GATE ECE 2016 Set 1

Our goal is to find an expression for $M^{-1}$ using the given fact that $M^4 = I$.

By definition, the inverse $M^{-1}$ is the matrix that satisfies $M \cdot M^{-1} = I$. Let's start with our given equation, $M^4 = I$, and multiply both sides by $M^{-1}$:
$M^4 M^{-1} = I M^{-1}$

Using the law of exponents ($M^a M^b = M^{a+b}$), the left side becomes $M^{4-1} = M^3$. The right side is simply $M^{-1}$. This tells us that $M^{-1} = M^3$.

To match the options, we need to introduce the natural number $k$. Since $M^4 = I$, it is also true that $(M^4)^k = I^k$, which means $M^{4k} = I$ for any natural number $k$. We can multiply any expression by $I$ without changing its value.

So, we can write $M^{-1} = M^3 = M^3 \cdot I = M^3 \cdot M^{4k} = M^{4k+3}$.

For a Poisson distribution with parameter $\lambda$, the mean is $\lambda$ and the second moment is given by the formula $\lambda^2 + \lambda$.

We are given that the second moment is 2, which allows us to set up an equation:
$\lambda^2 + \lambda = 2$
Rearranging this into a standard quadratic equation gives $\lambda^2 + \lambda - 2 = 0$. Factoring this equation, we get $(\lambda + 2)(\lambda - 1) = 0$.

The two possible solutions are $\lambda = 1$ and $\lambda = -2$. Since the mean $\lambda$ of a Poisson distribution must be positive, we discard the negative solution. Therefore, the mean of the random variable is 1.

Let's analyze the relationship between continuity and differentiability.

Statement R is a fundamental theorem of calculus: for a function to be differentiable at a point $x_0$, it must first be continuous at $x_0$. Thus, R is true. Differentiability is a stronger condition than continuity.

This means statement P, which claims the reverse (that continuity implies differentiability), must be false. A classic counterexample is the function $f(x) = |x|$, which is continuous at $x=0$ but not differentiable there due to the sharp corner.

Since we have a counterexample where a continuous function is not differentiable, statement Q is true. It correctly states that a continuous function may not be differentiable at a given point.

Solutions to the Laplace equation, $\nabla^2 f = 0$, are known as harmonic functions. These functions obey the mean value property, stating that the value of the function at any point is the average of its values on a sphere or circle centered at that point. A direct consequence of this is the maximum principle. If an interior point were a maximum, its value would be greater than all its neighbors, which would contradict the fact that it must be their average. The same logic applies to a minimum. Therefore, for any non-constant solution, the extreme values (maxima and minima) must lie on the boundary of the domain.

By the Fundamental Theorem of Calculus, the derivative of $F(x)$ is $f(x)$, so $F'(x) = f(x)$. This means the graph of $f(x)$ shows us the slope of the function $F(x)$ at every point.

From the given plot, we can see that $f(x)$ is negative for $x < 0$. This implies that the slope of $F(x)$ is negative in this region, so $F(x)$ must be a decreasing function.

Conversely, for $x > 0$, the plot shows that $f(x)$ is positive. This means the slope of $F(x)$ is positive, so $F(x)$ must be an increasing function.

A function that decreases for $x<0$ and increases for $x>0$ will have a local minimum at $x=0$.

An eigenfunction of a system is an input signal whose functional form is preserved at the output, only scaled by a constant factor. For any continuous-time LTI system, the output $y(t)$ is the convolution of the input $x(t)$ with the system's impulse response $h(t)$.

Let's test the input $x(t) = e^{j\omega_0 t}$. The output from convolution is:
$y(t) = \int_{-\infty }^{\infty } h(\tau)x(t-\tau)d\tau = \int_{-\infty }^{\infty } h(\tau)e^{j\omega_0(t-\tau)}d\tau$

By factoring out the term dependent on $t$, we get:
$y(t) = e^{j\omega_0 t} \int_{-\infty }^{\infty } h(\tau)e^{-j\omega_0\tau}d\tau$

The integral is the definition of the system's frequency response, $H(j\omega_0)$, which is a complex scalar for a given frequency $\omega_0$. The output is therefore $y(t) = H(j\omega_0) e^{j\omega_0 t}$, a scaled version of the input. This relationship holds for any LTI system.

