GATE ECE 2020

The dimension of the vector space $\mathbb{R}^4$ is 4. This single fact dictates the properties of any set of vectors within it.

A set of 6 vectors in a 4-dimensional space is always linearly dependent because it has more vectors than the dimension. This makes statement B true. The set is also not guaranteed to span $\mathbb{R}^4$; for example, all six vectors could lie in a 3-dimensional subspace. This makes statement A true.

For a set with exactly 4 vectors in $\mathbb{R}^4$, the conditions of spanning the space and being linearly independent are equivalent. If one is true, so is the other, and the set is a basis. Therefore, statement D is true.

Statement C is false because it makes the strong claim that any four of the vectors form a basis. This is not guaranteed, as the chosen subset could easily be linearly dependent (e.g., if $v_1 = 2v_2$).

Let's analyze the core definitions of vector field properties. A vector field $\vec{A}$ is defined as irrotational, or curl-free, if its curl is zero: $\nabla \times \vec{A} = 0$. Option C incorrectly claims the condition is $\nabla^2 \vec{A} = 0$. This equation, known as the vector Laplace equation, describes a different property and is not the test for an irrotational field.

The other statements are correct. A solenoidal field has zero divergence ($\nabla \cdot \vec{A} = 0$), the curl of a vector field is another vector field, and the "curl of the curl" identity is a standard formula in vector calculus.

To find the partial derivative with respect to $x$, we differentiate the function $f(x,y,z)$ by treating $y$ and $z$ as constant values.

Applying the chain rule to the first term and treating the second term as $x$ times a constant, the derivative is:
$\frac{\partial f}{\partial x} = e^{1-x \cos y} \cdot (-\cos y) + ze^{-1/(1+y^2)}$

Now, we evaluate this derivative at the specified point $(x,y,z)=(1,0,e)$ by substituting these values into our expression.
$\left. \frac{\partial f}{\partial x} \right|_{(1,0,e)} = e^{1-(1)\cos(0)}(-\cos 0) + e \cdot e^{-1/(1+0^2)}$

Since $\cos(0)=1$, the calculation simplifies to:
$e^{1-1}(-1) + e \cdot e^{-1} = e^0(-1) + 1 = -1 + 1 = 0$

To solve this second-order linear homogeneous differential equation, we begin by writing its characteristic (or auxiliary) equation. We replace $\frac{d^2y}{dx^2}$ with $m^2$, $\frac{dy}{dx}$ with $m$, and $y$ with $1$.

This gives us the quadratic equation: $m^2 - 6m + 9 = 0$.

Factoring the quadratic, we get $(m-3)^2 = 0$. This shows we have a repeated real root, $m_1 = m_2 = 3$.

For a repeated real root $m$, the general solution to the differential equation is $y = (C_1 + C_2x)e^{mx}$.

Substituting our root $m=3$ into this form gives the general solution: $y = (C_1 + C_2x)e^{3x}$.

The shift of the intrinsic Fermi level ($E_{F_i}$) from the exact center of the bandgap ($E_{mid}$) depends on the effective densities of states for electrons ($N_C$) and holes ($N_V$). This shift is calculated using the formula:
$\Delta E = E_{F_i} - E_{mid} = \frac{KT}{2}\ln\left(\frac{N_V}{N_C}\right)$

We are given that the thermal energy $KT$ is $26 \text{ meV}$. The problem also states that the density of states for holes is twice that for electrons, meaning $N_V = 2N_C$, so the ratio $\frac{N_V}{N_C} = 2$.

Substituting these values into the equation:
$\Delta E = \frac{26 \text{ meV}}{2}\ln(2) = 13 \text{ meV} \times 0.6931 \approx 9.01 \text{ meV}$

This circuit is a non-inverting Schmitt trigger. The output, $V_o$, will be at one of the saturation voltages, either $+5$ V or $-5$ V. The output switches states only when the voltage at the non-inverting input, $V^+$, crosses the voltage at the inverting input, which is $0$ V.

