GATE EE 2019

To find the inverse Laplace transform, we first want to simplify the expression $H(s)$ into a form that matches standard transform pairs. The denominator, $s^{2+}2s+1$, is a perfect square and can be factored into $(s+1)^2$.

Our strategy is to rewrite the numerator, $s+3$, in terms of the factor $(s+1)$, which gives us $(s+1)+2$. This allows us to split the fraction:
$H(s) = \frac{(s+1)+2}{(s+1)^2} = \frac{s+1}{(s+1)^2} + \frac{2}{(s+1)^2} = \frac{1}{s+1} + \frac{2}{(s+1)^2}$
Now we can take the inverse transform of each term separately. Using the standard transform properties $L^{-1}\left\{\frac{1}{s+a}\right\}=e^{-at}$ and $L^{-1}\left\{\frac{1}{(s+a)^2}\right\}=te^{-at}$, we get:
$h(t) = e^{-t} + 2te^{-t}$

A key property relates the eigenvalues of a matrix to the eigenvalues of its powers. If a matrix $M$ has an eigenvalue $\lambda$, then the matrix $M^2$ will have an eigenvalue of $\lambda^2$. This rule extends to any power.

We are given that the matrix $M$ has eigenvalues of $4$ and $9$.

To find the eigenvalues for $M^2$, we apply this property by squaring the original eigenvalues.

Therefore, the eigenvalues of $M^2$ are $4^2 = 16$ and $9^2 = 81$.

This partial differential equation relates the second derivative with respect to time ($t$) to the second derivatives with respect to spatial coordinates ($x, y$). The term $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}$ is the two-dimensional Laplacian operator, often written as $\nabla^2 u$.

Rearranging the given equation, we get $\frac{\partial^2 u}{\partial t^2} = c^2 \nabla^2 u$. This is the canonical form of the wave equation, which describes the propagation of waves with speed $c$. The second-order time derivative, $\frac{\partial^2 u}{\partial t^2}$, is the key feature that signifies a wave-like or oscillatory behavior.

In contrast, the heat equation involves a first-order time derivative (\frac{\partial u}{\partial t}$), while Laplace's and Poisson's equations ($\nabla^2 u=0$and$\nabla^2 u=f$, respectively) are time-independent, describing steady-state phenomena. Therefore, the given equation represents the two-dimensional wave equation.

A function is considered analytic in a region if it has a derivative at every point within that region. For rational functions, like the ones given, the only points where they are not analytic are the singularities, which are the values of $z$ that make the denominator zero.

We are interested in the region defined by $|z| \leq 1$, which is a closed disk of radius 1 centered at the origin. For a function to be analytic here, all its singularities must lie outside this disk.

The function $f(z) = \frac{z^{2-}1}{z+2}$ has a single singularity where its denominator is zero, at $z = -2$. We check if this point is inside our region by calculating its magnitude: $|-2| = 2$. Since $2 > 1$, the singularity is outside the disk, making the function analytic everywhere within the region $|z| \leq 1$.

To find the standard deviation, we first need to determine the variance. The variance of a random process is defined as its mean-square value minus the square of its mean value: Var(X) = E\[X^2]$ - (E[X])^2$.

We are given that the process is "zero-mean," which means $E[X] = 0$. The mean-square value is explicitly stated as E\[X^2]$ = \frac{kT}{C}$.

Substituting these values into the variance formula, we get:
$Var(X) = \frac{kT}{C} - (0)^2 = \frac{kT}{C}$

The standard deviation is simply the square root of the variance. Therefore, the standard deviation is $\sqrt{\frac{kT}{C}}$.

By substituting $a_1=0$ and $b_1=0$ into the equation, the transfer function becomes $H(s) = \frac{c_1}{a_2s^{2+}b_2s+c_2}$.

To determine the filter type, we analyze its response at extreme frequencies.

At very low frequencies (as $s \to 0$), the gain is $H(0) = \frac{c_1}{c_2}$. Since the coefficients are positive constants, the filter passes these frequencies with a finite gain.

At very high frequencies (as $s \to \infty$), the $s^2$ term in the denominator dominates, causing the denominator to approach infinity. This makes the overall gain $H(\infty )$ approach zero.

A filter that passes low frequencies and blocks high frequencies is a low-pass filter.

A fundamental principle of causality in LTI systems is that the impulse response, $h(t)$, must be zero for all negative time ($t < 0$). This means the system cannot respond to an impulse before it occurs.

