GATE ECE 2011

Power density of any point source decreases with distance i.e. the density decreases and area of cross-over increases with the product being constant.

Hence the phase velocity of wave along the line is,

The overall information rate is the product of how many samples are sent per second and how much information each individual sample carries.

First, let's determine the information content of a single sample. With 4 distinct and equally probable quantization levels, the information per sample (also known as entropy) is calculated as:
$H = \log_2(\text{Number of levels}) = \log_2(4) = 2$ bits/sample.

Next, the problem directly gives us the transmission rate of the samples:
$r = 2$ samples/second.

Finally, multiplying the sample rate by the information per sample gives the total information rate, $R$:
$R = r \times H = 2 \text{ samples/sec} \times 2 \text{ bits/sample} = 4$ bits/second.

To understand the open-loop transfer function, we can analyze the starting and ending points of the root locus branches on the given plot.

The branches of the root locus always start at the open-loop poles and end at the open-loop zeros or at infinity. From the plot, we can see one branch starting from $s=-2$ and ending at $s=-1$. This tells us there is a pole at $s=-2$ and a zero at $s=-1$.

Furthermore, at the location $s=-3$, two branches break away from the real axis. A point where multiple loci depart like this indicates a multiple-order pole. The two departing branches mean there is a double pole at $s=-3$.

Combining these observations, the system has one zero at $s=-1$, one simple pole at $s=-2$, and a double pole at $s=-3$. This configuration matches the transfer function $G(S)H(S) = k\frac{(s+1)}{(s+2)(s+3)^2}$.

First, let's examine causality. A system is causal if its impulse response, $h(n)$, is zero for all negative time, i.e., $h(n) = 0$ for $n < 0$. In this case, $h(n) = 2^n u(n-2)$. The unit step function $u(n-2)$ is zero for all $n < 2$. This means the impulse response is zero for all $n < 2$, which satisfies the condition for causality.

Next, we check for stability. A system is stable if its impulse response is absolutely summable, meaning $\sum_{n=-\infty }^{\infty } |h(n)|$ must be finite. For our system, the sum is:
$\sum_{n=-\infty }^{\infty } |2^n u(n-2)| = \sum_{n=2}^{\infty } |2^n| = 4 + 8 + 16 + \dots$
This is a geometric series whose terms are growing, so the sum diverges to infinity. Since the sum is not finite, the system is unstable.

A fundamental principle connects a system's step and impulse responses: the unit impulse response, $h(t)$, is the time derivative of the unit step response, $s(t)$.

Given the step response $s(t) = 1 - e^{-\alpha t}$, we can find the impulse response by differentiating:
$h(t) = \frac{d}{dt} s(t) = \frac{d}{dt} (1 - e^{-\alpha t})$

The derivative of the constant '1' is zero. Applying the chain rule to the second term gives us:
$h(t) = 0 - (-\alpha e^{-\alpha t}) = \alpha e^{-\alpha t}$

The logical expression for the output $Y$ is given by $Y = PQ + PR + QR$. For the output $Y$ to be '1', the OR condition means that at least one of the AND terms ($PQ$, $PR$, or $QR$) must be '1'. An AND term, such as $PQ$, evaluates to '1' only when both of its inputs are '1'. Therefore, for $Y$ to be '1', either ($P$ and $Q$ are '1'), or ($P$ and $R$ are '1'), or ($Q$ and $R$ are '1'). Fulfilling any of these conditions requires that at least two of the inputs $P$, $Q$, and $R$ must be '1'.

The parallel inductor and capacitor in the collector branch form a resonant tank circuit, which means this is a tuned amplifier. The voltage gain of a tuned amplifier is designed to be greatest at its specific resonant frequency, which is determined by the formula $\omega_0 = 1/\sqrt{LC}$. The question asks for the gain at an input frequency of 10 M rad/s. By calculating the resonant frequency using the given component values, we find it is also 10 M rad/s. Since the amplifier is operating at its resonant frequency, the gain magnitude is at its maximum.

Drift current is the flow of charge carriers, like electrons and holes, that are pushed or "pushed" by an external electric field. We can express the density of this current, $J$, with the formula:
$J = nqv_d$
Here, $n$ is the charge carrier concentration, $q$ is the charge of each carrier, and $v_d$ is their average drift velocity.

The crucial point is that this drift velocity, $v_d$, is directly caused by the electric field, $E$. Their relationship is defined by the carrier mobility, $\mu$, as $v_d = \mu E$.