Using superposition, the voltage at the non-inverting input is $V^+ = (V_i + V_o)/2$. For the output to switch, $V^+$ must cross $0$ V. If $V_o = +5$ V, switching requires $V^+ < 0$, which means $V_i < -5$ V. If $V_o = -5$ V, switching requires $V^+ > 0$, meaning $V_i > +5$ V.

The input signal is a sine wave with a peak amplitude of 1 V, so its value is always in the range $[-1 V, +1 V]$. Since the input voltage never reaches the required thresholds of $\pm 5$ V, the output will never switch. It will remain constant at either $+5$ V or $-5$ V.

To find the Thevenin voltage, $V_{TH}$, we can systematically simplify the circuit from left to right using source transformations.

First, convert the 1A current source and its parallel 1Ω resistor into a $1\text{V}$ ($1\text{A} \times 1\text{\Omega }$) voltage source in series with the 1Ω resistor. This new source adds in series with the existing 2V source and 2Ω resistor, creating a simplified branch with a total voltage of $1\text{V} + 2\text{V} = 3\text{V}$ and a total resistance of $1\text{\Omega } + 2\text{\Omega } = 3\text{\Omega }$.

Next, transform this 3V source and 3Ω series resistor back into a current source of $1\text{A}$ ($3\text{V}/3\text{\Omega }$) in parallel with a 3Ω resistor. Now combine this with the parallel 2A source and 2Ω resistor. The total current is $1\text{A} + 2\text{A} = 3\text{A}$, and the equivalent parallel resistance is $3\text{\Omega } || 2\text{\Omega } = 1.2\text{\Omega }$.

This leaves a simplified circuit of a 3A source in parallel with a 1.2Ω resistor. The voltage across this combination is $V = I \times R = 3\text{A} \times 1.2\text{\Omega } = 3.6\text{V}$. Since $V_{TH}$ is the open-circuit voltage, no current flows through the 4Ω resistor, so $V_{TH}$ is simply 3.6V.

The output of a 4-to-1 multiplexer is a sum-of-products expression, built by pairing each data input with its corresponding select line combination. With select lines $S_1 = P$ and $S_0 = Q$, the general output function is $F = \bar{P}\bar{Q}I_0 + \bar{P}QI_1 + P\bar{Q}I_2 + PQI_3$.

By substituting the specific data inputs from the diagram ($I_0=R, I_1=0, I_2=R, I_3=1$), we get:
$F = \bar{P}\bar{Q}(R) + \bar{P}Q(0) + P\bar{Q}(R) + PQ(1)$

This expression simplifies to $F = \bar{P}\bar{Q}R + P\bar{Q}R + PQ$.
We can factor out the common term $\bar{Q}R$ from the first two parts of the expression:
$F = (\bar{P} + P)\bar{Q}R + PQ$

Using the Boolean identity $\bar{P} + P = 1$, the function reduces to its final simplified form:
$F = \bar{Q}R + PQ$

This problem is an application of the Argument Principle. The number of encirclements of the origin in the $G(s)$-plane depends on the number of poles and zeros enclosed by the contour $\Gamma$ in the s-plane.

From the diagram, the clockwise contour $\Gamma$ encloses $Z=3$ zeros and $P=2$ poles. Each zero contributes one encirclement in the same direction as $\Gamma$ (clockwise). In contrast, each pole contributes one encirclement in the opposite direction (counter-clockwise).

This results in 3 clockwise encirclements from the zeros and 2 counter-clockwise encirclements from the poles. The net effect is $3 - 2 = 1$ encirclement in the clockwise direction.

Let's first calculate the probability of the complementary event, which is the entire block being decoded perfectly with no errors. The probability of a single bit being decoded correctly is $1-\alpha$. Since each bit's decoding is an independent event, the probability that all $N$ bits are correct is the product of their individual probabilities, which is $(1-\alpha)^N$.