When a periodic continuous-time signal is sampled, the resulting discrete-time signal is periodic only if the sampling process eventually repeats its pattern relative to the original signal's cycle. This happens when an integer number of sampling periods, let's call it $N T_s$, aligns perfectly with an integer number of the original signal's periods, say $k T$.

This condition can be written as the equation $N T_s = k T$.

By rearranging this equation, we find the core requirement for periodicity: $\frac{T}{T_s} = \frac{N}{k}$. This shows that the ratio of the signal's period to the sampling period must be a rational number (a fraction of two integers).

Looking at the options, the ratio $\frac{T}{T_s} = 1.2$ can be expressed as the fraction $\frac{12}{10}$ or $\frac{6}{5}$. Since this is a rational number, the sampled signal will be periodic. The other option, $\sqrt{2}$, is irrational.

The signal $x[n]$ is a sum of two right-sided exponential sequences, $a^n u[n]$ and $b^n u[n]$. The Z-transform of a right-sided sequence of the form $c^n u[n]$ has a region of convergence (ROC) given by $|z| > |c|$.

Therefore, the ROC for the first term is $|z| > |a|$, and the ROC for the second term is $|z| > |b|$. The ROC for the sum of these two terms is the intersection of their individual ROCs, as the Z-transform must converge for both components.

Given that $0 < |a| < |b|$, the region $|z| > |b|$ is more restrictive than $|z| > |a|$. The set of all $z$ satisfying $|z| > |b|$ is a subset of those satisfying $|z| > |a|$. Thus, their intersection is $|z| > |b|$.

A two-port network is defined as reciprocal if the ratio of the excitation at one port to the response at the other port remains the same when the ports are interchanged. For a network described by the transmission (ABCD) matrix, $T$, this physical property of reciprocity corresponds to a specific mathematical condition. The condition is that the determinant of the transmission matrix must be equal to 1. Given

, its determinant is calculated as $AD - BC$. Therefore, for the network to be reciprocal, we must have $AD - BC = 1$.

Let the input sinusoid be $x(t)$ with frequency $f_m = 33$ Hz. Its spectrum consists of impulses at $\pm 33$ Hz. This signal is sampled by a Dirac impulse train with a sampling frequency of $f_s = 46$ Hz.

In the frequency domain, sampling creates replicas of the original signal's spectrum shifted by integer multiples of the sampling frequency. The new spectral components are located at frequencies $f = n f_s \pm f_m$ for all integers $n$.

For $n=0$, we have the original component at $\pm 33$ Hz. For $n=1$, we generate new components at $f_s \pm f_m = 46 \pm 33$ Hz. This creates frequencies at $13$ Hz and $79$ Hz.

The resulting signal is passed through an ideal low-pass filter with a cutoff of $23$ Hz. This filter passes frequencies $|f| < 23$ Hz. Of the components we've calculated ($\pm 13$ Hz, $\pm 33$ Hz, $\pm 79$ Hz), only the $\pm 13$ Hz components pass through.

Therefore, the fundamental frequency of the output signal is 13 Hz.

The energy band diagram shows a new, discrete energy level located just below the conduction band ($E_c$). This level is known as a donor energy level. Such a level is created by doping an intrinsic semiconductor with pentavalent impurity atoms, which have five valence electrons (e.g., Phosphorus in Silicon). These atoms donate their extra electron, which requires very little energy (0.01 eV here) to move into the conduction band. This process vastly increases the concentration of free electrons, making them the majority charge carriers and forming an n-type semiconductor.

Reducing a MOSFET's channel length ($L$) introduces short-channel effects (SCEs), which alter its ideal characteristics.