Let's examine the impulse response $h(t) = 1 + e^{-at}u(t)$. This function is the sum of a constant term, $1$, and a decaying exponential term, $e^{-at}u(t)$.

For any time $t < 0$, the unit step function $u(t)$ is zero, which makes the term $e^{-at}u(t)$ equal to zero. However, the constant term $1$ exists for all values of $t$. Therefore, for $t < 0$, the total impulse response is $h(t) = 1 + 0 = 1$. Since $h(t)$ is not zero for $t<0$, the system is non-causal.

Voltage regulation is a measure of the change in a transformer's secondary voltage when it goes from having no load to being fully loaded. It's calculated as the voltage drop under load, expressed as a percentage of the no-load voltage.

The no-load secondary voltage ($V_{NL}$) is given by the transformer's rating, which is $100 \text{ V}$.
The secondary voltage under load ($V_{FL}$) is given as $95 \text{ V}$.

The formula for voltage regulation is:
$\text{Regulation} (\%) = \frac{V_{NL} - V_{FL}}{V_{NL}} \times 100\%$
Substituting the given values:
$\text{Regulation} = \frac{100 \text{ V} - 95 \text{ V}}{100 \text{ V}} \times 100\% = \frac{5}{100} \times 100\% = 5\%$

For a six-pulse converter, the AC input current is distorted and contains harmonics. The order of these harmonic components is determined by the formula $nk \pm 1$, where $n$ is the pulse number and $k$ is any positive integer.

Here, the pulse number is $n=6$. The lowest-order harmonics are found by setting $k=1$, which gives us the $6(1) \pm 1$, or 5th and 7th, harmonics.

The lowest of these is the 5th harmonic. Its frequency is the harmonic order multiplied by the fundamental source frequency. Therefore, the lowest harmonic frequency is $5 \times 50 \text{ Hz} = 250 \text{ Hz}$.

The supply voltage $V$ is approximately proportional to the product of frequency $f$ and the peak mutual magnetic flux $\Phi_m$ in the motor's core, as given by the EMF equation $V \approx E_1 = 4.44 f N \Phi_m$. When the supply voltage is reduced at a rated (constant) frequency, the mutual flux $\Phi_m$ must also decrease. The magnetizing reactance, $X_m$, is associated with this mutual flux path.

The iron core exhibits magnetic saturation, a non-linear property. At rated voltage, the core is typically operating near saturation. Reducing the voltage and thus the flux moves the operating point to a less saturated region of the B-H curve. In this region, the core's magnetic permeability ($\mu$) is higher. Since magnetizing inductance $L_m$ is proportional to permeability, and $X_m = 2 \pi f L_m$, the magnetizing reactance increases. The other parameters, such as resistances and leakage reactances, are largely determined by physical construction and are not significantly affected by voltage changes.

The problem states that the motor's load and line voltage remain constant. "Load" on a motor refers to the mechanical work it's doing, which corresponds to the real power ($P$) it consumes. The formula for three-phase real power is $P = \sqrt{3} V_L I_L \cos\phi$.

Since both the real power ($P$) and the line voltage ($V_L$) are constant, the product of the line current and the power factor, $I_L \cos\phi$, must also remain constant to satisfy the equation.

We can equate this product for the two different operating conditions:
$I_1 \cos\phi_1 = I_2 \cos\phi_2$

Plugging in the values from the problem:
$200 \, \text{A} \times 1.0 = I_2 \times 0.5$

Solving for the new line current ($I_2$) gives us $400 \, \text{A}$.

The total current in the circuit is the sum of a transient response (the DC offset) and a steady-state AC response. Because the inductor current cannot change instantaneously, the total current must be zero at the moment the switch is closed ($t=0$). This implies that the initial value of the transient current must be equal in magnitude and opposite in sign to the initial value of the steady-state current.

The DC offset is maximized when the steady-state current component, $I_m \sin(\omega t + \theta - \phi)$, would be at its maximum value at $t=0$. This occurs when the angle at $t=0$, which is $(\theta - \phi)$, is equal to $\pm 90^\circ$.

First, we calculate the impedance angle $\phi$:
$\phi = \tan^{-1}\left(\frac{\omega L}{R}\right) = \tan^{-1}\left(\frac{377 \times 10 \times 10^{-3}}{3.77}\right) = \tan^{-1}(1) = 45^\circ$.

