GATE ECE 2012

The small-signal parameter $r_{\pi}$ depends on the DC operating point of the transistor, not the time-varying signal component. The given base current is $i_b(t) = 1 + 0.1 \cos(10000 \pi t)$ mA. The DC component of this current, which sets the bias point, is $I_B = 1$ mA.

The formula for the small-signal input resistance is $r_{\pi} = \frac{V_T}{I_B}$.

At room temperature (300 K), the thermal voltage $V_T$ is approximately $25$ mV.

Substituting the values, we get:
$r_{\pi} = \frac{25 \text{ mV}}{1 \text{ mA}} = \frac{25 \times 10^{-3} \text{ V}}{1 \times 10^{-3} \text{ A}} = 25 \, \Omega$.

The mean-square value $E[X^2(t)]$ represents the total average power of the signal. For a real process, this power is calculated by integrating the given one-sided power spectral density (PSD) over all positive frequencies and dividing by $\pi$.

P = E\[X^2(t)]$ = \frac{1}{\pi}\int_0^\infty S_X(\omega)d\omega$

The integral is simply the area under the provided PSD plot. The area of the given shape between $\omega = 9 \times 10^3$ and $11 \times 10^3$ rad/s is 6400. Thus, the total power is E\[X^2(t)]$ = \frac{6400}{\pi}$.

The term $|E[X(t)]|$ is the magnitude of the signal's mean, also known as its DC component. A DC component would manifest as spectral power at $\omega=0$. Since the given PSD is zero for all frequencies below $9 \times 10^3$ rad/s, there is no power at $\omega=0$. Therefore, the mean of the process is zero, and $|E[X(t)]|=0$.

To find the maximum signaling rate without inter-symbol interference, we use the formula for the bandwidth of a baseband signal with a raised-cosine pulse. This relationship connects the bandwidth ($BW$), the symbol rate ($R_s$), and the roll-off factor ($\alpha$).

The formula is:
$BW = \frac{R_s}{2}(1 + \alpha)$

We are given a bandwidth of $3500$ Hz and an excess bandwidth of 75%, which means $\alpha = 0.75$. Substituting these values allows us to solve for the maximum symbol rate, $R_s$.

$3500 = \frac{R_s}{2}(1 + 0.75)$
$3500 = \frac{R_s}{2}(1.75)$
$R_s = \frac{2 \times 3500}{1.75} = 4000$ symbols/sec.

First, let's analyze the wave's propagation. The phase term $e^{j(\omega t+3x-4y)}$ indicates the incident wave travels in a direction where the $x$-component is negative and the $y$-component is positive. When this wave reflects off the conducting slab at $x=0$, the component of propagation normal to the surface (the $x$-component) reverses direction. The tangential component (the $y$-component) is unchanged. This results in a reflected wave phase term of $e^{j(\omega t-3x-4y)}$.

Next, we apply the boundary conditions for a perfect conductor. The total tangential electric field at the surface must be zero. This means the tangential component of the reflected E-field must be the exact opposite of the incident tangential E-field. The incident field's tangential part (in the y-z plane) is $(6\hat{a}_{y}+5\hat{a}_{z})$. Therefore, the reflected tangential part is $-(6\hat{a}_{y}+5\hat{a}_{z})$.

Similarly, for reflection from a perfect conductor, the normal component of the E-field also inverts. The incident field's normal component (along $\hat{a}_x$) is $8\hat{a}_{x}$. Upon reflection, this becomes $-8\hat{a}_{x}$.

Combining these reflected components gives the full amplitude of the reflected E-field: $-8\hat{a}_{x} - 6\hat{a}_{y} - 5\hat{a}_{z}$. Paired with the reflected phase term, we get the final expression for the reflected wave.

First, we determine the wave's frequency from its propagation constant, $k=25$ rad/m, found in the exponential term. In free space, the angular frequency is $\omega = kc$, where $c$ is the speed of light. This gives $\omega = 25 \times (3 \times 10^8) = 7.5 \times 10^9$ rad/s. The frequency is therefore $f = \omega / (2\pi) \approx 1.2$ GHz.

