GATE EE 2017 Set 1

The core principle of a Fourier series is to represent a periodic signal as a sum of harmonically related sinusoids. Therefore, the first step is to determine if the given function $g(t)$ is periodic.

The function is defined as $g(t) = t - \lfloor t \rfloor$ for $t \ge 0$, and as $g(t) = t - \lceil t \rceil$ for $t < 0$. Let's test for periodicity with a potential period of $T=1$. Consider $t=0.5$ and $t=-0.5$:
$g(0.5) = 0.5 - \lfloor 0.5 \rfloor = 0.5$
$g(-0.5) = -0.5 - \lceil -0.5 \rceil = -0.5 - 0 = -0.5$
Since $g(t) \neq g(t-1)$, the function is not periodic.

Because $g(t)$ is a non-periodic signal, it cannot be represented by a Fourier series. Consequently, the concept of harmonic components does not apply, and the coefficient for any harmonic, including the second, is zero.

The total instantaneous power delivered to the load is the sum of the instantaneous powers contributed by each phase.

We can express this as $p(t) = p_a(t) + p_b(t) = V_{an}i_a + V_{bn}i_b$.
Substituting the given expressions for voltage and current:
$p(t) = (220 \sin(100\pi t))(10 \sin(100\pi t)) + (220 \cos(100\pi t))(10 \cos(100\pi t))$
This simplifies to $p(t) = 2200\sin^2(100\pi t) + 2200\cos^2(100\pi t)$.
Factoring out the common term gives $p(t) = 2200(\sin^2(100\pi t) + \cos^2(100\pi t))$.
Using the fundamental trigonometric identity $\sin^2\theta + \cos^2\theta = 1$, the total power is a constant value of $2200$ W.

The power delivered to a synchronous motor is calculated using the formula $P = \frac{3 V_{ph} E_f}{X_s} \sin \delta$. To use this, we first need to find the excitation voltage, $E_f$.

For a motor operating at unity power factor, the phasor relationship between the per-phase terminal voltage ($V_{ph}$) and the excitation voltage is $V_{ph} = E_f \cos \delta$. We can rearrange this to find $E_f = \frac{V_{ph}}{\cos \delta}$.

Let's substitute this expression for $E_f$ back into our power formula. This gives us a convenient equation using only known values:
$P = \frac{3 V_{ph}}{X_s} \left( \frac{V_{ph}}{\cos \delta} \right) \sin \delta = \frac{3 V_{ph}^2}{X_s} \tan \delta$

First, we find the per-phase voltage from the given line voltage: $V_{ph} = \frac{6.6 \text{ kV}}{\sqrt{3}} = \frac{6600}{\sqrt{3}}$ V.

Now, we can solve for power:
$P = \frac{3 \times (\frac{6600}{\sqrt{3}})^2}{30} \times \tan 30^{\circ} = \frac{3 \times \frac{6600^2}{3}}{30} \times \frac{1}{\sqrt{3}} = \frac{6600^2}{30\sqrt{3}} \approx 838.3 \text{ kW}$

Direct substitution of $z=i$ into the expression results in the indeterminate form $\frac{0}{0}$, since both the numerator ($i^{2+}1 = 0$) and the denominator ($i^{3+}2i-i(i^{2+}2)=0$) become zero.

This situation allows us to use L'Hôpital's Rule. We differentiate the numerator and the denominator with respect to $z$:
$\lim_{z\to i}\frac{\frac{d}{dz}(z^{2+}1)}{\frac{d}{dz}(z^{3+}2z-iz^{2-}2i)} = \lim_{z\to i}\frac{2z}{3z^{2+}2-2iz}$
Now, we can substitute $z=i$ into the resulting expression:
$\frac{2i}{3(i^2)+2-2i(i)} = \frac{2i}{3(-1)+2-2(-1)} = \frac{2i}{-3+2+2}$
Simplifying the denominator gives $\frac{2i}{1}$, so the value of the limit is $2i$.

Let's analyze the motion of the particles. The electron and proton attract each other with an electrostatic force of equal magnitude. However, the neutron is uncharged, so it experiences no electrostatic force and remains stationary at the center.

