GATE ECE 2016 Set 3

There are two fundamental properties connecting a matrix to its eigenvalues. While the trace (sum of diagonal elements) equals the sum of eigenvalues, it doesn't help find $x$ here. Instead, we'll use the second property: the determinant of a matrix equals the product of its eigenvalues.

First, let's calculate the determinant of matrix $A$:
$\text{det}(A) = (\sigma)(\sigma) - (x)(\omega) = \sigma^2 - x\omega$
Next, we find the product of the given eigenvalues. Since they are a complex conjugate pair, their product is:
$(\sigma + j\omega)(\sigma - j\omega) = \sigma^2 - (j\omega)^2 = \sigma^2 - (j^2\omega^2) = \sigma^2 + \omega^2$
By equating the determinant with the product of the eigenvalues, we get:
$\sigma^2 - x\omega = \sigma^2 + \omega^2$
Subtracting $\sigma^2$ from both sides leaves $-x\omega = \omega^2$. Solving for $x$ gives us $x = -\omega$.

To find the residue at the pole $z=0$, we'll examine the Laurent series of the function centered at that point. We begin by using the well-known Maclaurin series expansion for $\sin(z)$:
$\sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots$
Now, we can find the series for $f(z)$ by dividing this expression by $z^2$:
$f(z) = \frac{1}{z^2} \left( z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots \right) = \frac{1}{z} - \frac{z}{3!} + \frac{z^3}{5!} - \cdots$
The residue at $z=0$ is defined as the coefficient of the $\frac{1}{z}$ term in this Laurent expansion. By inspection, this coefficient is 1.

Here's a clearer way to understand this problem.

For the first "head" to appear on the fifth toss, a specific sequence of events must unfold: the first four tosses must be "tails" (T), followed by a "head" (H) on the fifth toss. The sequence is T, T, T, T, H.

We are given the probability of a head, $P(H) = 0.3$. Since there are only two outcomes, the probability of a tail is $P(T) = 1 - P(H) = 1 - 0.3 = 0.7$.

Because the tosses are independent, we can find the probability of this specific sequence by multiplying the probabilities of each individual outcome.

The calculation is: $P(T) \times P(T) \times P(T) \times P(T) \times P(H)$.
This equals $(0.7)^4 \times 0.3 = 0.2401 \times 0.3 = 0.07203$.

This integral can be solved by finding the antiderivative using a reverse chain rule. We notice the integrand $\frac{1}{\sqrt{1-x}}$ looks related to the derivative of $\sqrt{1-x}$.

Let's check: the derivative of $\sqrt{1-x}$ is $\frac{1}{2\sqrt{1-x}} \cdot (-1) = \frac{-1}{2\sqrt{1-x}}$.
Our function is off by a constant factor of $-2$. This means the antiderivative of $\frac{1}{\sqrt{1-x}}$ is $-2\sqrt{1-x}$.

Now, we evaluate this antiderivative at the given limits of integration:
$[-2\sqrt{1-x}]_{0}^{1} = (-2\sqrt{1-1}) - (-2\sqrt{1-0})$
$= (-2\sqrt{0}) - (-2\sqrt{1}) = 0 - (-2) = 2$.

We are asked to perform a single step of the second-order Runge-Kutta (RK2) method to approximate $y(0.1)$ and compare it to the exact value. The governing equation is $f(x, y) = y' = y+2x-x^2$, with initial conditions $x_0=0$, $y_0=1$, and step-size $h=0.1$.

First, we calculate the slope estimate at the starting point:
$k_1 = h f(x_0, y_0) = 0.1(1 + 2(0) - 0^2) = 0.1$.

Next, we find the slope estimate at the end of the step interval:
$k_2 = h f(x_0+h, y_0+k_1) = 0.1 f(0.1, 1.1) = 0.1(1.1 + 2(0.1) - (0.1)^2) = 0.129$.

The RK2 approximation is the weighted average of these slopes:
$y(0.1) \approx y_1 = y_0 + \frac{1}{2}(k_1 + k_2) = 1 + \frac{1}{2}(0.1 + 0.129) = 1.1145$.

The exact solution at $x=0.1$ is $y(0.1) = (0.1)^2 + e^{0.1} \approx 0.01 + 1.1052 = 1.1152$.