To maximize the DC offset, we set $\theta - \phi = \pm 90^\circ$. Using the negative case:
$\theta - 45^\circ = -90^\circ$
$\theta = -45^\circ$

To find the steady-state value of a system's output, $y(t)$, from its Laplace transform, $Y(s)$, we use the Final Value Theorem. This theorem states that the steady-state value, $y_{ss}$, is the limit of $sY(s)$ as $s$ approaches zero, provided the system is stable.

The formula is $y_{ss} = \lim_{s \to 0} sY(s)$.

Let's apply this to the given function:
$y_{ss} = \lim_{s \to 0} s \cdot \frac{10}{s(s^{2+}s+100\sqrt{2})}$

The $s$ in the numerator cancels with the $s$ in the denominator's pole at the origin:
$y_{ss} = \lim_{s \to 0} \frac{10}{s^{2+}s+100\sqrt{2}}$

Now, we evaluate the limit by substituting $s=0$ into the simplified expression:
$y_{ss} = \frac{10}{0^{2+}0+100\sqrt{2}} = \frac{10}{100\sqrt{2}} = \frac{1}{10\sqrt{2}}$

To find where the Nyquist plot crosses the negative real axis, we must find the frequency ($\omega_{pc}$) at which the phase angle of $G(j\omega)$ is $-180^\circ$ (or $-\pi$ radians).

First, let's express the phase of $G(j\omega) = \frac{\pi e^{-0.25j\omega}}{j\omega}$. The phase angle is the sum of the angles of its components: $\angle G(j\omega) = \angle(e^{-0.25j\omega}) - \angle(j\omega) = -0.25\omega - \frac{\pi}{2}$.

Setting the phase angle to $-\pi$ to find the phase crossover frequency, $\omega_{pc}$:
$-0.25\omega_{pc} - \frac{\pi}{2} = -\pi \implies 0.25\omega_{pc} = \frac{\pi}{2} \implies \omega_{pc} = 2\pi$

Now, we calculate the magnitude of $G(j\omega)$ at this frequency:
$|G(j\omega_{pc})| = \left| \frac{\pi e^{-0.25j(2\pi)}}{j(2\pi)} \right| = \frac{\pi}{|j2\pi|} = \frac{\pi}{2\pi} = 0.5$

A magnitude of 0.5 at an angle of $-180^\circ$ means the plot passes through the point $(-0.5, j0)$ on the negative real axis.

To determine the range of $k$ for BIBO stability, we apply the Routh-Hurwitz criterion to the characteristic equation. This criterion requires all the elements in the first column of the Routh array to be positive for the system's poles to be in the left-half s-plane.

First, we construct the Routh array using the coefficients of $\Delta(s)$. The first column of this array is found to be: $1$, $3$, $\frac{8}{3}$, $\frac{8-9k}{3}$, and $k$.

For stability, every entry in this column must be positive. The first three terms ($1$, $3$, and $\frac{8}{3}$) are already positive. We only need to enforce this condition on the remaining two terms, which depend on $k$.

This gives us two inequalities:

1. $k > 0$
2. $\frac{8-9k}{3} > 0 \implies 8-9k > 0 \implies k < \frac{8}{9}$

Combining these two conditions, we find that the system is stable when $0 < k < \frac{8}{9}$.

For an NMOS transistor to conduct current, it must first be turned "on." This happens when the gate-source voltage ($V_{gs}$) is greater than the device's threshold voltage ($V_{th}$), creating a conductive channel. This gives us our first condition: $V_{gs} > V_{th}$.

To operate in the saturation region, this channel must be "pinched-off" at the drain end. This occurs when the drain-source voltage ($V_{ds}$) is sufficiently large. The specific condition for pinch-off is $V_{ds} \geq V_{gs} - V_{th}$. Both of these requirements must be met for the transistor to be biased in saturation.

A Current Controlled Current Source (CCCS) is a type of current amplifier. When negative feedback is applied to a current amplifier, its output impedance increases, making it a more ideal current source.

The closed-loop output impedance, $Z_{of}$, is calculated by multiplying the open-loop output impedance, $Z_o$, by a feedback factor. This factor is $(1 + A\beta)$, where $A\beta$ represents the loop gain.

The formula is:
$Z_{of} = Z_o (1 + A\beta)$

Given $Z_o = 100 \text{ k}\Omega$ and a loop gain $A\beta = 9$, we can find the new impedance:
$Z_{of} = 100 \text{ k}\Omega \times (1 + 9) = 100 \text{ k}\Omega \times 10 = 1000 \text{ k}\Omega$.