Next, we analyze the polarization. The vector term $(\hat{a}_{y}+j\hat{a}_{z})$ shows that the electric field has components in the $y$ and $z$ directions with equal amplitudes. The factor $j$ indicates that the $z$-component leads the $y$-component by a phase of $90^\circ$. Equal amplitudes and a $90^\circ$ phase difference define circular polarization.

To determine the handedness, let's observe the field's rotation at a fixed position like $x=0$. The time-domain field is $\vec{E}(0,t) = \text{Re}\{10(\hat{a}_{y}+j\hat{a}_{z})e^{j\omega t}\} = 10\cos(\omega t)\hat{a}_{y} - 10\sin(\omega t)\hat{a}_{z}$. As time increases from $t=0$, the vector rotates from the $+y$ axis toward the $-z$ axis. For a wave propagating in the $+x$ direction, this counter-clockwise rotation signifies left circular polarization.

The circuit shown is a clocked S-R flip-flop, where the inputs $A$ and $B$ function as the Set ($S$) and Reset ($R$) signals, respectively. The race-around condition is a specific problem that occurs in level-triggered J-K flip-flops when both $J$ and $K$ inputs are held high, causing the output to oscillate. This phenomenon is a consequence of the J-K flip-flop's internal design and does not apply to S-R flip-flops. While an S-R flip-flop has its own problematic "invalid" state when $S=R=1$, this is not the race-around condition.

The comparator's output, $Y$, is 1 only when the 2-bit number $A$ is strictly greater than the 2-bit number $B$. We can determine the total number of such cases by examining each possible value of $B$.

• If $B=00_2$, $A$ can be $01_2$, $10_2$, or $11_2$ (3 cases).
• If $B=01_2$, $A$ can be $10_2$ or $11_2$ (2 cases).
• If $B=10_2$, $A$ must be $11_2$ (1 case).
• If $B=11_2$, no value of $A$ is greater (0 cases).

Summing these up, the total number of combinations where $A > B$ is $3 + 2 + 1 + 0 = 6$.

To find the current in the circuit, we must find the operating point that satisfies both the external circuit's behavior and the diode's characteristics.

First, apply Kirchhoff's Voltage Law (KVL) around the loop:
$10 - (1000 \cdot i) - v = 0$

Next, we use the given i-v characteristic for the diode, which we can rearrange to express $v$ in terms of $i$:
$i = \frac{v-0.7}{500} \implies v = 500i + 0.7$

Now, substitute this expression for $v$ into our KVL equation:
$10 - 1000i - (500i + 0.7) = 0$

Solving for the current $i$:
$9.3 - 1500i = 0 \implies 1500i = 9.3 \implies i = \frac{9.3}{1500} = 0.0062 \text{ A}$

This corresponds to a current of $6.2$ mA.

This is an ideal circuit, which means there is no resistance ($R=0$). The time constant of a capacitor circuit is given by $\tau = RC$, which is therefore zero. This implies that any change in the circuit's state must happen instantaneously.

When the switch closes at $t=0$, charge redistributes from the pre-charged capacitor $C_1$ to the uncharged capacitor $C_2$. This causes the voltage across $C_2$ to jump from 0V to a new equilibrium voltage instantly. The relationship between current and voltage for a capacitor is $i(t) = C \frac{dv(t)}{dt}$. For the voltage to change in a discontinuous step, its derivative, $\frac{dv}{dt}$, must be infinite. A current that is infinite for an infinitesimally short duration is, by definition, an impulse function.

The average power in an AC circuit is the power dissipated by the resistive component. The imaginary part of the impedance (reactance) stores and releases energy but consumes no average power. From the impedance $Z = (4 - j3) \Omega$, we identify the resistance as $R = 4 \Omega$.

The given current has a peak amplitude of $I_m = 5$ A. To calculate average power, we must use the RMS (root mean square) value of the current, which is $I_{rms} = \frac{I_m}{\sqrt{2}} = \frac{5}{\sqrt{2}}$ A.

