GATE ECE 2014 Set 4

This problem asks for the sum of the series $\sum_{n=0}^{\infty } \frac{1}{n!}$. This series is a famous result from calculus, and its value can be found by recalling the Maclaurin series expansion for the exponential function, $e^x$.

The series for $e^x$ is given by:
$e^x = \sum_{n=0}^{\infty } \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$

By substituting the specific value $x=1$ into this general formula, we get:
$e^1 = \sum_{n=0}^{\infty } \frac{1^n}{n!} = \sum_{n=0}^{\infty } \frac{1}{n!}$

This matches the series in the question exactly. Therefore, the series converges to the value of $e^1$, which is simply $e$.

To find the magnitude of the gradient, we first need to compute the gradient vector, $\nabla f$. This vector is composed of the partial derivatives of the function $f(x,y,z)=x^{2}+3y^{2}+z^{3}$.
The gradient is $\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle = \langle 2x, 6y, 3z^2 \rangle$.

Next, we evaluate this gradient vector at the specific point $(1,1,1)$ by substituting these values for $x, y,$ and $z$:
$\nabla f(1,1,1) = \langle 2(1), 6(1), 3(1)^2 \rangle = \langle 2, 6, 3 \rangle$.

Finally, we calculate the magnitude (or length) of this resulting vector, which is the square root of the sum of the squares of its components:
$||\nabla f(1,1,1)|| = \sqrt{2^2 + 6^2 + 3^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$.

We are asked to find the expected value of the absolute value of a standard normal random variable, $E[|X|]$. For a standard normal distribution, the mean is $\mu=0$, the variance is $\sigma^2=1$, and the probability density function (PDF) is $f_X(x) = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}$.

The expectation is calculated by integrating $|x|$ against the PDF: $E[|X|] = \int_{-\infty }^{\infty } |x| \frac{1}{\sqrt{2\pi}}e^{-x^2/2} dx$.
Because the entire function inside the integral is even (symmetric about the y-axis), we can simplify the calculation by integrating over the positive half and multiplying by two: $E[|X|] = 2 \int_{0}^{\infty } x \frac{1}{\sqrt{2\pi}}e^{-x^2/2} dx$.

We can solve this integral using a substitution. Let $u = x^2/2$, which means $du = x dx$. The integral becomes:
E[|X|] = \frac{2}{\sqrt{2\pi}} \int_{0}^{\infty } e^{-u} du = \frac{2}{\sqrt{2\pi}} \[-e^{-u}]_0^{∞}$.

Evaluating the definite integral gives $\frac{2}{\sqrt{2\pi}} \left( -e^{-\infty } - (-e^0) \right) = \frac{2}{\sqrt{2\pi}}(0 - (-1)) = \frac{2}{\sqrt{2\pi}} = \sqrt{\frac{2}{\pi}}$.
This exact value is approximately $0.7979$.

This is a second-order linear homogeneous differential equation. We can solve it by examining its characteristic equation, which is obtained by letting $x = e^{mt}$:
$m^2 + 2m + 1 = 0$
This quadratic equation factors neatly into a perfect square:
$(m+1)^2 = 0$
This gives us a repeated real root, $m_1 = m_2 = -1$. For a repeated root $m$, the general form of the solution is $x(t) = c_1e^{mt} + c_2te^{mt}$. Substituting our value of $m=-1$ and using the constants $a$ and $b$, we arrive at the general solution:
$x(t) = ae^{-t} + bte^{-t}$

To find the directional derivative, we calculate the dot product of the gradient of the function and the unit direction vector, expressed as $D_{\vec{u}}f = \nabla f \cdot \vec{u}$.

First, we find the gradient of $f(x,y)=\frac{1}{\sqrt{2}}(x^2y+xy^2)$. The gradient vector is $\nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle = \langle \frac{2xy+y^2}{\sqrt{2}}, \frac{x^{2+}2xy}{\sqrt{2}} \rangle$.
At the point $(1,1)$, the gradient evaluates to $\nabla f(1,1) = \langle \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}} \rangle$.

The direction is given by a unit vector at an angle of $\frac{\pi}{4}$ with the y-axis. This vector is $\vec{u} = \langle \cos(\frac{\pi}{4}), \sin(\frac{\pi}{4}) \rangle = \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle$.