The problem asks for the probability that a block is "erroneous," meaning it has at least one error. This is the exact opposite of the block being entirely correct. Therefore, we can find the desired probability by subtracting the probability of a perfect block from 1.

Probability(erroneous block) = $1 - P(\text{perfect block}) = 1 - (1-\alpha)^N$.

The given impedance, $Z=jX$, is purely reactive, meaning its resistive component is zero. We first find the normalized impedance, $z=Z/Z_0 = 0+jx$, which also has a zero resistive part ($r=0$). The Smith chart is a graphical representation of the complex reflection coefficient, $\Gamma$.

The reflection coefficient is calculated using the formula $\Gamma = \frac{z-1}{z+1}$.
Substituting $z=jx$ for our purely reactive load, we get:
$\Gamma = \frac{jx-1}{jx+1}$

The magnitude of this complex number is $|\Gamma| = \frac{|jx-1|}{|jx+1|} = \frac{\sqrt{x^2 + (-1)^2}}{\sqrt{x^2 + 1^2}} = 1$.
A reflection coefficient with a magnitude of 1 corresponds to a circle of radius 1 centered at the origin $(0,0)$ on the Smith chart, which represents the chart's outer boundary.

To find the pole-zero plot, we first need the system's transfer function, $H(z)$. We begin by expanding the summation in the difference equation:
$y[n] = x[n] - x[n-1] + x[n-2] - x[n-3]$

Next, we apply the Z-transform to this equation, which gives us:
$Y(z) = X(z) - z^{-1}X(z) + z^{-2}X(z) - z^{-3}X(z)$

The transfer function $H(z)$ is the ratio $\frac{Y(z)}{X(z)}$. We can rewrite this as a rational function with positive powers of $z$:
$H(z) = 1 - z^{-1} + z^{-2} - z^{-3} = \frac{z^3 - z^2 + z - 1}{z^3}$

The poles are the roots of the denominator, $z^3=0$, which means there is a third-order pole at the origin, $z=0$. The zeros are the roots of the numerator, $z^3 - z^2 + z - 1 = (z-1)(z^{2+}1)$. Solving for the roots gives us three zeros at $z=1$, $z=j$, and $z=-j$.

To ensure maximum power is transferred to the load, the load impedance $Z_L$ must be equal to the Thevenin equivalent impedance $Z_{th}$ seen from the output terminals. For a two-port network with a source resistance $R_S$ connected at the input, the Thevenin impedance at the output is given by the standard formula:
$Z_{th} = Z_{22} - \frac{Z_{12}Z_{21}}{Z_{11} + R_S}$
Using the values from the given impedance matrix and the circuit ($R_S = 10 \Omega$):
$Z_{th} = 120 - \frac{60 \times 60}{40 + 10} = 120 - \frac{3600}{50}$
Solving this gives $Z_{th} = 120 - 72 = 48 \Omega$. Therefore, the required load impedance for maximum power transfer is $Z_L = 48 \Omega$.

To determine the inductor's value, we first analyze the circuit in the frequency domain using phasors. The voltage source $v(t)$ corresponds to the phasor $V = 200\angle 0^{\circ}$ V, and the current $i(t)$ corresponds to the phasor $I = 10\angle -45^{\circ}$ A.

Using Ohm's Law for AC circuits, we find the total impedance $Z$:
$Z = \frac{V}{I} = \frac{200\angle 0^{\circ}}{10\angle -45^{\circ}} = 20\angle 45^{\circ} \, \Omega$
To isolate the reactive component, we convert this impedance to rectangular form:
$Z = 20(\cos 45^{\circ} + j\sin 45^{\circ}) = 10\sqrt{2} + j10\sqrt{2} \, \Omega$
For this series RL circuit, the impedance is $Z = R + jX_L$. By comparison, the inductive reactance is $X_L = 10\sqrt{2} \, \Omega$. Since $X_L = \omega L$ and the angular frequency is $\omega = 5$ rad/s (from the source), we can solve for the inductance:
$L = \frac{X_L}{\omega} = \frac{10\sqrt{2}}{5} = 2\sqrt{2} \approx 2.83 \, \text{H}$

This circuit is a full-wave voltage doubler. In steady state, with ideal components, it produces a DC output voltage equal to twice the peak amplitude of the sinusoidal input voltage.

