GATE ECE 2017 Set 1

An instruction's execution time is determined by the number of T-states it requires and the duration of each T-state. A T-state is simply one period of the processor's clock.

First, let's calculate the duration of a single T-state by finding the inverse of the given clock frequency, $f = 5 \text{ MHz}$:
$T = \frac{1}{f} = \frac{1}{5 \times 10^6 \text{ Hz}} = 0.2 \times 10^{-6} \text{ s} = 0.2 \ \mu s$.

To find the number of T-states needed for the instruction, we divide the total execution time by the duration of one T-state:
Number of T-states $= \frac{\text{Total Execution Time}}{\text{Duration per T-state}} = \frac{1.4 \ \mu s}{0.2 \ \mu s} = 7$.

To determine the system's properties, let's analyze its causality and stability.

A system is causal if its output at any time $n$ depends only on the current and past inputs. For all values of $n$, this system's output $y[n]$ is computed from $x[n]$ and/or $x[n-1]$. Since it never depends on future inputs (like $x[n+1]$), the system is causal.

For stability, we check if a bounded input, say $|x[n]| \le M_x$, produces a bounded output. In the interval $0 \le n \le 10$, we have $|y[n]| = n|x[n]| \le 10 M_x$. Elsewhere, $|y[n]| = |x[n] - x[n-1]| \le |x[n]| + |x[n-1]| \le 2M_x$. In both cases, the output is finite. Thus, the system is stable.

To determine if the functions are linearly independent, we set their linear combination equal to zero and see what it implies about the coefficients. Let's consider the equation $c_1(1) + c_2(x) + c_3(x^2) = 0$.

For this equation to hold true for all values of $x$ within an interval, the polynomial on the left must be the zero polynomial. A non-zero polynomial can only have a finite number of roots, but here it must be zero for infinitely many points in the interval. This is only possible if all the coefficients are zero.

Therefore, we must have $c_1 = 0$, $c_2 = 0$, and $c_3 = 0$. This is the definition of linear independence. This reasoning is valid for any interval on the real line, including both $[-1, 0]$ and $[0, 1]$. Consequently, the functions are linearly independent on both intervals, making statements I and III true.

A key property of a matrix is that if the sum of the elements in every row is the same constant value, let's call it $k$, then $k$ is an eigenvalue of that matrix.

Let's examine the given matrix $A$. Calculating the sum of the elements in the first row gives us $1 + 2 + 3 + 4 + 5 = 15$. Because every row is just a cyclic permutation of the first row, the sum for every other row is also 15.

Based on the property mentioned above, this constant row sum, 15, must be an eigenvalue of $A$. Since the problem states that there is only one real eigenvalue, it must be $\lambda = 15$.

The propagation constant's real part, $\alpha$, dictates the wave's attenuation. From the given $\gamma = (0.1+j40)m^{-1}$, we can identify this attenuation constant as $\alpha = 0.1$ Nepers/meter. The voltage magnitude decays exponentially with distance $l$ according to $|V(l)| = |V_{\in}|e^{-\alpha l}$.

Attenuation in decibels per meter is calculated by finding the loss over a 1-meter length using the standard voltage ratio:
$A_{dB/m} = 20 \log_{10}\left(\frac{|V(0)|}{|V(1)|}\right) = 20 \log_{10}\left(\frac{|V_{\in}|}{|V_{\in}|e^{-\alpha(1)}}\right)$
This simplifies to $20 \log_{10}(e^\alpha)$. Substituting $\alpha=0.1$, the calculation is $20(0.1)\log_{10}(e) \approx 0.868$ dB/m.

To understand the behavior of Silicon ($Si$) in a Gallium Arsenide ($GaAs$) crystal, we must consider the valence electrons of each element. Gallium ($Ga$) is in Group III (3 valence electrons), Arsenic ($As$) is in Group V (5 valence electrons), and Silicon ($Si$) is in Group IV (4 valence electrons).

When a $Si$ atom replaces a $Ga$ atom, it introduces one extra valence electron ($4 - 3 = 1$). This extra electron is donated to the crystal, making $Si$ an n-type dopant (a donor).