Finally, we take the dot product of the gradient and the direction vector:
$D_{\vec{u}}f(1,1) = \langle \frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}} \rangle \cdot \langle \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \rangle = (\frac{3}{\sqrt{2}})(\frac{1}{\sqrt{2}}) + (\frac{3}{\sqrt{2}})(\frac{1}{\sqrt{2}}) = \frac{3}{2} + \frac{3}{2} = 3$.

The circuit diagram shows a two-port network. The output side features a dependent source, identified by the diamond symbol. The arrow inside the diamond indicates it is a current source. The value of this source is given as $A_i I_1$. This expression shows that the output current is dependent on, or controlled by, the input current $I_1$. Therefore, because an input current ($I_1$) controls an output current source ($A_i I_1$), the circuit represents a current controlled current source (CCCS). The resistor $R$ in parallel represents the output impedance of the source.

To find the current through $R_2$, we can simplify the circuit using source transformation. The current source on the left in parallel with $R_1$ becomes a $10 \text{ mA} \times 2 \text{ k}\Omega = 20 \text{ V}$ voltage source in series with $R_1$. Likewise, the current source on the right in parallel with $R_3$ becomes a $2 \text{ mA} \times 4 \text{ k}\Omega = 8 \text{ V}$ source in series with $R_3$.

This results in a single series loop. Let the current flowing through this loop be $I$. We can now apply Kirchhoff's Voltage Law (KVL). Summing the voltages around the loop, we get:
$20 \text{ V} + 8 \text{ V} = I(R_1 + R_2 + R_3 + R_4)$
Plugging in the resistance values yields $28 = I(2+1+4+3) \text{ k}\Omega$.
This simplifies to $28 = 10I$, which gives a current $I$ of $2.8 \text{ mA}$. Since this is a series circuit, this is the current flowing through resistor $R_2$.

To generate an electron-hole pair, an incoming photon must have an energy ($E_p$) at least equal to the material's band gap energy ($E_g$). This fundamental condition is written as $E_p \ge E_g$.

Since a photon's energy is inversely related to its wavelength ($\lambda$) by the formula $E_p = \frac{hc}{\lambda}$, the energy requirement sets a maximum limit on the wavelength. Any wavelength longer than this cutoff will not have enough energy.

This cutoff wavelength ($\lambda_c$) is found by setting the photon energy equal to the band gap: $\lambda_c = \frac{hc}{E_g}$.

Using the common and useful approximation $\lambda_c (\mu m) \approx \frac{1.24}{E_g (eV)}$, we can calculate the value for GaAs:
$\lambda_c \le \frac{1.24}{1.42 \text{ eV}} \approx 0.87 \mu m$

Therefore, the incident radiation must have a wavelength less than or equal to $0.87 \mu m$. The given range falls entirely within this requirement.

The intrinsic resistivity ($\rho_i$) of a semiconductor is the reciprocal of its intrinsic conductivity ($\sigma_i$). The conductivity is given by $\sigma_i = n_i e (\mu_n + \mu_p)$, where $n_i$ is the intrinsic carrier concentration. Therefore, $\rho_i = \frac{1}{n_i e (\mu_n + \mu_p)}$. The intrinsic carrier concentration $n_i$ is strongly dependent on temperature (T) and the band gap energy ($E_g$) as $n_i \propto \exp(-\frac{E_g}{2kT})$, where $k$ is the Boltzmann constant.

Substituting this into the resistivity equation, we get $\rho_i \propto \frac{1}{\exp(-E_g/2kT)} = \exp(\frac{E_g}{2kT})$. Taking the natural logarithm gives $ln(\rho_i) \approx \frac{E_g}{2k} (\frac{1}{T}) + C$, where C is a constant that includes the weaker temperature dependencies of mobility. This equation is in the form of a straight line, $y = mx + c$, for a plot of $ln(\rho_i)$ versus $1/T$. The slope of this line is $m = \frac{E_g}{2k}$. Since $k$ is a known constant, the slope can be directly used to estimate the band gap energy, $E_g$.

For intrinsic excitation to occur, the energy of an incoming photon must be at least equal to the semiconductor's bandgap energy, $E_g$. The cut-off wavelength, $\lambda_c$, represents the longest possible wavelength (and thus the minimum energy) that can cause this excitation.

