GATE ECE 2019

A function is analytic over the entire complex plane if it is "entire," meaning it has no singularities or points where it is not defined. We can test each option for such points.

The functions $\ln(z)$, $e^{1/z}$, and $\frac{1}{1-z}$ are not entire. The logarithm has a branch point at $z=0$. The function $e^{1/z}$ has an essential singularity at $z=0$. The rational function $\frac{1}{1-z}$ has a simple pole at $z=1$.

In contrast, $\cos(z)$ is a well-known entire function. Its definition, $\cos(z) = \frac{e^{iz} + e^{-iz}}{2}$, shows it is a combination of the entire function $e^z$. As a sum of entire functions, $\cos(z)$ is itself analytic everywhere.

We can determine the family of curves for each value of $n$ by solving the differential equation.

For $n=-1$, the equation becomes $\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{-1} = -\frac{y}{x}$. We can separate variables to get $\frac{1}{y}dy = -\frac{1}{x}dx$. Integrating both sides gives $\ln|y| = -\ln|x| + C$, which rearranges to $xy = k$. This is the equation for a family of hyperbolas.

For $n=1$, the equation is $\frac{dy}{dx} = -\frac{x}{y}$. Separating variables gives $y\,dy = -x\,dx$. Integrating yields $\frac{y^2}{2} = -\frac{x^2}{2} + C$. This simplifies to the familiar form $x^2 + y^2 = k$, which represents a family of circles.

The structure of $P(z)$ reveals two key symmetries for its zeros. First, because the original signal $h[n]$ is real-valued, the coefficients of $P(z)$ are also real. This means that any complex zeros must occur in conjugate pairs. If $z_0$ is a zero, then its conjugate $z_0^*$ must also be a zero.

Second, observe the definition $P(z) = H(z)H(1/z)$. If we evaluate this at $1/z$, we find $P(1/z) = H(1/z)H(z)$, which is identical to $P(z)$. This property implies that if $z_0$ is a zero, then its reciprocal $1/z_0$ is also a zero.

Given the zero $z_1 = \frac{1}{2}+\frac{1}{2}j$, we can find the other three zeros by applying these two rules.

1. The conjugate must be a zero: $z_2 = z_1^* = \frac{1}{2}-\frac{1}{2}j$.
2. The reciprocal must be a zero: $z_3 = \frac{1}{z_1} = \frac{1}{0.5+0.5j} = 1-j$.
3. The fourth zero is the conjugate of the third: $z_4 = z_3^* = 1+j$.

The complete set of four zeros is therefore $\{\frac{1}{2}+\frac{1}{2}j, \frac{1}{2}-\frac{1}{2}j, 1+j, 1-j\}$. This collection of points is correctly shown in plot D.

This problem is a direct application of the reciprocity theorem. This principle applies to any linear, bilateral network, like the resistive circuit shown. The theorem states that the ratio of the response (the measured current) to the excitation (the applied voltage) remains the same if we interchange their positions.

In the first case, the ratio is $\frac{\text{response}}{\text{excitation}} = \frac{1 \text{ A}}{5 \text{ V}}$.
In the second case, the positions are swapped, and the ratio is $\frac{I}{5 \text{ V}}$.

According to the reciprocity theorem, these two ratios must be equal.
$\frac{I}{5 \text{ V}} = \frac{1 \text{ A}}{5 \text{ V}}$
Solving for the unknown current $I$, we find that $I = 1 \text{ A}$.

The total response of the system to a unit step input, $u(t)$, is given by the inverse Laplace transform of $Y(s) = \frac{G(s)}{s}$.
Plugging in the given transfer function, we get:
$Y(s) = \frac{3-s}{s(s+1)(s+3)}$
To find the time-domain response $y(t)$, we use partial fraction expansion:
$Y(s) = \frac{A}{s} + \frac{B}{s+1} + \frac{C}{s+3}$
The total response $y(t)$ is the sum of the forced response and the natural response. The forced response is determined by the input signal's pole (here, the pole at $s=0$ from the step input), while the natural response comes from the system's own poles ($s=-1, s=-3$).

The forced response is the time-domain equivalent of the $\frac{A}{s}$ term. We can find the coefficient $A$ using the cover-up method:
$A = \left. sY(s) \right|_{s=0} = \left. \frac{3-s}{(s+1)(s+3)} \right|_{s=0} = \frac{3}{(1)(3)} = 1$
Thus, the forced response in the s-domain is $\frac{1}{s}$, and its inverse Laplace transform is $u(t)$. The other terms, associated with $B$ and $C$, form the transient natural response which decays to zero.