Conversely, when a $Si$ atom replaces an $As$ atom, it is deficient by one valence electron ($4 - 5 = -1$). This creates a "hole" that can accept an electron, making $Si$ a p-type dopant (an acceptor). Therefore, $Si$ behaves as an amphoteric dopant in $GaAs$.

To determine the rank of a matrix, we can convert it into row echelon form using elementary row operations. The rank is the number of non-zero rows in the resulting matrix.

Let's start with the given matrix

.

First, swapping the first and second rows ($R_1 \leftrightarrow R_2$) gives us a convenient leading 1. Next, we create zeros in the first column by performing $R_2 \to R_2 - 5R_1$ and $R_3 \to R_3 - 3R_1$.
$\begin{bmatrix} 1 & 0 & 2 \\ 5 & 10 & 10 \\ 3 & 6 & 6 \end{bmatrix} \to \begin{bmatrix} 1 & 0 & 2 \\ 0 & 10 & 0 \\ 0 & 6 & 0 \end{bmatrix}$
At this stage, we can see that the third row is a multiple of the second row ($R_3 = 0.6 R_2$). This dependency means one row can be eliminated. The operation $R_3 \to R_3 - 0.6R_2$ results in a zero row:
$\begin{bmatrix} 1 & 0 & 2 \\ 0 & 10 & 0 \\ 0 & 0 & 0 \end{bmatrix}$
Since the final matrix in echelon form has two non-zero rows, the rank of matrix M is 2.

The operating mode of a BJT is defined by the biasing of its two junctions. We can determine the bias by observing the minority carrier concentration at each junction.

Looking at the emitter-base junction ($J_E$), the graph shows the normalized excess minority carrier concentration, $\Delta n_E / n_{E0}$, is negative. This indicates a depletion of minority carriers from their equilibrium level, which is characteristic of a reverse-biased junction.

Conversely, at the collector-base junction ($J_C$), the graph shows a large positive excess concentration on both sides of the junction. This signifies a strong injection of minority carriers, the hallmark of a forward-biased junction.

With the emitter-base junction reverse biased and the collector-base junction forward biased, the transistor is operating in the inverse active mode.

In a Common Emitter amplifier, the small base-collector capacitance ($C_{\mu}$) acts as a feedback path between the output and input. Due to the Miller effect, this capacitance appears at the input as a much larger "Miller capacitance," given by $C_{\in(Miller)} \approx C_{\mu}(1 - A_v)$, where $A_v$ is the stage's voltage gain. This large effective input capacitance, along with the source resistance, forms a low-pass RC filter. The upper cutoff frequency is defined as $f_H = 1 / (2\pi R_{eq}C_{\in})$. A significantly larger $C_{\in}$ therefore causes a decrease in $f_H$, limiting the amplifier's high-frequency response.

The D-latch is transparent when its clock input, CLK2, is high. During this time, the output Q simply follows the data input D, which is CLK1. Thus, the output Q will be high only when both CLK1 and CLK2 are simultaneously high.

Let the clock period be $T_{CLK}$. CLK1 is high for the first half of the period, from $t=0$ to $T_{CLK}/2$. CLK2 is delayed by $T_{CLK}/5$, so it becomes high at $t=T_{CLK}/5$.

The output Q becomes high when CLK2 goes high at $T_{CLK}/5$ (since CLK1 is already high) and goes low when CLK1 goes low at $T_{CLK}/2$. The total time the output remains high is the difference between these two points:
$T_{high} = \frac{T_{CLK}}{2} - \frac{T_{CLK}}{5} = \frac{5T_{CLK} - 2T_{CLK}}{10} = \frac{3T_{CLK}}{10}$

The duty cycle is the ratio of the high duration to the total period, expressed as a percentage:
Duty Cycle = $\frac{T_{high}}{T_{CLK}} \times 100 = \frac{3T_{CLK}/10}{T_{CLK}} \times 100 = 30\%$

A phase-lag controller is described by a transfer function of the form $G_c(s) = K \frac{s+z_c}{s+p_c}$. This system introduces a pole at $s = -p_c$ and a zero at $s = -z_c$. For a controller to be a phase-lag compensator, its pole must be closer to the origin of the s-plane than its zero. Assuming both the pole and zero are on the negative real axis for stability, this means the magnitude of the pole's position is smaller than the zero's. This condition is written as $|p_c| < |z_c|$. On the pole-zero plot, the pole (x) will therefore be positioned to the right of the zero (o).