Now, we can find the average power using the formula $P = I_{rms}^2 R$:
$P = \left(\frac{5}{\sqrt{2}}\right)^2 \times 4 = \frac{25}{2} \times 4 = 50 \text{ W}$

This problem is a direct application of the frequency differentiation property of the Laplace transform. This property states that if the Laplace transform of $f(t)$ is $F(s)$, then the transform of $t f(t)$ is given by $L\{t f(t)\} = -\frac{d}{ds}F(s)$.

We are given the transform $F(s) = \frac{1}{s^{2+}s+1}$.

To find the solution, we must calculate the negative derivative of $F(s)$ with respect to $s$. Using the chain rule for differentiation:
$L\{t f(t)\} = -\frac{d}{ds}\left(\frac{1}{s^{2+}s+1}\right)$
$= - \left[ \frac{-1}{(s^{2+}s+1)^2} \cdot \frac{d}{ds}(s^{2+}s+1) \right]\$$
$= - \left[ \frac{-1}{(s^{2+}s+1)^2} \cdot (2s+1) \right]\$$
The two negative signs cancel, leading to the final answer:
$\frac{2s+1}{(s^{2+}s+1)^2}$

This is a first-order linear differential equation. First, let's arrange it into the standard form $\frac{dx}{dt} + P(t)x = Q(t)$ by dividing the entire equation by $t$. This gives us $\frac{dx}{dt} + \frac{1}{t}x = 1$.

The next step is to find the integrating factor, $I(t)$, which is calculated as $I(t) = e^{\int P(t) dt}$. Here, $P(t)=\frac{1}{t}$, so the integrating factor is $e^{\int \frac{1}{t} dt} = e^{\ln t} = t$.

Now, multiply the standard form of the equation by the integrating factor $t$. This simplifies the left side to the derivative of a product: $t\frac{dx}{dt} + x = t \implies \frac{d}{dt}(t \cdot x) = t$.

Integrating both sides with respect to $t$ gives us $t \cdot x = \int t \,dt = \frac{t^2}{2} + C$. The general solution for $x$ is thus $x = \frac{t}{2} + \frac{C}{t}$.

Finally, we use the initial condition $x(1) = 0.5$ to find the constant $C$. Substituting $t=1$ and $x=0.5$ into the general solution gives $0.5 = \frac{1}{2} + \frac{C}{1}$, which means $C=0$.

Therefore, the specific solution is $x = \frac{t}{2}$.

Let's analyze the loop containing the voltage source, capacitor $C_1$, and diode $D_1$. After an initial charging period, the capacitor $C_1$ will hold a steady DC voltage, which we'll call $V_{C1}$. By applying Kirchhoff's Voltage Law (KVL) to this loop, we can express the voltage across the diode, $v(t)$, as the input voltage minus the capacitor voltage: $v(t) = \cos(\omega t) - V_{C1}$.

The ideal diode $D_1$ has its cathode connected to ground. It will conduct and act as a short circuit if its anode voltage $v(t)$ tries to exceed its cathode voltage (0V). This means the diode "clamps" the voltage $v(t)$, preventing it from ever becoming positive. Therefore, the maximum possible value of $v(t)$ is 0V.

To find $V_{C1}$, we set the maximum value of our expression for $v(t)$ to zero. The maximum value of $\cos(\omega t)$ is 1, so $max(v(t)) = 1 - V_{C1}$. Setting this to zero gives $1 - V_{C1} = 0$, which means $V_{C1} = 1$V. Substituting this back into our KVL equation yields the final answer: $v(t) = \cos(\omega t) - 1$.

This circuit is a CMOS logic gate, where the output $Y$ is the inverse of the logic performed by the NMOS pull-down network. Let's analyze this network first.

The transistors controlled by inputs $A$ and $B$ are in parallel, which creates a path to ground if either $A$ or $B$ is high. This corresponds to a logical OR function, $A+B$. This parallel group is in series with the transistor for input $C$, which corresponds to a logical AND.