By substituting this expression for $v_d$ into our first equation, we get:
$J = nq(\mu E) = (nq\mu)E$
This final relationship demonstrates that drift current fundamentally depends on both the carrier concentration ($n$) and the strength of the electric field ($E$).

The primary function of a Zener diode is to provide a stable reference voltage. It is specifically engineered to operate in the reverse breakdown region to achieve this. When the reverse voltage across the diode reaches its specific breakdown voltage ($V_Z$), it begins to conduct significant current. Critically, in this region, the voltage across the diode remains almost perfectly constant despite large changes in the current flowing through it. This stable voltage is what makes it invaluable for voltage stabilization circuits.

This problem is best solved using AC steady-state analysis with phasors and impedances. The input signal $v_i(t) = V_p \cos(t/RC)$ tells us the angular frequency is $\omega = 1/(RC)$.

At this frequency, we can calculate the impedance of the components. The output voltage is across the parallel combination of a resistor and a capacitor. The impedance of this parallel section is $Z_p = R || \frac{1}{j\omega C} = \frac{R}{1+j\omega RC}$. Substituting $\omega=1/(RC)$, this simplifies to $Z_p = \frac{R}{1+j}$.

The total impedance of the circuit, $Z_{total}$, is the sum of the resistor, the series capacitor, and the parallel section $Z_p$.
$Z_{total} = R + \frac{1}{j\omega C} + Z_p = R + \frac{R}{j} + \frac{R}{1+j} = \frac{3R}{1+j}$.

The circuit acts as a voltage divider, so the ratio of the output to input voltage is the ratio of the parallel impedance to the total impedance:
$\frac{V_o}{V_i} = \frac{Z_p}{Z_{total}} = \frac{R/(1+j)}{3R/(1+j)} = \frac{1}{3}$.

Since this transfer function is a real, positive constant, the output signal is in phase with the input and scaled by a factor of $1/3$. Therefore, $v_o(t) = \frac{1}{3} v_i(t) = \frac{V_p}{3} \cos(t/RC)$.

This problem is a classic application of the Divergence Theorem, which relates a surface integral to a volume integral. The theorem states that for a vector field $\vec{F}$, $\oint \oint_S \vec{F} \cdot \hat{n} dS = \iiint_V (\nabla \cdot \vec{F}) dV$.

In this question, our vector field is $\vec{F} = 5\vec{r}$. Let's find its divergence. The divergence of the position vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is a well-known result: $\nabla \cdot \vec{r} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1+1+1=3$.

Therefore, the divergence of our field is $\nabla \cdot (5\vec{r}) = 5(\nabla \cdot \vec{r}) = 5(3) = 15$.

Substituting this constant value back into the volume integral from the Divergence Theorem gives us $\iiint_V 15 \, dV$.

This integral simplifies to $15 \iiint_V dV$, and since the integral of $dV$ over the entire volume is just the volume $V$, the final answer is $15V$.

The existence of modes in a rectangular waveguide is governed by specific rules for the indices $m$ and $n$. For Transverse Electric ($TE_{mn}$) modes, propagation is possible as long as at least one index is non-zero. This means modes like $TE_{10}$ and $TE_{01}$ can exist.

However, the condition for Transverse Magnetic ($TM_{mn}$) modes is more restrictive: both indices must be non-zero ($m \ge 1$ and $n \ge 1$). This is because the field equations for TM waves result in zero fields if either $m$ or $n$ is zero. Therefore, a mode designated as $TM_{10}$ has $n=0$, violating this condition, and thus cannot exist.

This differential equation describes exponential growth and can be solved using the method of separation of variables.

First, rearrange the equation to group all $y$ terms on one side and all $x$ terms on the other:
$\frac{dy}{y} = k\,dx$
Next, integrate both sides to find the general solution. This gives us $\ln(y) = kx + A$, where $A$ is the constant of integration.

We can find the value of $A$ by using the initial condition $y(0)=c$. Substituting $x=0$ and $y=c$ into our equation yields $\ln(c) = k(0) + A$, so $A = \ln(c)$.

Plugging this constant back into the solution gives $\ln(y) = kx + \ln(c)$. To solve for $y$, we can rearrange this to $\ln(y) - \ln(c) = kx$. Using the property of logarithms, this becomes $\ln\left(\frac{y}{c}\right) = kx$.

Finally, exponentiating both sides with base $e$ isolates the term $\frac{y}{c}$, resulting in $\frac{y}{c} = e^{kx}$. The explicit solution is therefore $y = ce^{kx}$.

Let's break down the best modulation system for each attribute.

