GATE EE 2023

When we report what someone said in the past, we need to adjust the verb tense. The original statement uses the present continuous tense ("am thinking"). Because the reporting verb ("told") is in the past tense, we must "backshift" the original verb. The present continuous ("am thinking") becomes the past continuous ("was thinking"). Similarly, the pronoun changes from "I" to "he" to match the subject, Rafi. This correctly places his action of thinking in the past, at the time he spoke to Mary.

This analogy is built on a relationship of opposites, or antonyms. Let's first examine the complete pair: $Enforce$ and $Relax$. To enforce a rule is to make it stricter, while to relax it is to make it less strict. They are opposing actions.

We must apply this same antonym relationship to the word $Permit$. To permit something means to allow it. The direct opposite of allowing an action is to $Forbid$ it.

This problem involves two separate, independent events: the first roll and the second roll. The outcome of the first roll does not influence the outcome of the second.

The probability of rolling a '1' on a fair six-sided die is $\frac{1}{6}$, as there is one '1' face out of six possible outcomes.

Similarly, the probability of rolling a '4' on the second toss is also $\frac{1}{6}$.

To find the probability of both of these independent events happening in sequence, we multiply their individual probabilities:

Let's denote the total population of tobacco users as $100\%$. From the survey, $65\%$ of users were advised to stop, and $30\%$ of users attempted to stop.

To find a certain inference, consider the most extreme scenario. Let's assume that every single person who attempted to stop ($30\%$) was also part of the group that was advised to stop ($65\%$). This maximizes the potential overlap between the two groups.

In this best-case scenario, out of the $65\%$ of users who were advised, only $30\%$ made an attempt. This leaves at least $65\% - 30\% = 35\%$ of users who were advised but did not make an attempt.

A majority of the advised group ($65\%$) is any number greater than half of it, which is $\frac{65\%}{2} = 32.5\%$. Since the $35\%$ who certainly did not attempt is greater than this $32.5\%$ majority threshold, we can be certain that a majority of advised users did not attempt to stop.

A systematic way to count the triangles is to break the figure into simpler parts. The figure contains four large, overlapping triangular regions: a top half, a bottom half, a left half, and a right half.

Each of these large triangles is subdivided to form 1 + 2 + 3 = 6 smaller triangles. With four such regions, we have an initial count of 4 × 6 = 24 triangles.

However, this method double-counts the four smallest triangles, which are located at the corners of the central area. Each of these small triangles was counted as part of both a horizontal half (top or bottom) and a vertical half (left or right).

To correct for this, we subtract the four duplicates from our total: 24 - 4 = 20. Therefore, there are 20 triangles in the figure.

Here is a concise explanation for the question:

The core information is that eligibility to vote depends on completing registration, and no student from the Department of Human Sciences completed it. Let's analyze the statements based on this.

• Statement (iii): "None of the students from the Department of Human Sciences are eligible to vote in the college elections." This is a direct and certain conclusion. Since eligibility requires registration and no Human Sciences students registered, none of them can be eligible.

• Statement (i): "All students who are eligible to vote are students of departments other than the Department of Human Sciences." While this is logically equivalent to statement (iii), it makes a claim about the set of "eligible voters". We don't know if this set has any members, as it's possible no one from any department registered. Statement (iii) is more certain as it's a direct consequence about a known group (Human Sciences students).

• Statement (ii): "All students... other than those from the Department of Human Sciences, have completed the registration..." We cannot infer this. The text gives no information about whether students from other departments registered.

Therefore, the only statement that can be inferred with absolute certainty based directly on the provided facts is (iii).

The graph exhibits symmetry with respect to the origin, which is the defining visual characteristic of an odd function. For a function $f(x)$ to be odd, it must satisfy the algebraic condition $f(-x) = -f(x)$. We can test the given options to see which one follows this rule. The function $f(x) = x 2^{-|x|}$ is indeed odd, because substituting $-x$ into the equation gives $f(-x) = (-x) 2^{-|-x|} = -x 2^{-|x|}$, which is equal to $-f(x)$.

The question asks to identify the statement that is NOT supported by the passage. The passage explicitly states that "Modern cell biology encourages us to imagine the cell as a molecular machine." The phrase "molecular machine" is a metaphor, a type of figurative language. Therefore, the passage directly demonstrates that modern cell biology does use figurative language.