Thus, the entire pull-down network's logic is $(A+B) \cdot C$. The final output $Y$ is the complement of this expression: $Y = \overline{(A+B) \cdot C}$.

To simplify, we apply De Morgan's laws. First, we break the primary AND operation: $Y = \overline{(A+B)} + \bar{C}$. Then, we break the OR operation: $\overline{(A+B)} = \bar{A}\bar{B}$. Substituting this back gives the final result: $Y = \bar{A}\bar{B} + \bar{C}$.

A key principle of information theory is that a source's entropy is maximized when its symbols are equally probable. The entropy function is a concave curve that peaks at the point of uniform distribution.

Initially, the first two symbols have equal probabilities, let's call them $p_1$ and $p_2$, so $p_1 = p_2$. The encoder then shifts these probabilities to $p_1' = p_1 + \varepsilon$ and $p_2' = p_2 - \varepsilon$. This action makes the probabilities unequal. Because you are moving away from the point of equality, you are moving down from the peak of the entropy function. Therefore, the overall entropy of the source decreases.

The characteristic impedance ($Z_0$) of a coaxial cable is determined by its physical dimensions and the dielectric material separating the conductors. We can calculate it using the approximate engineering formula:
$Z_0 = \frac{138}{\sqrt{\epsilon_r}} \log_{10}\left(\frac{D}{d}\right)$
Here, $D$ is the outer diameter (2.4 mm), $d$ is the inner diameter (1 mm), and $\epsilon_r$ is the relative permittivity of the dielectric (10.89).

Plugging the given values into the formula:
$Z_0 = \frac{138}{\sqrt{10.89}} \log_{10}\left(\frac{2.4}{1}\right)$
Since $\sqrt{10.89}$ is 3.3, the calculation becomes $\frac{138}{3.3} \times \log_{10}(2.4)$, which yields an impedance of approximately $15.89 \, \Omega$.

Directivity, $D$, is the ratio of the maximum radiation intensity, $U_{max}$, to the average intensity, which can be expressed as $D = \frac{4\pi U_{max}}{P_{rad}}$. The given radiation pattern is $F(\theta) = \cos^4\theta$. Since this is a normalized pattern, its maximum value is $U_{max} = F(0) = 1$.

We find the total radiated power, $P_{rad}$, by integrating the intensity over the specified hemisphere:
$P_{rad} = \int_{\phi=0}^{2\pi} \int_{\theta=0}^{\pi/2} \cos^4\theta \sin\theta \,d\theta \,d\phi$
Solving the integral gives $P_{rad} = 2\pi \left[-\frac{\cos^5\theta}{5}\right]_{0}^{\pi/2} = 2\pi(0 - (-\frac{1}{5})) = \frac{2\pi}{5}$.

Now, we can calculate the directivity:
$D = \frac{4\pi U_{max}}{P_{rad}} = \frac{4\pi(1)}{2\pi/5} = 10$
Finally, converting this value to decibels yields $D_{dB} = 10 \log_{10}(10) = 10$ dB.

To find the region of convergence (ROC) for the Z-transform of $x[n]$, we analyze its constituent parts. The signal is a sum of two components: a two-sided exponential, $(1/3)^{|n|}$, and a right-sided exponential, $-(1/2)^{n}u[n]$.

First, consider the two-sided term $(1/3)^{|n|}$. This signal is composed of a right-sided part, $(1/3)^n u[n]$, and a left-sided part. The right-sided part's transform converges for $|z| > 1/3$, while the left-sided part's transform converges for $|z| < 3$. Together, the ROC for $(1/3)^{|n|}$ is their intersection: $\frac{1}{3} < |z| < 3$.

Next, consider the right-sided term $(1/2)^{n}u[n]$. The ROC for any causal exponential signal of the form $a^n u[n]$ is $|z| > |a|$. In this case, with $a=1/2$, the ROC is $|z| > 1/2$.

The ROC of the overall signal $x[n]$ is the intersection of the ROCs of its individual components. We must find the region where both conditions, $\frac{1}{3} < |z| < 3$ and $|z| > 1/2$, are satisfied simultaneously. This common region is $\frac{1}{2} < |z| < 3$.