This threshold condition is described by the equation $E_g = \frac{hc}{\lambda_c}$, where $h$ is Planck's constant and $c$ is the speed of light. A highly useful approximation for this relationship is:
$\lambda_c (\mu\text{m}) = \frac{1.24}{E_g (\text{eV})}$
Using the given bandgap of $E_g = 1.1$ eV, we can calculate the cut-off wavelength:
$\lambda_c = \frac{1.24}{1.1} \approx 1.127 \text{ } \mu\text{m}$

An unbypassed emitter resistor, $R_E$, introduces negative feedback into the amplifier circuit. The total input impedance seen from the base is given by the expression $R_{\in} = \beta r_e + (1+\beta)R_E$. This formula clearly shows that the presence of $R_E$ adds a large term, $(1+\beta)R_E$, which significantly increases the overall input impedance. The voltage gain, $A_v$, is inversely related to this impedance. Since $R_E$ makes the input impedance larger, it simultaneously reduces the overall voltage gain of the amplifier.

For the output voltage $V_o$ to equal the input voltage $V_i$, no current can flow through the resistor $R$. This happens only when both diodes, D1 and D2, are in their non-conducting (OFF) state.

Let's determine the conditions for each diode to be OFF. A diode remains OFF as long as it is not forward-biased by at least its forward voltage drop of $0.7V$.

1. Diode D2: This diode turns ON if its anode voltage, $V_o$, becomes $0.7V$ greater than its cathode voltage, which is held at $2V$. So, D2 turns ON if $V_o > 2V + 0.7V = 2.7V$. To keep D2 OFF, we require $V_o \le 2.7V$.

2. Diode D1: This diode is set up to conduct when $V_o$ becomes sufficiently negative. It will turn ON when the voltage on its anode side exceeds the voltage on its cathode side by $0.7V$. Here, this occurs when $-1V > V_o + 0.7V$, or $V_o < -1.7V$. To keep D1 OFF, we require $V_o \ge -1.7V$.

For both diodes to be OFF, the output voltage must satisfy both conditions: $-1.7V \le V_o \le 2.7V$. Since this is the region where $V_o = V_i$, the range for the input voltage is $-1.7V < V_i < 2.7V$.

This circuit is a non-inverting Schmitt trigger with an RC timing network, which forms a monostable multivibrator. In its stable state, the capacitor $C_1$ is an open circuit, holding the non-inverting input $v_+$ at ground potential ($0V$). The inverting input $v_-$ is biased at $-2V$. Since $v_+ > v_-$, the op-amp's output $v_o$ is saturated at the positive supply, $+12V$.

A sufficiently large positive trigger pulse at $v_i$ raises $v_-$ above $v_+$, causing the output to switch to $-12V$. This is a temporary, quasi-stable state. During this time, capacitor $C_1$ charges through resistor $R_1$, causing the voltage $v_+$ to rise exponentially. The circuit returns to its stable state when $v_+$ reaches the reference voltage on the inverting input, $-2V$. This behavior of having one stable state and one triggered, timed, quasi-stable state defines a monostable multivibrator.

Let's consider the two effects of increasing the hold capacitor's value ($C$).

First, the acquisition time is the duration required to charge the capacitor to the input voltage. A larger capacitor needs more charge ($Q=CV$) to reach a given voltage. Assuming a fixed charging current, this means it takes a longer time to charge. Therefore, the acquisition time is directly proportional to the capacitance.

Second, the droop rate is the speed at which the stored voltage decays during the hold phase due to small leakage currents ($I_{leak}$). From the fundamental capacitor relationship $I = C \frac{dV}{dt}$, the rate of voltage change is $\frac{dV}{dt} = \frac{I_{leak}}{C}$. This shows that the droop rate is inversely proportional to capacitance, so a larger capacitor will have a slower voltage droop.

To find the expression for Y when C=0, we first analyze the circuit's full Boolean expression and then substitute the given condition.

The overall output is represented by $Y = \overline{\bar{C} \cdot (\overline{A+B}+C) + AB}$. By substituting $C=0$ and its complement $\bar{C}=1$ at the beginning, the expression becomes much simpler:
$Y = \overline{1 \cdot (\overline{A+B}+0) + AB} = \overline{(\overline{A+B}) + AB}$.