First, we must find the peak input voltage, $V_p$, from the given RMS value:
$V_p = V_{rms} \times \sqrt{2} = 230\sqrt{2}$ V.

The output voltage, $V_o$, is then twice this peak value.
$V_o = 2 \times V_p = 2 \times (230\sqrt{2}) \text{ V} \approx 650.54$ V.

This circuit is a classic non-inverting amplifier. The first step is to calculate the output voltage, $V_o$. The voltage gain is determined by the resistors, giving $V_o = V_i \left(1 + \frac{R_f}{R_1}\right)$. With $V_i = 2$ V and both resistors at $1 \text{ k}\Omega$, the output voltage is $V_o = 2 \text{ V} \left(1 + \frac{1\text{k}}{1\text{k}}\right) = 4$ V.

The current $I_o$ supplied by the op-amp splits into two branches: one through the feedback resistor and one through the load resistor. To find the current in the feedback path, we use the ideal op-amp property that the inverting input voltage matches the non-inverting input: $V_- = V_+ = 2$ V.

Applying Kirchhoff's Current Law at the output node, $I_o$ is the sum of these two branch currents:
$I_o = I_{\text{feedback}} + I_{\text{load}} = \frac{V_o - V_-}{1 \text{ k}\Omega} + \frac{V_o - 0}{1 \text{ k}\Omega}$
$I_o = \frac{4 \text{ V} - 2 \text{ V}}{1 \text{ k}\Omega} + \frac{4 \text{ V}}{1 \text{ k}\Omega} = 2 \text{ mA} + 4 \text{ mA} = 6 \text{ mA}$.

The number of address lines, which we'll call $n$, determines the total number of unique memory locations ($N$) that can be accessed. This relationship is defined by the formula $N = 2^n$.

We are given a memory bank of 16 K bytes. To find the required number of address lines, we first express this size as a power of 2, recalling that $1 \text{ K} = 2^{10}$ and $16 = 2^4$.

The total number of memory locations is:
$N = 16 \times 1 \text{ K} = 2^4 \times 2^{10} = 2^{14}$

By equating this with our formula, we have $2^n = 2^{14}$.
This means that $n=14$ address lines are needed to access all 16 K byte locations.

To find the analog output voltage, we first need to understand the digital input. The hexadecimal value (13A)₁₆ must be converted to its decimal equivalent:
$(13A)_{16} = 1 \times 16^2 + 3 \times 16^1 + 10 \times 16^0 = 256 + 48 + 10 = 314$.

Next, we determine the DAC's resolution, which is the smallest voltage step it can produce. For a 10-bit converter with a 10 V full-scale range, there are $2^{10}-1 = 1023$ steps. The voltage for each step is:
Resolution $= \frac{V_{FS}}{2^n - 1} = \frac{10 \text{ V}}{1023}$.

Finally, the output voltage is the decimal input value multiplied by the resolution:
$V_{out} = 314 \times \frac{10}{1023} \approx 3.069$ V.

The impedance characteristics along a transmission line repeat every half-wavelength ($\lambda/2$). Therefore, the input impedance for a line of length $3\lambda/4$ is equivalent to that of a line with length $(\frac{3\lambda}{4} - \frac{\lambda}{2}) = \lambda/4$.

For a quarter-wavelength line, the input impedance $Z_{\in}$ is given by the quarter-wave transformer equation:
$Z_{\in} = \frac{Z_0^2}{Z_L}$
Here, the characteristic impedance is $Z_0 = 50 \, \Omega$ and the load impedance is $Z_L = 400 \, \Omega$. Substituting these values gives:
$Z_{\in} = \frac{(50)^2}{400} = \frac{2500}{400} = 6.25 \, \Omega$

The entropy of a random variable, which measures its uncertainty, is maximized when all possible outcomes are equally likely. For the binary variable $X$, the two outcomes are $+2$ and $-2$.

