GATE ECE 2015 Set 2

The bilateral Laplace transform is found by integrating from $-\infty$ to $\infty$. For the given function $f(t)$, which is a rectangular pulse, the integral is non-zero only between $t=a$ and $t=b$.

The transform is calculated as:
$F(s) = \int_{-\infty }^{\infty } f(t)e^{-st} dt = \int_{a}^{b} (1)e^{-st} dt$

Now, we evaluate this straightforward integral:
$F(s) = \left[ \frac{e^{-st}}{-s} \right]_{t=a}^{t=b} = \frac{e^{-sb}}{-s} - \frac{e^{-sa}}{-s}$

Rearranging the terms gives the final result:
$F(s) = \frac{e^{-as} - e^{-bs}}{s}$

A matrix with complex entries will have all real eigenvalues if and only if it is a Hermitian matrix. A matrix $A$ is Hermitian if it is equal to its own conjugate transpose, a condition written as $A = A^*$, where $A^* = (\bar{A})^T$.

This condition implies that the element in the $i$-th row and $j$-th column must be the complex conjugate of the element in the $j$-th row and $i$-th column, i.e., $a_{ij} = \overline{a_{ji}}$.

For the given matrix, let's look at the element in the first row, second column: $a_{12} = 5+j$. The element in the second row, first column is $a_{21} = x$.

For the matrix to be Hermitian, we must have $a_{21} = \overline{a_{12}}$.
Therefore, $x = \overline{5+j}$, which means $x = 5-j$.

The core idea is to express the condition $f(z_1) = f(z_2)$ algebraically and simplify it to find a relationship between the coefficients. Let's start by setting the expressions for $f(z_1)$ and $f(z_2)$ equal to each other:
$\frac{az_1+b}{cz_1+d} = \frac{az_2+b}{cz_2+d}$
To eliminate the fractions, we cross-multiply, yielding $(az_1+b)(cz_2+d) = (az_2+b)(cz_1+d)$. After expanding both sides, we can cancel the common terms $acz_1z_2$ and $bd$, which simplifies the equation to $adz_1 + bcz_2 = adz_2 + bcz_1$.

By rearranging the terms to group $z_1$ and $z_2$, we get:
$bc(z_2 - z_1) = ad(z_2 - z_1)$
Since the condition holds for $z_1 \neq z_2$, the term $(z_2 - z_1)$ is non-zero, allowing us to divide both sides by it. This reveals the simple relationship $bc = ad$. Finally, we plug in the given values and solve for $d$:
$d = \frac{bc}{a} = \frac{4 \times 5}{2} = 10$

This is a first-order differential equation that can be solved using the method of separation of variables. First, we group the $x$ and $y$ terms on opposite sides of the equation.

$\frac{dy}{1+\cos(2y)} = \frac{dx}{1-\cos(2x)}$

Using the trigonometric double-angle identities $1+\cos(2\theta) = 2\cos^2(\theta)$ and $1-\cos(2\theta) = 2\sin^2(\theta)$, we can simplify the denominators.

$\frac{dy}{2\cos^2(y)} = \frac{dx}{2\sin^2(x)}$

This is equivalent to $\frac{1}{2}\sec^2(y) \, dy = \frac{1}{2}\csc^2(x) \, dx$. After cancelling the $\frac{1}{2}$ on both sides, we integrate.

$\int \sec^2(y) \, dy = \int \csc^2(x) \, dx$

Integrating yields $\tan(y) = -\cot(x) + c$, where $c$ is the constant of integration. Rearranging the terms gives the final general solution: $\tan(y) + \cot(x) = c$.

To determine the nature of the signal $x(t)$, we examine the symmetry properties of its Fourier series coefficients, $a_k$. A signal is purely real if and only if its coefficients exhibit conjugate symmetry, meaning $a_k = a_{-k}^*$.

This single condition implies two separate properties for the magnitude and phase:

1. The magnitude spectrum must be even: $|a_k| = |a_{-k}|$.
2. The phase spectrum must be odd: $\angle a_k = -\angle a_{-k}$.