Initially, with P=Q='0', both NAND gates have at least one '0' input, which forces both of their outputs to be '1'. Therefore, the starting state is X='1', Y='1'.

When the inputs change to P=Q='1', a race condition develops. Both gates initially have '1's on all their inputs (P,Y for the top; Q,X for the bottom) and thus try to switch their outputs to '0'. Because the propagation delays are unequal, one gate will "win" this race.

If the top gate is faster, its output X will become '0' first. This X='0' is fed back to the bottom gate, forcing its output Y to remain '1'. The stable state becomes X='0', Y='1'.

Conversely, if the bottom gate is faster, Y will become '0' first. This Y='0' forces the top gate's output X to remain '1'. The stable state becomes X='1', Y='0'.

To solve this, we first need to determine the total number of possible outcomes when throwing three dice. Since each die has 6 independent faces, the total number of combinations is $6 \times 6 \times 6 = 216$.

Next, we identify the number of favorable outcomes. We want the cases where all three dice show the same number. This occurs if we roll $(1,1,1)$, $(2,2,2)$, $(3,3,3)$, $(4,4,4)$, $(5,5,5)$, or $(6,6,6)$. There are exactly 6 such outcomes.

The probability is the ratio of favorable outcomes to the total number of outcomes.
$\text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{6}{216} = \frac{1}{36}$
Converting this fraction to a decimal, we get approximately $0.0277...$, which rounds to $0.028$ to three decimal places.

The problem provides two key properties of the signal $x(t)$. First, the condition $x(t) = -x(-t)$ defines an odd function. The Fourier series of an odd function contains only sine terms, which means the DC component $a_0$ and all cosine coefficients $a_n$ must be zero for all $n$.

Second, the condition $x(t) = -x(t - \pi/\omega_0)$ describes half-wave symmetry, as $\pi/\omega_0$ is half the fundamental period $T=2\pi/\omega_0$. A signal with half-wave symmetry has a Fourier series containing only odd harmonics. This means all coefficients for even-numbered harmonics ($n=2, 4, 6, ...$) are zero.

Combining these two properties, we know that all coefficients $a_n$ are zero due to the odd symmetry. Furthermore, due to the half-wave symmetry, the sine coefficients $b_n$ must be zero for all even values of $n$.

The steady-state error performance of a system is dictated by its "type," which corresponds to the number of pure integrators (poles at $s=0$) in the open-loop transfer function. For the given function $G(s)$, the system type is equal to the integer $p$.

We are given that the steady-state error ($e_{ss}$) is zero for a unit step input and finite for a unit ramp input. This specific behavior is the defining characteristic of a Type 1 system. Therefore, to meet these requirements, the system must have exactly one pole at the origin, which means the value of the parameter $p$ must be 1.

The built-in potential, $V_0$, across this $n^{+}-n$ junction is determined by the ratio of the electron concentrations in the two regions. The governing equation is $V_0 = V_T \ln(\frac{n_1}{n_2})$, where $V_T$ is the thermal voltage.

Assuming complete ionization in these n-type regions, the electron concentration is approximately equal to the donor doping concentration ($n \approx N_D$). We can therefore substitute the given doping levels into the formula.

Using the thermal voltage $V_T = kT/q = 0.025$ V, the calculation becomes:
$V_0 = 0.025 \text{ V} \times \ln\left(\frac{N_{D1}}{N_{D2}}\right) = 0.025 \text{ V} \times \ln\left(\frac{10^{18}}{10^{15}}\right)$
$V_0 = 0.025 \text{ V} \times \ln(1000) \approx 0.173 \text{ V}$

This circuit is a Schmitt trigger, where the reference voltage at the non-inverting input ($V_+$) determines the switching thresholds. This voltage is set by a divider network involving the output $V_{out}$ and the 3V source. The general expression for this reference voltage is:
$V_+ = \frac{V_{out} + 6}{3}$

The upper threshold voltage, $V_{UT}$, is the value of $V_+$ when the output is at its positive saturation limit of $+15$V.
$V_{UT} = \frac{15 + 6}{3} = 7 \text{V}$

The lower threshold voltage, $V_{LT}$, is the value of $V_+$ when the output is at its negative saturation limit of $-15$V.
$V_{LT} = \frac{-15 + 6}{3} = -3 \text{V}$

To find the angular frequency $\omega$ where the phase difference between $V_1$ and $V_2$ is $\pi/4$, we'll work with phasors. Let the current phasor be $I$.