To achieve maximum entropy, their probabilities must be equal:
$P(X=+2) = P(X=-2)$

Since there are only two outcomes, their probabilities must sum to 1. If they are equal, each must be $1/2$. We are given $P(X=+2) = \alpha$, therefore:
$\alpha = \frac{1}{2} = 0.5$

The stability of the system is governed by the characteristic equation, $1+G(s)H(s)=0$. Substituting the loop transfer function, we get $s^3 + 10s^2 + (16+K)s + 11K = 0$.

We use the Routh-Hurwitz criterion to find the value of $K$ for which the system is marginally stable. This occurs when a row of zeros appears in the Routh array, which happens if the first element in the $s^1$ row is zero.

The term for the $s^1$ row is $\frac{10(16+K)-1(11K)}{10}$.

Setting this term to zero to achieve marginal stability:
$10(16+K) - 11K = 0$
$160 + 10K - 11K = 0$
Solving for K, we find $K=160$.

The variance of the random variable $Y$ is defined as Var(Y) = E\[Y^2]$ - (E[Y])^2$. The input noise$W(t)$has a constant power spectral density (PSD), which implies it is a zero-mean process. Because the mean of$W(t)$is zero, the mean of$Y$is also zero,$E[Y]=0$, simplifying the variance calculation to just$Var(Y) = E[Y^2]$.

To find the mean-square value $E[Y^2]$, we use the autocorrelation function of the noise, $R_W(\tau)$, which is the inverse Fourier transform of its PSD. For a constant PSD of $S_W(f)=3$, the autocorrelation is $R_W(\tau) = 3\delta(\tau)$.

The mean-square value is then given by the double integral:
$E\$[Y^2]\$ = \int_{-\infty }^{\infty }\int_{-\infty }^{\infty } R_W(t_1-t_2)\phi(t_1)\phi(t_2)dt_1dt_2 = \int_{-\infty }^{\infty }\int_{-\infty }^{\infty } 3\delta(t_1-t_2)\phi(t_1)\phi(t_2)dt_1dt_2$
Using the sifting property of the delta function, this expression simplifies to $3\int_{-\infty }^{\infty }\phi^2(t)dt$. The integral term is simply the energy of the rectangular pulse $\phi(t)$.

The energy of $\phi(t)$ is calculated as $\int_{-\infty }^{\infty }\phi^2(t)dt = \int_{5}^{7} (1)^2 dt = 2$. Therefore, the variance is the noise power level multiplied by the signal energy: $Var(Y) = 3 \times 2 = 6$.

Let M and N be the outcomes of the first and second coin tosses, respectively. Since the coin is fair and the tosses are independent, there are four equally likely outcomes for the pair (M, N), each with a probability of $1/4$: (0, 0), (0, 1), (1, 0), and (1, 1).

The random variable $X$ is defined as the minimum of these two values, $X = \min(M, N)$. We calculate $X$ for each possible outcome:
$\min(0, 0) = 0$
$\min(0, 1) = 0$
$\min(1, 0) = 0$
$\min(1, 1) = 1$

From this, we can determine the probability distribution of $X$. The value $X=0$ occurs in three of the four cases, so $P(X=0) = 3/4$. The value $X=1$ occurs in one case, so $P(X=1) = 1/4$.

The expected value, $E(X)$, is the sum of each value of X multiplied by its corresponding probability:
$E(X) = \sum x \cdot P(X=x) = (0 \times \frac{3}{4}) + (1 \times \frac{1}{4}) = \frac{1}{4} = 0.25$.

For a system of equations to have a solution, the equations must be consistent. First, notice that the left side of the second equation, $2x_1+4x_2$, is exactly twice the left side of the first equation, $x_1+2x_2$. For this to be consistent, the same relationship must hold for the right-hand sides, meaning $b_2=2b_1$.

