GATE ECE 2014 Set 1

Let's examine the properties listed. Options A, B, and C are all standard, fundamental rules for the matrix transpose operation, and they are always true.

Option D, $MN = NM$, is the commutative property of multiplication. While this property is true for regular numbers (for example, $3 \times 5 = 5 \times 3$), it is famously not true for matrices. In matrix algebra, the order of multiplication is crucial. Swapping the order of two matrices $M$ and $N$ will, in most cases, yield a completely different result. Because $MN$ does not always equal $NM$, this property does not always hold.

The key is to focus on the population of children, not families. Let's consider a representative sample of just two families to reflect the society's makeup: one family has 1 child, and the other has 2 children.

In this sample, we have a total of $1 + 2 = 3$ children. The children who have a sibling are the two from the two-child family. The child from the single-child family has no siblings.

Therefore, if we pick one child at random from this group of three, there are 2 favorable outcomes (the children with a sibling) out of 3 possible outcomes. The probability is $\frac{2}{3}$, which is approximately $0.667$.

This integral can be evaluated using Cauchy's Integral Formula. The integrand's denominator, $z+2j$, indicates a simple pole at $z_0 = -2j$. Since $|-2j| = 2$, this pole is inside the circular contour $|z|=3$.

The formula is $\oint_C \frac{f(z)}{z-z_0} dz = 2\pi j \cdot f(z_0)$, where $f(z)$ is the analytic part of the integrand, which in this case is the numerator, $f(z) = z^2 - z + 4j$.

We now evaluate $f(z)$ at the pole $z_0 = -2j$:
$f(-2j) = (-2j)^2 - (-2j) + 4j = -4 + 2j + 4j = -4 + 6j$.

Substituting this into the formula gives the value of the integral:
$2\pi j (-4 + 6j) = -8\pi j + 12\pi j^2 = -12\pi - 8\pi j = -4\pi(3 + 2j)$.

Let $\lambda$ represent an eigenvalue of the matrix $A$. A core property of eigenvalues is that if $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$.

We are given the equation $A^2 = I$. The eigenvalues of the identity matrix $I$ are all equal to 1.

Therefore, we can equate the eigenvalues of both sides of the equation, which gives us $\lambda^2 = 1$.

Solving for $\lambda$ yields two possible values: $\lambda = 1$ and $\lambda = -1$.

The question asks for the positive eigenvalue, which is 1.

Since the three random variables $X_1, X_2,$ and $X_3$ are independent and identically distributed, they are perfectly symmetric. This means no single variable is inherently more likely to be larger or smaller than the others.

Think of it as a fair "competition" to see which variable gets the highest value. Each of the three variables has an equal opportunity to win. Therefore, the probability that any specific variable, such as $X_1$, is the largest is simply one out of the three possibilities. This gives a probability of $\frac{1}{3}$.

This question concerns the Maximum Power Transfer Theorem for AC circuits. Let the output impedance of the first section (source) be $Z_1 = R_1 + jX_1$ and the input impedance of the second section (load) be $Z_2 = R_2 + jX_2$.

The average power delivered to the load $Z_2$ is given by $P_2 = |I|^2 R_2$, where the current $I$ depends on the total impedance $Z_1 + Z_2$. To maximize this power, we must first cancel the total reactance of the circuit to maximize the current. This is achieved when the load reactance is the negative of the source reactance, i.e., $X_2 = -X_1$.

With the reactive components canceled, the circuit behaves resistively. For a purely resistive circuit, maximum power is transferred when the load resistance equals the source resistance, so $R_2 = R_1$.

Combining these two conditions gives the optimal load impedance: $Z_2 = R_2 + jX_2 = R_1 + j(-X_1) = R_1 - jX_1$. This relationship defines the complex conjugate of the source impedance $Z_1$. Therefore, for maximum power transfer, the load impedance must be the complex conjugate of the source impedance, $Z_2 = Z_1^*$.