The integral in question is $\int_C \text{grad } f \cdot d\mathbf{r}$. This is the line integral of a conservative vector field. By the Fundamental Theorem for Line Integrals, the value of this integral depends only on the function $f$ evaluated at the endpoints of the path, not the specific path taken between them.

The path starts at point $A = (-3, -3, 2)$ and ends at point $B = (2, 6, -1)$. The value of the integral is simply the change in the scalar function $f$ from the start point to the end point.

Value of integral = $f(B) - f(A) = f(2, 6, -1) - f(-3, -3, 2)$

We have the function $f = 2x^{3+}3y^{2+}4z$. Let's evaluate it at the endpoints:
$f(2, 6, -1) = 2(2)^3 + 3(6)^2 + 4(-1) = 16 + 108 - 4 = 120$
$f(-3, -3, 2) = 2(-3)^3 + 3(-3)^2 + 4(2) = -54 + 27 + 8 = -19$

Therefore, the integral's value is $120 - (-19) = 139$.

To solve for the current $I$, we can use nodal analysis. Let's label the voltage at the central node, where the three branches meet, as $V_x$.

By applying Kirchhoff's Current Law (KCL) at node $V_x$, we sum the currents leaving the node and set them to zero. This gives us the equation: $-I - 2 + \frac{V_x - 5I}{3} = 0$. The negative signs indicate that current $I$ and the 2 A source are flowing into the node.

Simplifying the KCL equation, we get $V_x - 8I = 6$. We need a second relationship between $V_x$ and $I$. Looking at the leftmost branch, Ohm's law gives us $I = \frac{20 - V_x}{2}$, which we can rearrange to $V_x = 20 - 2I$.

Now we substitute this expression for $V_x$ into our simplified KCL equation: $(20 - 2I) - 8I = 6$. Solving for $I$, we find that $10I = 14$, which means $I = 1.4$ A.

The capacitance of a coaxial cylinder is directly proportional to the relative permittivity of the dielectric material it contains. Let's denote the initial capacitance as $C_1$ and the capacitance from the geometry and vacuum permittivity as $C_0$.

Initially, the capacitor is filled with a dielectric of $\varepsilon_{r1}=2$. So, the capacitance is $C_1 = C_0 \cdot \varepsilon_{r1} = 2C_0$.

In the second scenario, the capacitor acts as a parallel combination of two capacitors. One occupies three-fourths of the volume with dielectric $\varepsilon_{r1}=2$, and the other occupies one-fourth with dielectric $\varepsilon_{r2}$. The new total capacitance, $C_2$, is the sum of these parts:
$C_2 = \frac{3}{4}(C_0 \cdot \varepsilon_{r1}) + \frac{1}{4}(C_0 \cdot \varepsilon_{r2})$

Substituting $\varepsilon_{r1}=2$, we get:
$C_2 = C_0 \left( \frac{3}{4} \cdot 2 + \frac{1}{4} \cdot \varepsilon_{r2} \right) = C_0 \left( \frac{3}{2} + \frac{\varepsilon_{r2}}{4} \right)$

We are given that the capacitance doubles, so $C_2 = 2C_1$.
$C_0 \left( \frac{3}{2} + \frac{\varepsilon_{r2}}{4} \right) = 2(2C_0)$

The $C_0$ term cancels, leaving:
$\frac{3}{2} + \frac{\varepsilon_{r2}}{4} = 4 \implies \frac{\varepsilon_{r2}}{4} = 4 - \frac{3}{2} = \frac{5}{2}$

Solving for $\varepsilon_{r2}$, we find $\varepsilon_{r2} = 10$.

The off-diagonal element of the $Y_{bus}$ matrix, $Y_{12}$, represents the negative of the total admittance connecting bus 1 and bus 2. From the given matrix, $Y_{12} = j20$ pu.

Therefore, the total admittance between the buses is $y_{total} = -Y_{12} = -j20$ pu.

Since the two transmission lines are identical and in parallel, their admittances add up. The admittance of a single line is half of the total admittance:
$y_{line} = \frac{y_{total}}{2} = \frac{-j20}{2} = -j10$ pu.

The series impedance of each line is the reciprocal of its admittance.
$z_{line} = \frac{1}{y_{line}} = \frac{1}{-j10} = j0.1$ pu.

The series reactance is the imaginary part of this impedance. Thus, the magnitude of the series reactance is $0.1$ pu.