Finally, the percentage difference is $\frac{|1.1152 - 1.1145|}{1.1152} \times 100\% \approx 0.062\%$.

First, we find the expression for the new signal, $y(t)$, by applying the time transformation $t \to (2t+5)$ to the original signal $x(t)$.
$y(t) = x(2t+5) = \cos(6\pi(2t+5)) + \sin(8\pi(2t+5))$

Simplifying the arguments of the cosine and sine functions, we get:
$y(t) = \cos(12\pi t + 30\pi) + \sin(16\pi t + 40\pi)$

The signal $y(t)$ is composed of two sinusoids. Their angular frequencies are $\omega_1 = 12\pi$ rad/s and $\omega_2 = 16\pi$ rad/s. To find the cyclic frequencies in Hz, we use the relation $f = \omega / (2\pi)$. This gives us $f_1 = 6$ Hz and $f_2 = 8$ Hz.

The maximum frequency component in $y(t)$ is therefore $f_{max} = 8$ Hz. The Nyquist sampling rate is defined as twice the maximum frequency of the signal.

Thus, the Nyquist rate is $f_s = 2 f_{max} = 2 \times 8 = 16$ samples/second.

This problem is a classic example of the convolution theorem. Let's analyze the operation in the frequency domain, where convolution becomes simple multiplication.

Let the base signal be $s(t) = \frac{\sin(t)}{\pi t}$. The question asks for the result of $s(t) * s(t)$. The Fourier transform of $s(t)$ is a perfect rectangular pulse, $S(\omega)$, which has a value of 1 for frequencies between -1 and 1, and is 0 everywhere else.

In the frequency domain, the convolution becomes the product $S(\omega) \cdot S(\omega)$. Since the rectangular pulse $S(\omega)$ only takes on values of 0 or 1, multiplying it by itself results in the exact same pulse.

Therefore, the product is still just $S(\omega)$. Transforming back to the time domain, we find that the resulting signal must be identical to the original signal, $s(t)$.

First, we find the z-transform of the given signal, $x[n]$. Using the time-shifting property of the z-transform, which states that the transform of $\delta[n-n_0]$ is $z^{-n_0}$, we get:
$X(z) = z^{-3} + 2z^{-5}$

The problem states that $y[n]$ has a z-transform $Y(z) = X(-z)$. To find $Y(z)$, we substitute $-z$ for every $z$ in the expression for $X(z)$:
$Y(z) = (-z)^{-3} + 2(-z)^{-5}$

Since the exponents -3 and -5 are odd numbers, we can factor out the negative sign:
$Y(z) = -(z^{-3}) - 2(z^{-5}) = -(z^{-3} + 2z^{-5})$

Notice that the expression in the parentheses is simply the original $X(z)$. Therefore, we have:
$Y(z) = -X(z)$

By the linearity property of the z-transform, this relationship in the z-domain corresponds to $y[n] = -x[n]$ in the time domain.

The input signal contains two distinct frequencies, so we can analyze the circuit's response to each frequency individually using the superposition principle.

First, let's consider the component with angular frequency $\omega = 200$ rad/s. In the series part of the circuit, the inductive reactance is $X_L = \omega L = (200)(0.25) = 50 \Omega$, and the capacitive reactance is $X_C = 1/(\omega C) = 1/(200 \cdot 100 \mu \text{F}) = 50 \Omega$. Since these reactances are equal, the series L-C branch is at resonance and acts as a short circuit. This makes the total series impedance zero, so the output voltage equals the input voltage for this component.

Next, for the component with $\omega = 500$ rad/s, we examine the parallel part of the circuit. The inductive reactance is $X_L = \omega L = (500)(0.4) = 200 \Omega$, and the capacitive reactance is $X_C = 1/(\omega C) = 1/(500 \cdot 10 \mu \text{F}) = 200 \Omega$. Because these reactances are equal for the parallel elements, this creates a parallel resonance condition, which acts as an open circuit (infinite impedance). With an open circuit in the shunt path, the output voltage must again equal the input voltage.

Since the circuit passes both frequency components without any change, the total output voltage $v_o(t)$ is simply the sum of the individual responses, which is identical to the input voltage $v_i(t)$.

