2 = I_1 \times 3\Omega$, which gives us $Z_{21} = \frac{V_2}{I_1} = 3\Omega$.

To find the total energy supplied by the source, we first determine the equivalent resistance ($R_{eq}$) the source "sees." The resistor network is a balanced Wheatstone bridge, so the central 5 Ω resistor carries no current and can be removed. This simplifies the circuit to two parallel branches, each with a resistance of $5+5=10 \Omega$. The equivalent resistance is then $R_{eq} = (10 \parallel 10) = 5 \Omega$.

The current $i(t)$ drawn from the source while charging the capacitor is given by $i(t) = \frac{V}{R_{eq}}e^{-t/(R_{eq}C)}$.
Plugging in the values, we get $i(t) = \frac{10 \text{ V}}{5 \Omega}e^{-t/(5 \Omega \cdot 1 \text{ F})} = 2e^{-0.2t}$ A.

The total energy $E$ from the source is the integral of power ($P=V \cdot i(t)$) from $t=0$ until steady state ($t \to \infty$):
$E = \int_{0}^{\infty } V \cdot i(t) dt = \int_{0}^{\infty } 10 \cdot (2e^{-0.2t}) dt = 20 \int_{0}^{\infty } e^{-0.2t} dt$
$E = 20 \left[ \frac{e^{-0.2t}}{-0.2} \right]_{0}^{\infty } = 20 \left[ 0 - \frac{1}{-0.2} \right] = 20(5) = 100 \text{ J}$

The diode $D_3$ is a freewheeling diode, which is designed to conduct when the voltage across the load attempts to go negative. This situation arises in circuits with inductive loads, where stored energy forces current to continue flowing.

However, this circuit features a purely resistive load. In a rectifier circuit with a resistive load, the output voltage is always composed of positive segments of the AC input voltage or is zero. The output voltage can never become negative.

Because the voltage across the load is always non-negative ($v_o \geq 0$), the freewheeling diode $D_3$ is never forward biased and thus never turns on. Consequently, no current flows through it. The RMS value of a current that is always zero is simply 0 A.

The size of the Jacobian matrix in a Newton-Raphson load flow analysis is determined by the number of unknown variables in the system. For a power system with $n$ total buses, including $n_{PV}$ PV buses and one slack bus, the dimension of the square Jacobian matrix is given by $(2n - n_{PV} - 2)$.

We are given that the Jacobian size is $100 \times 100$ and the number of PV buses is $n_{PV} = 20$. We can set up the equation for the dimension:
$100 = 2n - 20 - 2$
Simplifying the equation gives:
$100 = 2n - 22$
Solving for $n$, the total number of buses:
$122 = 2n \implies n = 61$

The speed of the magnetic field created by the stator, known as the synchronous speed ($N_s$), is the reference speed in the motor. We can calculate it as:
$N_s = \frac{120f}{P} = \frac{120 \times 50}{4} = 1500$ rpm

A fundamental concept in induction motors is that the rotor's magnetic field, although generated by the rotor, rotates at the synchronous speed with respect to the stationary stator. This is because the rotor field's speed relative to the rotor itself ($sN_s$) perfectly combines with the rotor's mechanical speed ($N = N_s(1-s)$) to match the stator's field speed.

Therefore, a stationary observer always sees the rotor flux moving at the synchronous speed, $N_s$.

To find this speed in mechanical rad/sec, we convert from rpm:
Speed = $N_s \times \frac{2\pi}{60} = 1500 \times \frac{2\pi}{60} = 50\pi \approx 157$ rad/sec.

To find the uncertainty in a calculated quantity that depends on multiple independent measurements, we must use the formula for propagation of uncertainty. This rule states that the total variance is the sum of the variances contributed by each measurement.

For a quantity $R$ that is a function of $R_1$ and $R_2$, the variance in $R$, denoted $\sigma_R^2$, is given by:
$\sigma_R^2 = \left(\frac{\partial R}{\partial R_1}\right)^2 \sigma_1^2 + \left(\frac{\partial R}{\partial R_2}\right)^2 \sigma_2^2$

Here, $\sigma_1$ and $\sigma_2$ are the standard deviations (uncertainties) of $R_1$ and $R_2$, respectively. The partial derivatives act as sensitivity factors, scaling the contribution of each uncertainty.

