GATE ECE 2010

For a real skew-symmetric matrix, its transpose is equal to its negative, $A^T = -A$. This property has a direct consequence on its eigenvalues. Let $\lambda$ be an eigenvalue with eigenvector $v$. A fundamental relationship we can derive is $\bar{\lambda} = -\lambda$, where $\bar{\lambda}$ is the complex conjugate of $\lambda$.

To understand why this is true, consider any complex number $\lambda = a + ib$. The condition $\bar{\lambda} = -\lambda$ means $a - ib = -(a + ib)$, which simplifies to $2a = 0$. This forces the real part of the eigenvalue, $a$, to be zero.

Therefore, any eigenvalue must be of the form $\lambda = ib$, which is either a purely imaginary number or zero (if $b=0$).

First, we identify the waveform's symmetry. Since the function is a mirror image across the vertical axis, it is an even function, meaning $f(t) = f(-t)$. A key property of Fourier series is that even functions are composed exclusively of cosine terms and a DC component; all sine terms ($b_n$) are zero.

Next, we determine the DC component, $a_0$, which is simply the average value of the function over one period $T$. We can see by inspection that the negative portion of the wave has a larger magnitude ($-2A$) than the positive portion ($A$), suggesting the average value will be negative.

To confirm this, we calculate the average value:
$a_0 = \frac{1}{T} \times (\text{Area under curve \in one period})$
$a_0 = \frac{1}{T} \left[ \underbrace{A \times \frac{T}{2}}_{\text{Positive Area}} + \underbrace{(-2A) \times \frac{T}{2}}_{\text{Negative Area}} \right] = \frac{1}{T} \left[ \frac{AT}{2} - AT \right] = -\frac{A}{2}$.
The DC component is indeed negative.

This problem involves a second-order linear homogeneous differential equation. The characteristic equation is $r^2 - 1/L^2 = 0$, which has roots $r = \pm 1/L$. This leads to the general solution $n(x) = C_1 e^{x/L} + C_2 e^{-x/L}$.

Now, we apply the boundary conditions to find the constants $C_1$ and $C_2$. For the condition $n(\infty ) = 0$ to be met, the coefficient of the term that grows infinitely with $x$ must be zero. Since $e^{x/L}$ explodes as $x \to \infty$, we must set $C_1 = 0$.

This simplifies our solution to $n(x) = C_2 e^{-x/L}$. Using the second condition, $n(0) = K$, we get $K = C_2 e^0$, which implies $C_2 = K$.

Substituting the values of the constants back gives the final solution: $n(x) = K e^{-x/L}$.

This circuit is a symmetric $\pi$-network, which allows us to find the Y-parameters by inspection. First, we convert the component resistances into admittances using the formula $Y = 1/R$. For each $0.5\,\Omega$ resistor, the admittance is $Y = 1/0.5\,\Omega = 2$ S.

The diagonal parameter $Y_{11}$ is the total admittance seen from port 1, which is the sum of the shunt and series branches connected to it: $Y_{11} = 2 + 2 = 4$ S. Due to the circuit's symmetry, $Y_{22}$ is identical. The off-diagonal parameters, $Y_{12}$ and $Y_{21}$, are the negative of the admittance connecting the two ports, so $Y_{12} = Y_{21} = -2$ S.

For a parallel RLC circuit, the bandwidth is given by the expression $BW = \frac{1}{RC}$. This formula directly shows that increasing the resistance $R$ will decrease the bandwidth, making statement A correct. It also shows the bandwidth is independent of the inductance $L$, making statement B correct.

At resonance, the reactive currents from the inductor and capacitor cancel each other out. This means the circuit behaves as a pure resistor, so the input impedance is a real quantity, confirming statement C. This condition corresponds to minimum total admittance ($Y = 1/R$). Since impedance is the reciprocal of admittance ($Z=1/Y$), the impedance at parallel resonance is at its maximum value ($Z=R$), not its minimum.