Observing the plots, the magnitude $|a_k|$ is clearly an even function, as it is a mirror image about the vertical axis. The phase plot shows that for non-zero coefficients like $k=\pm 2$ and $k=\pm 3$, the phase is $-\pi$. To check for odd symmetry, we test if $\angle a_k = -\angle a_{-k}$. For $k=2$, this means $-\pi = -(\angle a_{-2}) = -(-\pi) = \pi$. This identity holds true because phase angles of $\pi$ and $-\pi$ are equivalent. Since both symmetry conditions are met, the signal $x(t)$ must be purely real.

In a series RLC circuit, Kirchhoff's Voltage Law applies to the phasors, not the magnitudes. The relationship between the voltage magnitudes is described by the voltage triangle, which follows the Pythagorean theorem: $V_{source}^2 = V_R^2 + (V_L - V_C)^2$.

Plugging in the given values:
$100^2 = 80^2 + (40 - V_C)^2$

Now, we can solve for the net reactive voltage term:
$(40 - V_C)^2 = 100^2 - 80^2 = 10000 - 6400 = 3600$

Taking the square root of both sides gives $|40 - V_C| = 60$. This yields two possibilities: $V_C = -20$ V or $V_C = 100$ V. Since a voltage magnitude must be a positive value, the correct answer is $100$ V.

The total voltage $V_{ab}$ is a combination of a DC component and an AC component. To find its average value, we can use the principle of superposition. The average of the total voltage is the sum of the average from the DC source and the average from the AC source.

The AC source, $5\pi \sin(5000t)$, produces a purely sinusoidal voltage. By definition, the average value of any pure sinusoidal waveform over a full cycle is zero.

For the 5 V DC source, we analyze the circuit in its DC steady state. The capacitor acts as an open circuit, blocking DC current. With no current flowing, there is no voltage drop across the resistors. This leaves the potential at node 'a' equal to 5 V and the potential at node 'b' at 0 V. The DC component of $V_{ab}$ is therefore $5 \text{ V} - 0 \text{ V} = 5 \text{ V}$.

Summing the average values from both sources, the total average voltage is $5 \text{ V} + 0 \text{ V} = 5 \text{ V}$.

First, we characterize the network using its impedance matrix, $[Z]$. The port equations $V_1 = 6I_1 + 4I_2$ and $V_2 = 4I_1 + 6I_2$ define this matrix as

.

The admittance matrix, $[Y]$, is the inverse of the impedance matrix. To find $[Y] = [Z]^{-1}$, we first calculate the determinant, which is $\det(Z) = (6)(6) - (4)(4) = 20$.

The inverse matrix is then calculated as

.

Since the provided options only contain positive values, we take the magnitude of each element in the resulting matrix to match the answer format.

In a steady state, the rate at which charge carriers are generated is balanced by the rate at which they recombine. The excess concentration of minority carriers ($\Delta p$) is therefore the product of the generation rate ($G$) and the minority carrier lifetime ($\tau_p$).

We are given a generation rate of $G = 10^{20}$ pairs/(cm³·s) and a lifetime of $\tau_p = 1 \mu s = 10^{-6}$ s.

The steady-state excess hole concentration is calculated as:
$\Delta p = G \times \tau_p = (10^{20} \text{ cm}^{-3}\text{s}^{-1}) \times (10^{-6} \text{ s}) = 10^{14} \text{ cm}^{-3}$.

Under strong illumination, this excess concentration is much larger than the equilibrium concentration, so it represents the total hole concentration. Comparing this to the given form $10^x$, we find that $x=14$.

The silicon is doped with phosphorus, making it an n-type semiconductor where conductivity is determined by electrons. The formula for conductivity ($\sigma$) is $\sigma = nq\mu_n$.

Assuming full ionization, the electron concentration ($n$) equals the given doping concentration, $N_D = 10^{16} \text{ cm}^{-3}$.