Frequency Modulation (FM) is highly power-efficient. Its constant amplitude allows the transmitter to operate at full power, providing superior performance against noise.

For voice signals, Single-Sideband Suppressed Carrier (SSB-SC) offers the greatest bandwidth efficiency by transmitting only a single sideband without the carrier.

Conventional Amplitude Modulation (AM) is known for having the simplest receiver structure, requiring only a basic envelope detector for demodulation.

Finally, Vestigial Sideband (VSB) is a bandwidth-efficient method tailored for signals with significant low-frequency or DC components, like video, as it preserves this vital information while trimming bandwidth.

To understand the system's behavior, we begin by finding its transfer function, $H(s) = \frac{Y(s)}{X(s)}$. Taking the Laplace transform of the differential equation, assuming the system is initially relaxed, gives us $(100s^2 - 20s + 1)Y(s) = X(s)$.

The system's stability is dictated by the poles of its transfer function, which are the roots of the characteristic equation $100s^2 - 20s + 1 = 0$. Factoring this quadratic expression yields $(10s - 1)^2 = 0$.

This reveals a repeated pole at $s = +0.1$. Since this pole is located in the right-half of the s-plane (i.e., it has a positive real part), the system is unstable.

For a bounded input like a unit step, an unstable system's output will increase without limit. Among the given options, only waveform A depicts this type of unbounded growth.

The Nyquist plot traces the complex number $G(j\omega)$ as frequency $\omega$ goes from $0$ to $\infty$.

Let's find the starting point of the plot. At $\omega=0$, the function evaluates to $G(j0) = 5+j(0) = 5$. This is a point on the positive real axis.

Now, consider the behavior as $\omega \to \infty$. The real part of $G(j\omega)$ is constant at 5, while the imaginary part, $\omega$, increases without bound.

Therefore, the plot starts at the point (5, 0) and moves vertically upwards, extending to infinity as the imaginary component grows.

A function's Fourier series is composed of a DC offset ($a_0$), cosine terms, and sine terms. An even function, defined by the property $f(t) = f(-t)$, is inherently symmetric like a cosine wave. The coefficients for the sine terms in a Fourier series, denoted $b_n$, are calculated from the product of the function and a sine wave.

Since a sine wave is an odd function, the product of an even function $f(t)$ and an odd $\sin(n\omega_0 t)$ results in an odd function. Integrating any odd function over a symmetric period always yields zero. Consequently, all sine coefficients ($b_n$) for an even function are zero, and its series expansion is constructed entirely from the DC and cosine terms.

For the output $Y$ to be 1, the inputs to the AND gate must both be 1, meaning $Q_1=1$ and $Q_2=1$. Let's analyze the conditions required to reach this state.

For the second flip-flop's output $Q_2$ to become 1, its input $D_2$ must have been 1 before the clock pulse. The circuit shows that $D_2$ is connected to $\bar{Q_1}$. This implies that in the state prior to the clock pulse, $\bar{Q_1}$ was 1, which means $Q_1$ was 0.

So, the clock pulse caused $Q_1$ to transition from 0 to 1. The output of the first flip-flop, $Q_1$, follows its input, the Data line. Therefore, for $Q_1$ to change from 0 to 1, the Data input must have changed from 0 to 1.

To understand the function of this logic circuit, let's analyze its output, F, for every possible combination of the inputs P and Q.

We can summarize the behavior in a truth table. The output F is high (1) only when the inputs P and Q are different from each other (one is 0 and the other is 1). When the inputs are the same (both 0 or both 1), the output F is low (0). This pattern, where the output is true only when an odd number of inputs is true, is the definition of the Exclusive OR (XOR) operation.

This relationship is expressed by the logical equation $F = P \oplus Q$.

To determine the filter type, we can analyze the circuit's behavior at extreme frequencies.

For very low frequencies, approaching DC ($\omega \to 0$), the inductor's impedance ($X_L = \omega L$) becomes zero. It acts like a short circuit, connecting the non-inverting input directly to ground. This forces the output voltage $V_o$ to be zero.

For very high frequencies ($\omega \to \infty$), the inductor's impedance becomes infinite, and it acts as an open circuit. The entire input current $I_i$ is forced to flow through the resistor $R_1$, creating a non-zero voltage at the non-inverting input, which then appears at the output.

Since the circuit passes high frequencies but blocks low frequencies, it functions as a high-pass filter.