The voltage $V_1$ is across a pure 1Ω resistor, so its phasor is $V_1 = I \cdot 1$. This means $V_1$ is in phase with the current $I$.

The voltage $V_2$ is across a series combination of a 1Ω resistor and a 1H inductor. The impedance of this combination is $Z_2 = R + j\omega L = 1 + j\omega$. The voltage phasor is $V_2 = I \cdot Z_2$. The phase of $V_2$ leads the current's phase by the angle of the impedance, $\angle Z_2$.

The phase difference between $V_2$ and $V_1$ is therefore the phase angle of $Z_2$, which is $\arg(Z_2) = \arctan(\frac{\omega}{1})$.
We are given this phase difference is $\pi/4$. So, we set $\arctan(\omega) = \pi/4$.
Solving for the angular frequency gives $\omega = \tan(\pi/4) = 1$ rad/s.

To prevent symbols from interfering with each other, the overall pulse shape must satisfy the Nyquist criterion for zero inter-symbol interference (ISI). In the frequency domain, this criterion states that the sum of the pulse's spectrum, $P(f)$, and all its replicas shifted by the symbol rate, $R_s$, must be constant.

Given the symbol rate is 2000 symbols/second, the frequency shift is $R_s = 2$ kHz. The condition is:
$\sum_{n=-\infty }^{\infty } P(f - nR_s) = \text{constant}$
We must check which spectrum, when shifted by multiples of 2 kHz and summed, produces a flat line.

Only the spectrum in option (B) satisfies this. Its triangular roll-off regions are designed with a special symmetry. For instance, in the range from 0.8 kHz to 1.2 kHz, the decreasing tail of the original spectrum $P(f)$ adds with the increasing tail of the shifted spectrum $P(f-2)$ to maintain a constant sum of 1. This holds true across all frequencies, ensuring zero ISI.

The asymptotic slope of a system's Bode magnitude plot at high frequencies ($\omega \rightarrow \infty$) is directly related to the difference between the degree of the denominator polynomial ($q$) and the numerator polynomial ($p$).

Each pole adds -20 dB/decade to the slope and each zero adds +20 dB/decade. The net high-frequency slope is therefore $(p - q) \times 20$ dB/decade.

Given a final slope of -60 dB/decade, we can set up the equation:
$(p - q) \times 20 = -60$

Solving for the difference gives $p - q = -3$, which means $q - p = 3$. This tells us the system must have three more poles than zeros. Reviewing the options, we seek the pair $(p, q)$ that satisfies this condition.

A transconductance amplifier is designed to be a voltage-controlled current source, producing an output current $i_{out}$ proportional to an input voltage $v_{\in}$. For the input, to accurately measure the source's voltage without loading it down, the amplifier must draw minimal current. This is achieved with a very high input resistance ($R_{\in} \to \infty$). For the output, to act as an ideal current source, it must deliver a constant current to the load, regardless of the load's resistance. A very high output resistance ($R_{out} \to \infty$) ensures the amplifier's generated current is forced into the load rather than being shunted internally.

Mutual information, $I(X_1; X_2)$, quantifies the amount of information one random variable contains about another. A key piece of information given is that $X_1$ and $X_2$ are independent. By definition, two independent variables have no statistical relationship; knowing the value of one provides no information whatsoever about the value of the other. Therefore, the information they "mutually" share is zero. The given means and variances are distractors, as the independence of the variables is the only fact needed to determine the answer.

The fundamental condition for BIBO stability is that the Region of Convergence (ROC) of the system's transfer function, $H(s)$, must include the imaginary axis ($j\omega$ axis).

Statement I is false. A common misconception is that poles in the right-half plane (RHP) always imply instability. However, this is only true for causal systems. A non-causal system with an RHP pole can have an ROC that is a left-half plane or a vertical strip, which can include the $j\omega$ axis, making the system stable.