From the graph, we find the electron mobility ($\mu_n$) corresponding to this concentration by locating $10^{16}$ on the x-axis and reading the value from the solid "Electron" curve, which is $\mu_n = 1200 \text{ cm}^2/(\text{V}\cdot\text{s})$.

Substituting these values, we calculate the conductivity:
$\sigma = (10^{16}) \times (1.6 \times 10^{-19}) \times (1200) = 1.92 \text{ S cm}^{-1}$.

A clamping circuit adds a DC level to an AC signal. During the negative half-cycle of the input signal V, the diode is forward-biased, allowing the capacitor C to charge rapidly to the peak voltage. For the remainder of the cycle, the diode is reverse-biased, and the capacitor attempts to discharge through the resistor R. For the circuit to function as a clamper, the voltage across the capacitor must remain nearly constant. This requires the capacitor to discharge very slowly. The time constant for discharge is $\tau = RC$. To ensure minimal discharge over the signal's period T, the time constant must be much larger than the period, so the condition is $RC \gg T$.

First, let's analyze the circuit with the switch in position B. In this configuration, the op-amp acts as an inverting summing amplifier. The output voltage, $V_{0B}$, is the inverted sum of the two inputs, each weighted by the ratio of the feedback resistor to the input resistor.
$V_{0B} = -\left(\frac{1 \text{ k}\Omega}{1 \text{ k}\Omega} \cdot 5\text{V} + \frac{1 \text{ k}\Omega}{1 \text{ k}\Omega} \cdot 1\text{V}\right) = -6\text{ V}$
Now, with the switch in position A, the circuit becomes a difference amplifier. We can find the output $V_{0A}$ using superposition. The contribution from the 5V inverting input is $-5\text{V}$. The non-inverting terminal voltage $V_+$ is determined by a voltage divider: $V_+ = 1\text{V} \left(\frac{1 \text{ k}\Omega}{1 \text{ k}\Omega + 1 \text{ k}\Omega}\right) = 0.5\text{V}$. The gain for this non-inverting input is $(1 + \frac{1k}{1k}) = 2$. Its contribution to the output is $0.5\text{V} \times 2 = 1\text{V}$.
The total output is the sum of these contributions: $V_{0A} = -5\text{V} + 1\text{V} = -4\text{V}$.
Finally, the ratio of the two output voltages is:
$\frac{V_{0B}}{V_{0A}} = \frac{-6\text{ V}}{-4\text{ V}} = 1.5$

This circuit is a non-inverting Schmitt trigger, which uses positive feedback to create two switching thresholds for the input voltage, $v_{\in}$. The difference between these thresholds is the hysteresis width ($V_H$). The upper threshold voltage is $V_{TH} = -L_- \frac{R_1}{R_2}$ and the lower is $V_{TL} = -L_+ \frac{R_1}{R_2}$.

The hysteresis width is therefore $V_H = V_{TH} - V_{TL} = (L_+ - L_-) \frac{R_1}{R_2}$.
Given the op-amp saturation levels $L_+ = +5 \text{ V}$ and $L_- = -5 \text{ V}$, the difference is $10 \text{ V}$.
We can now plug in the known values to find $R_1$:
$500 \text{ mV} = (5 \text{ V} - (-5 \text{ V})) \frac{R_1}{20 \text{ k}\Omega} = 10 \text{ V} \cdot \frac{R_1}{20 \text{ k}\Omega}$
This simplifies to $0.5 \text{ V} = \frac{R_1}{2 \text{ k}\Omega}$.
Solving for $R_1$ gives $R_1 = 0.5 \times 2 \text{ k}\Omega = 1 \text{ k}\Omega$.

To find the required gates, let's work backward from the desired output expression, $Y = AB + \bar{C}\bar{D}$.

The final gate, G2, combines two signals to produce $Y$. The expression for $Y$ is a logical OR (sum) of two terms. The lower input to G2 comes from a NOR gate, which transforms inputs C and D into $\overline{C+D}$. By De Morgan's theorem, this is equivalent to $\bar{C}\bar{D}$. For the final output to be $AB + \bar{C}\bar{D}$, G2 must therefore be an OR gate.