Statement Q is incorrect: Output resistance is given by $r_o \approx \frac{1}{\lambda I_D}$. As channel length ($L$) reduces, Channel Length Modulation (CLM) becomes more severe, increasing the parameter $\lambda$ ($\lambda \propto 1/L$). Consequently, the output resistance $r_o$ decreases, not increases.

Statement R is incorrect: As $L$ reduces, the source and drain depletion regions share more of the bulk charge with the gate. This "charge sharing" effect reduces the gate voltage required to form the channel, thus decreasing the threshold voltage ($V_T$). This phenomenon is known as $V_T$ roll-off.

Statements P and S are correct. Reducing $L$ enhances Drain-Induced Barrier Lowering (DIBL), increasing OFF-state current, and it also reduces carrier transit time, which increases the ON current. Therefore, the incorrect statements are Q and R.

The op-amp, in its negative feedback configuration, works to keep its input terminals at the same voltage. The Zener diode fixes the voltage at the inverting input to $V_{CC} - V_{ref}$, so the emitter voltage at the non-inverting input must also be $V_E = V_{CC} - V_{ref}$.

This stable emitter voltage sets the voltage drop across the resistor $R$ to $V_{CC} - V_E = V_{ref}$.

Using Ohm's law, we can find the emitter current flowing through this resistor: $I_E = \frac{V_{ref}}{R}$.

The load current, $I_O$, is the transistor's collector current, $I_C$. These two currents are related by the common-base gain, $\alpha$, such that $I_C = \alpha I_E$.

Substituting our expression for $I_E$ and expressing $\alpha$ in terms of $\beta$ ($\alpha = \frac{\beta}{\beta+1}$), we get the final load current: $I_O = (\frac{\beta}{\beta+1}) \frac{V_{ref}}{R}$.

The op-amp is configured as a comparator, which compares the input signal $V_i$ with a fixed reference voltage. This reference voltage is set by the voltage divider at the inverting terminal:
$V_{ref} = 10 \text{ V} \times \frac{2 \text{ k}\Omega}{8 \text{ k}\Omega + 2 \text{ k}\Omega} = 2 \text{ V}$
The LED will glow when the transistor turns ON, which happens when the op-amp's output goes high. This occurs whenever the input voltage at the non-inverting terminal is greater than the reference voltage, i.e., when $V_i > 2 \text{ V}$. By inspecting the input waveform, we can see that the signal $V_i$ exceeds the 2 V threshold three times. Therefore, the LED glows 3 times.

This circuit is a Wien-bridge oscillator, which generates sinusoidal waveforms. For oscillations to begin, the loop gain must be slightly greater than one. In this configuration, this means the amplifier's gain, set by the negative feedback network, must be slightly above 3.

Initially, for small output voltages, the diodes are off and the gain is $1 + (22.1 \text{ k}\Omega / 10 \text{ k}\Omega) = 3.21$. This allows the oscillations to grow in amplitude. As the output voltage increases, the diodes begin to conduct during the peaks of the waveform. This action effectively reduces the feedback resistance, which in turn lowers the amplifier's gain to below 3. This gain-limiting effect prevents the op-amp from saturating and stabilizes the output at a constant, undistorted sinusoidal amplitude.

A frequency synthesizer uses a Phase Locked Loop (PLL) to generate an output frequency, $f_{out}$, that is a multiple of a reference frequency, $f_{ref}$. In a locked state, the frequencies at both inputs of the phase detector are equal. One input is $f_{ref}$, and the other is the feedback frequency, which is the synthesizer output divided by the counter's division factor, $N$. This gives the locking condition $f_{ref} = \frac{f_{out}}{N}$.