Now, we can simplify this result using De Morgan's laws. Applying the law to the main expression gives $Y = \overline{(\overline{A+B})} \cdot \overline{(AB)}$. The double negation cancels out, and applying De Morgan's law a second time yields $Y = (A+B)(\bar{A}+\bar{B})$.

Expanding this product gives $Y = A\bar{A} + A\bar{B} + \bar{A}B + B\bar{B}$. Since a variable ANDed with its complement is zero (e.g., $A\bar{A}=0$), the expression reduces to the final answer: $Y = A\bar{B} + \bar{A}B$.

This circuit implements a three-input CMOS NAND gate. The inputs to the gate are $A$, $B$, and $\bar{C}$. The function of a NAND gate is to perform an AND operation on its inputs and then invert the result.

So, the output $Y$ is the logical NAND of $A$, $B$, and $\bar{C}$. This can be written as the Boolean expression $Y = \overline{A \cdot B \cdot \bar{C}}$.

By applying De Morgan's theorem, we can transform this expression. The theorem allows us to distribute the negation, changing the operation from AND to OR: $Y = \bar{A} + \bar{B} + \overline{(\bar{C})}$.

Finally, simplifying the double negation on the $C$ term ($\overline{\bar{C}} = C$) gives the final output expression: $Y = \bar{A} + \bar{B} + C$.

To find the value of $A$, we can directly calculate the Fourier transform of the signal $x[n] = (\frac{2}{3})^{n} u[n+3]$ using its definition. The presence of the unit step $u[n+3]$ means the signal is non-zero only for $n \ge -3$.

The Discrete-Time Fourier Transform (DTFT) is given by:
$X(e^{j\omega}) = \sum_{n=-3}^{\infty } \left(\frac{2}{3}\right)^{n} e^{-j\omega n}$
To evaluate this sum, we can use a change of variables to start the index at zero. Let $m = n+3$, which means $n = m-3$.
$X(e^{j\omega}) = \sum_{m=0}^{\infty } \left(\frac{2}{3}\right)^{m-3} e^{-j\omega(m-3)} = \left(\frac{2}{3}\right)^{-3} e^{j3\omega} \sum_{m=0}^{\infty } \left(\frac{2}{3} e^{-j\omega}\right)^m$
The summation is an infinite geometric series, which converges to $\frac{1}{1 - \frac{2}{3}e^{-j\omega}}$. The constant factor $(\frac{2}{3})^{-3}$ simplifies to $(\frac{3}{2})^3 = 3.375$. Substituting $\omega = 2\pi f$, we get:
$X(f) = \frac{3.375 e^{j3(2\pi f)}}{1 - \frac{2}{3}e^{-j2\pi f}} = \frac{3.375 e^{j6\pi f}}{1 - \frac{2}{3}e^{-j2\pi f}}$
By comparing our result with the given form $\frac{Ae^{j6\pi f}}{1-(\frac{2}{3})e^{-j2 \pi f}}$, we can see that $A=3.375$.

The frequency response $H(f) = e^{-j4\pi f}$ describes a system that does not change the magnitude of the input signal but introduces a linear phase shift. Such a system causes a pure time delay.

To determine the impulse response $h(t)$, we can express the frequency response in terms of angular frequency, $\omega = 2\pi f$. Substituting $f = \omega/(2\pi)$ gives $H(\omega) = e^{-j4\pi (\omega/2\pi)} = e^{-j2\omega}$.

From Fourier transform properties, we know that a phase shift of $e^{-j\omega t_0}$ in the frequency domain corresponds to a time-domain impulse response of $h(t) = \delta(t-t_0)$. Comparing this to our $H(\omega)$, we see the time delay is $t_0=2$, making the impulse response $h(t) = \delta(t-2)$.

The output $y(t)$ is the convolution of the input $x(t)$ with the impulse response $h(t)$. Convolving any signal with a shifted delta function results in a shifted version of that signal:
$y(t) = x(t) * h(t) = x(t) * \delta(t-2) = x(t-2)$.

We are asked to find the total sum of the output sequence $y[n]$, which results from the self-convolution of $x[n]=0.5^{n}u[n]$. A powerful property of linear time-invariant systems states that the sum of the output sequence is equal to the product of the sums of the input sequence and the impulse response. Here, $x[n]$ acts as both.