The number of poles and zeros can be determined by analyzing the slope changes in the Bode gain plot. Each pole contributes a $-20$ dB/decade change to the slope, while each zero contributes a $+20$ dB/decade change. By tracking the slope changes at each corner frequency, we can count the poles and zeros.

The slope changes by $-20$ at $10^1$ Hz (1 pole), $-40$ at $10^2$ Hz (2 poles), $+20$ at $10^3$ Hz (1 zero), $+40$ at $10^4$ Hz (2 zeros), $-40$ at $10^5$ Hz (2 poles), and $-20$ at $10^6$ Hz (1 pole). Summing these contributions gives a total of $N_p = 1+2+2+1=6$ poles and $N_z = 1+2=3$ zeros within the given frequency range.

The condition $\Phi_{ms}=0V$ and the absence of oxide charges indicate an ideal MOS capacitor. For such a device, the flat-band condition, where there is no band bending in the semiconductor, occurs precisely at a gate voltage of $V_G = 0V$. This corresponds to point Q.

Applying a large positive voltage ($V_G > 0$) repels majority carriers and forms an inversion layer, leading to the minimum high-frequency capacitance at point R. Conversely, a negative gate voltage ($V_G < 0$) attracts majority carriers to the semiconductor surface, creating an accumulation layer. This condition yields the maximum capacitance, equal to the oxide capacitance ($C_{ox}$), as seen at point P.

Therefore, the flat-band, inversion, and accumulation regions are represented by points Q, R, and P, respectively.

The electric field $\vec{E}$ from an infinite line with linear charge density $Q$ is purely radial, with magnitude $E = \frac{Q}{2\pi\varepsilon_0 \rho}$. Because the field is radial, it is parallel to the flat top, bottom, and side surfaces of the quarter-cylinder, so only the curved surface contributes to the flux.

On this curved surface, the electric field vector $\vec{E}$ is always parallel to the area element vector $d\vec{a}$. The flux integral is therefore:
$\Phi_E = \int E \cdot da = \int_0^H \int_0^{\pi/2} \left( \frac{Q}{2\pi\varepsilon_0 \rho} \right) (\rho \, d\phi \, dz)$
The radius $\rho$ cancels, leaving a simple integral over height $z$ and angle $\phi$.
$\Phi_E = \frac{Q}{2\pi\varepsilon_0} \int_0^H dz \int_0^{\pi/2} d\phi = \frac{Q}{2\pi\varepsilon_0} (H) (\frac{\pi}{2}) = \frac{HQ}{4\varepsilon_0}$

This question asks us to correctly assemble Maxwell's equations, which form the foundation of classical electromagnetism. Let's match each term in List I with its corresponding source term in List II.

1. Gauss's law for electricity, $\nabla \cdot \vec{D} = \rho$, states that the divergence of the electric displacement field is equal to the electric charge density. This matches 1 with Q.
2. Faraday's law of induction, $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$, shows that a time-varying magnetic field creates a curling electric field. This matches 2 with R.
3. Gauss's law for magnetism, $\nabla \cdot \vec{B} = 0$, reflects the fact that there are no magnetic monopoles, so the magnetic field is always divergenceless. This matches 3 with P.
4. The Ampere-Maxwell law, $\nabla \times \vec{H} = \vec{J} + \frac{\partial \vec{D}}{\partial t}$, shows that a curling magnetic field is produced by both conduction current density ($\vec{J}$) and displacement current. This matches 4 with S.

Increasing the width of the pMOS transistor makes it electrically stronger than the nMOS. This means the pull-up network is now more powerful than the pull-down network. To make the inverter switch, the input voltage must rise to a higher level to enable the nMOS to overcome the stronger pMOS. This shifts the inverter's switching threshold ($V_M$) to a higher voltage, closer to $V_{DD}$.

The LOW noise margin is approximately $NM_L \approx V_M$, so as $V_M$ increases, $NM_L$ also increases. The HIGH noise margin is approximately $NM_H \approx V_{DD} - V_M$. Since $V_M$ is now a larger value, the difference decreases, causing $NM_H$ to decrease.

Let's analyze the circuit's behavior based on the enable signal, EN.