To determine the short-circuit level, we first find the equivalent per-unit (p.u.) reactance of the five parallel alternators. Since they are identical, we can use a common base of 5 MVA. The total reactance is the individual reactance divided by the number of alternators.
$X_{eq} = \frac{0.25}{5} = 0.05 \text{ p.u.}$
The per-unit short-circuit current is the reciprocal of this equivalent reactance, assuming a pre-fault voltage of 1 p.u.
$I_{SC} = \frac{1}{X_{eq}} = \frac{1}{0.05} = 20 \text{ p.u.}$
Finally, the short-circuit MVA is found by multiplying the per-unit short-circuit current by the base MVA.
$\text{Short-Circuit MVA} = I_{SC} \times \text{MVA}_{base} = 20 \times 5 = 100 \text{ MVA}$

First, let's find the secondary current ($I_{sec}$) during the fault. Since the fault is 20 times the rated primary current, the secondary current will be 20 times the CT's rated secondary current of 5 A.
$I_{sec} = 20 \times 5 \text{ A} = 100 \text{ A}$

Next, we use Ohm's Law to find the voltage ($V_{sec}$) developed across the total secondary circuit impedance.
$V_{sec} = I_{sec} \times Z_{total} = 100 \text{ A} \times 0.01 \, \Omega = 1 \text{ V}$

Finally, the VA output is simply the product of the secondary voltage and current.
$\text{VA Output} = V_{sec} \times I_{sec} = 1 \text{ V} \times 100 \text{ A} = 100 \text{ VA}$

The rank of a matrix is the number of linearly independent rows, which we can find by reducing the matrix to its row echelon form and counting the non-zero rows.

Let's begin by swapping the first and second rows to get a leading 1 in the top-left position.
$M = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix} \xrightarrow{R_1 \leftrightarrow R_2} \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{bmatrix}$
Next, we'll create zeros below the leading 1. We perform the operation $R_3 \rightarrow R_3 - R_1$.
$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & -1 \end{bmatrix}$
Finally, to complete the echelon form, we use the second row to create a zero in the third row with the operation $R_3 \rightarrow R_3 - R_2$.
$\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & -2 \end{bmatrix}$
The resulting matrix has three non-zero rows. Therefore, the rank of the matrix is 3.

The fundamental RMS component of the output voltage ($V_{o1,rms}$) for this unipolar PWM scheme is given by the expression $V_{o1,rms} = \frac{2\sqrt{2}V_s}{\pi}\sin(d)$, where $V_s$ is the DC input voltage and $2d$ is the pulse width.

We are given the condition that this output must be 75% of the DC voltage, so $V_{o1,rms} = 0.75 V_s$.

Equating the two expressions for $V_{o1,rms}$ and cancelling $V_s$ from both sides, we get:
$\frac{2\sqrt{2}}{\pi}\sin(d) = 0.75$

Solving for $d$ gives us:
$d = \arcsin\left(\frac{0.75 \pi}{2\sqrt{2}}\right) \approx 56.44^\circ$

The total required pulse width is $2d$, which is $2 \times 56.44^\circ \approx 112.9^\circ$.

The core property of an inverse matrix is that when multiplied by the original matrix, it yields the identity matrix, $I$. Therefore, we know that $M^{-1}M = I$.

Let's substitute the given forms of $M$ and $M^{-1}$ into this equation:
$M^{-1}M = \begin{bmatrix} u_1^T \\ u_2^T \end{bmatrix} [v_1 \;\; v_2] = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Performing the block matrix multiplication gives us a new 2x2 matrix:
$\begin{bmatrix} u_1^T v_1 & u_1^T v_2 \\ u_2^T v_1 & u_2^T v_2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

By equating the corresponding elements of these matrices, we find that $u_1^T v_1 = 1$ and $u_2^T v_2 = 1$. This validates Statement 1. We also find that $u_1^T v_2 = 0$ and $u_2^T v_1 = 0$, which validates Statement 2. Thus, both statements are correct.

This integral is best solved using Cauchy's Residue Theorem. The function $f(z) = \frac{z^{3+}z^{2+}8}{z+2}$ has a simple pole where the denominator is zero, which is at $z = -2$. This pole is located inside the circular contour $|z|=5$.

To find the residue at this simple pole, we calculate the limit:
$\text{Res}(f, -2) = \lim_{z \to -2} (z+2) \frac{z^{3+}z^{2+}8}{z+2}$
This simplifies to evaluating the numerator at $z = -2$:
$(-2)^3 + (-2)^2 + 8 = -8 + 4 + 8 = 4$.