To solve for the unknown currents, we can apply Kirchhoff's Current Law (KCL) at each junction, or node. KCL states that the total current entering a node must equal the total current leaving it.

Let's proceed node by node to find the currents we need:

1. At node B, currents $i_1$ and $i_4$ flow in, and $i_2$ flows out. So, $i_2 = i_1 + i_4 = 2 A + (-1 A) = 1 A$.
2. Now, at node A, currents $i_2$ and $i_5$ flow in, and $i_3$ flows out. This gives us $i_3 = i_2 + i_5 = 1 A + (-4 A) = -3 A$.
3. Finally, at node C, currents $i_3$ and $i_6$ flow in, and $i_1$ flows out. The equation is $i_3 + i_6 = i_1$.
   We can now isolate and solve for $i_6$: $i_6 = i_1 - i_3 = 2 A - (-3 A) = 5 A$.

The responsivity ($R$) of a photodiode measures how much photocurrent ($I_p$) it generates for a given amount of incident optical power ($P_{\in}$). The units of responsivity, Amps per Watt (A/W), tell us the relationship directly.

To find the photocurrent, we multiply the responsivity by the incident power:
$I_p = R \times P_{\in}$

Plugging in the given values:
$I_p = (0.8 \text{ A/W}) \times (10 \mu\text{W})$
$I_p = 8 \mu\text{A}$

To solve this, we use the method of assumption and verification. The Schottky diode $D_2$ has a lower forward voltage ($0.3 \text{ V}$) than the PN diode $D_1$ ($0.7 \text{ V}$), making it more likely to conduct. Let's assume $D_2$ is ON and $D_1$ is OFF.

With $D_1$ acting as an open circuit, the current $I$ flows in a single loop. Applying Kirchhoff's Voltage Law (KVL) to this loop gives:
$10 \text{ V} = I \cdot (1000 \Omega + 20 \Omega) + 0.3 \text{ V}$
Solving for the current, we get $I = \frac{9.7 \text{ V}}{1020 \Omega} \approx 9.5 \text{ mA}$. Since $I$ is positive, our assumption that $D_2$ is ON is valid.

Next, we verify the state of $D_1$. The voltage across $D_1$ is the voltage at the node after the 1kΩ resistor. This voltage is the source voltage minus the drop across the 1kΩ resistor:
$V_{D1} = 10 \text{ V} - (9.5 \text{ mA} \times 1 \text{ k}\Omega) = 10 \text{ V} - 9.5 \text{ V} = 0.5 \text{ V}$.

Since the voltage across $D_1$ ($0.5 \text{ V}$) is less than its turn-on voltage of $0.7 \text{ V}$, our assumption that $D_1$ is OFF is also correct. Thus, $D_1$ is OFF and $D_2$ is ON.

The purpose of the gate voltage in an n-channel MOSFET is to attract electrons to the silicon surface to form a conductive channel. The minimum voltage required to do this is the threshold voltage, $V_{th}$.

Fixed positive charges trapped in the gate oxide create their own electric field. This field independently attracts electrons toward the silicon-oxide interface, even without any gate voltage applied. This effect assists the gate's primary function. Consequently, a smaller external positive voltage is needed on the gate to accumulate enough electrons to form the channel, leading to a decrease in $V_{th}$.

A current buffer is designed to replicate an input current at its output, effectively isolating the source from the load.

1. Input Side: To sense the input current accurately, the buffer must draw as much current as possible from the source. By the current divider rule, this requires the buffer's input impedance ($Z_{\in}$) to be much lower than the source's impedance. An ideal current input has $Z_{\in} = 0$, so a good buffer has a low input impedance.

2. Output Side: The buffer must deliver this current to the load, acting as an ideal current source. An ideal current source has infinite internal impedance to ensure the output current is independent of the load impedance ($Z_L$). Therefore, a good buffer must have a high output impedance ($Z_{out} \gg Z_L$) to force the current through the load.