The synthesized output frequency is therefore given by the formula $f_{out} = N \times f_{ref}$.
With the reference input $f_{ref} = 5$ kHz, we can calculate the output for each switch position:

• Position 1 (N=2): $f_{out} = 2 \times 5 \text{ kHz} = 10 \text{ kHz}$
• Position 2 (N=4): $f_{out} = 4 \times 5 \text{ kHz} = 20 \text{ kHz}$
• Position 3 (N=8): $f_{out} = 8 \times 5 \text{ kHz} = 40 \text{ kHz}$
• Position 4 (N=16): $f_{out} = 16 \times 5 \text{ kHz} = 80 \text{ kHz}$
  As the switch moves sequentially from 1 to 4, the corresponding frequencies are 10 kHz, 20 kHz, 40 kHz, and 80 kHz.

The output of the circuit, $Y$, can be represented by the Boolean expression $Y = ABC \oplus AB \oplus BC$. We can simplify this by tackling the XOR operations in stages.

First, let's simplify the sub-expression $ABC \oplus AB$. Using the XOR definition ($P \oplus Q = \overline{P}Q + P\overline{Q}$), this term simplifies neatly to $AB\overline{C}$.

This reduces the overall expression for the output to $Y = (AB\overline{C}) \oplus BC$. Applying the XOR definition again to this expression gives $\overline{(AB\overline{C})} \cdot BC + (AB\overline{C}) \cdot \overline{BC}$, which simplifies to $Y = BC + AB\overline{C}$.

Finally, we factor out the common variable $B$, which gives $Y = B(C + A\overline{C})$. Using the Boolean absorption law ($X + \overline{X}Y = X+Y$), the term $C + A\overline{C}$ simplifies to $C+A$.

Therefore, the final simplified expression for the output is $Y = B(A+C)$.

This circuit configures a CMOS inverter with negative feedback. The 10kΩ resistor connects the output node ($V_{out}$) to the common gate input. Since the gate current of MOSFETs is practically zero, there is no voltage drop across the resistor. This forces the input voltage to be equal to the output voltage, i.e., $V_{\in} = V_{out}$. On the voltage transfer characteristic (VTC) of an inverter, the point where $V_{\in} = V_{out}$ is, by definition, the switching threshold voltage ($V_M$). The circuit will stabilize at this DC operating point, where both the PMOS and NMOS transistors are in the saturation region. Thus, the output voltage $V_{out}$ is the switching threshold of the inverter.

The Routh-Hurwitz stability criterion provides three key rules based on the elements of the first column in the Routh array.

First, for a system to be stable (X), all roots of its characteristic equation must lie in the left-half of the s-plane. This condition is met when all elements in the first column are positive (P).

Second, if there is any change in sign between the elements of the first column (R), it signifies that there are roots in the right-half of the s-plane, making the system unstable (Y).

Finally, a special case occurs when an element in the first column is zero (Q). This prevents the completion of the array, so the standard test is inconclusive and is said to break down (Z).

The Nyquist stability criterion is governed by the formula $N = P - Z$. In this equation, $P$ represents the number of open-loop poles in the right-half of the s-plane, and $Z$ is the number of closed-loop poles in that same region. The term $N$ counts the number of times the Nyquist plot encircles the critical point $(-1, j0)$, with counterclockwise encirclements being positive.

For a closed-loop system to be stable, it cannot have any poles in the right-half s-plane, which mathematically means we must have $Z = 0$. Substituting this requirement into the criterion gives the condition for stability: $N = P$. Therefore, a stable system requires the Nyquist plot to encircle $(-1, j0)$ counterclockwise exactly $P$ times.

To solve this, we use the formula for the bandwidth of a baseband digital signal that uses raised-cosine pulse shaping. This bandwidth depends on both the bit rate ($R_b$) and the filter's roll-off factor ($\alpha$).

The formula is:
$\text{Bandwidth} = \frac{R_b}{2} (1 + \alpha)$

Here, the term $\frac{R_b}{2}$ represents the minimum possible (Nyquist) bandwidth. The roll-off factor accounts for the additional "excess" bandwidth required.