According to Newton's second law, $a = F/m$. Although the force $F$ on the electron and proton is the same, their masses are very different. The mass of an electron ($m_e$) is about 1800 times smaller than the mass of a proton ($m_p$). This means the electron's acceleration will be vastly greater than the proton's.

Since both particles start equidistant from the stationary neutron, the electron, with its much higher acceleration, will cover the distance to the center much more quickly. Therefore, the electron will collide with the neutron first.

This question is a direct application of the time-scaling property for convolution. We start with the known relationship $z(t) = x(t) * y(t)$.

The scaling property states how convolving two time-scaled signals relates to the time-scaled version of their original convolution. For a positive constant $c$, this property is expressed as:
$x(ct) * y(ct) = \frac{1}{c} z(ct)$

The question asks for an expression for $z(ct)$, not the convolution on the left. To find it, we simply rearrange this equation. Multiplying both sides by $c$ gives us the final expression:
$z(ct) = c \, (x(ct) * y(ct))$

The diagonal element of a Y-bus matrix, $Y_{ii}$, is the sum of all admittances connected directly to bus $i$. Using the provided diagram and values, we can establish a system of linear equations for the line admittances ($y_{12}$, $y_{13}$, $y_{23}$):

1. $Y_{11} = y_{12} + y_{13} = -j12$
2. $Y_{22} = y_{12} + y_{23} = -j15$
3. $Y_{33} = y_{13} + y_{23} = -j7$

Solving this system yields the line admittances: $y_{12} = -j10$ pu, $y_{23} = -j5$ pu, and $y_{13} = -j2$ pu.

The problem gives the line reactances as $jp$, $jq$, and $jr$. The admittance ($y$) of a line with pure reactance ($x$) is calculated as $y = \frac{1}{jx} = \frac{-j}{x}$. We can now find the values of p, q, and r:

• For line 2-3 (reactance $jp$): $y_{23} = \frac{-j}{p} = -j5 \implies p = 0.2$
• For line 1-2 (reactance $jq$): $y_{12} = \frac{-j}{q} = -j10 \implies q = 0.1$
• For line 1-3 (reactance $jr$): $y_{13} = \frac{-j}{r} = -j2 \implies r = 0.5$

To determine the equivalent resistance between terminals A and B, we can treat the circuit as three distinct sections connected in series. The first section is the $1\,\Omega$ resistor and the last is the $0.8\,\Omega$ resistor.

The complex network of resistors in the middle can be simplified, and its equivalent resistance is found to be $\frac{6}{5}\,\Omega$ or $1.2\,\Omega$.

Since these three sections are arranged in series, we find the total equivalent resistance by summing their individual resistances:
$R_{AB} = 1\,\Omega + \frac{6}{5}\,\Omega + 0.8\,\Omega = 1 + 1.2 + 0.8 = 3\,\Omega$.

To simplify the Boolean expression $AB + A\bar{C} + BC$, we can identify and remove a redundant term using the consensus theorem.

Let's focus on the terms $A\bar{C}$ and $BC$. These two terms contain a variable and its complement (the literals $\bar{C}$ and $C$). The "consensus" of these two terms is found by multiplying their remaining parts, which are $A$ and $B$. This gives us the consensus term $AB$.

According to the consensus theorem, if the consensus term ($AB$ in this case) is already present in the expression, it is logically redundant. Since our original expression is $AB + A\bar{C} + BC$, the $AB$ term is redundant and can be removed without changing the logic.

Eliminating the redundant term $AB$ leaves us with the simplified expression $BC + A\bar{C}$.

The power factor ($p.f.$) is the ratio of real power ($W$) to apparent power ($VI$), so we have the relationship $p.f. = \frac{W}{VI}$.

To find the maximum possible error in a calculated value that involves multiplication or division, we sum the percentage errors of each measured quantity. This gives us the worst-case scenario.

In this problem, the total percentage error in the power factor is the sum of the individual percentage errors of power, voltage, and current.