First, when EN=0, the top NAND gate's output is \overline{D \cdot 0} = 1. A high signal on the PMOS gate turns it OFF. The bottom NOR gate's inputs are D and \overline{EN}=1, so its output is \overline{D+1} = 0. A low signal on the NMOS gate turns it OFF. Since both transistors are off, the output F is disconnected, creating a high-impedance (Hi-Z) state.

Next, when EN=1, the NAND gate's output becomes \overline{D \cdot 1} = \bar{D}. The NOR gate's inputs are now D and \overline{EN}=0, making its output \overline{D+0} = \bar{D}. If D=1, \bar{D}=0 turns ON the PMOS, setting F=1. If D=0, \bar{D}=1 turns ON the NMOS, setting F=0. Therefore, when enabled, the output F simply follows the input D.

To understand this circuit, let's analyze its behavior for all possible input combinations of $A$ and $B$.

The output $F$ will be HIGH (1) only when both inputs are identical. If $A=0$ and $B=0$, the two series PMOS transistors (top) turn on, connecting $F$ to the power supply Vdd. If $A=1$ and $B=1$, both NMOS transistors (bottom) turn on, passing the HIGH logic level from the inputs to the output $F$.

Conversely, if the inputs are different ($A=0, B=1$ or $A=1, B=0$), one of the NMOS transistors will connect the output $F$ to the input that is LOW (0), pulling $F$ down to ground. This behavior, where the output is HIGH if and only if the inputs are equal, is the definition of an XNOR gate.

Let's evaluate the integral by first simplifying the expression inside. We can rewrite the integrand as:
$\left(z+\frac{1}{z}\right)^2 = \frac{(z^{2+}1)^2}{z^2}$.

The integral now has the form $I = \frac{1}{2 \pi j} \oint_{|z|=1} \frac{(z^{2+}1)^2}{z^2} dz$. This perfectly matches Cauchy's Integral Formula for derivatives, which states that $\frac{1}{2 \pi j} \oint_C \frac{f(z)}{(z-z_0)^{n+1}} dz = \frac{f^{(n)}(z_0)}{n!}$.

For our integral, we can identify $f(z) = (z^{2+}1)^2$, the pole $z_0 = 0$ (which is inside the unit circle), and an exponent of $2$, meaning $n+1=2$ or $n=1$.

Therefore, the value of the integral is $\frac{f'(0)}{1!}$. The derivative is $f'(z) = \frac{d}{dz}(z^{2+}1)^2 = 2(z^{2+}1)(2z) = 4z(z^{2+}1)$. Evaluating this at $z=0$ gives $f'(0) = 4(0)(0^{2+}1) = 0$, which is our final answer.

The given matrix is an upper triangular matrix, as all entries below the main diagonal are zero.

A fundamental property of triangular matrices is that their eigenvalues are simply the entries on the main diagonal. For this matrix, the diagonal entries are $2, 1, 3,$ and $2$.

The set of eigenvalues is $\{2, 1, 3, 2\}$. To find the number of distinct eigenvalues, we identify the unique values in this set, which are $\{1, 2, 3\}$.

By counting the elements in this set, we find that there are 3 distinct eigenvalues.

The key to solving this is the linearity of expectation. This property allows us to rewrite the given equations. The expression $E[2X+Y]=0$ becomes $2E[X]+E[Y]=0$, and similarly, $E[X+2Y]=33$ becomes $E[X]+2E[Y]=33$.

We now have a system of two linear equations. Instead of solving for $E[X]$ and $E[Y]$ individually, notice what happens when we add the two equations together:
$(2E[X]+E[Y]) + (E[X]+2E[Y]) = 0 + 33$

Combining like terms gives us $3E[X]+3E[Y]=33$.
Finally, dividing the entire equation by 3 yields the desired sum: $E[X]+E[Y]=11$.

The integral $\int \frac{\sin x}{x} dx$ cannot be solved using elementary functions, which indicates we should change the order of integration. The original bounds are $0 \le y \le \pi$ and $y \le x \le \pi$. Sketching this triangular region allows us to reverse the order, giving new bounds of $0 \le x \le \pi$ and $0 \le y \le x$. The integral becomes:
$I = \int_{0}^{\pi} \int_{0}^{x} \frac{\sin x}{x} dy \, dx$
Now, we can perform the inner integration with respect to $y$:
$I = \int_{0}^{\pi} \left[ y \frac{\sin x}{x} \right]_{y=0}^{y=x} dx = \int_{0}^{\pi} \left( x \frac{\sin x}{x} \right) dx = \int_{0}^{\pi} \sin x \, dx$
Evaluating this simplified integral is straightforward:
$\int_{0}^{\pi} \sin x \, dx = [-\cos x]_{0}^{\pi} = (-\cos \pi) - (-\cos 0) = 1 - (-1) = 2$