The voltage at which a diode "turns on" and current begins to flow rapidly is known as the knee voltage. This voltage is directly related to the semiconductor's band gap energy, $E_g$. To overcome the diode's internal potential barrier, charge carriers must be given an amount of energy proportional to the band gap. Therefore, a larger band gap requires a higher turn-on voltage.

From the graph, we can observe the order of the knee voltages: $V_{knee, X} < V_{knee, Y} < V_{knee, Z}$. This directly implies that the corresponding band gap energies follow the same increasing order: $E_{gX} < E_{gY} < E_{gZ}$.

To understand the state of this MOS device, let's first identify the semiconductor type. In the bulk region (far to the right), the Fermi level ($E_{FS}$) is above the intrinsic Fermi level ($E_i$), which means the substrate is n-type.

Now, look at the surface, near the SiO₂ interface. The energy bands are bent downwards. This downward bending is so significant that the intrinsic Fermi level ($E_i$) has crossed over and is now positioned above the semiconductor's Fermi level ($E_{FS}$).

This condition, where $E_{FS}$ is below $E_i$ at the surface, indicates that holes (minority carriers in the n-type bulk) have become the majority carriers in this region. When the surface carrier type is opposite to the bulk carrier type, the MOS is in inversion.

The efficiency of a solar cell ($\eta$) is the ratio of its maximum electrical power output ($P_{m}$) to the total incident light power ($P_{\in}$).

First, we calculate the total incident power by multiplying the given power density by the cell's area:
$P_{\in} = (100 \text{ mW/cm}^2) \times (3 \text{ cm}^2) = 300 \text{ mW} = 0.3 \text{ W}$.

Next, we find the maximum output power using the fill factor ($FF$) and the values from the I-V curve, where the open-circuit voltage $V_{oc} = 0.5 \text{ V}$ and the short-circuit current $I_{sc} = 180 \text{ mA}$.
$P_{m} = FF \times V_{oc} \times I_{sc} = 0.7 \times 0.5 \text{ V} \times (180 \times 10^{-3} \text{ A}) = 0.063 \text{ W}$.

Finally, we calculate the efficiency:
$\eta = \frac{P_{m}}{P_{\in}} = \frac{0.063 \text{ W}}{0.3 \text{ W}} = 0.21$, which corresponds to an efficiency of $21\%$.

During the initial positive half-cycle of the AC source, the ideal diode D1 becomes forward-biased, allowing the top capacitor to charge. At the same time, any tendency for the output voltage $V_o$ to rise above ground potential will forward-bias the ideal diode D2. This clamps the output node, forcing $V_o=0V$.

This continues until the input reaches its peak of $10 \mathrm{V}$. At this instant, the top capacitor is fully charged to $10 \mathrm{V}$. Since $V_o$ is held at $0 \mathrm{V}$, the bottom capacitor remains uncharged. For any subsequent time, the input voltage is less than or equal to $10 \mathrm{V}$, so D1 becomes permanently reverse-biased. With D1 off and D2 still preventing any positive voltage, the output $V_o$ has no source of charge and remains at a steady state of $0 \mathrm{V}$.

To find the DC voltage $V_{C2}$, we first analyze the circuit around transistor Q1. Assuming Q1 operates with its collector-base voltage at zero ($V_{CB1}=0$), its collector-emitter voltage becomes equal to its base-emitter voltage, so $V_{CE1} = V_{BE1} = 0.7$ V.

Next, we apply Kirchhoff's Voltage Law (KVL) to the input loop of transistor Q2 to find its base current, $I_{B2}$. The loop equation is $-2.5 + V_{CE1} + 0.7 + I_{B2}(10 \text{ k}\Omega) + 1 = 0$. Substituting $V_{CE1} = 0.7$ V and solving, we get $I_{B2} = 0.01$ mA.

The collector current of Q2, $I_{C2}$, is found by multiplying the base current by the current gain $\beta_2$. Based on the original calculation, $\beta_2 = 50$, so $I_{C2} = \beta_2 I_{B2} = 50 \times 0.01 \text{ mA} = 0.5$ mA.