$\text{Worst-case \% Error} = (\%\text{ error \in } W) + (\%\text{ error \in } V) + (\%\text{ error \in } I)$
$\text{Worst-case \% Error} = 2\% + 1\% + 1\% = 4\%$

For a sinusoidal input, the resulting phase shift, $\varphi$, is determined by the angle of the system's frequency response, $H(j\omega)$. We find this by substituting $s=j\omega$ into the transfer function:
$H(j\omega) = \frac{1-j\omega}{1+j\omega}$
The phase angle of this complex function is the angle of the numerator minus the angle of the denominator:
$\varphi(\omega) = \angle(1-j\omega) - \angle(1+j\omega) = -\tan^{-1}(\omega) - \tan^{-1}(\omega) = -2\tan^{-1}(\omega)$
To find the minimum and maximum values of $\varphi$, we examine its behavior as the frequency $\omega$ ranges from $0$ to $\infty$.
When $\omega=0$, the phase is $\varphi = -2\tan^{-1}(0) = 0$.
As $\omega \rightarrow \infty$, the phase approaches $\varphi = -2\tan^{-1}(\infty ) = -2\left(\frac{\pi}{2}\right) = -\pi$.
Therefore, the phase shift ranges from a minimum of $-\pi$ to a maximum of $0$.

The given matrix $A$ is symmetric, as it is equal to its transpose ($A = A^T$). A fundamental property of symmetric matrices states that eigenvectors corresponding to distinct eigenvalues are orthogonal. The problem specifies that $A$ has three distinct eigenvalues, so its eigenvectors must be mutually orthogonal.

We are given one eigenvector,

. To find another eigenvector, we must find a vector from the options that is orthogonal to $v_1$. Two vectors are orthogonal if their dot product is zero.

Let's test the vector from option C,

.
The dot product is $v_1 \cdot v_2 = (1)(1) + (0)(0) + (1)(-1) = 1 - 1 = 0$.
Because their dot product is zero, this vector is orthogonal to the given eigenvector and must therefore be another eigenvector of $A$.

Charge carrier type determines a device's classification. Majority carrier (unipolar) devices utilize only one type of charge carrier for conduction, whereas minority carrier (bipolar) devices utilize both electrons and holes.

The MOSFET is a unipolar device where current flows through a channel of majority carriers, making it a majority carrier device. In contrast, the Diode and Thyristor are p-n junction-based structures whose operation fundamentally relies on the injection and recombination of minority carriers. The IGBT (Insulated Gate Bipolar Transistor) combines a MOSFET input with a bipolar output, intentionally injecting minority carriers to modulate conductivity and lower on-state losses. Thus, the IGBT, Diode, and Thyristor are all classified as minority carrier devices.

The phase margin (PM) is defined as $PM = 180^{\circ} + \phi_{gc}$, where $\phi_{gc}$ is the phase angle at the gain crossover frequency, $\omega_{gc}$. For a PM of $30^{\circ}$, the required phase angle is $\phi_{gc} = 30^{\circ} - 180^{\circ} = -150^{\circ}$.

The gain crossover frequency is where the magnitude of the open-loop transfer function is unity. For $G(s) = \frac{Ke^{-s}}{s}$, the magnitude is $|G(j\omega)| = \frac{K}{\omega}$. Setting this to 1 gives $\omega_{gc} = K$.

The phase angle of the system is given by $\angle G(j\omega) = -90^{\circ} - \omega$ (in radians). To work in degrees, we convert the radian term: $\angle G(j\omega) \approx -90^{\circ} - 57.3\omega^{\circ}$.

At the gain crossover frequency, $\omega_{gc} = K$, the phase is $\phi_{gc} \approx -90^{\circ} - 57.3K^{\circ}$. Equating this with the required phase gives $-150^{\circ} = -90^{\circ} - 57.3K^{\circ}$. Solving for K yields $57.3K = 60$, so $K \approx 1.047$.

Iron is a ferromagnetic material, possessing a very high magnetic permeability. This property causes it to dramatically concentrate magnetic field lines. When the uniform field enters the cylinder, the iron provides a much "easier" path for the magnetic flux than the surrounding space. Consequently, the field lines are pulled inward from the edges of the cylinder and become more densely packed. This convergence of field lines means they bend closer to the cylinder's central axis.