Given a bit rate $R_b = 56 \text{ kbps}$ and a roll-off factor $\alpha = 0.25$, we can find the required transmission bandwidth:
$\text{Bandwidth} = \frac{56}{2} (1 + 0.25) = 28 \times 1.25 = 35 \text{ kHz}$

The relationship between the signal frequency ($f_s$), image frequency ($f_{si}$), and intermediate frequency ($f_{IF}$) is given by $f_{si} = f_s + 2f_{IF}$. We need to choose an $f_{IF}$ such that for any $f_s$ in the 58-68 MHz band, its corresponding $f_{si}$ falls outside this band.

Since the image frequency is always higher than the signal frequency, we must ensure it lies above the band's upper edge. The most restrictive, or "worst-case," scenario is when the desired signal is at its lowest frequency, $f_s = 58$ MHz, as its image will be closest to the operating band.

To prevent this image from falling into the band, it must be greater than the highest frequency in the band:
$f_{si} > 68 \text{ MHz}$
Substituting the worst-case values into this condition gives:
$58 \text{ MHz} + 2f_{IF} > 68 \text{ MHz}$
Solving for $f_{IF}$, we find $2f_{IF} > 10 \text{ MHz}$, which means $f_{IF} > 5 \text{ MHz}$. Thus, the minimum required intermediate frequency is 5 MHz.

The provided expression is a standard representation of an amplitude-modulated (AM) signal in the frequency domain. It consists of a carrier frequency and two sidebands. The general form is $s(t) = A_c \cos(2\pi f_c t) + \frac{A_c \mu}{2} \cos(2\pi(f_c - f_m)t) + \frac{A_c \mu}{2} \cos(2\pi(f_c + f_m)t)$.

In the given signal, $s(t) = 5 \cos(1600\pi t) + 20 \cos(1800\pi t) + 5 \cos(2000 \pi t)$, the term with the central frequency, $20 \cos(1800\pi t)$, represents the carrier. From this, we can identify the carrier amplitude as $A_c = 20$.

The other two terms are the sidebands, and their amplitudes are both 5. The amplitude of each sideband is given by the formula $\frac{A_c \mu}{2}$.

By equating the known sideband amplitude to this formula, we get $\frac{A_c \mu}{2} = 5$. Substituting $A_c = 20$, the equation becomes $\frac{20 \mu}{2} = 5$, which simplifies to $10\mu = 5$. Solving for the modulation index gives $\mu = 0.5$.

To ensure the electric flux density $\vec{D}$ is zero at a radius of 10 m, Gauss's Law requires the total charge enclosed within a 10 m Gaussian sphere to be zero. Since this sphere encloses all three charged shells, the sum of the charges on each shell must equal zero: $Q_1 + Q_2 + Q_3 = 0$.

The charge on each shell is its surface charge density times its surface area, $Q = \rho_s \cdot 4\pi r^2$. Setting the total charge to zero yields:
$\rho_{s1}(4\pi r_1^2) + \rho_{s2}(4\pi r_2^2) + \rho_{s3}(4\pi r_3^2) = 0$

We can cancel the common $4\pi$ term and substitute the known values:
$(20)(2^2) + (-4)(4^2) + \rho_s(8^2) = 0$
$80 - 64 + 64\rho_s = 0$
$16 + 64\rho_s = 0 \implies \rho_s = -16/64 = -0.25 \text{ nC/m}^2$.

The primary line constants (R, L, G, C) can be determined by cleverly combining the expressions for the propagation constant, $\gamma = \sqrt{(R+j\omega L)(G+j\omega C)}$, and the characteristic impedance, $Z_0 = \sqrt{\frac{R+j\omega L}{G+j\omega C}}$.

Multiplying $\gamma$ and $Z_0$ isolates the series impedance per unit length, $R+j\omega L$.
$\gamma Z_0 = (2+j5)(50) = 100 + j250$
From this, we equate the real parts to get $R=100 \, \Omega/m$ and the imaginary parts to get $\omega L = 250$, so $L = \frac{250}{10^6} = 250 \, \mu H/m$.