The mobility of electrons in bulk silicon at room temperature is high, typically around $1350 cm^{2}/V-s$. However, in a MOSFET, the electrons are confined to the inversion layer, a very thin region at the silicon-silicon dioxide (Si-SiO2) interface. The strong vertical electric field from the gate forces the electrons to travel along this interface. This leads to a significant additional scattering mechanism known as surface scattering, caused by interface roughness and trapped charges. This surface scattering severely degrades the mobility, reducing it to a value much lower than the bulk mobility. Therefore, a value of $150 cm^{2}/V-s$ is a realistic value for electron mobility in the inversion layer.

The quality of the gate oxide is paramount for transistor performance. Dry oxidation, which involves reacting silicon with pure oxygen gas at high temperatures, is used for this critical step. Although this process is significantly slower than wet oxidation, its slow growth rate provides the precise control needed to create the ultra-thin, uniform layers required in modern CMOS devices. This meticulous process results in a denser, higher-quality oxide with fewer defects and a superior silicon-to-oxide interface, which is essential for ensuring device reliability and minimizing electrical leakage.

First, we determine the reference current ($I_{REF}$) flowing through the left branch into the diode-connected transistor $Q_1$. Applying KVL from the ground rail to the -10 V supply, the voltage across the resistor is the total span minus the base-emitter drop.
$I_{REF} = \frac{(0\text{ V} - (-10\text{ V})) - V_{BE1}}{9.3 \text{ k}\Omega}$
Assuming a standard silicon BJT forward voltage of $V_{BE1} \approx 0.7$ V, the reference current is:
$I_{REF} = \frac{10\text{ V} - 0.7\text{ V}}{9.3 \text{ k}\Omega} = \frac{9.3\text{ V}}{9.3 \text{ k}\Omega} = 1 \text{ mA}$
In a current mirror, the ratio of the collector currents is proportional to the ratio of the transistors' emitter areas. Since $Q_1$ and $Q_2$ share the same base-emitter voltage, we have:
$\frac{I_0}{I_{REF}} = \frac{\text{Area}(Q_2)}{\text{Area}(Q_1)}$
Given that Area($Q_1$) is half of Area($Q_2$), this ratio is 2. Therefore, the output current $I_0$ is twice the reference current.
$I_0 = 2 \times I_{REF} = 2 \times 1 \text{ mA} = 2 \text{ mA}$

Removing the emitter bypass capacitor $C_E$ means the resistor $R_E$ is no longer shorted out for AC signals. This introduces the resistor into the emitter's signal path, creating a form of negative feedback called emitter degeneration. This feedback dramatically increases the impedance looking into the transistor's base to $r_{\pi} + (1+\beta)R_{E}$, which in turn increases the total input resistance $R_{i}$ of the amplifier. The same feedback effect reduces the voltage gain, as the gain magnitude $|A_v|$ changes from approximately $g_{m}R_C$ to the smaller value of $\frac{g_{m}R_C}{1+g_{m}R_E}$.

This circuit is a classic inverting amplifier configuration. With an ideal op-amp, the non-inverting (+) input being grounded forces the inverting (-) input to be at a "virtual ground," meaning its potential is also $0$V.

We can find the gain by applying Kirchhoff's Current Law at this virtual ground node. The current entering through $R_1$ is $V_{\in}/R_1$, and the current leaving through the feedback resistor $R_2$ is $-V_{out}/R_2$. As no current enters the ideal op-amp, these two currents must be equal:
$\frac{V_{\in}}{R_1} = -\frac{V_{out}}{R_2}$
Solving for the voltage gain, $A_V = V_{out}/V_{\in}$, yields $-\frac{R_2}{R_1}$. The resistor $R_3$ is merely a load on the output and does not influence the gain equation for an ideal op-amp.

Let's break down the equivalences between the gates in Column A and the descriptions in Column B.

First, the NOR and NAND gates are direct applications of De Morgan's theorems.