Therefore, Statement II is true. A system can be non-causal, have a pole in the RHP, and still be BIBO stable, as its ROC can be properly defined to encompass the imaginary axis.

Differential Pulse Code Modulation (DPCM) is a predictive coding technique used for signal compression. It capitalizes on the fact that many signals have high correlation between adjacent samples. Instead of quantizing the actual signal sample, $m[n]$, DPCM first creates a prediction of that sample, $\hat{m}[n]$, based on past values. The core of the method is to then find the difference, or prediction error, $e[n] = m[n] - \hat{m}[n]$. This error signal, which is typically much smaller in magnitude than the original signal, is what gets quantized and transmitted.

The capacity of a wireless link depends on the received signal power, which is described by the Friis transmission equation. This equation states that received power ($P_r$) is proportional to the product of the antenna effective areas and inversely proportional to the square of the distance.

$P_r \propto \frac{A_{et} A_{er}}{d^2}$

In this scenario, both antenna areas are doubled, increasing the numerator by a factor of $2 \times 2 = 4$. Simultaneously, the distance is doubled, which increases the denominator ($d^2$) by a factor of $2^2 = 4$.

The factor of 4 increase in the numerator is perfectly cancelled by the factor of 4 increase in the denominator. As a result, the received power $P_r$ remains the same. Since capacity is a function of the signal-to-noise ratio, and both the signal ($P_r$) and noise (unchanged) are constant, the maximum capacity does not change.

To solve this using the Newton-Raphson method, we first need to find the root of the function $f(x) = x^3 + x - 1$. The derivative is $f'(x) = 3x^2 + 1$. The iterative formula is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$.

Starting with the initial guess $x_0 = 1$, the first iteration is:
$x_1 = 1 - \frac{f(1)}{f'(1)} = 1 - \frac{1^{3+}1-1}{3(1)^{2+}1} = 1 - \frac{1}{4} = 0.75$.

For the second iteration, we use the result from the first, $x_1=0.75$:
$x_2 = 0.75 - \frac{f(0.75)}{f'(0.75)} = 0.75 - \frac{(0.75)^{3+}0.75-1}{3(0.75)^{2+}1} = 0.75 - \frac{0.171875}{2.6875} \approx 0.686$.

After two iterations, the solution is approximately $0.686$.

For two FSK signals to be uncorrelated (orthogonal), their frequency separation must be an integer multiple of half the bit rate. The frequencies of the given signals are $f_0 = \frac{20000\pi}{2\pi} = 10$ kHz and $f_1 = \frac{22000\pi}{2\pi} = 11$ kHz.

The condition for orthogonality is $|f_1 - f_0| = \frac{n}{2T}$, where $n$ is a positive integer and $T$ is the bit duration.
Substituting the known values, we have $11000 - 10000 = \frac{n}{2T}$, which simplifies to $1000 = \frac{n}{2T}$.

We can now solve for the bit duration: $T = \frac{n}{2000}$ seconds.
To find the smallest positive value for $T$, we must use the smallest positive integer for $n$, which is $n=1$.
This gives $T = \frac{1}{2000} \text{ s} = 0.0005 \text{ s}$, which is equivalent to 0.5 ms.

To determine the midband voltage gain, we begin with a DC analysis to find the transistor's operating point. The Thevenin equivalent voltage at the base is found using the voltage divider rule: $V_{Th} = \frac{12 \text{ V} \times 47 \text{ k}\Omega}{73 \text{ k}\Omega + 47 \text{ k}\Omega} = 4.7 \text{ V}$. With negligible base current, we can find the collector current, $I_C = \frac{V_{Th}-V_{BE}}{R_E} = \frac{4.7 \text{ V} - 0.7 \text{ V}}{2 \text{ k}\Omega} = 2 \text{ mA}$.

Next, we calculate the transconductance, $g_m = \frac{I_C}{V_T} = \frac{2 \text{ mA}}{25 \text{ mV}} = 80 \text{ mS}$. For the midband AC gain, the bypass capacitor $C_E$ effectively shorts the emitter to ground. The total AC resistance at the collector is $R'_L = R_C || R_L = 2 \text{ k}\Omega || 8 \text{ k}\Omega = 1.6 \text{ k}\Omega$. The voltage gain is then $A_V = -g_m R'_L = -80 \text{ mS} \times 1.6 \text{ k}\Omega = -128$. Therefore, the magnitude of the gain is $|A_V|=128$.