We are asked to find the conditional probability $P(Z > 2 | Z > 1)$. The definition of conditional probability states $P(A|B) = \frac{P(A \cap B)}{P(B)}$. Applying this, we get:
$P(Z > 2 | Z > 1) = \frac{P(Z > 2 \text{ and } Z > 1)}{P(Z > 1)}$
If a variable is greater than 2, it is automatically greater than 1, so the intersection simplifies to just $P(Z > 2)$. The expression becomes $\frac{P(Z > 2)}{P(Z > 1)}$.

The probability $P(Z > x)$ is the complement of the given CDF, $F_Z(x)$. Thus, $P(Z > x) = 1 - P(Z \leq x) = 1 - (1-e^{-x}) = e^{-x}$.

Substituting this into our fraction gives:
$\frac{e^{-2}}{e^{-1}} = e^{-2 - (-1)} = e^{-1}$
This result demonstrates the memoryless property of the exponential distribution. Calculating the value, $e^{-1} \approx 0.3678$, which rounds to $0.37$.

To find the fundamental period of a signal that is a sum of sinusoids, we must find the lowest frequency that is a common divisor of all the individual frequencies. First, let's identify the angular frequencies ($\omega$) of the periodic components in the signal $f(t)$:
$\omega_1 = \pi$, $\omega_2 = \frac{2\pi}{3}$, and $\omega_3 = \frac{\pi}{2}$.

The fundamental angular frequency, $\omega_0$, is the greatest common divisor (GCD) of these frequencies.
$\omega_0 = \text{GCD}\left(\pi, \frac{2\pi}{3}, \frac{\pi}{2}\right) = \pi \cdot \text{GCD}\left(1, \frac{2}{3}, \frac{1}{2}\right) = \pi \cdot \frac{1}{6} = \frac{\pi}{6}$ rad/s.

The fundamental time period, $T_0$, is then calculated from the fundamental angular frequency.
$T_0 = \frac{2\pi}{\omega_0} = \frac{2\pi}{\pi/6} = 12$ seconds.
The constant term (1) and the phase shift ($\pi/4$) do not affect the signal's periodicity.

The instantaneous frequency $f_i(t)$ of a phase-modulated signal is determined by the carrier frequency and the rate of change (slope) of the message signal $m(t)$. The relationship is given by:
$f_i(t) = f_c + \frac{k}{2\pi}\frac{dm(t)}{dt}$

Substituting the given values $f_c = 50$ kHz and $k=10\pi$, our specific formula becomes:
$f_i(t) = 50 \text{ kHz} + 5\frac{dm(t)}{dt}$

From the graph of $m(t)$, we find the maximum and minimum slopes. The maximum slope is $\frac{1 - (-1)}{1 \text{ ms}} = 2000 \text{ s}^{-1}$, or 2 kHz. The minimum slope is $\frac{-1 - 1}{2 \text{ ms}} = -1000 \text{ s}^{-1}$, or -1 kHz.

Using these slopes, we find the maximum and minimum instantaneous frequencies:
$f_{max} = 50 \text{ kHz} + 5(2 \text{ kHz}) = 60 \text{ kHz}$
$f_{min} = 50 \text{ kHz} + 5(-1 \text{ kHz}) = 45 \text{ kHz}$

The ratio of the minimum to the maximum instantaneous frequency is therefore $\frac{45}{60} = 0.75$.

The radiation resistance ($R_{rad}$) of a small dipole antenna is proportional to the square of its length ($l$). This relationship stems from the formula $R_{rad} = 80 \pi^2 (\frac{l}{\lambda})^2$, so we can state that $R_{rad} \propto l^2$.

If the length $l$ is changed by 1%, the new length $l_{new}$ becomes $1.01$ times the original length $l_{old}$.

Due to the squared relationship, the new resistance $R_{new}$ will be related to the old resistance $R_{old}$ by the square of this factor:
$\frac{R_{new}}{R_{old}} = \left(\frac{l_{new}}{l_{old}}\right)^2 = (1.01)^2 = 1.0201$.

To find the percentage change in resistance, we calculate $(1.0201 - 1) \times 100\%$, which results in a $2.01\%$ increase.

Let's analyze the circuit's behavior for each half of the input square wave's period.