The function $f(X, Y, Z) = \sum(2, 3, 4, 5)$ is defined by a set of minterms. To find the prime implicants, we simplify the expression by grouping adjacent minterms using Boolean algebra.

First, we group minterms 2 ($\bar{X}Y\bar{Z}$) and 3 ($\bar{X}YZ$). This pair simplifies to $\bar{X}Y(\bar{Z}+Z) = \bar{X}Y$.

Next, we group minterms 4 ($X\bar{Y}\bar{Z}$) and 5 ($X\bar{Y}Z$). This pair simplifies to $X\bar{Y}(\bar{Z}+Z) = X\bar{Y}$.

These two resulting terms, $\bar{X}Y$ and $X\bar{Y}$, represent the largest possible groupings of the original minterms and cannot be simplified further. Therefore, they are the prime implicants.

The steady-state output of a system to a sinusoidal input is determined by the system's response at the input frequency. The system's transfer function, $G(s)$, has a numerator term $(s^{2+}9)$, which means the system has zeros at $s=\pm j3$. These are known as transmission zeros.

An input signal of $\sin(\omega t)$ introduces poles into the response at $s=\pm j\omega$. These poles are responsible for the sustained sinusoidal output.

If the input frequency $\omega$ matches the frequency of the system's zeros on the imaginary axis, the system completely blocks the signal. Here, choosing $\omega=3$ rad/s means the input's poles at $s=\pm j3$ are cancelled by the system's zeros. This pole-zero cancellation eliminates the sinusoidal component from the output, causing the steady-state response to be zero.

To find the impedance between two nodes, we can connect a 1 V test source ($V_{\in}$) across them and calculate the current ($I_{\in}$) that flows from the source. The impedance is then the ratio $Z_{\in} = V_{\in} / I_{\in}$.

Applying this method, the analysis of the circuit yields two key results. First, the control current $i_b$ is found by relating it to the 1 V source:
$i_b = \frac{1 \text{ V}}{10 \text{ k}\Omega} = 10^{-4} \text{ A}$

Second, applying Kirchhoff's laws at the input node establishes a relationship between the total current $I_{\in}$ and the control current $i_b$, which simplifies to $I_{\in} = 200 i_b$.

By substituting the value of $i_b$ into this expression, we can find the total current:
$I_{\in} = 200 \times (10^{-4} \text{ A}) = 0.02 \text{ A}$

Finally, the impedance is calculated as:
$Z_{\in} = \frac{V_{\in}}{I_{\in}} = \frac{1 \text{ V}}{0.02 \text{ A}} = 50 \Omega$

Let $i_1$ be the current through the inductor and $i$ be the current through the resistor. By applying Kirchhoff's Current Law (KCL) at the node connecting the current source, inductor, and resistor, we can establish the relationship $i_1 = i + 1$.

Next, we apply Kirchhoff's Voltage Law (KVL) around the loop containing the resistor and the inductor. The two $1V$ sources oppose each other, resulting in a net voltage of zero from them in the loop. This leaves us with the voltage drops across the resistor and inductor summing to zero: $i(1) + i_1(j1) = 0$.

Now, substitute the KCL expression into the KVL equation: $i + (i+1)j = 0$.
Solving for $i$, we get $i + ij + j = 0$, which simplifies to $i(1+j) = -j$, so $i = \frac{-j}{1+j}$.

Finally, substitute this result back into our initial KCL equation to find the inductor current $i_1$:
$i_1 = 1 + i = 1 + \frac{-j}{1+j} = \frac{(1+j)-j}{1+j} = \frac{1}{1+j}$ A.

By Cauchy's Residue Theorem, the value of the expression is the sum of the residues of $f(z)$ at the poles located inside the contour $C$.

The function $f(z)=\frac{1}{z+1}-\frac{2}{z+3}$ has two simple poles at $z=-1$ and $z=-3$. The contour $C$ is described by $|z+1|=1$, which is a circle of radius 1 centered at $z=-1$.