This means the top input to G2 must be $AB$. This signal is the output of gate G1, whose inputs are $\bar{A}$ and $\bar{B}$. We need a gate that produces $AB$ from $\bar{A}$ and $\bar{B}$. If G1 is a NOR gate, its output is $\overline{\bar{A} + \bar{B}}$. Applying De Morgan's law simplifies this to $\overline{\bar{A}} \cdot \overline{\bar{B}}$, which is exactly $AB$. Therefore, G1 must be a NOR gate.

To identify the correct instruction, let's examine what each one does and its effect on the accumulator (Register A).

The MOV B,M instruction copies the byte from the memory location pointed to by the HL register pair into register B. This operation does not involve the accumulator. The PCHL instruction loads the contents of the HL pair into the Program Counter, effectively causing a jump. The RNZ instruction is a conditional return from a subroutine that depends on the Zero flag, not the accumulator's content.

In contrast, SBI BEH stands for "Subtract Immediate with Borrow". This arithmetic instruction subtracts both the immediate data byte ($BE_{16}$) and the state of the carry flag from the value currently in the accumulator. The result of this subtraction is then stored back into the accumulator, changing its content.

The counter's active-low synchronous $\overline{CLEAR}$ input is controlled by a NAND gate whose inputs are $Q_B$ and $Q_C$. This means the clear signal is asserted (goes low) when both $Q_B=1$ and $Q_C=1$. As the counter increments, the first state that satisfies this condition is $Q_D Q_C Q_B Q_A = 0110_2$, which is decimal 6.

Because the clear is synchronous, the counter remains in state 6 for one full clock cycle. On the next rising clock edge, the active clear signal causes the counter to reset to 0 instead of advancing to 7. The counter therefore cycles through the states 0, 1, 2, 3, 4, 5, and 6. This constitutes a total of 7 unique states, making it a mod-7 counter.

The region of convergence (RoC) includes all complex values of $s$ for which the bilateral Laplace transform integral converges. Since the signal $f(t)$ is non-zero only over the finite interval $[T_1, T_2]$, the transform simplifies to $F(s) = \int_{T_1}^{T_2} f(t)e^{-st}dt$.

For this integral to converge, its value must be finite. We know $|f(t)|$ is bounded. The other term, $e^{-st}$, will also always be finite for any finite $s$ and any time $t$ within the finite interval $[T_1, T_2]$. The product of two bounded functions is also bounded. Integrating a bounded function over a finite interval always results in a finite value. Therefore, the integral converges for any finite value of $s$, making the RoC the entire s-plane.

The given relationship $y[n]=\sum_{m=0}^{n}x[m]$ describes a running sum or accumulator. In the time domain, this is equivalent to the convolution of the input signal $x[n]$ with the unit step function $u[n]$.

In the z-domain, convolution becomes multiplication. Taking the z-transform of both sides gives $Y(z) = X(z)U(z)$, where $U(z)=\frac{z}{z-1}$.

We can now find $X(z)$ by rearranging the equation: $X(z) = Y(z) \frac{1}{U(z)} = Y(z) \frac{z-1}{z}$.
Substituting the given $Y(z)$: $X(z) = \frac{2}{z(z-1)^2} \cdot \frac{z-1}{z} = \frac{2}{z^2(z-1)}$.

To find the time-domain signal $x[n]$, we can rewrite $X(z)$ as $X(z) = 2z^{-3} \left( \frac{z}{z-1} \right)$. This form helps us recognize the inverse transform. The term $\frac{z}{z-1}$ corresponds to $u[n]$, and the $z^{-3}$ factor indicates a time delay of 3 samples.

Therefore, the inverse z-transform is $x[n] = 2u[n-3]$. This signal is a scaled unit step that starts at $n=3$. For any time index less than 3, the signal is zero. Thus, the value at $n=2$ is $x[2] = 2u[2-3] = 2u[-1] = 0$.

The fundamental operations available for combining transfer function blocks are cascading (multiplication) and summing/differencing (addition/subtraction). Let's examine the options based on these rules.