In this problem, the uncertainties are given as $\Delta R_1$ and $\Delta R_2$, which correspond to $\sigma_1$ and $\sigma_2$. Taking the square root to find the standard deviation $\sigma_R$ gives the final expression for the combined error.

The magnitude of the characteristic impedance, $|Z_c|$, for a nominal-$\pi$ network is calculated using the formula $|Z_c| = \sqrt{|Z|/|Y|}$, where $|Z|$ is the magnitude of the series impedance and $|Y|$ is the magnitude of the total shunt admittance.

We are given that the series impedance magnitude is $|Z| = 100 \ \Omega$.

The total shunt admittance $Y$ is the sum of the admittances of the two parallel capacitors. The admittance of a single capacitor with reactance $X$ is $j/X$. Therefore, the total shunt admittance is $Y = j/X + j/X = j(2/X)$. The magnitude is $|Y| = 2/X$.

Given $X = 3300 \ \Omega$, we find $|Y| = 2/3300 \ S$.

Now, we can calculate the magnitude of the characteristic impedance:
$|Z_c| = \sqrt{\frac{100}{2/3300}} = \sqrt{\frac{100 \times 3300}{2}} = \sqrt{165000} \approx 406.2 \ \Omega$.

The frequency response of a digital filter is evaluated on the unit circle, where the point $z=1$ corresponds to the lowest frequency (DC) and $z=-1$ corresponds to the highest frequency (Nyquist). A zero placed on the unit circle forces the filter's gain to be zero at that specific frequency.

Observing the plots, we see that all three systems, P, Q, and R, have zeros located at both $z=1$ and $z=-1$. By simultaneously blocking the lowest and highest frequencies, these systems allow only a band of frequencies in between to pass. Therefore, all three systems are band-pass filters.

The mean square value of a periodic signal is its average power, which is found by averaging the square of its amplitude, $[f(t)]^2$, over one complete cycle.

First, let's find the period ($T$) of the waveform from the graph. The signal pattern repeats every 4 units (e.g., from $t=-0.3$ to $t=3.7$), so $T=4$.

Next, we calculate the energy over one period by integrating $[f(t)]^2$. For a piecewise constant signal, this simplifies to summing the squared amplitudes multiplied by their durations:

• The signal is $f(t)=4$ for a duration of $1$ (from $t=-0.3$ to $t=0.7$).
• The signal is $f(t)=-2$ for a duration of $2$ (from $t=0.7$ to $t=2.7$).

The mean square value is this total energy divided by the period:
$\text{Mean Square Value} = \frac{1}{T} \left[ (4)^2 \cdot (1) + (-2)^2 \cdot (2) \right]\$$
$= \frac{1}{4} [16 \times 1 + 4 \times 2] = \frac{16+8}{4} = 6$

Lissajous patterns reveal the frequency ratio of the two input signals. The ratio of the vertical frequency ($f_y$) to the horizontal frequency ($f_x$) is directly proportional to the ratio of the number of horizontal tangencies to the number of vertical tangencies.

This relationship is given by the formula:
$\frac{f_y}{f_x} = \frac{\text{Number of horizontal tangencies}}{\text{Number of vertical tangencies}}$
We are given $f_x = 3$ kHz, 3 horizontal tangencies, and 2 vertical tangencies. Plugging these values into the equation:
$\frac{f_y}{3 \text{ kHz}} = \frac{3}{2}$
Solving for $f_y$ yields $f_y = \frac{3}{2} \times 3 \text{ kHz} = 4.5 \text{ kHz}$.

Let's analyze the given equations. The first equation, $y^2 - 2y + 1 = x$, can be factored as a perfect square: $(y-1)^2 = x$. Taking the square root of both sides gives us $y-1 = \sqrt{x}$, which rearranges to $y = 1 + \sqrt{x}$.

Now, we can substitute this expression for $y$ into the second equation, $\sqrt{x} + y = 5$. This gives us $\sqrt{x} + (1 + \sqrt{x}) = 5$.

Combining terms, we have $2\sqrt{x} + 1 = 5$, which simplifies to $2\sqrt{x} = 4$, or $\sqrt{x} = 2$. Squaring both sides yields $x=4$.

With $x=4$, we can find $y$ using our earlier relationship: $y = 1 + \sqrt{4} = 1 + 2 = 3$.