For a silicon PN junction operating with a constant forward current, the forward voltage has a negative temperature coefficient. This means the voltage required to maintain the current decreases as the temperature rises. The rate of this change is a well-known characteristic of silicon diodes.

The forward voltage $V$ changes with temperature $T$ at a rate of approximately:
$\frac{dV}{dT} \approx -2.5 \text{ mV}/^{\circ}\text{C}$
For a temperature increase of $\Delta T = 10^{\circ}\text{C}$, the change in voltage is:
$\Delta V = \frac{dV}{dT} \times \Delta T \approx (-2.5 \text{ mV}/^{\circ}\text{C}) \times (10^{\circ}\text{C}) = -25 \text{ mV}$
Thus, the forward bias voltage decreases by 25 mV.

To find the Norton equivalent current ($I_N$), we must calculate the short-circuit current ($I_{sc}$) that flows between terminals P and Q.

The total source current of $16 \angle 0^{\circ}$ A divides between the two parallel branches. We can find the current in the branch containing the short circuit by applying the current divider rule.

$I_N = I_{sc} = 16 \angle 0^{\circ} \times \frac{25}{25 + (15 + j30)} = \frac{400}{40 + j30}$

To simplify, we first convert the denominator into polar form: $40 + j30 \Omega = 50 \angle 36.87^{\circ} \Omega$. Now we can perform the division:

$I_N = \frac{400 \angle 0^{\circ}}{50 \angle 36.87^{\circ}} = 8 \angle -36.87^{\circ} \text{ A}$

Finally, we convert this polar form to the required rectangular form to get the answer:

$I_N = 8(\cos(-36.87^{\circ}) + j\sin(-36.87^{\circ})) = (6.4 - j4.8) \text{ A}$

According to the maximum power transfer theorem, the load resistance ($R_L$) will receive maximum power when it is equal to the Thévenin equivalent resistance ($R_{Th}$) of the source circuit.

To find $R_{Th}$, we calculate the resistance seen from the terminals of $R_L$ after deactivating all independent sources. This means we replace the 20V voltage source with a short circuit and the 2A current source with an open circuit.

By analyzing the simplified resistive network, the equivalent resistance is found to be $15 \, \Omega$. Therefore, for maximum power to be delivered to the load, $R_L$ must be equal to $R_{Th}$, which is $15 \, \Omega$.

To solve this integral, we can use the Cauchy-Goursat theorem, which states that the integral of a function over a closed path is zero if the function is analytic everywhere inside that path.

First, let's find the singularities (poles) of the integrand $f(z) = \frac{-3z+4}{z^{2}+4z+5}$ by finding the roots of the denominator.
Setting $z^{2}+4z+5 = 0$ and using the quadratic formula gives us the poles:
$z = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm 2i}{2} = -2 \pm i$.

Next, we determine if these poles are inside the contour $c$, which is the unit circle $|z|=1$. We calculate the magnitude (distance from the origin) of the poles:
$|-2 \pm i| = \sqrt{(-2)^2 + (\pm 1)^2} = \sqrt{4+1} = \sqrt{5}$.

Since $\sqrt{5} \approx 2.236$, which is greater than 1, both poles lie outside the unit circle. This means the function $f(z)$ is analytic everywhere inside and on the contour $|z|=1$. Therefore, by the Cauchy-Goursat theorem, the integral is zero.

To find the magnetic field $\vec{H}_2$ in Region-2, we'll use the boundary conditions for magnetostatics at the interface ($x=0$), examining the normal and tangential components separately.

First, the normal component of the magnetic flux density $\vec{B}$ is continuous, meaning $B_{n1} = B_{n2}$. Since $\vec{B} = \mu\vec{H}$ and the normal direction is $\hat{u}_x$, this implies $\mu_1 H_{x1} = \mu_2 H_{x2}$. Given $\mu_{r1}=1$ and $\mu_{r2}=2$, and $H_{x1}=3$ A/m, we find $(\mu_0 \cdot 1)(3) = (\mu_0 \cdot 2)H_{x2}$, which gives $H_{x2} = 1.5$ A/m.

Next, the tangential component of the magnetic field intensity $\vec{H}$ is discontinuous in the presence of a surface current. The boundary condition is $\vec{H}_{t1} - \vec{H}_{t2} = \vec{J}_s \times \hat{a}_n$, where $\hat{a}_n$ is the unit normal vector pointing from Region 2 to Region 1 (i.e., $\hat{a}_n = -\hat{u}_x$). The tangential component of $\vec{H}_1$ is $\vec{H}_{t1} = 30\hat{u}_y$.