From the input voltage $100 \cos(3t)$, we can identify the angular frequency as $\omega = 3$ rad/s. The voltage $V_1$ is across the impedance $Z_1 = 4\Omega + j\omega(1\text{H}) = (4+j3)\Omega$. Similarly, the voltage $V_2$ is across the impedance $Z_2 = 5\Omega + \frac{1}{j\omega(1/36\text{F})} = (5-j12)\Omega$.

Since the two impedance sections are in series, the same current, let's call it $I$, flows through both. According to Ohm's law, the voltage amplitudes are $|V_1| = |I||Z_1|$ and $|V_2| = |I||Z_2|$.

The ratio of the voltage amplitudes is therefore the ratio of the impedance magnitudes:
$\frac{|V_2|}{|V_1|} = \frac{|I||Z_2|}{|I||Z_1|} = \frac{|Z_2|}{|Z_1|}$
Calculating the magnitudes, we get $|Z_1| = \sqrt{4^2 + 3^2} = 5\Omega$ and $|Z_2| = \sqrt{5^2 + (-12)^2} = 13\Omega$.

Finally, the ratio is $\frac{13}{5} = 2.6$.

This first-order differential equation can be solved by making a substitution to simplify the expression on the right-hand side. Let's define a new variable $v = x+y-1$. By differentiating this with respect to $x$, we find $\frac{dv}{dx} = 1 + \frac{dy}{dx}$, which allows us to replace $\frac{dy}{dx}$ with $\frac{dv}{dx}-1$.

Substituting both $v$ and this new derivative expression into the original equation transforms it into $\frac{dv}{dx} - 1 = v^2$. This is now a separable equation, which we can rearrange to $\frac{dv}{v^{2+}1} = dx$. Integrating both sides gives $\int \frac{1}{v^{2+}1} dv = \int dx$, which results in $\tan^{-1}(v) = x+c$.

To get our final answer in terms of $x$ and $y$, we substitute back $v=x+y-1$, yielding $\tan^{-1}(x+y-1) = x+c$. Finally, we solve for $y$ by taking the tangent of both sides, which gives $x+y-1 = \tan(x+c)$, or $y = 1-x+\tan(x+c)$.

The root locus traces the path of the closed-loop poles as the gain $K$ varies. The locus starts at the open-loop poles, which are the roots of $s^{2-}3s+2=0$, namely $s=1$ and $s=2$. Since both poles are in the right-half plane (RHP), the system is initially unstable for small $K$.

The locus branches terminate at the open-loop zeros, which are the roots of $s^{2+}2s+2=0$, namely $s=-1 \pm j$. These zeros are in the stable left-half plane (LHP). For the system to become stable, the root locus must cross the imaginary axis into the LHP.

We are given that this crossover occurs at $K=1.5$. This means for $K>1.5$, the poles move from the imaginary axis into the LHP, making the closed-loop system stable.

To evaluate this line integral, we first need a parametric representation of the path C. For the straight line segment from A(0,2,1) to B(4,1,-1), we can define the path as $\vec{r}(t) = (1-t)A + tB$ for $t \in [0, 1]$. This gives us the equations $x=4t$, $y=2-t$, and $z=1-2t$.

From these, we find the differentials: $dx = 4dt$, $dy = -dt$, and $dz = -2dt$.

Now, we substitute these expressions for $x, y, z$ and their differentials into the integral, converting it into a definite integral with respect to $t$:
$I = \int_{0}^{1} [2(1-2t)(4) + 2(2-t)(-1) + 2(4t)(-2)] dt$

Simplifying the integrand leads to:
$I = \int_{0}^{1} (8-16t - 4+2t - 16t) dt = \int_{0}^{1} (4 - 30t) dt$

Evaluating this definite integral gives the final value:
I = \[4t - 15t^2]_0^1 = (4(1) - 15(1)^2) - (0) = -11$

The junction capacitance ($C$) of a reverse-biased p-n diode is modeled like a parallel-plate capacitor, $C = \frac{\varepsilon A}{W}$, where $W$ is the depletion width. This relationship means that the capacitance is inversely proportional to the depletion width, $C \propto \frac{1}{W}$.