First, let's find the sum of the sequence $x[n]$:
$S_x = \sum_{n=-\infty }^{\infty } x[n] = \sum_{n=0}^{\infty } (0.5)^n$
This is a standard geometric series with a ratio of $0.5$, which sums to $\frac{1}{1-0.5} = 2$.

Since $y[n] = x[n] * x[n]$, the sum of the output sequence $y[n]$ is the square of the sum of the sequence $x[n]$:
$\sum_{n=-\infty }^{\infty } y[n] = \left( \sum_{n=-\infty }^{\infty } x[n] \right) \left( \sum_{n=-\infty }^{\infty } x[n] \right) = (S_x)^2$
Therefore, the final sum is $2^2 = 4$.

An all-pole system of $4^{th}$ order is characterized by a transfer function with four poles and no zeros. The high-frequency slope of a system's Bode magnitude plot is dictated by the number of poles and zeros. A fundamental principle of Bode plots is that each pole contributes a slope of $-20$ dB/decade to the high-frequency asymptote. Since this system has four poles, their effects are cumulative. Therefore, the total slope at high frequencies is the sum of the individual contributions, yielding $4 \times (-20 \text{ dB/decade}) = -80 \text{ dB/decade}$.

To determine the natural frequency, we first need to find the overall closed-loop transfer function of the system. This is a unity negative feedback system, so the transfer function $T(s)$ is given by the formula $T(s) = \frac{G(s)}{1 + G(s)}$.

Substituting the given forward path gain, $G(s) = \frac{4}{s(s+4)}$, we get:
$T(s) = \frac{\frac{4}{s(s+4)}}{1 + \frac{4}{s(s+4)}} = \frac{4}{s(s+4) + 4} = \frac{4}{s^2 + 4s + 4}$

Next, we compare the denominator of our result to the standard form of a second-order system's characteristic equation: $s^2 + 2\zeta\omega_n s + \omega_n^2$.

By matching the terms, we can see that $\omega_n^2 = 4$.
Taking the square root gives the undamped natural frequency: $\omega_n = 2$ rad/s.

This problem describes the counting of events that occur independently and at a constant average rate over a fixed interval. This is the definition of a Poisson process. The key characteristics are:

1. Independence: The arrival of one call does not influence the arrival of another.
2. Counting: We are interested in the total number of calls ($k$), which is a discrete count.
3. Fixed Interval: The count is measured over a fixed duration of time ($t$).

The probability distribution for the number of events $k$ in a fixed interval for a Poisson process is the Poisson distribution. Its probability mass function is given by $P(X=k) = \frac{\mu^k e^{-\mu}}{k!}$, where $\mu = \lambda t$ is the average number of calls in the interval, with $\lambda$ being the average arrival rate.

In a DSB-FC AM system, the ratio of the total sideband power ($P_{SB}$) to the carrier power ($P_C$) is determined by the modulation index, $\mu$. The specific relationship is:

$\frac{P_{SB}}{P_C} = \frac{\mu^2}{2}$

This formula shows that the power ratio is directly proportional to the square of the modulation index ($R \propto \mu^2$).

Let's say our initial modulation index is $\mu_1$. If we double it, the new index becomes $\mu_2 = 2\mu_1$. To find the new power ratio, we can compare the two scenarios:

$\frac{\text{New Ratio}}{\text{Old Ratio}} = \frac{\mu_2^2 / 2}{\mu_1^2 / 2} = \frac{(2\mu_1)^2}{\mu_1^2} = \frac{4\mu_1^2}{\mu_1^2} = 4$

Thus, doubling the modulation index increases the ratio of sideband power to carrier power by a factor of 4.

In the far-field region of a radiating antenna, the electric field strength ($E$) is inversely proportional to the distance ($r$). Since the distance is doubling from 1 km to 2 km, the electric field will be halved. The new electric field at 2 km is therefore $E_{2\text{km}} = \frac{12 \text{ mV/m}}{2} = 6 \text{ mV/m}$.

The average power density ($P_{\text{avg}}$) is related to the electric field and the intrinsic impedance of free space ($\eta$) by the formula $P_{\text{avg}} = \frac{E^2}{2\eta}$.