This immediately eliminates options B and D. We now test the second condition in option A, which suggests a linear dependency among equations (i), (iii), and (iv). Let's see if the same combination of the left-hand sides simplifies to zero. We check the expression $6(x_1+2x_2) - 3(3x_1+7x_2) + (3x_1+9x_2)$:
$(6x_1+12x_2) - (9x_1+21x_2) + (3x_1+9x_2)$
$(6-9+3)x_1 + (12-21+9)x_2 = 0x_1 + 0x_2 = 0$
Since this combination of the equations' left sides always equals zero, the corresponding combination of the right sides must also be zero for a solution to exist. Thus, we must have $6b_1-3b_3+b_4=0$.

This equation can be solved by the method of separation of variables. By rearranging the terms to group $x$ and $y$ on opposite sides, we get $\frac{1}{y-1} dy = x dx$.

Integrating both sides yields the general solution:
$\int \frac{1}{y-1} dy = \int x dx$
This results in the family of solutions described by $\ln|y-1| = \frac{1}{2}x^2 + C$.

Additionally, we must check for constant (equilibrium) solutions where $\frac{dy}{dx}=0$. In the original equation, this occurs if $(y-1)x = 0$ for all $x$. This condition is satisfied if $y-1=0$, which gives the particular constant solution $y=1$.

To determine the current $I$, we can model it as the sum of contributions from both voltage sources. The direction of the current $I$ as shown in the diagram means it is equal to the negative sum of the currents notionally supplied by each source through the impedance $Z$.

This gives us the relationship $I = -(I_1 + I_2)$. We can express the component currents using Ohm's Law as $I_1 = V_1/Z$ and $I_2 = V_2/Z$.

Combining these terms, we get the expression for the total current:
$I = -\left[ \frac{120\angle -90^{\circ}}{80-j35} + \frac{120\angle -30^{\circ}}{80-j35} \right]$

Since the impedance $Z$ is a common denominator, we can sum the voltage phasors first:
$I = - \frac{120\angle -90^{\circ} + 120\angle -30^{\circ}}{80 - j35}$

Evaluating this complex expression by performing the vector addition in the numerator and then the division yields the final answer, $I \approx 2.38 \angle 143.7^{\circ}$ A.

First, we find the discrete-time signal $x[n]$ by substituting the sampling time $t=n/400$ into the continuous-time signal $x(t)$.
$x[n] = \cos\left(200\pi \frac{n}{400}\right) = \cos\left(\frac{\pi n}{2}\right)$
For $n=0, 1, \dots, 7$, the 8-point sequence is $x[n] = \{1, 0, -1, 0, 1, 0, -1, 0\}$.

We can see that $x[n]$ is formed by interleaving zeros into a shorter 4-point sequence, $y[n] = \{1, -1, 1, -1\}$. A key property of the DFT states that the $2N$-point DFT of such a zero-interleaved sequence is simply two back-to-back copies of the $N$-point DFT of the original shorter sequence.

Let's find the 4-point DFT, $Y[k]$, of $y[n]$. By inspection or calculation, the only non-zero term is $Y[2] = \sum_{n=0}^{3} y[n] e^{-j\pi n} = 1(-1)^0 - 1(-1)^1 + 1(-1)^2 - 1(-1)^3 = 1+1+1+1=4$. Thus, the 4-point DFT is $Y[k] = \{0, 0, 4, 0\}$.

Applying the property, the 8-point DFT of $x[n]$ is formed by repeating $Y[k]$ twice:
$X[k] = \{ Y[0], Y[1], Y[2], Y[3], Y[0], Y[1], Y[2], Y[3] \} = \{0, 0, 4, 0, 0, 0, 4, 0\}$
This result shows that only the components $X[2]$ and $X[6]$ are non-zero.