To classify the feedback, we analyze how the output is sampled and how the feedback signal is mixed with the input.

First, the feedback network, represented by $R_F$, connects directly to the output node. This parallel (or shunt) connection at the output is used to sense the output voltage, $V_o$. This is therefore voltage sampling.

Next, at the input, the feedback current through $R_F$ combines at a node with the input signal current, $i_{\in}$. Since the currents are being summed, this is current mixing (or shunt mixing).

Because the amplifier samples the output voltage and mixes a current at the input, the topology is voltage-current feedback.

This circuit is an active first-order low-pass filter. Its cutoff frequency ($f_c$) is determined by the resistor and capacitor in the feedback loop. The standard formula relating these components is $f_c = \frac{1}{2\pi R_2 C}$.

To find the value of $R_2$ that achieves the desired cutoff frequency, we can algebraically rearrange this equation:
$R_2 = \frac{1}{2\pi f_c C}$

Now, we substitute the given values, $f_c = 5 \text{ kHz}$ and $C = 10 \text{ nF}$:
$R_2 = \frac{1}{2\pi (5 \times 10^3 \text{ Hz})(10 \times 10^{-9} \text{ F})} \approx 3183 \, \Omega$

This result is equivalent to approximately $3.18 \text{ k}\Omega$.

This circuit uses NMOS transistors in a pass-transistor logic configuration. An NMOS transistor is a "poor" passer of a high voltage signal. The output at the source terminal, $V_{out}$, can only rise until the gate-to-source voltage, $V_{GS}$, drops to the threshold voltage, $V_T$. At this point, the transistor turns off, and the output voltage can't rise any further.

This condition is described by the equation $V_{GS} = V_G - V_{out} = V_T$. Rearranging for the output voltage gives us the maximum possible output: $V_{out} = V_G - V_T$.

In this problem, all three transistors have their gates connected to $5V$ ($V_G=5V$) and a threshold voltage of $V_T=1V$. Although the inputs are different, the output voltage at each node is clamped by this threshold drop. Thus, for nodes P, Q, and R, the output voltage is calculated as:
$V_P = V_Q = V_R = V_G - V_T = 5V - 1V = 4V$.

Let's simplify the expression by breaking it into its two main parts connected by the 'OR' (+) operation.

First, we'll expand the term $(X+Y)(X+\bar{Y})$ to $XX+X\bar{Y}+YX+Y\bar{Y}$. Based on Boolean identities, this becomes $X+X\bar{Y}+XY+0$. We can factor this to $X(1+\bar{Y}+Y)$, which simplifies to just $X$.

Next, let's simplify the second term, $\overline{(X\bar{Y}+\bar{X})}$. Using De Morgan's law, this becomes $\overline{(X\bar{Y})} \cdot \overline{(\bar{X})}$. This expression simplifies further to $(\bar{X}+Y) \cdot X$.

Distributing the $X$ in $(\bar{X}+Y)X$ gives us $\bar{X}X+YX$. Since $\bar{X}X=0$, this part simplifies to $XY$.

Finally, we combine our simplified parts: the first part became $X$ and the second became $XY$. The full expression is now $X+XY$. Applying the absorption law ($A+AB=A$), the final result is $X$.

The circuit shown is an asynchronous ripple counter where the output of one flip-flop clocks the next. Since the J and K inputs of each flip-flop are tied high, they are in toggle mode. In this mode, each flip-flop divides its input clock frequency by two.

The output $Q_3$ is taken from the fourth flip-flop in the cascade (assuming a zero-based index: $Q_0, Q_1, Q_2, Q_3$). The frequency is successively halved at each stage. Thus, the total frequency division at the $Q_3$ output is $2^4 = 16$.