Similarly, dividing $\gamma$ by $Z_0$ isolates the shunt admittance per unit length, $G+j\omega C$.
$\frac{\gamma}{Z_0} = \frac{2+j5}{50} = 0.04 + j0.1$
This yields $G=0.04 \, S/m$ and $\omega C = 0.1$, so $C = \frac{0.1}{10^6} = 0.1 \, \mu F/m$.

The circular domain of integration makes polar coordinates the ideal approach. We'll use the substitution $x=r\cos\theta$, $y=r\sin\theta$, and $dx dy = r dr d\theta$. For the disk $x^{2+}y^2 \leq 4$, our new limits of integration are $0 \le r \le 2$ and $0 \le \theta \le 2\pi$.

The integral becomes $\frac{1}{2\pi} \int_{0}^{2\pi} \int_{0}^{2} (r\cos\theta + r\sin\theta + 10)r \,dr\,d\theta$.

Let's evaluate the inner integral with respect to $r$:
$\int_{0}^{2} (r^2(\cos\theta+\sin\theta) + 10r) \,dr = \left[ \frac{r^3}{3}(\cos\theta+\sin\theta) + 5r^2 \right]_{0}^{2} = \frac{8}{3}(\cos\theta+\sin\theta) + 20$

Now, we integrate this result with respect to $\theta$:
$\frac{1}{2\pi} \int_{0}^{2\pi} \left( \frac{8}{3}(\cos\theta+\sin\theta) + 20 \right) d\theta$

The integral of $\cos\theta+\sin\theta$ over a full period $[0, 2\pi]$ is zero. We are left with the integral of the constant term: $\frac{1}{2\pi} \int_{0}^{2\pi} 20 \, d\theta = \frac{1}{2\pi} [20\theta]_{0}^{2\pi} = \frac{20(2\pi)}{2\pi} = 20$.

To find $x[12]$, we need to calculate the matrix

raised to the 12th power. A direct calculation would be tedious, so we use a more elegant approach with the Cayley-Hamilton theorem. The characteristic equation of $A$ is $\lambda^2 - \lambda - 1 = 0$, which means the matrix itself satisfies $A^2 - A - I = 0$, or $A^2 = A + I$.

This powerful identity allows us to express any high power of $A$ as a linear combination of $A$ and $I$. Through successive squaring ($A^4 = (A^2)^2$, etc.), we can efficiently determine that $A^{12} = 144A + 89I$.

Substituting the matrices for $A$ and $I$ gives:

.

Finally, we use this result in the original equation for $n=12$:

.

From this, we see that $x[12] = 233$.

This problem is a direct application of Cauchy's Integral Formula for derivatives, which states that $\oint_C \frac{f(z)}{(z-z_0)^{n+1}} dz = \frac{2\pi j}{n!} f^{(n)}(z_0)$.

The integrand $\frac{\sin z}{(z-2\pi j)^{3}}$ has a single pole of order $n+1=3$ at $z_0 = 2\pi j$ inside the contour. We can identify $f(z) = \sin z$ and $n=2$.

First, we find the second derivative of $f(z)$: $f'(z) = \cos z$ and $f''(z) = -\sin z$.
Evaluating this at the pole gives $f''(2\pi j) = -\sin(2\pi j)$.

Now, substitute these values into the formula for the full expression:
Value $= -\frac{1}{2\pi} \left( \frac{2\pi j}{2!} f''(2\pi j) \right) = -\frac{1}{2\pi} \left( \pi j (-\sin(2\pi j)) \right) = \frac{j}{2} \sin(2\pi j)$.

Using the identity $\sin(ix) = i \sinh(x)$, we get $\frac{j}{2} (i \sinh(2\pi)) = \frac{i^2}{2} \sinh(2\pi) = -\frac{1}{2}\sinh(2\pi)$, which is approximately $-133.87$.

To find the volume of this region, we integrate the cylindrical volume element, $dV = \rho \, d\rho \, d\varphi \, dz$, over the specified bounds.

The volume integral is set up as $V = \int_{3}^{4.5} \int_{\pi/8}^{\pi/4} \int_{3}^{5} \rho \, d\rho \, d\varphi \, dz$.