To realize the function $\frac{G_{1}(s)}{G_{2}(s)}$, you would need to implement division by $G_{2}(s)$, which requires a block with the transfer function $1/G_{2}(s)$. Creating an inverse of a given block is not a standard elementary operation like cascading or summing.

On the other hand, the other options are realizable. $G_{1}(s)G_{1}(s)$ is achieved by cascading two $G_{1}(s)$ blocks. The expressions in C and D simplify to $1 \pm G_{1}(s)G_{2}(s)$. These can be constructed by first cascading $G_{1}(s)$ and $G_{2}(s)$ to get $G_{1}(s)G_{2}(s)$, and then summing or differencing this result with a unity gain path, which represents the constant '1'.

Transfer function

First, we need to find the characteristic equation of the system. For a unity negative feedback system, the closed-loop transfer function is $T(s) = \frac{G(s)}{1+G(s)}$. Substituting the given open-loop function gives $T(s) = \frac{K}{s^{2+}10s+K}$.

The denominator of this function, $s^{2+}10s+K=0$, is the system's characteristic equation. We can compare this to the standard second-order characteristic equation, $s^{2+}2\xi\omega_n s+\omega_n^2=0$.

By matching the coefficients, we find two relationships: $2\xi\omega_n = 10$ and $\omega_n^2 = K$.

We are given a damping ratio of $\xi = 0.25$. Plugging this into the first relationship gives $2(0.25)\omega_n = 10$, which we can solve to find the natural frequency, $\omega_n = 20$ rad/s.

Finally, we use the second relationship to calculate the gain: $K = \omega_n^2 = (20)^2 = 400$.

For a sinusoidal signal, the Signal-to-Quantization-Noise Ratio (SQNR) is related to the number of quantization bits, $n$, by the well-known approximation:
$SQNR_{dB} \approx 1.8 + 6n$
We are given that the SQNR is $31.8$ dB. By substituting this value into the formula, we can solve for the number of bits:
$31.8 = 1.8 + 6n$
$30 = 6n \implies n = 5 \text{ bits}$
The total number of quantization levels, $L$, is determined by the number of bits according to the relationship $L = 2^n$.
Therefore, the number of levels is $L = 2^5 = 32$.

The input signal, $x(t) = \cos(10\pi t+\frac{\pi}{4})$, is a cosine wave with a frequency of 5 Hz. When we sample it at $f_s = 15$ Hz, we create replicas of its spectrum at integer multiples of 15 Hz. This process, known as aliasing, generates a new spectral component at $5 \text{ Hz} + 15 \text{ Hz} = 20$ Hz.

The filter has an impulse response corresponding to a band-pass filter centered at a frequency of $\frac{40\pi}{2\pi} = 20$ Hz. This filter is designed to select frequencies only in a narrow band around $\pm 20$ Hz.

When the sampled signal passes through this filter, only the aliased component at 20 Hz is allowed through. The output is a new cosine wave at this 20 Hz frequency. Its amplitude is scaled by $\frac{f_s}{2} = \frac{15}{2}$, and its phase is the sum of the original signal's phase (\frac{\pi}{4}$) and the filter's phase shift ($-\frac{\pi}{2}$), resulting \in a final phase of$-\frac{\pi}{4}$. This gives the output signal$\frac{15}{2}\cos(40\pi t - \frac{\pi}{4})$.

In a source-free region, there is no net electric charge. This physical condition requires the electrostatic potential, $\varphi$, to satisfy Laplace's equation: $\nabla^2 \varphi = 0$. The term $\nabla^2 \varphi$ represents the sum of the second partial derivatives of the potential.

For the given potential, $\varphi = 2x^{2}+y^{2}+cz^{2}$, we find these derivatives:
$\frac{\partial^2 \varphi}{\partial x^2} = 4, \quad \frac{\partial^2 \varphi}{\partial y^2} = 2, \quad \frac{\partial^2 \varphi}{\partial z^2} = 2c$
To satisfy Laplace's equation, the sum of these terms must be zero:
$4 + 2 + 2c = 0$
This simplifies to $6 + 2c = 0$. Solving for the constant $c$, we find that $c = -3$.