Finally, the collector voltage $V_{C2}$ is the voltage across the 1 kΩ collector resistor, calculated with Ohm's law: $V_{C2} = I_{C2} \times 1 \text{ k}\Omega = 0.5 \text{ mA} \times 1 \text{ k}\Omega = 0.5$ V.

This circuit shows a 555 timer configured as an astable multivibrator, designed to generate a continuous square wave. The frequency of this oscillation depends on the external resistors, $R_A$ and $R_B$, and the capacitor, $C$.

We calculate the frequency, $f$, using the standard formula for this circuit:
$f = \frac{1}{0.69(R_A + 2R_B)C}$
Substituting the given component values:
$f = \frac{1}{0.69 \left( 2.2 \times 10^3 \, \Omega + 2 \times 4.7 \times 10^3 \, \Omega \right) \left( 0.022 \times 10^{-6} \, \text{F} \right)}$
$f = \frac{1}{0.69 \left( 11.6 \times 10^3 \, \Omega \right) \left( 0.022 \times 10^{-6} \, \text{F} \right)} \approx 5681.8 \, \text{Hz}$
Therefore, the frequency of oscillation is approximately $5.68$ kHz.

The RLC instruction performs a "Rotate Accumulator Left" operation. This means every bit in the accumulator shifts one position to the left. The most significant bit (MSB) is unique: it is copied into the carry flag (CY) and also wraps around to become the new least significant bit (LSB).

Initially, the accumulator holds $A7_H = (10100111)_2$. The MSB is 1.

When RLC is executed, this MSB of 1 sets the carry flag, so $CY=1$. The bits shift left, and this same MSB of 1 circles back to the LSB position.

The new accumulator value becomes $(01001111)_2$. Converting this binary value to hexadecimal gives $4F_H$. Therefore, the final contents are $4F_H$ in the accumulator and 1 in the carry flag.

This circuit consists of two NMOS transistors, which act as switches that are closed (ON) when their gate voltage is high (logic 1).

Let's analyze the circuit's behavior based on input $B$.
If $B=0$, then $\bar{B}=1$. The bottom transistor turns on, connecting the output $Y$ directly to ground. This forces $Y=0$, regardless of the value of $A$.

If $B=1$, then $\bar{B}=0$. The top transistor turns on, connecting the output $Y$ to the input $A$. The bottom transistor is off. Therefore, the output is $Y=A$.

Combining these conditions, the output $Y$ can only be 1 if $B=1$ AND $A=1$. This behavior matches the truth table for a logical AND gate.

The Boolean expression for a 2-input XOR gate is $Y = A \oplus B = A\bar{B} + \bar{A}B$. This expression can be implemented using only NAND gates. The standard minimum-gate circuit is constructed by first creating an intermediate signal $P = \overline{A \cdot B}$ using one NAND gate. This signal is then used as a common input to two more NAND gates, whose other inputs are $A$ and $B$, respectively. This produces $Q = \overline{A \cdot P}$ and $R = \overline{B \cdot P}$. The outputs $Q$ and $R$ are then fed into a final, fourth NAND gate. The resulting output, $Y = \overline{Q \cdot R} = \overline{(\overline{A \cdot \overline{AB}}) \cdot (\overline{B \cdot \overline{AB}})}$, simplifies algebraically to $A \oplus B$. This well-known configuration uses exactly 4 gates.

To find the overall gain, we can simplify the block diagram in steps. The core of the system can be seen as an equivalent forward path, which we'll call $G_{eq}(s)$, placed within a larger unity feedback loop.

This equivalent forward path is given by $G_{eq}(s) = \frac{G_1 G_2}{1+G_1 H_1}$. This term represents the cascaded gains $G_1$ and $G_2$ over the inner feedback loop involving $H_1$.

Now, we apply the standard formula for a unity feedback system, $G = \frac{G_{eq}}{1+G_{eq}}$:
$G = \frac{\frac{G_1 G_2}{1+G_1 H_1}}{1 + \frac{G_1 G_2}{1+G_1 H_1}}$

To simplify this complex fraction, we multiply the numerator and denominator by $(1+G_1 H_1)$:
$G = \frac{G_1 G_2}{(1+G_1 H_1) + G_1 G_2} = \frac{G_1 G_2}{1+G_1 G_2 + G_1 H_1}$

To determine the steady-state error, we first find the expression for the error signal, $E(s)$. For a unity feedback system, this is given by $E(s) = \frac{R(s)}{1+G(s)}$. The input is a unit step, so its Laplace transform is $R(s) = \frac{1}{s}$.