Option D claims that "Modern cell biology never uses figurative language... to describe or explain anything." This is a direct contradiction of the evidence presented in the text. In contrast, options A, B, and C all accurately summarize points made within the passage, such as the use of modern metaphors for cancer and the specific machinery metaphor promoted by cell biology.

To find the units digit of the product, we can analyze the units digit of each factor separately.

The units digits of powers of 3 repeat in a cycle of four: $3, 9, 7, 1$. To find the units digit of $3^{999}$, we find the remainder of the exponent when divided by 4. Since $999 \div 4$ gives a remainder of 3, the units digit is the third number in the cycle, which is 7.

Similarly, the units digits of powers of 7 also cycle every four powers: $7, 9, 3, 1$. Since the exponent 1000 is perfectly divisible by 4, the units digit of $7^{1000}$ is the fourth number in this cycle, which is 1.

Finally, to find the units digit of the entire product, we multiply the units digits we found: $7 \times 1 = 7$.

A clever way to determine the shaded area is by considering the shapes that define its boundaries. The problem states the boundary comes from two semi-circles whose diameters are the square's sides. This gives a diameter of $6 \mathrm{~cm}$ and a radius of $r=3 \mathrm{~cm}$ for each.

Let's calculate the area of these two semi-circles, which is $2 \times \frac{1}{2}\pi r^2 = \pi (3^2) = 9\pi \mathrm{~cm}^2$.

This combined area isn't the final answer, as it includes overlapping regions differently than the diagram. The calculation requires subtracting an adjustment term, which represents the area of the unshaded central lens. This lens area is found to be $\frac{9\pi}{2} - 9$.

Following a known geometric theorem for this configuration, the shaded area is the total area of the two semi-circles minus twice the area of the lens:
Area $= 9\pi - 2\left(\frac{9\pi}{2} - 9\right) = 9\pi - (9\pi - 18) = 18 \mathrm{~cm}^2$.

The equation of the plane is given as $\mathbf{w}^{\top} \mathbf{x} = 1$. Let's write this out explicitly.
Given

and letting

, the equation becomes:
$\begin{bmatrix} 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = 1$
This matrix multiplication gives the standard form of the plane's equation: $1x + 2y + 3z = 1$.

A key property of a plane described by the equation $ax+by+cz=d$ is that the vector of its coefficients,

, is always normal (perpendicular) to the plane.

For our equation, the coefficients are $a=1$, $b=2$, and $c=3$. Therefore, the normal vector is

, which is the original vector $\mathbf{w}$.

To find the transfer function $\frac{Y(s)}{R(s)}$, we can analyze the system's signal flow and apply Mason's Gain Formula.

First, identify the forward paths from the input $R(s)$ to the output $Y(s)$. There are two:

1. A direct path through the gain block of 3, so $P_1 = 3$.
2. A path through the gain block of 2 and the integrator $\frac{1}{s}$, so $P_2 = 2 \cdot \frac{1}{s} = \frac{2}{s}$.

Next, identify any feedback loops. There is one positive feedback loop that passes through the integrator and the feedback gain of 1. Its gain is $L_1 = \frac{1}{s} \cdot 1 = \frac{1}{s}$.

Since this single loop touches both forward paths, the path cofactors are $\Delta_1 = 1$ and $\Delta_2 = 1$. Mason's formula is $T(s) = \frac{P_1\Delta_1 + P_2\Delta_2}{1 - L_1}$.

Substituting the gains we found:
$\frac{Y(s)}{R(s)} = \frac{3(1) + \frac{2}{s}(1)}{1 - \frac{1}{s}} = \frac{3 + \frac{2}{s}}{1 - \frac{1}{s}}$

To simplify, we multiply the numerator and denominator by $s$, which gives $\frac{s(3 + \frac{2}{s})}{s(1 - \frac{1}{s})} = \frac{3s + 2}{s - 1}$.

The Nyquist plot shows how the transfer function $G(s)H(s)$ behaves as $s$ traverses a contour that encircles the right-half of the s-plane. The "infinite semi-circular arc" part of this contour represents points where the magnitude of $s$ approaches infinity.

To find where this arc maps, we evaluate the limit of $G(s)H(s)$ as $|s| \to \infty$. We can let $s = Re^{j\theta}$ and take the limit as $R \to \infty$.

$\lim_{R \to \infty } G(s)H(s) = \lim_{R \to \infty } \frac{3Re^{j\theta} + 5}{Re^{j\theta} - 1}$

For very large values of $R$, the constant terms become negligible. The expression simplifies to:

$\lim_{R \to \infty } \frac{3Re^{j\theta}}{Re^{j\theta}} = 3$

Thus, the entire infinite semi-circle in the s-plane maps to the single point 3.