First, when $V_s = +8$ V, the positive voltage at the diode's cathode reverse-biases it. The diode acts as an open circuit, creating a simple voltage divider. The load voltage is thus $V_L = V_s \frac{R_2}{R_1 + R_2} = 8 \text{ V} \times \frac{50\Omega}{50\Omega + 50\Omega} = 4$ V.

Next, when $V_s = -10$ V, the diode becomes forward-biased. As an ideal diode, it acts like a short circuit, effectively bypassing resistor $R_1$. This connects the source directly across the load, so $V_L = V_s = -10$ V.

The output $V_L$ is 4 V for half the period and -10 V for the other half. The average value is the sum of these voltage levels weighted by their duration, divided by the total period:
$V_{L, \text{avg}} = \frac{(4 \text{ V} \times T/2) + (-10 \text{ V} \times T/2)}{T} = \frac{4 - 10}{2} = -3$ V.

This circuit is a synchronous counter. By tracing its state transitions for the outputs $(Q_2, Q_1)$, we can see it follows the sequence $00 \rightarrow 01 \rightarrow 10$ before repeating back to $00$. The counter cycles through three distinct states, which means it is a modulo-3 (MOD-3) counter.

A MOD-N counter functions as a frequency divider, where the output frequency is the input clock frequency divided by N. For this MOD-3 counter, the frequency at the output $Q_2$ is one-third of the clock frequency.

$f_{Q_2} = \frac{f_{\text{clk}}}{3} = \frac{12 \text{ kHz}}{3} = 4 \text{ kHz}$

We can solve this using the Mean Value Theorem (MVT). The MVT states that for any two points, there's a point in between where the instantaneous rate of change equals the average rate of change. Let's apply this to the interval from $-1$ to any other point $x$.

According to the MVT, there exists a number $c$ between $-1$ and $x$ such that:
$\frac{f(x) - f(-1)}{x - (-1)} = f'(c)$
Given that $f(-1)=0$, this simplifies to $\frac{f(x)}{x+1} = f'(c)$.

The original explanation's logic relies on the bound $|f'(x)| \leq 2$. This means $|f'(c)| \leq 2$.
Taking the absolute value of our MVT result gives $|\frac{f(x)}{x+1}| = |f'(c)|$.
Combining these, we get $|\frac{f(x)}{x+1}| \leq 2$, which means $|f(x)| \leq 2|x+1|$.
This inequality is equivalent to $-2|x+1| \leq f(x) \leq 2|x+1|$. The right-hand side of this compound inequality confirms that $f(x) \leq 2|x+1|$ must be true.

This problem is a classic application of Green's Theorem, which connects a line integral around a closed curve to a double integral over the region it encloses. The theorem states $\oint_C (Pdx + Qdy) = \iint_R (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) dA$.

From the given integral $\int_C (xdy - ydx)$, we identify $P = -y$ and $Q = x$. Taking the partial derivatives, we find $\frac{\partial Q}{\partial x} = 1$ and $\frac{\partial P}{\partial y} = -1$.

Substituting these into Green's Theorem, the integrand becomes $(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) = (1 - (-1)) = 2$.
The line integral is therefore equivalent to $\iint_R 2 \,dA$, which simplifies to $2 \times (\text{Area of R})$.

The area of the region R is the sum of the area of the $3 \times 2$ rectangle and the semi-circle of radius 1.
Area(R) = $(3 \times 2) + \frac{1}{2}\pi(1)^2 = 6 + \frac{\pi}{2}$.
Finally, the value of the line integral is $2 \times (6 + \frac{\pi}{2}) = 12 + \pi$.

First, we can determine the expression for $X[1]$ by tracing the paths in the signal-flow graph:
$X[1] = a_1(x[0] - x[3]) + a_2(x[1] - x[4]) + a_3(x[2] - x[5])$

Next, we write the standard 6-point DFT definition for the $k=1$ frequency bin:
$X[1] = \sum_{n=0}^{5} x[n]W_6^n = x[0]W_6^0 + x[1]W_6^1 + x[2]W_6^2 + x[3]W_6^3 + x[4]W_6^4 + x[5]W_6^5$

To match the structure of the flow graph, we use the property $W_N^{n+N/2} = -W_N^n$. For $N=6$, this means $W_6^{n+3} = -W_6^n$. Applying this, we get $W_6^3=-W_6^0$, $W_6^4=-W_6^1$, and $W_6^5=-W_6^2$. Substituting these into the DFT equation and regrouping terms gives:
$X[1] = (x[0] - x[3])W_6^0 + (x[1] - x[4])W_6^1 + (x[2] - x[5])W_6^2$

By comparing this expression with the one derived from the graph, we find that $a_1=W_6^0=1$, $a_2=W_6$, and $a_3=W_6^2$.