Substituting the given transfer functions:
$E(s) = \frac{1/s}{1 + \frac{2}{s(s+1)}} = \frac{1/s}{\frac{s(s+1)+2}{s(s+1)}} = \frac{s+1}{s^{2+}s+2}$

Next, we apply the Final Value Theorem, $e_{ss} = \lim_{s \to 0} s E(s)$, to find the steady-state error.
$e_{ss} = \lim_{s \to 0} s \left( \frac{s+1}{s^{2+}s+2} \right)$

Evaluating the limit by substituting $s=0$ gives the final answer:
$e_{ss} = \frac{0(0+1)}{0^{2+}0+2} = \frac{0}{2} = 0$

In a superheterodyne receiver, an incoming signal mixes with a local oscillator ($f_{lo}$) to create a fixed intermediate frequency ($IF$). The image frequency ($f_{si}$) is an unwanted signal that can also produce this same $IF$.

The desired signal and its image are always on opposite sides of the local oscillator frequency. Since the problem states the received signal's frequency is greater than $f_{lo}$, the image frequency must be less than $f_{lo}$.

This means we find the image frequency by subtracting the $IF$ from the $f_{lo}$:
$f_{si} = f_{lo} - IF$

Given $f_{lo} = 3.5 \text{ GHz} = 3500 \text{ MHz}$ and $IF = 15 \text{ MHz}$:
$f_{si} = 3500 \text{ MHz} - 15 \text{ MHz} = 3485 \text{ MHz}$

First, we determine the sampling rate. Since the signal is bandlimited to 100 Hz, the Nyquist sampling rate is $f_s = 2 \times 100 = 200$ samples/sec.

Next, we find the probabilities for all four symbols. We're given $p_1 = p_4 = 0.125 = 1/8$. Since probabilities must sum to 1 and $p_2=p_3$, we can solve for them: $p_2 = p_3 = (1 - p_1 - p_4)/2 = (1 - 2 \times 1/8)/2 = 3/8$.

The average information per symbol, or entropy ($H$), is calculated as $H = \sum_{i=1}^{4} p_i \log_2(1/p_i)$.
$H = 2 \times \frac{1}{8}\log_2(8) + 2 \times \frac{3}{8}\log_2(\frac{8}{3}) \approx 1.811$ bits/sample.

Finally, the total information rate ($R$) is the product of the entropy and the sampling rate:
$R = H \times f_s = 1.811 \text{ bits/sample} \times 200 \text{ samples/sec} \approx 362.2$ bits/sec.

To find the output of the matched filter, we must convolve the input signal, $P(t)$, with the filter's impulse response, $h(t)$. For a matched filter, the impulse response is a time-reversed and delayed version of the input signal, defined as $h(t) = P(T_s - t)$.

In this case, the input $P(t)$ is a rectangular pulse. Because a rectangular pulse is symmetric about its midpoint, its time-reversed and shifted version, $h(t)$, is identical to the original pulse, $P(t)$.

Therefore, the output $y(t)$ is the convolution of the rectangular pulse with itself: $y(t) = P(t) * P(t)$. The convolution of two identical rectangular pulses of duration $T_s$ results in a triangular pulse of duration $2T_s$. The peak value of this triangle is 1, occurring at $t = T_s$.

When a wave reflects from a perfect conductor, the electric field at the conductor's surface is inverted, undergoing a phase shift of $\pi$ radians. This preserves the absolute direction of the electric field vector's rotation in space (e.g., a clockwise rotation remains clockwise). However, the wave's direction of propagation is reversed. The handedness of circular polarization is defined by the E-field's rotation relative to its propagation direction. Since the propagation direction flips but the field's spatial rotation does not, the defined handedness must be reversed. Therefore, the incident right-handed wave becomes a left-handed wave upon reflection.

Faraday's law of induction describes how a magnetic field that changes over time can create an electric field. This is the core principle of electromagnetic induction. The equation that mathematically represents this law in its differential (or "point") form is $\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$.