For a large reverse bias voltage ($V_R$), the depletion width is given by $W \approx \sqrt{\frac{2 \varepsilon V_{R}}{e}\left(\frac{1}{N_{A}}+\frac{1}{N_{D}}\right)}$. This shows that $W$ is proportional to $\sqrt{\frac{1}{N_A} + \frac{1}{N_D}}$.

The ratio of the capacitances can be found from the inverse ratio of the depletion widths:
$\frac{C_{2}}{C_{1}} = \frac{W_{1}}{W_{2}} = \sqrt{\frac{\frac{1}{N_{A1}} + \frac{1}{N_{D1}}}{\frac{1}{N_{A2}} + \frac{1}{N_{D2}}}}$

For Diode 1, $N_{A1} = N_{D1} = 10^{14}$ cm$^{-3}$. For Diode 2, $N_{A2} = N_{D2} = 10^{16}$ cm$^{-3}$.
Plugging in these values:
$\frac{C_{2}}{C_{1}} = \sqrt{\frac{\frac{1}{10^{14}} + \frac{1}{10^{14}}}{\frac{1}{10^{16}} + \frac{1}{10^{16}}}} = \sqrt{\frac{2/10^{14}}{2/10^{16}}} = \sqrt{\frac{10^{16}}{10^{14}}} = \sqrt{100} = 10$.

A circularly polarized wave starts with a $\frac{\pi}{2}$ phase difference between its orthogonal electric field components. For this wave to become linearly polarized, the components must be in phase or exactly out of phase, meaning their phase difference must become a multiple of $\pi$. The minimum distance corresponds to the smallest change, which is an additional phase shift of $\frac{\pi}{2}$ to reach a total difference of $\pi$.

This extra phase shift accumulates because the different refractive indices, $n_x$ and $n_y$, cause the components to travel at different phase velocities. Over a distance $z$, the accumulated phase difference is $\Delta\phi = |k_y - k_x|z$, where $k = \frac{2\pi n}{\lambda_0}$ is the wave number.

To find the minimum distance $z_{min}$, we set this accumulated phase difference to $\frac{\pi}{2}$:
$\left| \frac{2\pi n_y}{\lambda_0} - \frac{2\pi n_x}{\lambda_0} \right| z_{min} = \frac{\pi}{2}$
Solving for $z_{min}$ yields:
$z_{min} = \frac{\lambda_0}{4|n_y - n_x|}$
Plugging in the given values:
$z_{min} = \frac{1.5 \times 10^{-6} \, \text{m}}{4(1.5001 - 1.5000)} = \frac{1.5 \times 10^{-6} \, \text{m}}{4 \times 10^{-4}} = 0.00375 \, \text{m}$
Converting to the required units, the distance is $0.375$ cm.

We are given the signal $x[n] = \{1, 2, 1\}$. Since $h[0]=1$ and the signal is non-zero only for $n=0, 1, 2$, we can denote it as $h[n] = \{1, h[1], h[2]\}$. We'll use the given convolution outputs to find the unknown values $h[1]$ and $h[2]$.

The convolution formula $y[n] = x[n] * h[n]$ gives us:
$y[1] = x[0]h[1] + x[1]h[0] = (1)h[1] + (2)(1) = h[1] + 2$.
Given $y[1]=3$, we find $h[1]+2 = 3$, which means $h[1]=1$.

Next, we use $y[2]$:
$y[2] = x[0]h[2] + x[1]h[1] + x[2]h[0] = (1)h[2] + (2)h[1] + (1)(1) = h[2] + 2h[1] + 1$.
Given $y[2]=4$ and our result $h[1]=1$, we have $h[2] + 2(1) + 1 = 4$, which gives $h[2]=1$.

Now that we know $h[n]=\{1, 1, 1\}$, we can find $y[3]$ and $y[4]$:
$y[3] = x[1]h[2] + x[2]h[1] = (2)(1) + (1)(1) = 3$.
$y[4] = x[2]h[2] = (1)(1) = 1$.

Finally, the value of the expression is $10y[3] + y[4] = 10(3) + 1 = 31$.