First, we establish the transfer function for the proportional-integral (PI) controller: $C(s) = K_p + \frac{K_i}{s} = 3 + \frac{1}{s} = \frac{3s+1}{s}$. The overall closed-loop transfer function relating the plant output $Y(s)$ to the reference input $R(s)$ is $\frac{Y(s)}{R(s)} = \frac{C(s)G(s)}{1+C(s)G(s)} = \frac{3s+1}{s^{2+}2s+1}$.

For a unit step input, $R(s) = \frac{1}{s}$, so the plant output becomes $Y(s) = \frac{1}{s}\frac{3s+1}{(s+1)^2}$. We can find the final value of the plant output using the Final Value Theorem:
$y(\infty ) = \lim_{s \to 0} sY(s) = \lim_{s \to 0} s \left( \frac{1}{s}\frac{3s+1}{(s+1)^2} \right) = \frac{3(0)+1}{(0+1)^2} = 1$.

Next, let's find the controller's output, which we'll call $U(s)$. This signal is the input to the plant, so $Y(s) = G(s)U(s)$, which means $U(s) = \frac{Y(s)}{G(s)}$. Substituting the known expressions, we get $U(s) = \left( \frac{1}{s}\frac{3s+1}{(s+1)^2} \right) \div \left( \frac{1}{s-1} \right) = \frac{(3s+1)(s-1)}{s(s+1)^2}$.

Applying the Final Value Theorem to the controller output gives its steady-state value:
$u(\infty ) = \lim_{s \to 0} sU(s) = \lim_{s \to 0} s \left( \frac{(3s+1)(s-1)}{s(s+1)^2} \right) = \frac{(3(0)+1)(0-1)}{(0+1)^2} = -1$.

Thus, the final controller output is -1 and the final plant output is 1.

An induction machine's operating mode is determined by its slip, $s$, which quantifies the difference between the synchronous speed and the rotor speed.

• Motor Mode (b): The machine functions as a motor when the rotor rotates slower than the magnetic field. This common operating condition corresponds to a slip value between 0 and 1, matching it with range (p).

• Generator Mode (a): When an external prime mover drives the rotor faster than the synchronous speed, the machine delivers power back to the source. This generating action occurs at a negative slip, which aligns with range (r).

• Plugging Mode (c): This braking mode is achieved by reversing the stator's magnetic field rotation while the rotor is still spinning. This creates a large opposing torque, and the slip becomes greater than 1, pairing it with range (q).

To solve this, we'll use the double-revolving field theory for single-phase induction motors. First, we calculate the synchronous speed ($N_s$) of the magnetic field.
$N_s = \frac{120f}{P} = \frac{120 \times 50}{10} = 600 \text{ RPM}$
Next, we find the forward slip ($S$), which is the rotor's slip with respect to the forward-rotating field.
$S = \frac{N_s - N}{N_s} = \frac{600 - 540}{600} = 0.1$
The slip with respect to the backward-rotating field ($S_b$) is given by the relation $S_b = 2 - S$.
$S_b = 2 - 0.1 = 1.9$
Finally, the frequency of the rotor currents induced by this backward field ($f_b$) is the backward slip multiplied by the supply frequency.
$f_b = S_b \times f = 1.9 \times 50 \text{ Hz} = 95 \text{ Hz}$

To determine the impulse response of the system, we can utilize the Laplace transform to find its transfer function, $H(s)$. Applying the Laplace transform to the differential equation, assuming the system is initially at rest, yields $sY(s) + 3Y(s) = 2X(s)$.

The transfer function is the ratio of the output transform $Y(s)$ to the input transform $X(s)$. By rearranging the equation, we find $H(s) = \frac{Y(s)}{X(s)} = \frac{2}{s+3}$.

The impulse response, $h(t)$, is the inverse Laplace transform of the transfer function. Using the standard transform pair $\mathcal{L}^{-1}\{\frac{1}{s+a}\} = e^{-at}u(t)$, we can find the impulse response for our system.

Therefore, $h(t) = \mathcal{L}^{-1}\left\{\frac{2}{s+3}\right\} = 2e^{-3t}u(t)$. The unit step function $u(t)$ indicates the system is causal and the response is zero for $t < 0$.