Let's break down this equation. The term on the right, $-\frac{\partial \vec{B}}{\partial t}$, represents the rate at which the magnetic field $\vec{B}$ is changing. The term on the left, $\nabla \times \vec{E}$, is the curl of the electric field $\vec{E}$. A non-zero curl signifies a circulating field, which is precisely what's needed to induce a voltage around a closed loop.

To solve this second-order linear homogeneous differential equation, we first find its characteristic equation: $m^2 + 12m + 36 = 0$. This equation factors into $(m+6)^2 = 0$, revealing a repeated real root of $m = -6$. For a repeated root, the general solution takes the form $y(x) = (C_1 + C_2x)e^{-6x}$.

We use the initial conditions to find the constants $C_1$ and $C_2$. Applying $y(0)=3$ gives $3 = (C_1 + C_2 \cdot 0)e^0$, which simplifies to $C_1=3$.

Next, we find the derivative of the general solution: $y'(x) = C_2e^{-6x} - 6(C_1 + C_2x)e^{-6x}$. Applying the second condition, $y'(0)=-36$, yields $-36 = C_2e^0 - 6(C_1 + 0)e^0$, or $-36 = C_2 - 6C_1$.

Substituting our value for $C_1$, we get $-36 = C_2 - 6(3)$, which simplifies to $C_2 = -18$. Plugging the determined constants back into the general solution gives our particular solution: $y(x) = (3 - 18x)e^{-6x}$.

To express the vector $u$ in terms of the new basis vectors, we need to find the scalar coefficients $a, b,$ and $c$ that satisfy the equation $u = a e_1 + b e_2 + c e_3$.

Substituting the given vectors, we have:
$(4, 3, -3) = a(1, 0, 2) + b(0, 1, 0) + c(-2, 0, 1)$

This single vector equation is equivalent to a system of three linear equations, one for each component:

1. (x-component) $a - 2c = 4$
2. (y-component) $b = 3$
3. (z-component) $2a + c = -3$

From the y-component, we immediately find that $b=3$. Solving the first and third equations simultaneously gives $a = -2/5$ and $c = -11/5$.

Therefore, the vector $u$ can be written as $u = -\frac{2}{5} e_1 + 3 e_2 - \frac{11}{5} e_3$.

To find the volume, we integrate the height of the solid, given by the plane $z = 6 - x - y$, over the triangular region in the xy-plane. This triangular base is bounded by the lines $y=0$, $x=3$, and $y = \frac{2}{3}x$. We can express this region with the limits $0 \le x \le 3$ and $0 \le y \le \frac{2}{3}x$.

This setup gives us the double integral for the volume:
$V = \int_{0}^{3} \int_{0}^{2x/3} (6 - x - y) \, dy \, dx$
First, we integrate with respect to $y$:
$V = \int_{0}^{3} \left[6y - xy - \frac{y^2}{2}\right]_{y=0}^{y=2x/3} \, dx = \int_{0}^{3} \left(4x - \frac{2x^2}{3} - \frac{2x^2}{9}\right) \, dx$
After combining terms, the integral simplifies to:
$V = \int_{0}^{3} \left(4x - \frac{8x^2}{9}\right) \, dx$
Finally, integrating with respect to $x$ gives the volume:
$V = \left[2x^2 - \frac{8x^3}{27}\right]_{0}^{3} = \left(2(3)^2 - \frac{8(3)^3}{27}\right) - 0 = 18 - 8 = 10$

This problem is a classic application of Cauchy's Integral theorems. The value of the integral depends on whether the singularity at $z_0=2$ is inside the contour $c$.

(i) When the point $z_0 = 2$ is inside the contour $c$, we use Cauchy's Integral Formula. The formula states that for a function $f(z)$ that is analytic inside the contour, $\frac{1}{2\pi j}\oint_c \frac{f(z)}{z-z_0}dz = f(z_0)$. In our case, $f(z) = e^z$, so the integral evaluates to $f(2) = e^2 \approx 7.39$.

(ii) When the point $z_0 = 2$ is outside the contour $c$, the entire function being integrated, $\frac{e^z}{z-2}$, is analytic everywhere inside the contour. According to Cauchy's Integral Theorem, the integral of any analytic function along a closed path is zero. Therefore, the value of the integral is 0.