• A NOR gate has the function $F = \overline{X+Y}$. By De Morgan's law, this is equivalent to $F = \bar{X} \cdot \bar{Y}$, which describes an AND gate with inverted inputs. This is an INVERT-AND gate. (Q → 2)
• A NAND gate has the function $F = \overline{X \cdot Y}$. The equivalent expression is $F = \bar{X} + \bar{Y}$, describing an OR gate with inverted inputs. This is an INVERT-OR gate. (R → 1)

Next, we analyze the XOR and XNOR gates based on their circuit implementations.

• An XNOR gate is the complement of an XOR gate. Its function can be written as $S = \overline{X\bar{Y}+\bar{X}Y}$. This expression represents a NOR gate whose inputs are the outputs of two AND gates (for the terms $X\bar{Y}$ and $\bar{X}Y$). This structure is described as AND-NOR. (S → 3)
• An XOR gate can be constructed using only NOR gates. The resulting circuit is a network where a final NOR gate takes inputs from other NOR gates. This is aptly called a NOR-NOR gate. (P → 4)

To achieve an output of $F=1$, the final 3-input XNOR gate must receive an even number of '1' inputs. Let's trace the circuit with the inputs $A=0$, $B=0$, and $C=1$.

First, we calculate the intermediate values $X$ and $Y$.
The top gate is an XOR gate, so $X = A \oplus B = 0 \oplus 0 = 0$.
The gate below it is an XNOR gate, so $Y = \overline{A \oplus B} = \overline{0 \oplus 0} = 1$.

Now, we look at the inputs to the final XNOR gate: $(X, Y, C) = (0, 1, 1)$.
This set of inputs has two '1's. Since two is an even number, the output of the XNOR gate is indeed $F=1$.

To find the address range for the device at Y5, we must determine which address bits are fixed by the chip-select logic. The 3-to-8 decoder activates its Y5 output only when its select inputs, CBA, are set to 101 (the binary representation of 5).

The circuit connects these decoder inputs (C, B, A) and its enable pins to the upper address lines. This means that to select Y5, the upper address bits must have a specific, fixed pattern. The resulting pattern for bits $A_{15}$ through $A_8$ is 0010 1101. In hexadecimal, 0010 is 2 and 1101 is D.

The lower 8 address bits, $A_7$ through $A_0$, are not involved in selecting the device, so they are free to vary. This allows them to range from 0000 0000 (00H) to 1111 1111 (FFH). Therefore, the complete address range for Y5 is from 2D00H to 2DFFH.

To find the discrete-time signal $x[n]$, we can determine the inverse z-transform of $X(z)$ by inspection. The key is to recognize the fundamental z-transform pair for a shifted unit impulse: a term $z^{k}$ in the z-domain corresponds to an impulse at time $n=-k$, represented as $\delta[n+k]$.

We apply this principle to each term in the given expression for $X(z)$:

• The term $5z^2$ has a power of $k=2$, so its inverse transform is $5\delta[n+2]$.
• The term $3$, which can be written as $3z^0$, has a power of $k=0$, giving us $3\delta[n]$.
• The term $4z^{-1}$ has a power of $k=-1$, resulting in $4\delta[n-1]$.

Summing these individual components gives the complete time-domain signal $x[n]$.

For two systems connected in cascade, the overall impulse response is the convolution of their individual responses: $h[n] = h_1[n] * h_2[n]$. This calculation is often simpler in the Z-domain, where convolution in time becomes multiplication.

First, we find the Z-transform of each impulse response.
For $h_1[n]=\delta[n-1]$, the Z-transform is $H_1(z) = z^{-1}$.
For $h_2[n]=\delta[n-2]$, the Z-transform is $H_2(z) = z^{-2}$.

The overall transfer function, $H(z)$, is the product of the individual transfer functions:
$H(z) = H_1(z) H_2(z) = (z^{-1})(z^{-2}) = z^{-3}$.