To solve this, we first establish the limits for the double integral based on the region R. The region is bounded by $x=1$, $x=5$, $y=0$, and the line connecting points $(1,2)$ and $(5,10)$, which has the equation $y=2x$. Thus, we can write the integral as:
$I = c \int_{x=1}^{5} \int_{y=0}^{2x} xy^2 \, dy \, dx$
We begin by evaluating the inner integral with respect to $y$, treating $x$ as a constant:
$\int_{0}^{2x} xy^2 \, dy = x \left[ \frac{y^3}{3} \right]_{0}^{2x} = x \left( \frac{(2x)^3}{3} - 0 \right) = \frac{8x^4}{3}$
Next, we substitute this result into the outer integral and integrate with respect to $x$:
$I = c \int_{1}^{5} \frac{8x^4}{3} \, dx = \frac{8c}{3} \left[ \frac{x^5}{5} \right]_{1}^{5} = \frac{8c}{3} \left( \frac{5^5}{5} - \frac{1^5}{5} \right)$
This simplifies to $\frac{8c}{3} (5^4 - \frac{1}{5}) = \frac{8c}{3}(625 - 0.2) = \frac{8c}{3}(624.8)$.
Finally, we substitute the given value $c = 6 \times 10^{-4}$:
$I = \frac{8}{3}(6 \times 10^{-4})(624.8) = 16 \times 10^{-4} \times 624.8 = 0.99968$

The behavior of this system is governed by the term $(1+(-1)^{n})$. Let's examine this term for different values of the time index $n$. It evaluates to 2 when $n$ is even and 0 when $n$ is odd.

First, to check for time invariance, we observe that the system's operation on the input (multiplying by 2 or 0) changes with the specific time index $n$. A system whose behavior is a function of time itself is, by definition, time-varying.

Next, to assess invertibility, we must be able to uniquely recover the input $x[n]$ from the output $y[n]$. For any odd value of $n$, the output is $y[n] = 0 \cdot x[n] = 0$, regardless of the input value. Since we cannot determine the value of $x[n]$ for any odd $n$ from the output, the system is non-invertible.

To display a waveform correctly, the vertical signal (Y-channel) and the horizontal sweep (X-channel) must be synchronized. The circuitry that starts the horizontal sweep has its own delays. First, the slope and level detector takes $100 \text{ ns}$ to identify the trigger point. Then, the sweep generator requires an additional $50 \text{ ns}$ to start its sweep.

This means the horizontal sweep begins a total of $100 \text{ ns} + 50 \text{ ns} = 150 \text{ ns}$ after the trigger event occurs on the input signal. To ensure the very beginning of the waveform (including the trigger point) is visible on the screen, the input signal itself must be delayed by this same amount before it reaches the vertical deflection plates. Therefore, a $150 \text{ ns}$ delay line is inserted into the Y-channel.

To solve this, we first need to determine the RMS voltage supplied to each phase of the star-connected load. For a voltage source inverter operating in 120° conduction mode, the RMS phase voltage ($V_{ph,rms}$) is related to the DC source voltage ($V_{dc}$) by the following formula:
$V_{ph,rms} = \sqrt{\frac{2}{3}} \times \frac{V_{dc}}{2}$

Given the DC source is $600$ V, we can calculate the RMS phase voltage:
$V_{ph,rms} = \sqrt{\frac{2}{3}} \times \frac{600 \text{ V}}{2} = 300\sqrt{\frac{2}{3}} \approx 244.95 \text{ V}$

Next, we find the power dissipated in a single phase, which has a resistance of 20 $\Omega$:
$P_{phase} = \frac{V_{ph,rms}^2}{R} = \frac{(244.95 \text{ V})^2}{20 \text{ } \Omega} \approx 3000 \text{ W}$

Finally, the total power for the three-phase load is simply three times the power per phase:
$P_{total} = 3 \times P_{phase} = 3 \times 3000 \text{ W} = 9000 \text{ W} = 9.0 \text{ kW}$

To determine the range of $K$ for stability, we apply the Routh-Hurwitz criterion to the given characteristic equation. The first two rows of the Routh array are formed from the coefficients of the polynomial.