First, determine the system's frequency response by substituting $s=j\omega$ into the transfer function, yielding $H(j\omega) = e^{j\omega} + e^{-j\omega} = 2\cos(\omega)$. An LTI system scales each sinusoidal input component by the value of the frequency response at that component's frequency.

For the first input term, $2\cos(\frac{2\pi t}{3})$, the frequency is $\omega_1 = \frac{2\pi}{3}$. The system gain is $H(j\frac{2\pi}{3}) = 2\cos(\frac{2\pi}{3}) = -1$.
For the second input term, $-\cos(\pi t)$, the frequency is $\omega_2 = \pi$. The gain is $H(j\pi) = 2\cos(\pi) = -2$.

The total output signal is the sum of the individual responses:
$y(t) = (-1)\left[2\cos\left(\frac{2\pi t}{3}\right)\right] + (-2)\left[-\cos(\pi t)\right] = 2\cos(\pi t) - 2\cos\left(\frac{2\pi t}{3}\right)$.

To find the Fourier coefficients, we express $y(t)$ in terms of its fundamental frequency, $\omega_0 = \text{GCD}(\pi, \frac{2\pi}{3}) = \frac{\pi}{3}$. This gives $y(t) = 2\cos(3\omega_0 t) - 2\cos(2\omega_0 t)$. Using Euler's formula, $\cos(x) = \frac{e^{jx}+e^{-jx}}{2}$, the output becomes $y(t) = (e^{j3\omega_0 t} + e^{-j3\omega_0 t}) - (e^{j2\omega_0 t} + e^{-j2\omega_0 t})$. By comparing this to the exponential Fourier series definition, $y(t) = \sum_{k=-\infty }^{\infty } C_k e^{jk\omega_0 t}$, we can see the coefficient of the $k=3$ term, $C_3$, is 1.

The signal $x[n] = 2^{|n|}$ is a two-sided sequence, so we must consider its causal and anti-causal parts separately.

The causal part, for $n \ge 0$, is $2^n u[n]$. The region of convergence (ROC) for a right-sided exponential sequence like this is the area outside a circle, so its ROC is $|z| > 2$.

The anti-causal part, for $n < 0$, is $2^{-n} u[-n-1] = (1/2)^n u[-n-1]$. The ROC for a left-sided sequence is the area inside a circle, so its ROC is $|z| < 1/2$.

The overall ROC is the intersection of these two regions. However, there are no values of $z$ for which the magnitude $|z|$ is simultaneously greater than 2 and less than 1/2. Thus, the z-transform does not converge, and the ROC does not exist.

First, we find the initial conditions at $t=0^-$. In the DC steady state, capacitors act as open circuits. The circuit becomes a simple loop with a total resistance of $R = 1\Omega + 2\Omega + 3\Omega = 6\Omega$. The current flowing from the source is $I = 12\text{V} / 6\Omega = 2\text{A}$.

The voltages across the capacitors are determined by the node voltages. The voltage at the node after the $1\Omega$ resistor is $12\text{V} - (2\text{A} \times 1\Omega) = 10\text{V}$. The voltage at the node between the $2\Omega$ and $3\Omega$ resistors is $2\text{A} \times 3\Omega = 6\text{V}$. Therefore, the initial capacitor voltages are $V_{3F}(0^-) = 10\text{V} - 6\text{V} = 4\text{V}$ and $V_{2F}(0^-) = 6\text{V} - 0\text{V} = 6\text{V}$.

At $t=0^+$, the $3\Omega$ resistor is an open circuit, and the capacitor voltages remain constant. Let's call the central node C and the node to its right D. The capacitor voltages fix these node voltages: $V_C(0^+) = V_{2F} = 6\text{V}$, and the node before the capacitors (A) has $V_A - V_C = 4\text{V}$, so $V_A = 10\text{V}$.

Because the path through the $3\Omega$ resistor is open, KCL at node D gives $\frac{V_A-V_D}{2\Omega} + \frac{V_C-V_D}{2\Omega} = 0$. Substituting our known voltages gives $\frac{10 - V_D}{2} + \frac{6 - V_D}{2} = 0$, which solves to $V_D = 8\text{V}$.