The input signal $x[n]$ is composed of complex sinusoids at frequencies $\omega = \pm\frac{\pi}{2}$. For an LTI system to produce a zero output, its frequency response, $H(e^{j\omega})$, must be zero at the frequencies present in the input.

First, let's find the frequency response of the three-tap filter $h[n]=\{1, a, b\}$.
$H(e^{j\omega}) = \sum_{n=0}^{2} h[n]e^{-j\omega n} = 1 + ae^{-j\omega} + be^{-j2\omega}$

Now, we enforce the condition that the response is zero at $\omega = \frac{\pi}{2}$:
$H(e^{j\pi/2}) = 1 + ae^{-j\pi/2} + be^{-j2\pi/2} = 1 + a(-j) + b(-1) = 0$

This gives the complex equation $(1-b) - ja = 0$. For this equation to hold true, both the real and imaginary parts must be zero.
Real part: $1 - b = 0 \implies b = 1$
Imaginary part: $-a = 0 \implies a = 0$
Therefore, the filter taps are $a=0$ and $b=1$.

To find the steady-state current, we first analyze the circuit in the frequency domain. The voltage source $v(t)=2\sin(1000t)$ corresponds to a phasor $V=2\angle0^\circ$ V and angular frequency $\omega=1000$ rad/s. The impedance of the capacitors is $Z_C = \frac{1}{j\omega C} = \frac{1}{j(1000)(10^{-6})} = -j1 \text{ k}\Omega$. Thus, the capacitive reactance $X_C$ is $1 \text{ k}\Omega$, which is equal to the given resistance $R$.

The circuit is a symmetrical bridge structure. The special condition $R=X_C$ means the bridge is balanced, which allows the entire network to be simplified to an equivalent admittance $Y_{eq} = \frac{3}{2R} + j\frac{1}{2X_C}$. Substituting the component values, we get $Y_{eq} = (\frac{3}{2 \cdot 1000} + j\frac{1}{2 \cdot 1000}) = (1.5 + j0.5)$ mS.

The current phasor $I$ is the product of the voltage phasor and the equivalent admittance: $I = V \cdot Y_{eq} = (2\angle0^\circ) \cdot (1.5 + j0.5) \text{ mS} = (3+j1)$ mA.

Finally, we convert this phasor back to the time domain. For a sine voltage reference, a current phasor $A+jB$ corresponds to $i(t) = A\sin(\omega t) + B\cos(\omega t)$. Therefore, the steady-state current is $i(t) = 3\sin(1000t) + \cos(1000t)$ mA.

First, we find the system's output, $C(s)$, by multiplying the transfer function $G(s)$ by the input $R(s)$, giving $C(s) = \frac{1}{s(s+1)^2}$. To analyze the output in the time domain, we take the inverse Laplace transform of $C(s)$, which can be done using partial fraction expansion. This yields the time-domain response $c(t) = (1 - e^{-t} - te^{-t})u(t)$.

The steady-state value is the limit of the output as time goes to infinity, $\lim_{t \to \infty } c(t) = 1$. We are looking for the time $t$ when the output reaches 94% of this value. Therefore, we set up the equation $c(t) = 0.94$ and solve for $t$:
$1 - e^{-t} - te^{-t} = 0.94$
Solving this transcendental equation, for instance by iterating through the given options, we find that $t \approx 4.50$ seconds.

To determine the transfer function $H(s) = Y(s)/X(s)$, we can analyze the system by writing algebraic equations for the signals at key nodes. Let's label the signal at the output of the first summing junction as $E_1(s)$ and the signal at the input of the final $1/s$ block as $E_2(s)$.

From the block diagram, we can establish three key relationships:

1. At the first summer: $E_1(s) = X(s) - E_2(s) - Y(s)$
2. At the parallel blocks: $E_2(s) = (s + \frac{1}{s})E_1(s)$
3. At the output: $Y(s) = \frac{1}{s}E_2(s)$

By substituting the third equation into the other two to express them in terms of $Y(s)$, and then combining them to eliminate $E_1(s)$, we can find a single equation that relates the input $X(s)$ to the output $Y(s)$. Solving this equation for the ratio $\frac{Y(s)}{X(s)}$ yields the system's transfer function.