To determine the state equation, we derive two first-order differential equations for the state variables: the capacitor voltage $V$ and the inductor current $i$.

First, apply Kirchhoff's Current Law (KCL) at the central node (above the capacitor). The current $i$ entering from the inductor splits into three paths: through the capacitor ($0.25 \frac{dV}{dt}$), through the 1 Ω resistor ($V/1$), and into the current source $i_2$ (upwards, so it's $-i_2$ leaving the node). This gives $i = 0.25\frac{dV}{dt} + V - i_2$. Rearranging for $\frac{dV}{dt}$ yields $\frac{dV}{dt} = -4V + 4i + 4i_2$.

Next, apply Kirchhoff's Voltage Law (KVL) to the leftmost loop. First, transform the current source $i_1$ and parallel 2 Ω resistor into a voltage source $V_s = 2i_1$ in series with a 2 Ω resistor. The KVL equation for the loop is $2i_1 - 2i - 0.5\frac{di}{dt} - V = 0$. Solving for $\frac{di}{dt}$ gives $\frac{di}{dt} = -2V - 4i + 4i_1$.

Finally, we express these two equations in matrix form:
$\frac{d}{dt} \begin{bmatrix} V \\ i \end{bmatrix} = \begin{bmatrix} -4 & 4 \\ -2 & -4 \end{bmatrix} \begin{bmatrix} V \\ i \end{bmatrix} + \begin{bmatrix} 0 & 4 \\ 4 & 0 \end{bmatrix} \begin{bmatrix} i_1 \\ i_2 \end{bmatrix}$

For a one-sided abrupt pn-junction, the plot of $1/C_D^2$ versus the applied voltage $V$ is a straight line. The equation for this line is given by:
$\frac{1}{C_D^2} = \frac{2}{q \epsilon A^2 N_D} (V_{bi} - V)$
where $V_{bi}$ is the built-in potential and the other terms are constants for the diode. We are given a point on this line: when $V = -0.2$ V, the capacitance $C_D = 50$ pF. This gives a y-value of $1/C_D^2 = 1/(50 \times 10^{-12})^2 = 4 \times 10^{20}$ F$^{-2}$.

The line intercepts the voltage axis where $1/C_D^2 = 0$, which occurs at $V = V_{bi}$. This gives us a second point, $(V_{bi}, 0)$. The slope is then calculated using these two points:
$\text{Slope} = \frac{\Delta(1/C_D^2)}{\Delta V} = \frac{4 \times 10^{20} - 0}{-0.2 - V_{bi}}$
Since the value of the built-in potential $V_{bi}$ is unknown, the slope cannot be calculated from the information provided.

To find the depletion charge per unit area, $Q'_d$, we can rearrange the standard equation for the threshold voltage of a MOS capacitor. The equation is $V_{TH} = \phi_{ms} - \frac{Q'_{ox}}{C'_{ox}} - \frac{Q'_d}{C'_{ox}} + 2\phi_{Fp}$.

From the problem statement and diagram, we identify the necessary values. The work function difference is $\phi_{ms} = -0.93$ V, and the Fermi potential is $\phi_{Fp} = 0.3$ V. We are also given $V_{TH} = -0.16$ V and told there is no oxide charge, so $Q'_{ox} = 0$.

Substituting these values into the threshold voltage equation gives:
$-0.16 = -0.93 - 0 - \frac{Q'_d}{C'_{ox}} + 2(0.3)$

Solving for the term containing the depletion charge, we find $\frac{Q'_d}{C'_{ox}} = -0.93 + 0.60 + 0.16 = -0.17$ V.
Finally, using the given oxide capacitance per unit area, $C'_{ox} = 100 \text{ nF/cm}^2$, we calculate $Q'_d = -0.17 \times (100 \times 10^{-9} \text{ F/cm}^2) = -1.7 \times 10^{-8} \text{ C/cm}^2$. The magnitude of this charge is $1.7 \times 10^{-8} \text{ C/cm}^2$.