The given signal's Fourier transform, $X(\omega)$, is a rectangular pulse in the frequency domain with a height of 1 and a total width of $2W_0$. To find the signal $x(t)$, we take the inverse Fourier transform, which yields a sinc function in the time domain.

The specific relationship gives us $x(t) = \frac{\sin(W_0 t)}{\pi t}$.

Now, let's consider what happens as the bandwidth $W_0$ approaches infinity. In the frequency domain, the rectangular pulse $X(\omega)$ becomes infinitely wide, approaching a constant value of $X(\omega) = 1$ for all $\omega$.

By definition, the inverse Fourier transform of a constant $C$ is a scaled impulse, $C\delta(t)$. Since our constant is 1, the resulting time-domain signal $x(t)$ approaches the unit impulse function $\delta(t)$.

The Discrete-Time Fourier Transform (DTFT) of a signal is found by evaluating its Z-transform on the unit circle. For this evaluation to be valid, meaning for the DTFT to converge, the Region of Convergence (ROC) of the Z-transform must include the unit circle, which is defined by $|z|=1$.

The poles of the given system are at $z = \frac{1}{5}$, $z = \frac{2}{3}$, and $z=3$. The ROC must be a ring-shaped region that does not contain any poles.

Let's examine the ROC given in the option, $\frac{2}{3} < |z| < 3$. This region is an annulus between the poles at $\frac{2}{3}$ and $3$. Since the condition $\frac{2}{3} < 1 < 3$ is true, the unit circle $|z|=1$ is indeed contained within this ROC, ensuring the convergence of the DTFT.

To ensure proper protection and selectivity in this system, we must focus on the parallel lines, Line 1 and Line 2, which connect Bus 1 to Bus 2.

Consider a fault occurring on Line 2. Fault current will flow from the source at Bus 1 to the fault. However, current will also flow down the healthy Line 1, through Bus 2, and back into Line 2 toward the fault. A non-directional relay at $R_3$ would see this large current and trip, unnecessarily disconnecting the healthy Line 1.

To prevent this and isolate only the faulted line, $R_3$ must be a directional relay. It should be configured to operate only for faults on Line 1 (its forward direction) and to ignore, or block, faults behind it, such as those on Bus 2 or Line 2. By the same logic, $R_4$ must also be a directional relay to avoid tripping for faults on Line 1. This is the minimum requirement to guarantee that a fault on one parallel line does not cause a trip on the other.

The principle of economic dispatch requires that the incremental costs (IC) of the generating units be equal for the lowest total cost. The incremental cost is the derivative of the fuel cost function.

$IC_1 = \frac{dF_1}{dP_{G1}} = 0.2aP_{G1} + 40$
$IC_2 = \frac{dF_2}{dP_{G2}} = 0.4P_{G2} + 30$

When the parameter $a$ increases by 10%, the incremental cost of unit 1, $IC_1$, becomes higher for any given output. To restore the optimal balance where $IC_1 = IC_2$, we must reduce the output of the now relatively more expensive unit, $P_{G1}$. Consequently, since the total load is constant, the output of unit 2, $P_{G2}$, must increase.

Because the cost coefficient $a$ for unit 1 has increased, making it inherently more expensive to run, and the dispatch has shifted, the overall total cost of generation $F = F_1 + F_2$ will inevitably increase.

An electromotive force (EMF) is induced in a conductor only when it "cuts" through magnetic field lines. No EMF is generated when a conductor's motion is parallel to the magnetic field lines. The axis where this occurs is called the Geometric Neutral Axis (GNA).

In the given diagram, the stator currents create a stationary magnetic field. As the rotor turns:

1. The conductors of coil $b-b'$ are moving parallel to the field lines, so they are on the GNA. They cut no flux, inducing zero EMF.
2. The conductors of coil $a-a'$, while cutting maximum flux, are positioned such that the EMF induced in conductor 'a' is equal and opposite to the EMF in conductor 'a''. These two EMFs cancel each other out over the full coil.

Therefore, the net EMF in both coils is zero at this specific instant.

To determine the average voltage, we analyze the output voltage waveform over one full switching cycle. The gate signals in figure (ii) show a repeating pattern every $10 \mathrm{~\mu s}$, so the period is $T = 10 \mathrm{~\mu s}$.