To understand the wave's polarization, we first examine its components from the given phasor expression. The electric field has an x-component with magnitude $1$ and a y-component with magnitude $|j4| = 4$. Because the component magnitudes are unequal, the polarization must be elliptical, not circular.

Next, we determine the handedness. The complex number $j$ is equivalent to $e^{j\pi/2}$, which signifies a phase lead of $\pi/2$ for the y-component relative to the x-component. The term $e^{-j0.2z}$ in the exponent indicates that the wave is propagating in the positive z-direction.

A standard convention for a wave traveling in the $+z$ direction is that if the y-component leads the x-component in phase, the polarization is left-handed. Combining these findings-elliptical shape and left-handed rotation-we conclude the wave has left-handed elliptical polarization.

This problem can be solved by applying the Laplace transform to the differential equation. Let's first rearrange the equation to $\frac{dx}{dt} + 0.2x = 10$. Transforming each term yields $[sX(s) - x(0)] + 0.2X(s) = \frac{10}{s}$.

Next, we substitute the initial condition $x(0)=1$ and solve for $X(s)$:
$sX(s) - 1 + 0.2X(s) = \frac{10}{s}$
$X(s)(s+0.2) = \frac{10}{s} + 1 = \frac{10+s}{s}$
$X(s) = \frac{s+10}{s(s+0.2)}$

To find the inverse transform, we first use partial fraction decomposition to break down $X(s)$:
$X(s) = \frac{50}{s} - \frac{49}{s+0.2}$

Finally, applying the inverse Laplace transform term-by-term gives the solution in the time domain:
$x(t) = 50 - 49e^{-0.2t}$

This integral can be solved elegantly using Fourier transform properties, specifically Parseval's theorem. We can express the integral as $12 \int_{-\infty }^{\infty } x(t)y(t)dt$, where the integrand is a product of two functions, $x(t) = \cos(2\pi t)$ and $y(t) = \frac{\sin(4\pi t)}{4\pi t}$.

Parseval's theorem allows us to evaluate this integral in the frequency domain:
$\int_{-\infty }^{\infty } x(t)y(t)dt = \frac{1}{2\pi}\int_{-\infty }^{\infty } X(\omega)Y(\omega)d\omega$.

The Fourier transform of $x(t)$ is $X(\omega) = \pi[\delta(\omega - 2\pi) + \delta(\omega + 2\pi)]$, a pair of impulses. The transform of the sinc function $y(t)$ is a rectangular pulse, $Y(\omega)$, with an amplitude of $\frac{1}{4}$ for frequencies $|\omega| < 4\pi$.

The impulses in $X(\omega)$ sample the rectangular pulse $Y(\omega)$ at $\omega = \pm 2\pi$. Since both points are within the pulse's range, the value of the integral (without the constant 12) is $\frac{1}{2\pi} [\pi Y(2\pi) + \pi Y(-2\pi)] = \frac{1}{2} [\frac{1}{4} + \frac{1}{4}] = \frac{1}{4}$.

Finally, multiplying by the leading constant gives the total value: $12 \times \frac{1}{4} = 3$.

To evaluate this integral, we parameterize the unit circle $C$ using $z = e^{i\theta}$. For this parameterization, the real part is $Re\{z\} = \cos\theta$ and the differential is $dz = ie^{i\theta}d\theta$. The contour integral over $C$ becomes a definite integral for $\theta$ from $0$ to $2\pi$.

Substituting these into the given expression, we get:
$\frac{1}{2\pi i}\int_{0}^{2\pi} (\cos\theta)(ie^{i\theta}) d\theta = \frac{1}{2\pi}\int_{0}^{2\pi} \cos\theta \, e^{i\theta} d\theta$

Using Euler's formula, $e^{i\theta} = \cos\theta + i\sin\theta$, the integrand becomes $\cos\theta(\cos\theta + i\sin\theta) = \cos^2\theta + i\sin\theta\cos\theta$.