The resulting frequency at $Q_3$ is calculated as:
$f_{Q3} = \frac{f_{clock}}{16} = \frac{1 \text{ MHz}}{16} = \frac{1000 \text{ kHz}}{16} = 62.5 \text{ kHz}$

where m is the smallest integer that converts $\frac{2 \pi}{\omega_{0}}$ into a integer value. $\therefore \quad N=\frac{2 \pi}{\pi^{2}} \cdot m=\frac{2}{\pi} \cdot m$ So, there exists no such integer value of m which could make the N integer, so the system is not periodic.

Multiplication in time domain = convolution in frequency domain. $x_{1}(t) \cdot x_{2}(t) \leftrightarrow X_{1}(\omega) * X_{2}(\omega)$ So, highest frequency component contained by the convolved signal $z(t)=1500 \mathrm{Hz}$ $\therefore$ Nyquist rate $=2 \times 1500$ $=3000 \mathrm{Hz}=3 \mathrm{kHz}$

We know that Y(s)=H(s) X(s)=3\left[\frac{1}{s}-\frac{e^{-3 s}}{s}\right]\ \cdot \frac{5}{s}$ Steady state output

Given that, $\quad G(s)=\frac{K}{(s+2)(s-1)}$ Using root locus method, the break point can be obtain as

To have, both the poles at the same directions

The stability of the closed-loop system depends on the Nyquist plot of $G(s)$ encircling the critical point, which is located at $-1/k$. The provided plot is for $G(s)$ itself, corresponding to a gain of $k=1$.

As we increase the positive gain $k$, the critical point $-1/k$ moves along the negative real axis from $-1$ towards the origin. The given Nyquist plot clearly intersects the negative real axis somewhere between $-1$ and the origin.

For a sufficiently large value of $k$, the critical point $-1/k$ will cross this intersection point and become enclosed by the Nyquist contour. According to the Nyquist stability criterion, this encirclement ($N \neq 0$) will introduce right-half plane poles in the closed-loop system, making it unstable.

In a CDMA system, the number of chips in a signature sequence, denoted by $N$, determines the dimensionality of the signal space. Each user's unique signature sequence can be represented as a vector in this $N$-dimensional space.

For users to communicate without interfering with one another, their signature sequences must be mutually orthogonal. A key principle from linear algebra is that an $N$-dimensional space can accommodate a maximum of $N$ mutually orthogonal vectors.

Given that the number of chips is $N=8$, we are working in an 8-dimensional space. Consequently, we can define at most 8 signature sequences that are all mutually orthogonal to each other, allowing for a maximum of 8 users.

The capacity $C$ of a Binary Symmetric Channel (BSC) is calculated using the formula $C = 1 - H(p)$, where $p$ is the crossover probability and $H(p)$ is the binary entropy function.

The entropy is given by $H(p) = -p \log_2(p) - (1-p) \log_2(1-p)$.
For a crossover probability of $p = 0.5$, the channel is maximally noisy. The entropy is:
$H(0.5) = -0.5 \log_2(0.5) - (1-0.5) \log_2(0.5) = -0.5(-1) - 0.5(-1) = 1$.
Therefore, the capacity is $C = 1 - H(0.5) = 1 - 1 = 0$.

A capacity of zero means no information can be transmitted. This is because when $p=0.5$, the output is completely random and statistically independent of the input.

Here's a clearer, more insightful way to explain the solution:

The relationship between incident ($a$) and reflected ($b$) waves in a two-port network is defined by the S-parameter equations:
$b_1 = S_{11}a_1 + S_{12}a_2$
$b_2 = S_{21}a_1 + S_{22}a_2$

When port-2 is short-circuited, its reflection coefficient is $\Gamma_L = -1$. This means the wave incident on port-2 from the termination, $a_2$, is the negative of the wave emerging from the network's port-2, $b_2$. Thus, we have the crucial condition $a_2 = -b_2$.