Finally, we find the inverse Z-transform of $H(z)$ to get the overall impulse response in the time domain, which is $h[n] = \delta[n-3]$.

The fundamental building block of a Radix-2 FFT algorithm, known as the "butterfly," is designed for computational efficiency. A butterfly takes two complex inputs, let's call them $a$ and $b$, to produce two outputs, $a'$ and $b'$. The operations are $a' = a + (W_N^k \cdot b)$ and $b' = a - (W_N^k \cdot b)$.

Notice that the complex multiplication involving the twiddle factor, $(W_N^k \cdot b)$, is performed only once. The resulting product is then used for two separate complex additions (one addition and one subtraction) to compute both outputs. Therefore, each butterfly computation requires just one complex multiplication.

Let's trace the signals through the diagram. The signal leaving the first summing junction, let's call it $P(s)$, gets sent down two parallel paths that are summed together to form the feedback signal. This feedback signal is $P(s)\frac{1}{s+1} - P(s)\frac{1}{s+1} = 0$.

Since the feedback signal is zero, the first summer's output is simply $P(s) = R(s) - 0 = R(s)$.

The system's output, $Y(s)$, is produced by passing $P(s)$ through the upper block. So, $Y(s) = P(s) \cdot \frac{1}{s+1}$.

Substituting $P(s) = R(s)$, we get $Y(s) = R(s) \cdot \frac{1}{s+1}$.

Therefore, the overall transfer function is $\frac{Y(s)}{R(s)} = \frac{1}{s+1}$.

For a linear system with a sinusoidal input, the output will be a sinusoid at the same frequency but with a shifted phase and scaled amplitude. The system's frequency response, $G(j\omega)$, defines this transformation.

First, determine the phase shift introduced by the system from the given signals. The angular frequency is $\omega=2$ rad/s. The phase shift is the output phase minus the input phase:
$\phi = \phi_{out} - \phi_{\in} = (-\frac{\pi}{3}) - (-\frac{\pi}{2}) = \frac{\pi}{6}$ radians, or $30^\circ$.

Next, find the theoretical phase angle of the transfer function $G(s)=\frac{s}{s+p}$ by substituting $s=j\omega$:
$\angle G(j\omega) = \angle(j\omega) - \angle(p+j\omega) = 90^\circ - \tan^{-1}(\frac{\omega}{p})$.

Equating the observed phase shift to the system's phase angle at $\omega=2$:
$30^\circ = 90^\circ - \tan^{-1}(\frac{2}{p})$

Solving for $p$ yields $\tan^{-1}(\frac{2}{p}) = 60^\circ$. Therefore, $\frac{2}{p} = \tan(60^\circ) = \sqrt{3}$, which gives $p = \frac{2}{\sqrt{3}}$.

The general form of the transfer function is determined by its corner frequencies and gain. The plot shows an initial low-frequency gain of 0 dB, which means the DC gain constant $K=1$ since $20\log_{10}(K) = 0$.

The slope increases by +20 dB/decade at $\omega = 0.1$ rad/s, indicating a zero at that frequency. This corresponds to the term $(1+s/0.1)$ in the numerator.

The slope then decreases back to 0 dB/decade at $\omega=10$ rad/s, indicating a pole. This corresponds to the term $(1+s/10)$ in the denominator.

Combining these elements, the transfer function is $H(s) = \frac{1+s/0.1}{1+s/10}$. Simplifying this expression gives $H(s) = \frac{1+10s}{1+0.1s}$, which is equivalent to $\frac{10s+1}{0.1s+1}$.

The standard expression for a conventional AM signal is $s(t) = A_c[1 + k_a m(t)]\cos(2\pi f_c t)$. For no over-modulation, the modulation index, $\mu$, must be less than or equal to 1. The modulation index is defined as $\mu = k_a A_m$, where $A_m$ is the peak amplitude of the message signal $m(t)$.