This contour encloses the pole at $z=-1$ (its center) but excludes the pole at $z=-3$, since the distance from the center is $|-3 - (-1)| = 2$, which is greater than the radius.

Therefore, we only need to find the residue of $f(z)$ at $z=-1$. For this calculation, it's helpful to combine the terms of $f(z) = \frac{(z+3)-2(z+1)}{(z+1)(z+3)} = \frac{-z+1}{(z+1)(z+3)}$. The residue at the simple pole $z=-1$ is:
$\text{Res}(f, -1) = \lim_{z \to -1} (z+1) \frac{-z+1}{(z+1)(z+3)} = \frac{-(-1)+1}{-1+3} = \frac{2}{2} = 1$
The value of the integral is simply this residue, which is 1.

Since $X$ and $Y$ are independent and uniformly distributed, we can solve this using geometric probability. The sample space of all possible $(X,Y)$ pairs forms a square in the $xy$-plane defined by $-1 \le x \le 1$ and $-1 \le y \le 1$. The total area of this square is $2 \times 2 = 4$.

For the event $\max(X,Y) < 1/2$ to occur, it's necessary that both $X < 1/2$ and $Y < 1/2$. This defines our favorable region.

This favorable region is a smaller square where $x$ and $y$ are both restricted to the interval $[-1, 1/2)$. The side length of this square is $1/2 - (-1) = 3/2$, making its area $(3/2) \times (3/2) = 9/4$.

The probability is the ratio of the favorable area to the total sample space area: $\frac{9/4}{4} = \frac{9}{16}$.

First, we identify $x = \sqrt{-1}$ as the imaginary unit, $i$. The problem therefore asks us to find the value of $i^i$.

To raise an imaginary number to an imaginary power, we should express the base in its exponential form. Using Euler's formula, the number $i$ can be written as $e^{i\pi/2}$, since it lies at an angle of $\pi/2$ on the complex plane.

Now we can substitute this into our expression:
$i^i = (e^{i\pi/2})^i$

Using the property of exponents that $(a^b)^c = a^{bc}$, we multiply the powers:
$e^{(i\pi/2) \cdot i} = e^{i^2\pi/2}$

Finally, since $i^2 = -1$, the value simplifies to $e^{-\pi/2}$.

First, let's determine the volume of the transistor's source region. The volume is the area multiplied by the depth, which is $V = (1 \ \mu\text{m})^2 \times (1 \ \mu\text{m}) = 1 \ \mu\text{m}^3$. In cubic centimeters, this volume is $V = 10^{-12} \ \text{cm}^3$.

The source is heavily doped n-type silicon, so the electron concentration is approximately equal to the donor density, $n \approx N_D = 10^{19} / \text{cm}^3$. We can find the concentration of holes, which are the minority carriers, using the law of mass action: $p = n_i^2 / n$.

Substituting the given values, the hole concentration is $p = (10^{10} \text{cm}^{-3})^2 / (10^{19} \text{cm}^{-3}) = 10 \ \text{holes}/\text{cm}^3$.

To find the total number of holes, we multiply this concentration by the source volume:
Number of holes = $p \times V = (10 \ \text{holes}/\text{cm}^3) \times (10^{-12} \ \text{cm}^3) = 10^{-11}$.

This result represents the expected number of holes. Since this value is an extremely small fraction, the probability of finding even a single hole in this tiny volume is negligible. Therefore, the most likely number of holes is 0.

In a coherent BPSK receiver, a phase error of $\phi_e$ between the incoming carrier and the local oscillator reduces the desired signal component at the correlator output by a factor of $cos(\phi_e)$, while the noise variance remains unchanged. The signal amplitude at the decision stage becomes $\sqrt{E}cos(\phi_e)$ instead of $\sqrt{E}$. The noise variance at the correlator output remains $\sigma_{n}^{2} = N_{0}/2$.