Substituting this condition into the second equation gives $-a_2 = S_{21}a_1 + S_{22}a_2$. We can rearrange this to express $a_2$ in terms of $a_1$: $a_2 = \frac{-S_{21}a_1}{1+S_{22}}$.

Finally, we substitute this expression for $a_2$ into the first equation and solve for the ratio $\frac{b_1}{a_1}$, which represents the new $S_{11}$ parameter of the resulting one-port network:
$\frac{b_1}{a_1} = S_{11} + S_{12}\left(\frac{-S_{21}}{1+S_{22}}\right) = \frac{S_{11}(1+S_{22}) - S_{12}S_{21}}{1+S_{22}} = \frac{S_{11}+S_{11}S_{22}-S_{12}S_{21}}{1+S_{22}}$.

This problem is solved using the method of images. The infinite grounded metal plate can be replaced by a single "image charge" which, together with the original charge, satisfies the zero potential boundary condition on the plane of the plate. For a charge $+q$ at a distance $d$ from the plate, the image charge is $-q$ located at a distance $d$ behind the plate.

The force on the charge $+q$ is simply the Coulomb force exerted by the image charge $-q$. The distance between the original charge $+q$ and the image charge $-q$ is $d + d = 2d$. Since the charges have opposite signs, the force is attractive, meaning it is directed towards the plate.

The magnitude of this attractive force is given by Coulomb's law in a medium with permittivity $\varepsilon$:
$F = \frac{1}{4\pi \varepsilon} \frac{|(+q)(-q)|}{(2d)^2}$
$F = \frac{1}{4\pi \varepsilon} \frac{q^2}{4d^2} = \frac{q^2}{16\pi \varepsilon d^2}$
Thus, the force is $\frac{q^2}{16\pi \varepsilon d^2}$ directed towards the plate.

To determine the Taylor series for this function, we can combine the well-known series for $\sin x$ and $\cos x$.

First, recall the initial terms of their Maclaurin series (a Taylor series centered at 0):
$\sin x = x - \frac{x^3}{3!} + \dots$
$\cos x = 1 - \frac{x^2}{2!} + \dots$

Next, substitute these into the expression $3 \sin x + 2 \cos x$:
$3\left(x - \frac{x^3}{3!} + \dots\right) + 2\left(1 - \frac{x^2}{2!} + \dots\right)$

Distribute the coefficients and simplify the factorials ($3! = 6$, $2! = 2$):
$(3x - \frac{3x^3}{6} + \dots) + (2 - \frac{2x^2}{2} + \dots)$

Finally, combine the terms in order of increasing powers of $x$ to get the resulting series:
$2 + 3x - x^2 - \frac{x^3}{2} + \dots$

Let $G(j\omega)$ be the Fourier transform of $g(t)$, so $G(j\omega) = \omega e^{-2\omega^2}$. We are asked to find $\int_{-\infty }^{+\infty }y(t)dt$, which is equivalent to finding the Fourier transform of $y(t)$, denoted $Y(j\omega)$, evaluated at $\omega=0$.

The function $y(t)$ is the running integral of $g(t)$, which corresponds to the convolution $y(t) = g(t) * u(t)$, where $u(t)$ is the unit step function. Using the convolution property, this becomes $Y(j\omega) = G(j\omega)U(j\omega)$ in the frequency domain.

Substituting the known transforms $G(j\omega) = \omega e^{-2\omega^2}$ and $U(j\omega) = \frac{1}{j\omega} + \pi\delta(\omega)$, we get:
$Y(j\omega) = (\omega e^{-2\omega^2}) \left(\frac{1}{j\omega} + \pi\delta(\omega)\right) = \frac{e^{-2\omega^2}}{j} + \omega e^{-2\omega^2}\pi\delta(\omega)$.

Finally, we evaluate this at $\omega=0$. The first term becomes $\frac{e^0}{j} = \frac{1}{j}$. The second term becomes zero because the function $\omega e^{-2\omega^2}\pi$ is zero at $\omega=0$. Therefore, the integral is $Y(j0) = \frac{1}{j} = -j$.