Given $m(t)=2\cos(2\pi f_{m}t)$, the peak amplitude is $A_m = 2$.

Now, let's analyze option (C). We can factor the expression to match the standard form:
$x(t) = A_{c}\cos(2\pi f_{c}t) + \frac{A_{c}}{4}m(t)\cos(2\pi f_{c}t) = A_c[1 + \frac{1}{4}m(t)]\cos(2\pi f_{c}t)$

From this form, we can see that the constant $k_a = 1/4$. We can now calculate the modulation index:
$\mu = k_a A_m = \frac{1}{4} \times 2 = 0.5$

Since $\mu = 0.5$, which is less than 1, this represents a conventional AM signal without over-modulation.

The given signal, $x(t)$, is an angle-modulated wave. A key characteristic of all angle-modulated signals (like FM and PM) is that their amplitude remains constant, regardless of the complexity of the phase term. The information is encoded in the phase, not the amplitude.

From the equation, we can identify the constant carrier amplitude as $A_c = 6$ V.

The average power of any sinusoidal signal with a constant amplitude $A_c$ is given by the formula $P_{avg} = \frac{A_c^2}{2R}$. Assuming a standard load of $R=1 \Omega$, this simplifies to $P_{avg} = \frac{A_c^2}{2}$.

Substituting the value of our amplitude:
$P_{avg} = \frac{6^2}{2} = \frac{36}{2} = 18$ W.

A two-port network is reciprocal if its scattering matrix is symmetric, meaning $S_{12} = S_{21}$. In this case, both $S_{12}$ and $S_{21}$ are $0.9\angle 90^{\circ}$, so the network is indeed reciprocal.

For a network to be lossless, power must be conserved. This requires the sum of the squared magnitudes of the S-parameters in each column to equal 1. Let's check the first column:
$|S_{11}|^2 + |S_{21}|^2 = (0.2)^2 + (0.9)^2 = 0.04 + 0.81 = 0.85$.

Since this sum is not equal to 1, the network does not conserve power and is therefore not lossless.

For a transmission line to be distortionless, the ratio of its primary parameters must satisfy the condition $R/L = G/C$. This allows us to establish a key relationship for this problem.

The characteristic impedance is defined as $Z_0 = \sqrt{L/C}$. Due to the distortionless condition, we can substitute $L/C = R/G$, which gives us $Z_0 = \sqrt{R/G}$.

The attenuation constant is given by $\alpha = \sqrt{RG}$. We can rearrange our expression for $Z_0$ to solve for $\sqrt{G}$, yielding $\sqrt{G} = \sqrt{R}/Z_0$.

By substituting this into the equation for $\alpha$, we find a simple formula in terms of our known values:
$\alpha = \sqrt{R} \cdot \sqrt{G} = \sqrt{R} \cdot \left(\frac{\sqrt{R}}{Z_0}\right) = \frac{R}{Z_0}$.

Plugging in the given values, we get:
$\alpha = \frac{0.1 \, \Omega/\text{m}}{50 \, \Omega} = 0.002$ Np/m.

The impulse response, $h(t)$, of a filter matched to a signal, $s(t)$, is its time-reversed and delayed version. The mathematical relationship is given by $h(t) = s(T-t)$, where $T$ is the duration of the pulse.

We can find the shape of $h(t)$ in two simple steps:

1. Time Reversal: First, flip the signal $s(t)$ about the vertical axis ($t=0$) to obtain $s(-t)$. The original rectangular pulse from $t=0$ to $t=T$ now exists from $t=-T$ to $t=0$.
2. Time Shift: Next, shift the reversed signal $s(-t)$ to the right by $T$ to get $s(-(t-T)) = s(T-t)$. Shifting the pulse from the interval $[-T, 0]$ right by $T$ moves it to the interval $[0, T]$.

Therefore, the resulting impulse response $h(t)$ is a rectangular pulse from $t=0$ to $t=T$, which is identical to the original pulse $s(t)$.