Finally, we calculate the required expression, $x + \sqrt{y}$, which is $4 + \sqrt{3}$. This is approximately $4 + 1.732 = 5.732$.

The power factor of a synchronous motor is controlled by the DC field excitation current. When a synchronous motor operates at a leading power factor, it is said to be "over-excited." This means the field winding is energized to a point where the internally generated excitation voltage, $E_f$, becomes greater than the applied terminal voltage, $V_t$. Therefore, for a motor running with a leading power factor, the relationship is $E_f > V_t$.

The amount of peak overshoot ($M_p$) in a second-order system is inversely dependent on the damping ratio ($\xi$). A system with less damping will oscillate more and have a higher overshoot.

Let's compare each option to the standard form $\frac{\omega_n^2}{s^2 + 2\xi\omega_n s + \omega_n^2}$.
For all the given systems, the natural frequency is constant: $\omega_n^2 = 100$, so $\omega_n = 10$.

The middle term's coefficient is $2\xi\omega_n = 20\xi$. To maximize the overshoot, we need to find the system with the minimum $\xi$. This corresponds to the system with the smallest coefficient for the $s$ term. Option C has the term $5s$, giving the smallest damping ratio of $\xi = 5/20 = 0.25$.

The positive sequence reactance ($X_L$) of a transmission line is determined by its inductance. The formula for inductive reactance per unit length is $X_L = 2 \pi f L$, where the inductance per meter is given by $L = 2 \times 10^{-7} \ln(D_{eq}/GMR)$.

For this symmetrical equilateral spacing, the equivalent distance between conductors, $D_{eq}$, is simply the given side length of 1 m. The Geometric Mean Radius (GMR) is provided as 0.01 m.

First, let's calculate the reactance in Ohms per meter using the given frequency of 50 Hz:
$X_L = 2 \pi (50) \times 2 \times 10^{-7} \ln \left(\frac{1}{0.01}\right) \approx 2.893 \times 10^{-4} \, \Omega/\text{m}$

Finally, to express the reactance in the required units of $\Omega$/km, we multiply this result by 1000:
$X_L = 2.893 \times 10^{-4} \times 1000 \approx 0.289 \, \Omega/\text{km}$

When two generators operate in parallel, they must run at the same system frequency. The governor speed regulation (droop) dictates how each unit's frequency changes with its power output. We can model this as a linear relationship for each generator, starting from a common no-load frequency (assumed 50 Hz).

Let $P_1$ and $P_2$ be the loads on the 400 MW and 300 MW units, respectively.
For the 400 MW unit, a 4% droop causes a $0.04 \times 50 = 2$ Hz frequency drop at full load. Its operating frequency $f$ is $f = 50 - \frac{2}{400}P_1$.
For the 300 MW unit, a 6% droop causes a $0.06 \times 50 = 3$ Hz drop at full load. Its frequency is $f = 50 - \frac{3}{300}P_2$.

Since the frequencies must be equal and $P_1 + P_2 = 600$, we can set the equations equal and substitute $P_2 = 600 - P_1$:
$50 - \frac{2}{400}P_1 = 50 - \frac{3}{300}(600 - P_1)$
$\frac{1}{200}P_1 = \frac{1}{100}(600 - P_1)$
Solving for $P_1$ gives $P_1 = 2(600 - P_1)$, which simplifies to $3P_1 = 1200$.
Thus, the load on the larger unit is $P_1 = 400$ MW.

To find the Thevenin voltage, $V_{ab}$, we can simplify the circuit and apply nodal analysis. Imagine the circuit is reduced to a single node, with voltage $V_{ab}$, where the networks to its left and right have been simplified into equivalent branches.

By applying Kirchhoff's Current Law (KCL) at this node and summing the currents flowing out of it, we get the following relationship:
$\frac{V_{ab} - (-30 \text{ V})}{15 \text{ \Omega }} + \frac{V_{ab} - 8 \text{ V}}{5 \text{ \Omega }} = 0$
This equation models the simplified left and right branches connected to our node. To solve for $V_{ab}$, we can multiply the entire equation by 15:
$(V_{ab} + 30) + 3(V_{ab} - 8) = 0$
Simplifying this gives $4V_{ab} + 6 = 0$, which yields $V_{ab} = -1.5 \text{ V}$.