To find the volume, we need to calculate the double integral of the function $z(x,y) = x+y$ over the given triangular domain. The boundaries of the triangle, $0\leq x\leq 12$ and $0\leq y\leq x$, define the limits of our integration.

We set up the iterated integral as $V = \int_{0}^{12} \int_{0}^{x} (x+y) \,dy\,dx$.

First, we evaluate the inner integral with respect to $y$, treating $x$ as a constant:
$\int_{0}^{x} (x+y) \,dy = \left[ xy + \frac{y^2}{2} \right]_{y=0}^{y=x} = x(x) + \frac{x^2}{2} = \frac{3}{2}x^2$.

Now, we integrate this resulting expression with respect to $x$ over its range:
V = \int_{0}^{12} \frac{3}{2}x^2 \,dx = \frac{3}{2} \left[ \frac{x^3}{3} \right]_{0}^{12} = \frac{1}{2} \[x^3]_{0}^{12}$.

Finally, plugging in the limits for $x$ gives the volume:
$V = \frac{1}{2} (12^3 - 0^3) = \frac{1728}{2} = 864$.

We want to find the non-negative value of $\alpha$ for which the given matrix $P$ is singular, meaning its determinant is zero. The matrix is $P = I_6 + \alpha J_6$, where $I_6$ is the identity matrix and $J_6$ is the anti-diagonal matrix (ones on the anti-diagonal, zeros elsewhere).

The determinant of this matrix can be calculated elegantly. By reordering rows and columns to group pairs $(i, 7-i)$, the matrix becomes block-diagonal with three identical $2 \times 2$ blocks of the form

.

The determinant of each block is $1 \cdot 1 - \alpha \cdot \alpha = 1 - \alpha^2$. Since the determinant of a block-diagonal matrix is the product of the determinants of its blocks, we have $\det(P) = (1 - \alpha^2)^3$.
For $\det(P) = 0$, we must have $(1 - \alpha^2)^3 = 0$, which simplifies to $1 - \alpha^2 = 0$. As $\alpha$ is a non-negative real number, the only solution is $\alpha = 1$.

This problem requires a Y-to-$\Delta$ (or Star-to-Delta) transformation. We are given a Y-network with two arms of $10 \, \Omega$ and a third arm of $11 \, \Omega$. Let's label these resistances $R_1 = 10 \, \Omega$, $R_2 = 10 \, \Omega$, and $R_3 = 11 \, \Omega$.

The resistances of the equivalent $\Delta$-network are found using the standard conversion formula. Due to the symmetry of the two $10 \, \Omega$ resistors, two of the three $\Delta$ resistances will be identical. These are calculated as:
$R_a = 10 + 11 + \frac{10 \times 11}{10} = 32 \, \Omega$

The third, unique resistance is formed across the two $10 \, \Omega$ arms:
$R_c = 10 + 10 + \frac{10 \times 10}{11} = 20 + 9.09 \approx 29.09 \, \Omega$

Comparing the three calculated resistances ($32 \, \Omega$, $32 \, \Omega$, and $29.09 \, \Omega$), we find that the lowest value is approximately $29.09 \, \Omega$.

A better way to understand this problem is to calculate the total complex power by summing the complex power of each load.

For the first load, the real power is $P_1 = 10$ kW and the power factor is 0.8 leading. This means the reactive power is capacitive (negative). The complex power is $S_1 = 10 - j\left(10 \tan(\arccos 0.8)\right) = 10 - j7.5$ kVA.

For the second load, the apparent power is $|S_2| = 10$ kVA and the power factor is 0.8 lagging. This means the reactive power is inductive (positive). The complex power is $S_2 = 10(\cos\phi + j\sin\phi) = 10(0.8 + j0.6) = 8 + j6$ kVA.