The integral separates into two parts:
$\frac{1}{2\pi} \left( \int_{0}^{2\pi} \cos^2\theta \,d\theta + i \int_{0}^{2\pi} \sin\theta\cos\theta \,d\theta \right)$

Evaluating the standard definite integrals, we find that $\int_{0}^{2\pi} \cos^2\theta \,d\theta = \pi$ and $\int_{0}^{2\pi} \sin\theta\cos\theta \,d\theta = 0$.

Plugging these values back, we arrive at the final result: $\frac{1}{2\pi}(\pi + i \cdot 0) = \frac{1}{2} = 0.5$.

To find the expected number of tosses, $E[X]$, we calculate a weighted average. We sum each possible number of tosses ($x$) multiplied by the probability of stopping at that exact number, $P(X=x)$.

The soonest we can stop is after 2 tosses, with the sequence HH. The probability of this is $(\frac{1}{2}) \times (\frac{1}{2}) = \frac{1}{4}$. This gives the first term in our sum: $2 \times \frac{1}{4}$.

Considering all possible stopping points ($x=2, 3, 4, \dots$) generates an infinite series for the expectation. Based on the probabilities for each value of $x$, this series is:
$E[X] = 2\left(\frac{1}{4}\right) + 3\left(\frac{1}{8}\right) + 4\left(\frac{1}{16}\right) + \cdots = \sum_{k=2}^{\infty } \frac{k}{2^k}$
The sum of this arithmetico-geometric series converges to 1.5.

Resonance in this parallel circuit occurs when the total admittance is purely real, meaning its imaginary part (susceptance) is zero. The total admittance $Y_{eq}$ is the sum of the admittances of the capacitor branch ($Y_C$) and the inductor-resistor branch ($Y_{LR}$).

$Y_{eq} = Y_C + Y_{LR} = j\omega C + \frac{1}{R+j\omega L}$

To isolate the imaginary part, we rationalize the second term:
$Y_{eq} = j\omega C + \frac{R-j\omega L}{R^{2+}\omega^2 L^2} = \left(\frac{R}{R^{2+}\omega^2 L^2}\right) + j\left(\omega C - \frac{\omega L}{R^{2+}\omega^2 L^2}\right)$

Setting the imaginary part to zero gives the resonance condition:
$\omega C - \frac{\omega L}{R^{2+}\omega^2 L^2} = 0 \implies \omega^2 = \frac{1}{LC} - \frac{R^2}{L^2}$

Solving for the angular frequency $\omega$ yields $\omega = \sqrt{\frac{L-R^2C}{L^2C}} = \frac{1}{\sqrt{LC}}\sqrt{1-\frac{R^2C}{L}}$.

The resonant frequency is $f = \frac{\omega}{2\pi}$, which directly gives the final expression.

To find the Norton equivalent resistance, $R_N$, we can apply a test voltage source, $V_0$, across terminals a-b and calculate the resulting current, $I_0$. The resistance is then given by $R_N = V_0 / I_0$.

With $V_0$ applied across the terminals, the controlling current $I$ through the $4\Omega$ resistor is $I = V_0 / 4$. Consequently, the voltage of the dependent source becomes $4I = 4(V_0/4) = V_0$.

Now, we apply Kirchhoff's Current Law (KCL) at terminal 'a'. The total current $I_0$ leaving the test source is the sum of the currents flowing through the three branches:
$I_0 = \frac{V_0 - 4I}{2\Omega} + \frac{V_0}{2\Omega} + \frac{V_0}{4\Omega}$

Substituting $4I = V_0$, the equation simplifies to:
$I_0 = \frac{V_0 - V_0}{2} + \frac{V_0}{2} + \frac{V_0}{4} = 0 + \frac{2V_0 + V_0}{4} = \frac{3V_0}{4}$

Finally, we calculate the Norton resistance:
$R_N = \frac{V_0}{I_0} = \frac{V_0}{3V_0/4} = \frac{4}{3} \approx 1.33 \, \Omega$