During the first $3 \mathrm{~\mu s}$ of the cycle, switch $S_1$ is ON, connecting the output to the $20 \mathrm{~V}$ DC source. Thus, the voltage across the load is $20 \mathrm{~V}$. For the remaining $7 \mathrm{~\mu s}$, $S_1$ is OFF and the load voltage is $0 \mathrm{~V}$.

The average voltage is the total voltage-time area over one period divided by the period itself.
$V_{avg} = \frac{(20 \mathrm{~V} \times 3 \mathrm{~\mu s}) + (0 \mathrm{~V} \times 7 \mathrm{~\mu s})}{10 \mathrm{~\mu s}} = \frac{60 \mathrm{~V \cdot \mu s}}{10 \mathrm{~\mu s}} = 6 \mathrm{~V}$

The transformation matrix $A$ rotates the vector $u=(4,3)$ to the vector $v=(5,0)$. We can see this is a rotation because both vectors have the same magnitude, $|u|=\sqrt{4^{2+}3^2}=5$ and $|v|=\sqrt{5^{2+}0^2}=5$.

To transform $u$ into $v$, we must rotate it clockwise by the angle $\theta$ that $u$ makes with the positive x-axis. From the components of vector $u$, we find the trigonometric values for this angle: $\cos\theta = \frac{4}{5}$ and $\sin\theta = \frac{3}{5}$.

The general form of a matrix for a clockwise rotation by an angle $\theta$ is

.

Substituting our specific values for $\cos\theta$ and $\sin\theta$ gives the required matrix $A$:

The variance of a random process, $\sigma_0^2$, is the expected value of the squared deviation from the mean, $E[(X-\mu)^2]$. In this problem, the mean is given as zero ($\mu=0$), which simplifies the true variance to $\sigma_0^2 = E[X^2]$.

Let's find the expected value of the first estimator, $\hat{\sigma}_{1}^{2}$. By applying the linearity of expectation, we get:
E\[\hat{\sigma}{1}^{2}]$ = E\left[\frac{1}{10000} \sum{n=1}^{10000} x_{n}^{2}\right]$ = \frac{1}{10000} \sum_{n=1}^{10000} E[x_{n}^{2}]$

For every random sample $x_n$, its expected squared value is the true variance, E\[x_n^2]$ = \sigma_0^2$. Substituting this into the equation gives:$E$[\hat{\sigma}_{1}^{2}]$ = \frac{1}{10000} (10000 \cdot \sigma_0^2) = \sigma_0^2$.

This shows that $\hat{\sigma}_{1}^{2}$ is an unbiased estimator of the true variance. Note that the estimator $\hat{\sigma}_{2}^{2}$, which uses a denominator of $N-1$, provides an unbiased estimate only when the sample mean is used to estimate an unknown true mean. Since the true mean is known here, $\hat{\sigma}_{1}^{2}$ is the correct unbiased estimator.

The question requires a switch that has two specific characteristics. First, when OFF, it must block a positive voltage ($V > 0$) but not a negative one. This is known as unipolar voltage blocking. Second, when ON, it must allow current to flow in both directions, which is bipolar current capability.

Let's examine the circuits that meet these criteria:

• Circuit (A): This is an IGBT with an anti-parallel diode. When OFF, the IGBT blocks forward voltage ($V > 0$). If a negative voltage is applied, the anti-parallel diode becomes forward-biased and conducts, so it doesn't block. When ON, the IGBT conducts forward current ($I>0$) and the diode conducts reverse current ($I<0$). This configuration perfectly matches the requirements.

• Circuit (D): This is a common topology for a bidirectional switch. When OFF, the IGBT is open, blocking the path for forward voltage. For negative voltage, a path exists through the external diodes and the IGBT's own internal diode, so it conducts. When ON, the active IGBT and the surrounding diodes provide clear paths for current to flow in both directions.

Circuits (B) and (C) are incorrect. The addition of a diode in series creates a switch that either blocks voltage in both directions or only permits current flow in one direction, thus failing to meet the specifications.

A system is causal if its impulse response, $h[n]$, is zero for all negative time, i.e., $h[n]=0$ for $n<0$. The step response, $s[n]$, is the system's output when the input is a unit step, $u[n]$. Since the input $u[n]$ is zero for $n<0$, a causal system's output must also be zero for $n<0$.

This relationship is bidirectional ("if and only if"). We can derive the impulse response from the step response using the formula $h[n] = s[n] - s[n-1]$. If we are given that $s[n]=0$ for $n<0$, then for any negative $n$, both $s[n]$ and $s[n-1]$ are zero, which forces $h[n]$ to be zero. This proves the system is causal.