The total complex power is the sum of the individual powers:
$S_{total} = S_1 + S_2 = (10 - j7.5) + (8 + j6) = (18 - j1.5)$ kVA.

The root-mean-square (RMS) value is found by taking the square root of the average of the squared signal over one full period, $T$. The signal can be described in two parts: a ramp $x(t) = \frac{2t}{T}$ for the first half-period ($0 \le t \le T/2$), and zero for the second half.

First, let's find the mean of the squared signal, $\langle x^2 \rangle$:
$\langle x^2 \rangle = \frac{1}{T} \int_0^T x(t)^2 dt = \frac{1}{T} \left( \int_0^{T/2} \left(\frac{2t}{T}\right)^2 dt + \int_{T/2}^T 0 \, dt \right)$
Since the second integral is zero, we only need to solve the first one:
$\langle x^2 \rangle = \frac{1}{T} \int_0^{T/2} \frac{4t^2}{T^2} dt = \frac{4}{T^3} \left[ \frac{t^3}{3} \right]_0^{T/2} = \frac{4}{3T^3} \left( \frac{T^3}{8} \right) = \frac{1}{6}$
The RMS value is the square root of this result, so $x_{rms} = \sqrt{1/6} \approx 0.408$.

For a series RLC circuit to have a critically damped response, its damping ratio, $\xi$, must be equal to 1. The damping ratio is related to the component values by the formula $\xi = \frac{R}{2} \sqrt{\frac{C}{L}}$.

Setting $\xi=1$, we can solve for the required capacitance:
$1 = \frac{R}{2} \sqrt{\frac{C}{L}} \implies \sqrt{C} = \frac{2}{R} \sqrt{L} \implies C = L \left(\frac{2}{R}\right)^2$

Substituting the given values, which are $R=40 \, \Omega$ and $L=4 \, \text{H}$ (based on the original explanation's calculation, not the provided image):
$C = 4 \left(\frac{2}{40}\right)^2 = 4 \left(\frac{1}{20}\right)^2 = \frac{4}{400} = 0.01 \, \text{F}$

Converting from Farads to milliFarads, the required capacitance is $10$ mF.

To find the transconductance ($g_m$), we must first calculate the DC collector current ($I_C$) flowing through the BJT. The relationship is described by the diode equation:
$I_C = I_S e^{V_{BE}/V_T}$
Using the provided values, where $V_T = kT/q = 25$ mV:
$I_C = (10^{-13} \text{ A}) \times e^{(0.7 \text{ V})/(0.025 \text{ V})} \approx 0.144 \text{ mA}$
Now, transconductance is defined as the ratio of the collector current to the thermal voltage. Plugging in the values we have:
$g_m = \frac{I_C}{V_T} = \frac{0.144 \text{ mA}}{0.025 \text{ V}} = 5.76 \text{ mA/V}$

To find the electron concentration at the edge of the p-side under forward bias, we first determine the equilibrium minority carrier concentration, $n_{p0}$. Using the mass-action law and the given doping ($N_A$) and intrinsic concentration ($n_i$, using the standard value for silicon), we have:
$n_{p0} = \frac{n_i^2}{N_A} = \frac{(1.5 \times 10^{10} \text{ cm}^{-3})^2}{1 \times 10^{16} \text{ cm}^{-3}} = 2.25 \times 10^4 \text{ cm}^{-3}$

A forward bias, $V$, exponentially increases this concentration at the depletion edge. This relationship is described by the law of the junction:
$n_p = n_{p0} \exp\left(\frac{V}{V_T}\right)$, where $V_T = \frac{kT}{q}$ is the thermal voltage.

Plugging in the given values for the applied voltage and thermal voltage:
$n_p = (2.25 \times 10^4) \exp\left(\frac{0.3 \text{ V}}{0.026 \text{ V}}\right) \approx 2.3 \times 10^9 \text{ cm}^{-3}$