The time-average power density, also known as the Poynting vector's magnitude, is given by $P_{avg} = \frac{E_0^2}{2\eta}$, where $E_0$ is the electric field amplitude and $\eta$ is the medium's intrinsic impedance.

First, we must calculate the impedance of this specific dielectric. For a nonmagnetic ($\mu = \mu_0$) medium with relative permittivity $\epsilon_r=4$, the impedance is:
$\eta = \sqrt{\frac{\mu}{\epsilon}} = \sqrt{\frac{\mu_0}{4\epsilon_0}} = \frac{1}{2}\sqrt{\frac{\mu_0}{\epsilon_0}}$
Recognizing that the impedance of free space is $\eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} \approx 120\pi \, \Omega$, the impedance of our medium is simply $\eta = \frac{\eta_0}{2} = 60\pi \, \Omega$.

Now, we can substitute the given electric field amplitude ($E_0 = 1$ V/m) and our calculated impedance into the power density formula:
$P_{avg} = \frac{(1)^2}{2(60\pi)} = \frac{1}{120\pi} \, \text{W/m}^2$

To find the extrema of the function, we first need to express $y$ explicitly in terms of $x$. Taking the natural logarithm of both sides of $e^y = x^{1/x}$ gives us $y = \ln(x^{1/x})$, which simplifies to $y = \frac{\ln x}{x}$.

Next, we find the critical points by taking the first derivative and setting it to zero. Using the quotient rule, the derivative is $\frac{dy}{dx} = \frac{(1/x)(x) - (\ln x)(1)}{x^2} = \frac{1 - \ln x}{x^2}$. Setting this to zero, we find that a critical point exists when $1 - \ln x = 0$, which occurs at $x = e$.

To classify this point, we apply the second derivative test. Differentiating again, we get $\frac{d^2y}{dx^2} = \frac{2\ln x - 3}{x^3}$. Evaluating this at our critical point $x=e$, we find $\frac{d^2y}{dx^2}\Big|_{x=e} = \frac{2\ln(e) - 3}{e^3} = \frac{2-3}{e^3} = -\frac{1}{e^3}$.

Since the second derivative at $x=e$ is negative, the function $y$ has a maximum at this point.

We are tossing a coin four times. For the number of heads to be greater than the number of tails, we must have one of two scenarios: (1) 3 heads and 1 tail, or (2) 4 heads and 0 tails.

Let's count the outcomes for each scenario:

1. 4 Heads (HHHH): There is only one way this can happen.
2. 3 Heads and 1 Tail: The single tail can appear on the first, second, third, or fourth toss (THHH, HTHH, HHTH, HHHT). This gives us 4 different outcomes.

In total, there are $1 + 4 = 5$ favorable outcomes. The probability of any specific sequence of four tosses is $(\frac{1}{2})^4 = \frac{1}{16}$. Since there are 5 distinct sequences that satisfy our condition, the total probability is $5 \times \frac{1}{16} = \frac{5}{16}$.

To solve for the closed-loop line integral, we break the path into four straight segments and sum the integral for each part. The dot product of the vector field with the differential path element is $\vec{A} \cdot \vec{dl} = xy\,dx + x^2\,dy$.

1. Path P→Q: Here, $y=1$ and $dy=0$, so the integral is $\int_{1/\sqrt{3}}^{2/\sqrt{3}} x\,dx = \frac{1}{2}$.
2. Path Q→R: Along this vertical path, $x=2/\sqrt{3}$ and $dx=0$. The integral is $\int_{1}^{3} (\frac{2}{\sqrt{3}})^2\,dy = \frac{4}{3} \int_1^3 dy = \frac{8}{3}$.
3. Path R→S: Now $y=3$ and $dy=0$, giving $\int_{2/\sqrt{3}}^{1/\sqrt{3}} 3x\,dx = -\frac{3}{2}$. Note the integration limits are from right to left.
4. Path S→P: Finally, $x=1/\sqrt{3}$ and $dx=0$. The integral is $\int_{3}^{1} (\frac{1}{\sqrt{3}})^2\,dy = \frac{1}{3}\int_3^1 dy = -\frac{2}{3}$.