First, the input signal $x(t)$ is modulated by $\cos(1000\pi t)$. In the frequency domain, this corresponds to shifting the spectrum $X(\omega)$ to the left and right by the modulation frequency of $1000\pi$ rad/s. The resulting spectrum $Z(\omega)$ will thus contain frequencies up to $(1000\pi + 1000\pi) = 2000\pi$ rad/s.

Next, this modulated signal is passed through a system with the impulse response $h(t) = \frac{\sin(1500\pi t)}{\pi t}$. This function describes an ideal low-pass filter with a sharp cutoff frequency at $\omega_c = 1500\pi$ rad/s.

The filter passes frequencies from $Z(\omega)$ up to its cutoff, so the maximum frequency component in the final output signal $y(t)$ is $\omega_m = 1500\pi$ rad/s. In Hertz, this is $f_m = \frac{\omega_m}{2\pi} = \frac{1500\pi}{2\pi} = 750$ Hz.

To reconstruct the signal perfectly from its samples, the Nyquist sampling theorem requires the sampling rate to be at least twice the maximum frequency. Therefore, the minimum sampling rate is $f_s = 2 \times f_m = 2 \times 750 = 1500$ samples/s.

First, we analyze the motor's performance at its rated condition of 10 kW. Since rotational losses are neglected, this rated power is the same as the developed power, $P_{dev} = E_b I_a = (V - I_a R_a)I_a = 10,000$ W. Solving the resulting quadratic equation, $10000 = (220 - 0.02I_a)I_a$, gives a rated armature current of $I_{a,rated} \approx 45.75$ A.

Next, we calculate the torque produced at this rated operating point. The angular speed is $\omega = 2\pi N/60$, so the rated torque is $T_{rated} = P_{dev} / \omega = 10000 / (2\pi \cdot 900/60) \approx 106.15$ Nm.

For a separately excited DC motor with constant field excitation, the electromagnetic torque is directly proportional to the armature current ($T \propto I_a$).

We can now use this proportionality to find the armature current required to produce the specified torque of 70 Nm:
$I_{a,new} = I_{a,rated} \times \frac{T_{new}}{T_{rated}} = 45.75 \text{ A} \times \frac{70 \text{ Nm}}{106.15 \text{ Nm}} \approx 30.16$ A.

The overall impulse response $h(n)$ of a system in cascade is the convolution of its individual components. Thus, we have $h(n) = h_1(n) * h_2(n) = \{\overset{1}{\uparrow},-1\} * \{\overset{1}{\uparrow},1\} = \{\overset{1}{\uparrow}, 0, -1\}$.

To find the unknown input $x(n)$, it is most convenient to work in the Z-domain, where the convolution property states $Y(z) = X(z)H(z)$. This can be rearranged to solve for the input's Z-transform: $X(z) = Y(z) / H(z)$.

First, we find the Z-transforms of the overall impulse response and the given output sequence:
$H(z) = 1 - z^{-2}$
$Y(z) = 1+2z^{-1}+z^{-2}-z^{-3}-2z^{-4}-z^{-5}$

Performing polynomial long division gives the Z-transform of the input:
$X(z) = \frac{1+2z^{-1}+z^{-2}-z^{-3}-2z^{-4}-z^{-5}}{1-z^{-2}} = 1+2z^{-1}+2z^{-2}+z^{-3}$

By inspecting the coefficients of $X(z)$, we find the input sequence is $x(n)=\{\overset{1}{\uparrow},2,2,1\}$.

To solve for the ratio $\frac{R_B}{R}$, we'll write and solve two Kirchhoff's Voltage Law (KVL) equations for the circuit.

First, let's analyze the collector-emitter loop. The KVL equation is $V_{CC} = (I_B+I_C)4R + V_{CE} + I_E R$. We are given $V_{CE} = \frac{V_{CC}}{2}=5V$, and we know $I_E = I_B+I_C = (\beta+1)I_B$. With $\beta=29$, this becomes $10 = 30I_B(4R) + 5 + 30I_B(R)$, which simplifies to $5 = 150I_B R$.

Next, we write the KVL equation for the loop passing through the base: $V_{CC} = (I_B+I_C)4R + I_B R_B + V_{BE} + I_E R$. Substituting the given values $V_{BE}=0.7V$ and $\beta=29$, we get $10 = 30I_B(4R) + I_B R_B + 0.7 + 30I_B(R)$. This simplifies to $9.3 = 150I_B R + I_B R_B$.