Plugging in the values, we get $30\hat{u}_y - \vec{H}_{t2} = (10\hat{u}_y) \times (-\hat{u}_x) = 10\hat{u}_z$. Solving for the tangential part of $\vec{H}_2$ gives $\vec{H}_{t2} = 30\hat{u}_y - 10\hat{u}_z$.

Finally, combining the normal and tangential components gives the total field in Region-2:
$\vec{H}_2 = H_{x2}\hat{u}_x + \vec{H}_{t2} = 1.5\hat{u}_x + 30\hat{u}_y - 10\hat{u}_z$ A/m.

First, we find the magnitude of the reflection coefficient, $|\Gamma|$, from the given Voltage Standing Wave Ratio (VSWR). The relationship is $VSWR = \frac{1+|\Gamma|}{1-|\Gamma|}$. With a VSWR of 5, we solve for $|\Gamma|$ and find it to be $\frac{2}{3}$.

Next, we determine the phase of the reflection coefficient. We are told the first voltage maximum is at a distance of $\frac{\lambda}{4}$ from the load. On a transmission line, voltage maxima and minima are always separated by $\frac{\lambda}{4}$. This means a voltage minimum must be located precisely at the load itself.

A voltage minimum at the load signifies that the load impedance $Z_L$ is purely resistive and less than the characteristic impedance $Z_0$. This physical condition corresponds to a reflection coefficient that is a real, negative number. Therefore, the complete reflection coefficient is $\Gamma = -|\Gamma| = -\frac{2}{3}$.

Finally, we use the formula relating the reflection coefficient to impedance, $\Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}$. Plugging in the known values gives $-\frac{2}{3} = \frac{Z_L - 50}{Z_L + 50}$. Solving this equation for the load impedance $Z_L$ yields a value of $10 \, \Omega$.

To find the output power spectral density (PSD), we relate it to the input PSD via the system's transfer function, $H(f)$, using the formula $S_Y(f) = |H(f)|^2 S_X(f)$.

First, we find $H(f)$ by taking the Fourier transform of the system's differential equation, $y(t) = \frac{d}{dt}x(t) - x(t)$. This gives $Y(f) = j2\pi f X(f) - X(f)$, so the transfer function is $H(f) = j2\pi f - 1$. The squared magnitude is $|H(f)|^2 = |j2\pi f - 1|^2 = (2\pi f)^2 + (-1)^2 = 4\pi^2 f^2 + 1$.

Next, we find the input PSD, $S_X(f)$, by taking the Fourier transform of the given autocorrelation function, $R_X(\tau) = e^{-\pi\tau^2}$. A key property of the Gaussian function is that it is its own Fourier transform, which means $S_X(f) = e^{-\pi f^2}$.

Finally, we combine these results to find the output PSD: $S_Y(f) = (4\pi^2 f^2 + 1)e^{-\pi f^2}$.

Let's trace the states of the 3-stage Johnson counter, which begins at an initial state of $Q_2Q_1Q_0 = 000$. A Johnson counter is a shift register where the input is the inverted output of the final stage. This "twisted" feedback generates the state sequence: $000 \rightarrow 100 \rightarrow 110 \rightarrow 111 \rightarrow 011 \rightarrow 001$, before cycling back to $000$. The D/A converter transforms these binary numbers into a proportional analog voltage. Converting our sequence to decimal gives voltage levels proportional to $0, 4, 6, 7, 3, 1$, and then repeating from $0$. This specific staircase pattern of rising and falling voltage levels corresponds to waveform A.

To design the required synchronous counter, we must determine the logic for inputs $D_A$ and $D_B$ based on the present state outputs, $Q_A$ and $Q_B$. In a D flip-flop, the input value $D$ becomes the output $Q$ after the next clock pulse. Therefore, the required D inputs are simply the values of the desired next state.

Let's organize the given sequence into a state transition table:
| Present State ($Q_B Q_A$) | Next State ($Q_{B,next} Q_{A,next}$) |
|---|---|
| 0 0 | 1 1 |
| 0 1 | 1 0 |
| 1 0 | 0 0 |
| 1 1 | 0 1 |

From this table, we can derive the expressions for $D_A = Q_{A,next}$ and $D_B = Q_{B,next}$.
Looking at the column for $D_B$ (which is $Q_{B,next}$), we see its value is always the opposite of the present state's $Q_B$. Thus, $D_B=\bar{Q_B}$.
For $D_A$ (which is $Q_{A,next}$), the value is 1 for present states 00 and 11. This means $D_A$ is 1 only when $Q_A$ and $Q_B$ are equal, which is the definition of the XNOR logic function: $D_A = (Q_A Q_B + \bar{Q_A} \bar{Q_B})$.