Summing the results from all four paths gives the final answer: $\frac{1}{2} + \frac{8}{3} - \frac{3}{2} - \frac{2}{3} = (\frac{1}{2} - \frac{3}{2}) + (\frac{8}{3} - \frac{2}{3}) = -1 + 2 = 1$.

To find the residues of the function $X(z)$, we first identify its poles by finding the roots of the denominator. The function has simple poles at $z=0$, $z=1$, and $z=2$.

The residue at a simple pole $z_0$ can be calculated using the formula $\text{Res}(z_0) = \lim_{z \to z_0} (z - z_0)X(z)$.

Applying this formula to each pole:
The residue at $z=0$ is $\lim_{z \to 0} z \frac{1-2z}{z(z-1)(z-2)} = \frac{1-0}{(-1)(-2)} = \frac{1}{2}$.
The residue at $z=1$ is $\lim_{z \to 1} (z-1) \frac{1-2z}{z(z-1)(z-2)} = \frac{1-2}{1(1-2)} = \frac{-1}{-1} = 1$.
The residue at $z=2$ is $\lim_{z \to 2} (z-2) \frac{1-2z}{z(z-1)(z-2)} = \frac{1-4}{2(2-1)} = -\frac{3}{2}$.

Euler's method approximates a solution by taking sequential steps, using the tangent line at the start of each interval. The iterative formula is $y_{n+1} = y_n + h \cdot f(x_n, y_n)$, where $h$ is the step size and $\frac{dy}{dx} = f(x, y)$.

For this problem, $f(x, y) = x+y$, our initial condition is $(x_0, y_0) = (0, 0)$, and the step size is $h=0.1$. We need three steps to reach $x=0.3$.

1. Step 1 (to find y(0.1)):
   $y(0.1) \approx y(0) + h \cdot (x_0 + y_0) = 0 + 0.1(0 + 0) = 0$.

2. Step 2 (to find y(0.2)): Using the new point $(0.1, 0)$,
   $y(0.2) \approx y(0.1) + h \cdot (x_1 + y_1) = 0 + 0.1(0.1 + 0) = 0.01$.

3. Step 3 (to find y(0.3)): Using the new point $(0.2, 0.01)$,
   $y(0.3) \approx y(0.2) + h \cdot (x_2 + y_2) = 0.01 + 0.1(0.2 + 0.01) = 0.01 + 0.021 = 0.031$.

Therefore, the value of $y(0.3)$ is approximately $0.031$.

To determine the value of $K$, we can use the Final Value Theorem of Laplace transforms. This theorem states that for a stable system, the long-term behavior of a function $f(t)$ can be found from its transform $F(s)$ using the relation: $\lim_{t\rightarrow \infty }f(t)=\lim_{s\rightarrow 0} s F(s)$.

We are given that $\lim_{t\rightarrow \infty }f(t)=1$ and $F(s) = \frac{3s+1}{s^{3}+4s^{2}+(K-3)s}$.
Applying the theorem, we get:
$1 = \lim_{s\rightarrow 0} s \left[ \frac{3s+1}{s^{3}+4s^{2}+(K-3)s} \right]$

To evaluate the limit, we can factor an $s$ from the denominator and cancel it:
$1 = \lim_{s\rightarrow 0} \frac{s(3s+1)}{s(s^{2}+4s+K-3)} = \lim_{s\rightarrow 0} \frac{3s+1}{s^{2}+4s+K-3}$

Now, substitute $s=0$ into the expression:
$1 = \frac{3(0)+1}{0^{2+}4(0)+K-3} = \frac{1}{K-3}$

Solving the equation $1 = \frac{1}{K-3}$ gives us $K-3 = 1$, which means $K=4$.