Since the limits are all constants and the integrand is separable, we can evaluate this as a product of three single-variable integrals:
$V = \left( \int_{3}^{5} \rho \, d\rho \right) \left( \int_{\pi/8}^{\pi/4} d\varphi \right) \left( \int_{3}^{4.5} dz \right)$
Evaluating each definite integral gives us:
$V = \left[ \frac{\rho^2}{2} \right]_{3}^{5} \cdot \left[ \varphi \right]_{\pi/8}^{\pi/4} \cdot \left[ z \right]_{3}^{4.5}$
$V = \left( \frac{25-9}{2} \right) \cdot \left( \frac{\pi}{4} - \frac{\pi}{8} \right) \cdot (4.5 - 3)$
Multiplying these results yields the final volume: $V = (8) \cdot (\frac{\pi}{8}) \cdot (1.5) = 1.5\pi \approx 4.712$.

The Laplace transform of a periodic signal is found by transforming one period of the signal and then dividing by $1-e^{-sT}$. The first period, $f_1(t)$, is a single pulse that is 1 from $t=0$ to $t=T/2$.

First, we find the transform of this single pulse:
$L\{f_1(t)\} = \int_{0}^{T/2} (1)e^{-st} dt = \left[ \frac{e^{-st}}{-s} \right]_0^{T/2} = \frac{1 - e^{-sT/2}}{s}$.

Now, we apply the formula for a periodic signal:
$F(s) = \frac{L\{f_1(t)\}}{1-e^{-sT}} = \frac{(1 - e^{-sT/2})/s}{1-e^{-sT}}$.

To simplify, we factor the denominator using the difference of squares formula, $a^{2-}b^2=(a-b)(a+b)$, where $a=1$ and $b=e^{-sT/2}$:
$1-e^{-sT} = (1-e^{-sT/2})(1+e^{-sT/2})$.

Substituting this back and canceling the common term $(1-e^{-sT/2})$ gives the final answer:
$F(s) = \frac{1 - e^{-sT/2}}{s(1-e^{-sT/2})(1+e^{-sT/2})} = \frac{1}{s(1+e^{-sT/2})}$.

A circuit built from linear resistors, inductors, and capacitors is a Linear Time-Invariant (LTI) system. The input is a sum of three sinusoids at frequencies $\omega_0$, $2\omega_0$, and $3\omega_0$. Due to the principle of superposition, which applies to all linear systems, the total output is the sum of the outputs produced by each input frequency component individually.

A fundamental property of LTI systems is that they cannot create new frequencies. A sinusoidal input at frequency $f$ will produce a steady-state output at that same frequency $f$. The system only alters the signal's amplitude and phase. Thus, the output must also be a sum of sinusoids at $\omega_0$, $2\omega_0$, and $3\omega_0$, each with its own modified amplitude $b_k$ and phase shift $\phi_k$.

The system described is a first-order low-pass filter with a transfer function $H(s) = \frac{1}{1+sT}$. We can find the time response for each excitation by finding the inverse Laplace transform of the output signal $Y(s) = H(s)R(s)$, where $R(s)$ is the Laplace transform of the input.

For an impulse input (X), $R(s)=1$, the output is $Y(s) = \frac{1}{1+sT}$. Its inverse transform is an exponential decay, $y(t) \propto e^{-t/T}$, matching response R.

For a unit step input (Y), $R(s)=1/s$, the output is $Y(s) = \frac{1}{s(1+sT)}$. The inverse transform gives the system's step response, $y(t) = 1-e^{-t/T}$, which matches P.

For a ramp input (Z), $R(s)=1/s^2$, the output is $Y(s) = \frac{1}{s^2(1+sT)}$. The inverse transform yields the ramp response, $y(t) = t - T(1-e^{-t/T})$, matching Q.

Therefore, the correct pairings are X $\rightarrow$ R, Y $\rightarrow$ P, and Z $\rightarrow$ Q.