The Residue Theorem states that the integral is $2\pi j$ times the sum of the enclosed residues. Therefore, the integral's value is $2\pi j \times (4) = 8\pi j$.

To find the Fourier coefficients $a_1$ and $b_1$, we use the standard formulas for a function with period $2\pi$. Since $f(t)$ is zero on the interval $(\pi, 2\pi)$, we only need to integrate from $0$ to $\pi$.

First, let's find the cosine coefficient, $a_1$:
$a_1 = \frac{1}{\pi} \int_0^{2\pi} f(t) \cos(t) \, dt = \frac{1}{\pi} \int_0^{\pi} A \sin(t) \cos(t) \, dt$.
Using the identity $\sin(t)\cos(t) = \frac{1}{2}\sin(2t)$, the integral becomes $\frac{A}{2\pi} \int_0^{\pi} \sin(2t) \, dt$. The integral of $\sin(2t)$ over this interval (which is one full period of $\sin(2t)$) is zero, so $a_1 = 0$.

Next, we find the sine coefficient, $b_1$:
$b_1 = \frac{1}{\pi} \int_0^{2\pi} f(t) \sin(t) \, dt = \frac{1}{\pi} \int_0^{\pi} A \sin(t) \sin(t) \, dt = \frac{A}{\pi} \int_0^{\pi} \sin^2(t) \, dt$.
We use the power-reduction formula $\sin^2(t) = \frac{1}{2}(1 - \cos(2t))$.
This gives $b_1 = \frac{A}{2\pi} \int_0^{\pi} (1 - \cos(2t)) \, dt = \frac{A}{2\pi} \left[ t - \frac{\sin(2t)}{2} \right]_0^{\pi} = \frac{A}{2\pi}[\pi] = \frac{A}{2}$.

Let's break down the Bode plot to understand the transfer function, $G(s)$. The plot's initial slope is $-20$ dB/decade, which tells us there is one pole at the origin. At the corner frequency $\omega=1$, the slope steepens to $-40$ dB/decade, and at $\omega=20$, it steepens again to $-60$ dB/decade. Each $-20$ dB/decade change in slope adds another pole.

This means the system has a total of three poles and no finite zeros. Therefore, Statement I, which claims there is one zero, is false.

Now, let's consider the phase angle for Statement II. As the frequency $\omega$ approaches infinity, each of the three poles contributes a phase lag of $-90^{\circ}$. The total phase angle is the sum of these individual contributions. Thus, the limiting phase is $3 \times (-90^{\circ}) = -270^{\circ}$. Converting this to radians gives us $-\frac{3\pi}{2}$, which makes Statement II true.

A phase-lead compensator is designed to add positive phase shift to a system. The transfer function given is $D(s)=\frac{3\left ( s+\frac{1}{3T} \right )}{\left ( s+\frac{1}{T} \right )}$. This fits the general form $K \frac{s+z}{s+p}$, where $z$ is the zero frequency and $p$ is the pole frequency.

From the function, we can see the zero is at $z = \frac{1}{3T}$ and the pole is at $p = \frac{1}{T}$.

For any first-order phase-lead compensator, the frequency of maximum phase lead, denoted as $\omega_m$, occurs at the geometric mean of the pole and zero frequencies.

Therefore, we can calculate this frequency using the formula $\omega_m = \sqrt{z \cdot p}$.

Substituting our values for $z$ and $p$, we get:
$\omega_m = \sqrt{\left(\frac{1}{3T}\right) \left(\frac{1}{T}\right)} = \sqrt{\frac{1}{3T^2}}$

To find the damping ratio ($\xi$) and natural frequency ($\omega_n$), we first need to determine the system's characteristic equation. This is found by calculating the determinant of $(sI - A)$, where $A$ is the state matrix.

The characteristic equation is

.
This simplifies to $s^2 + 2\beta s + \alpha = 0$.

Now, we 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 can see that $\omega_n^2 = \alpha$, so the natural frequency is $\omega_n = \sqrt{\alpha}$.
Similarly, matching the middle terms gives $2\xi\omega_n = 2\beta$. Substituting $\omega_n = \sqrt{\alpha}$ yields $2\xi\sqrt{\alpha} = 2\beta$, which solves to $\xi = \frac{\beta}{\sqrt{\alpha}}$.

To convert the instrument into a voltmeter that can measure up to 100 V, a high resistance ($R_s$) must be added in series. The goal is that when 100 V is applied across the combination, the current flowing through it is exactly the full-scale deflection current, $I_g = 10 \text{ mA} = 0.01 \text{ A}$.