Plugging in the known values:
$P_{\text{avg}} = \frac{(6 \times 10^{-3} \text{ V/m})^2}{2 \times 120\pi \, \Omega} = \frac{36 \times 10^{-6}}{240\pi} \approx 47.7 \times 10^{-9} \text{ W/m}^2$

This result is equivalent to $47.7 \text{ nW/m}^2$.

Let's break down the radiation patterns of these different sources.

An ideal point electromagnetic source is a theoretical concept that radiates energy equally in all directions. This uniform, non-directional radiation is called isotropic.

A dish antenna uses a parabolic reflector to concentrate energy into a tight, narrow beam. This characteristic makes it highly directional, ideal for focusing signals over long distances.

A Yagi-Uda antenna, like those once common for TV reception, is designed to radiate strongly in a single direction along the axis of the antenna elements. This is known as an end-fire radiation pattern.

To solve this second-order linear homogeneous differential equation, we begin by examining its characteristic equation: $m^2 + 4m + 4 = 0$. This equation factors into $(m+2)^2 = 0$, which yields a repeated real root of $m = -2$. For repeated roots, the general solution takes the form $y(x) = (C_1 + C_2x)e^{-2x}$.

Next, we apply the initial conditions to determine the constants $C_1$ and $C_2$. Using $y(0)=1$, we get $1 = (C_1 + 0)e^0$, so $C_1 = 1$. The derivative of our solution is $y'(x) = -2(C_1 + C_2x)e^{-2x} + C_2e^{-2x}$. Using $y'(0)=1$, we have $1 = -2C_1 + C_2$. Substituting $C_1=1$, we find $1 = -2(1) + C_2$, which gives $C_2=3$.

Our particular solution is therefore $y(x) = (1+3x)e^{-2x}$. Finally, we evaluate this function at $x=1$:
$y(1) = (1 + 3(1))e^{-2(1)} = 4e^{-2} \approx 4(0.1353) \approx 0.541$.

This is a conditional probability problem. We need to find the probability that the loss occurred at the second post-office, out of the total probability that a loss occurred at all.

First, let's find the probability that the parcel is lost by the second post-office. This requires it to successfully pass the first (probability $1 - \frac{1}{5} = \frac{4}{5}$) and then be lost by the second. This combined probability is $\frac{4}{5} \times \frac{1}{5} = \frac{4}{25}$.

Next, we calculate the total probability that the parcel is lost anywhere. This is the sum of it being lost at the first office (\frac{1}{5}$) and being lost at the second (\frac{4}{25}$). The total probability of loss is $\frac{1}{5} + \frac{4}{25} = \frac{9}{25}$.

Finally, the conditional probability is the ratio of these two values: $\frac{\text{P(Lost at 2nd)}}{\text{P(Lost at all)}} = \frac{4/25}{9/25} = \frac{4}{9}$, which is approximately $0.444$.

To find the Laplace transform of $g(t) = t \cdot f(t)$, we use the frequency differentiation property. This property states that if $F(s)$ is the Laplace transform of $f(t)$, then the transform of $t \cdot f(t)$ is given by $-\frac{d}{ds}F(s)$.

We are given $F(s) = \frac{1}{s^{2+}s+1}$. Let's first compute its derivative with respect to $s$. Using the chain rule on $F(s) = (s^{2+}s+1)^{-1}$:
$\frac{dF(s)}{ds} = -1(s^{2+}s+1)^{-2} \cdot \frac{d}{ds}(s^{2+}s+1) = \frac{-(2s+1)}{(s^{2+}s+1)^2}$
Finally, we apply the property by taking the negative of this derivative:
$L\{t \cdot f(t)\} = -\frac{dF(s)}{ds} = - \left( \frac{-(2s+1)}{(s^{2+}s+1)^2} \right) = \frac{2s+1}{(s^{2+}s+1)^2}$

Let the sides of the right triangle be $a$ and $b$, with hypotenuse $h$. We are given that the sum of the hypotenuse and a side is constant, which we can write as $h+a=K$. Our goal is to maximize the area, $A = \frac{1}{2}ab$.

First, we express the area as a function of a single variable, $a$. From the Pythagorean theorem, $b = \sqrt{h^2 - a^2}$. Substituting $h=K-a$ from our constraint, we get $b = \sqrt{(K-a)^2 - a^2} = \sqrt{K^{2-}2aK}$. This gives the area function $A(a) = \frac{1}{2}a\sqrt{K^{2-}2aK}$.