The two nMOS transistors are connected in series, meaning their drain currents must be equal. We can first confirm that both operate in the saturation region. For a transistor to be saturated, $V_{DS} \ge V_{GS} - V_T$. This condition holds for both the upper transistor ($6-V_x \ge (5-V_x)-1$) and the lower transistor ($V_x \ge V_x-1$), as both inequalities simplify to true statements ($6 \ge 4$ and $0 \ge -1$).

The saturation current is proportional to $\frac{W}{L}(V_{GS}-V_T)^2$. We can now equate the expressions for the two currents:
$(\frac{W}{L})_{upper} (V_{GS,upper} - V_T)^2 = (\frac{W}{L})_{lower} (V_{GS,lower} - V_T)^2$
Plugging in the given values and circuit voltages ($V_{GS,upper} = 5-V_x$, $V_{GS,lower} = V_x$):
$4 \cdot ((5 - V_x) - 1)^2 = 1 \cdot (V_x - 1)^2$
$4(4 - V_x)^2 = (V_x - 1)^2$
Taking the square root of both sides gives $2(4-V_x) = V_x-1$. Solving this simple linear equation, we find $8 - 2V_x = V_x-1$, which rearranges to $3V_x = 9$, yielding $V_x = 3$ V.

To find the output of an LTI system, we can simplify the convolution process by working in the Laplace domain. The output transform $Y(s)$ is the product of the input transform $X(s)$ and the system's transfer function $H(s)$.

First, we transform the input $x(t) = e^{-2t}u(t) + \delta(t-6)$ to get $X(s) = \frac{1}{s+2} + e^{-6s}$. The impulse response $h(t) = u(t)$ transforms into $H(s) = \frac{1}{s}$.

Next, we multiply them to find the output transform: $Y(s) = X(s)H(s) = \left(\frac{1}{s+2} + e^{-6s}\right)\frac{1}{s} = \frac{1}{s(s+2)} + \frac{e^{-6s}}{s}$.

To transform this back to the time domain, we use partial fraction expansion on the first term: $\frac{1}{s(s+2)} = \frac{0.5}{s} - \frac{0.5}{s+2}$.

Finally, taking the inverse Laplace transform of the full expression $Y(s) = \frac{0.5}{s} - \frac{0.5}{s+2} + \frac{e^{-6s}}{s}$ term by term gives us the output signal $y(t) = 0.5u(t) - 0.5e^{-2t}u(t) + u(t-6)$, which can be rewritten as $y(t) = 0.5(1-e^{-2t})u(t) + u(t-6)$.

To determine the collector current ($I_C$), we must first find the common-emitter current gain, $\beta$, from the given common-base current gain, $\alpha$. The relationship between them is $\beta = \frac{\alpha}{1-\alpha}$.
Plugging in the given value, we get $\beta = \frac{0.98}{1-0.98} = \frac{0.98}{0.02} = 49$.

Now, we can use the complete equation for collector current in the common-emitter configuration, which accounts for the leakage current $I_{CO}$:
$I_C = \beta I_B + (1+\beta)I_{CO}$

Substituting all the known values into this equation:
$I_C = (49)(20 \mu A) + (1+49)(0.6 \mu A)$
$I_C = 980 \mu A + (50)(0.6 \mu A) = 980 \mu A + 30 \mu A$
Thus, the total collector current is $I_C = 1010 \mu A$, which is equivalent to $1.01$ mA.

To find the initial and final values of $f(t)$ from its Laplace transform $F(s)$, we apply the Initial and Final Value Theorems.

The initial value, $f(0)$, is given by the limit $\lim_{s \to \infty } sF(s)$.
$\lim_{s \to \infty } s \cdot \frac{2(s+1)}{s^{2+}4s+7} = \lim_{s \to \infty } \frac{2s^{2+}2s}{s^{2+}4s+7}$
For large $s$, the limit is the ratio of the highest-degree coefficients, so the initial value is $\frac{2}{1} = 2$.

The final value, $\lim_{t \to \infty } f(t)$, is given by the limit $\lim_{s \to 0} sF(s)$, provided the system is stable.
$\lim_{s \to 0} s \cdot \frac{2(s+1)}{s^{2+}4s+7} = 0 \cdot \frac{2(0+1)}{0+0+7} = 0$
Thus, the initial value is 2 and the final value is 0.