The probability of error for BPSK with a phase error is given by:
$P_{e} = Q\left(\frac{\text{signal amplitude}}{\text{noise standard deviation}}\right) = Q\left(\frac{\sqrt{E}cos(\phi_e)}{\sqrt{N_{0}/2}}\right)$

Given the phase error is $\phi_e = 45^{\circ}$, we have $cos(45^{\circ}) = \frac{1}{\sqrt{2}}$. Substituting this value into the error probability formula:
$P_{e} = Q\left(\frac{\sqrt{E}(1/\sqrt{2})}{\sqrt{N_{0}/2}}\right) = Q\left(\sqrt{\frac{E/2}{N_{0}/2}}\right)$

Simplifying the expression inside the Q-function gives the final result:
$P_{e} = Q\left(\sqrt{\frac{E}{N_{0}}}\right)$

The $100\,\Omega$ transmission line acts as a quarter-wave transformer, as its characteristic impedance is the geometric mean of the loads it is matching: $\sqrt{50\,\Omega \times 200\,\Omega} = 100\,\Omega$. The length $L$ of such a transformer must be an odd multiple of a quarter-wavelength, a condition that must be met for both frequencies.

The wavelengths are $\lambda_1 = c/f_1 \approx 0.7$ m for $f_1 = 429$ MHz, and $\lambda_2 = c/f_2 = 0.3$ m for $f_2 = 1$ GHz. For the length $L$ to be an odd multiple of both $\lambda_1/4$ and $\lambda_2/4$, we must have $L=m_1\frac{\lambda_1}{4}$ and $L=m_2\frac{\lambda_2}{4}$ for some odd integers $m_1$ and $m_2$.

This implies the ratio of these odd integers is $\frac{m_1}{m_2} = \frac{\lambda_2}{\lambda_1} \approx \frac{0.3}{0.7} = \frac{3}{7}$. The simplest solution uses $(m_1, m_2)=(3,7)$, giving a length of $L = 3 \times (0.7/4) = 0.525$ m. As this is not an option, we find the next solution by using the next pair of odd integers in the same ratio, which is $(9,21)$. This yields a length of $L = 9 \times (0.7/4) = 1.575$ m, which is approximately $1.58$ m.

Let's first test for time-invariance. If we delay the input by $t_0$ to get $x(t-t_0)$, the new output is $\int_{-\infty }^{t} x(\tau-t_0)\cos(3\tau) d\tau$. This is not equal to the original output delayed by $t_0$, which is $y(t-t_0) = \int_{-\infty }^{t-t_0} x(\tau)\cos(3\tau) d\tau$. Because the system's behavior depends on the fixed $\cos(3\tau)$ term, it is not time-invariant.

Next, we test for stability by checking if a bounded input can produce an unbounded output. Let's choose the bounded input $x(t) = \cos(3t)$. The output becomes $y(t) = \int_{-\infty }^{t} \cos^2(3\tau) d\tau$. Since the integrand $\cos^2(3\tau)$ is always non-negative, its integral will increase indefinitely as $t \to \infty$. This means a bounded input produces an unbounded output, so the system is not stable.

To find the conditions for oscillation, we first determine the system's characteristic equation, which is $1+G(s)H(s)=0$. This yields $s^{3}+a s^{2}+(2+K) s+(1+K)=0$.

For the system to exhibit sustained oscillations, a row in its Routh-Hurwitz array must be zero. Setting the $s^1$ row's element to zero gives the condition for marginal stability: $a(2+K) - (1+K) = 0$.

The frequency of these oscillations, $\omega$, is found from the auxiliary equation, which is formed from the $s^2$ row of the array: $as^2 + (1+K) = 0$. Substituting $s=j\omega$ gives $-a\omega^2 + (1+K) = 0$.

We are given $\omega = 2$ rad/s. Plugging this into the auxiliary equation results in $-4a + (1+K) = 0$, or $1+K = 4a$. Now we solve this simultaneously with the first condition. Substituting $1+K=4a$ into $a(2+K) = 1+K$ gives $a(2+K) = 4a$. Since $a \neq 0$, we can divide by $a$ to get $2+K=4$, which means $K=2$. Finally, substituting $K=2$ back into $1+K = 4a$ yields $3=4a$, so $a=0.75$.