To find the maximum area, we take the derivative of $A(a)$ with respect to $a$ and set it to zero. Solving $\frac{dA}{da}=0$ yields the condition $K=3a$.

Now we can relate the sides $h$ and $a$. Substituting $K=3a$ back into the original constraint $h+a=K$ gives $h+a=3a$, which simplifies to $h=2a$.

Finally, the angle $\theta$ between side $a$ and the hypotenuse $h$ is defined by $\cos(\theta) = \frac{a}{h}$. Using our finding that $h=2a$, we get $\cos(\theta) = \frac{a}{2a} = \frac{1}{2}$, which means the angle for maximum area is $\theta = 60^{\circ}$.

To solve for the frequency $\omega$, we first determine the circuit's transfer function, $H(j\omega) = Y/V_{\in}$, using phasor analysis. The input voltage source $v_{\in}(t) = \sin(\omega t)$ is represented by the phasor $V_{\in} = 1\angle0^\circ$.

Let $V_A$ be the phasor voltage at the node connecting the resistor and the capacitors. By applying Kirchhoff's Current Law (KCL) at this node, we sum the currents leaving it:
$\frac{V_A - V_{\in}}{R} + \frac{V_A}{1/(j\omega C)} + \frac{V_A}{2/(j\omega C)} = 0$
With $V_{\in}=1$, we can rearrange and solve for $V_A$: $V_A\left(\frac{1}{R} + j\omega C + \frac{j\omega C}{2}\right) = \frac{1}{R}$, which gives $V_A = \frac{2}{2+3j\omega RC}$.

The output voltage $Y$ is taken across one of the two identical series capacitors. By the voltage divider rule, $Y$ is half of $V_A$, so $Y = \frac{V_A}{2} = \frac{1}{2+3j\omega RC}$.

The amplitude $|A(\omega)|$ is the magnitude of this complex transfer function: $|A(\omega)| = \left| \frac{1}{2+3j\omega RC} \right| = \frac{1}{\sqrt{2^{2+}(3\omega RC)^2}}$.
Given that $|A(\omega)| = 0.25 = 1/4$, we can set up the equation:
$\frac{1}{4} = \frac{1}{\sqrt{4+9\omega^2(RC)^2}}$
Squaring both sides gives $16 = 4+9\omega^2(RC)^2$. Solving for $\omega$ yields $\omega = \frac{2}{\sqrt{3}RC}$.

As time approaches infinity, the circuit settles into a DC steady state. In this state, the inductor behaves as a short circuit (a wire), and the current source provides a constant 10 A. The voltage $v_o$ is thus the voltage of the entire top wire relative to the bottom wire.

Applying Kirchhoff's Current Law (KCL) at the top node yields:
$-10 + \frac{v_o - 2i_x}{5} + i_x = 0$
From Ohm's law on the right-hand resistor, we have the relationship $v_o = 5i_x$. Substituting this into the KCL equation allows us to solve for $i_x$:
$-10 + \frac{5i_x - 2i_x}{5} + i_x = 0 \implies \frac{8}{5}i_x = 10 \implies i_x = \frac{50}{8} \text{ A}$
Therefore, the steady-state voltage is $v_o = 5i_x = 5 \left(\frac{50}{8}\right) = 31.25$ V.

The key to solving this problem is to cleverly decompose the circuit. We can think of the initial $2R$ resistor as two resistors, $R+R$, connected in series. This reveals that the entire circuit is equivalent to a single resistor $R$ in series with a standard infinite ladder network, where every resistor is $R$.

Let the resistance of this standard infinite ladder be $R_{ladder}$. The total resistance is then $R_e = R + R_{ladder}$.

For the infinite ladder, adding another repeating section doesn't change its total resistance. This allows us to write a self-referential equation: $R_{ladder} = R + (R \parallel R_{ladder}) = R + \frac{R \cdot R_{ladder}}{R+R_{ladder}}$. This simplifies to the quadratic equation $R_{ladder}^2 - R \cdot R_{ladder} - R^2 = 0$, whose positive solution is $R_{ladder} = R\left(\frac{1+\sqrt{5}}{2}\right)$.