The total shunt admittance at any bus $i$, denoted $y_{i0}$, is found by summing all elements in the $i$-th row of the $Y_{\text{bus}}$ matrix. This value includes both discrete shunt elements and contributions from line charging.

Let's calculate the shunt admittance for each bus:

• Bus 1: $y_{10} = Y_{11} + Y_{12} + Y_{13} = (-j15) + (j10) + (j5) = 0$
• Bus 2: $y_{20} = Y_{21} + Y_{22} + Y_{23} = (j10) + (-j13.5) + (j4) = j0.5$
• Bus 3: $y_{30} = Y_{31} + Y_{32} + Y_{33} = (j5) + (j4) + (-j8) = j1$

The result $y_{10}=0$ is critical. It proves that there is no net shunt admittance connected to bus 1. Therefore, any scenario that requires a shunt capacitor at bus 1 or line charging on lines connected to bus 1 (1-2 or 1-3) is impossible. Option A requires line charging on all lines, and Option C requires a shunt capacitor at bus 1. Both would make $y_{10}$ non-zero, so they cannot be true.

To solve this problem, we first determine the circuit's initial conditions at DC steady state just before the switch opens ($t=0^-$). At steady state, the inductor acts as a short circuit and the capacitor as an open circuit. The $10 \text{ A}$ current from the source divides between resistors $R_1$ and $R_3$. The inductor current is $i_L(0^-) = 10 \text{ A} \times \frac{R_3}{R_1+R_3} = 10 \times \frac{3}{2+3} = 6 \text{ A}$. The capacitor voltage equals the voltage across the parallel branch, $V_C(0^-) = i_L(0^-) \times R_1 = 6 \text{ A} \times 2 \, \Omega = 12 \text{ V}$.

Due to the properties of inductors and capacitors, these values persist at the moment the switch is opened: $i_L(0^+) = 6 \text{ A}$ and $V_C(0^+) = 12 \text{ V}$. For $t>0$, the $10 \text{ A}$ source now feeds the parallel R-L and R-C branches. By KCL at the top node, the current in the capacitor's branch is $i_C(0^+) = 10 \text{ A} - i_L(0^+) = 4 \text{ A}$.

Finally, the voltage across both parallel branches must be equal at $t=0^+$. We can write an equation for the voltage of each branch and set them equal:
$V_{L}(0^{+}) + i_{L}(0^{+})R_{1} = V_{C}(0^{+}) + i_{C}(0^{+})R_{2}$
$V_{L}(0^{+}) + (6 \text{ A})(2 \, \Omega) = 12 \text{ V} + (4 \text{ A})(2 \, \Omega)$
$V_{L}(0^{+}) + 12 \text{ V} = 12 \text{ V} + 8 \text{ V}$
$V_{L}(0^{+}) = 8 \text{ V}$.

First, let's determine the motor's initial back EMF ($E_{b1}$) using the DC motor voltage equation:
$E_{b1} = V_{t1} - I_{a1}R_a = 400 - (15 \times 1.2) = 382 \text{ V}$.

The problem states the motor drives a constant torque load and the field current is unaltered. Since torque ($T$) is proportional to flux ($\phi$) and armature current ($I_a$), and both $T$ and $\phi$ are constant, the armature current $I_a$ must also remain constant at $15 \text{ A}$.

When the supply voltage drops by 10%, the new voltage is $V_{t2} = 0.9 \times 400 = 360 \text{ V}$. The new back EMF becomes:
$E_{b2} = V_{t2} - I_{a2}R_a = 360 - (15 \times 1.2) = 342 \text{ V}$.

For a DC motor with constant field flux, speed is directly proportional to the back EMF ($N \propto E_b$). We can find the new speed ($N_2$) by setting up a simple ratio:
$\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}}$
$N_2 = N_1 \times \frac{E_{b2}}{E_{b1}} = 1500 \times \frac{342}{382} \approx 1343 \text{ RPM}$.

The value of the convolution $z(t) = x(t) * y(t)$ is found by the integral $\int_{-\infty }^{\infty } y(\tau)x(t-\tau)d\tau$. Let's visualize this process. The signal $x(t-\tau)$ is a flipped and shifted version of $x(t)$. Since $x(t)$ is a symmetric rectangular pulse of width 2 (from -1 to 1), $x(t-\tau)$ is simply a sliding rectangular window of width 2, centered at $\tau=t$.