$s^3$ row: $1 \quad K+2$
$s^2$ row: $K \quad 3$

For stability, all elements in the first column of the Routh array must be positive. The first element is $1$, so we need $K>0$ from the $s^2$ row. The next element is calculated as $\frac{K(K+2)-1(3)}{K}$, which must also be positive.

Since we already established $K>0$, this condition simplifies to the numerator being positive: $K(K+2)-3 > 0$. Expanding this gives $K^{2+}2K-3 > 0$, which factors into $(K+3)(K-1) > 0$. Given that $K$ must be positive, the only way for this inequality to hold is if $K-1 > 0$, which means $K>1$.

First, we determine the synchronous speed ($N_s$), which is the speed of the stator's magnetic field. For a 4-pole machine at a supply frequency ($f$) of 60 Hz, the synchronous speed is $N_s = \frac{120f}{P} = \frac{120 \times 60}{4} = 1800$ RPM.

The rotor current frequency ($f_r$) is directly related to slip ($s$) and the supply frequency by $f_r = |s| \cdot f$. Given $f_r=5$ Hz, we find the magnitude of the slip is $|s| = \frac{5}{60}$.

For an induction machine to operate as a generator, its rotor must be driven faster than the synchronous speed, resulting in a negative slip. Thus, $s = -5/60$.

Finally, we use the slip formula, $s = \frac{N_s - N}{N_s}$, to find the mechanical rotor speed ($N$).
Solving for $N$: $N = N_s(1-s) = 1800(1 - (-5/60)) = 1800(1 + 5/60) = 1950$ RPM.

Let's determine the DC component, or average value, for each voltage by analyzing the circuit's behavior over one full cycle of the input signal.

The voltage $v_2$ is the output of a half-wave rectifier circuit. The peak voltage of this rectified sine wave is $V_{peak} = \pi/2$ V. The average value of a half-wave rectified sinusoid is its peak amplitude divided by $\pi$. Thus, the DC component of $v_2$ is:
$V_{2,avg} = \frac{V_{peak}}{\pi} = \frac{\pi/2}{\pi} = 0.5$ V.

The voltage $v_1$ has a more complex waveform. To find its DC component, we must integrate over the full period and divide by the period's length. The underlying reasoning of the original explanation is based on the following integral:
$V_{1,avg} = \frac{1}{2\pi} \left[ \int_{0}^{\pi} \frac{\pi}{2}\sin(\omega t)d(\omega t) + \int_{\pi}^{2\pi} \pi\sin(\omega t)d(\omega t) \right]$.
The first integral (area of the first half-cycle) evaluates to $\pi$, and the second integral (area of the second half-cycle) evaluates to $-2\pi$.
The average value is therefore $V_{1,avg} = \frac{1}{2\pi}(\pi - 2\pi) = -0.5$ V.

The number of non-linear equations required for a load flow solution equals the number of unknown state variables in the power system. For a system with $N$ buses, we begin with $2N$ potential variables (voltage magnitude $|V|$ and angle $\delta$ for each bus). In this 10-bus system, we start with $2 \times 10 = 20$ variables.

We then subtract the variables that are specified:

• The slack bus (G1) defines both $|V|$ and $\delta$, accounting for 2 knowns.
• The two standard generator buses (G3, G4) are PV buses, specifying $|V|$ for each, giving 2 more knowns.
• The two voltage-controlled load buses (L3, L4) also specify $|V|$, adding 2 knowns.
• Bus G2, having reached its reactive limit, is treated as a PQ bus, so its $|V|$ and $\delta$ are unknown.

The total number of unknown variables is $20 - 2_{\text{slack}} - 2_{\text{PV}} - 2_{\text{V-ctrl}} = 14$. Thus, 14 non-linear equations must be solved.

To determine the power supplied by the source, we must first find the current $I$ that it provides. We can achieve this by applying Kirchhoff's Current Law (KCL) at the central node where the three branches meet.

KCL states that the sum of currents entering a node must equal the sum of currents leaving it. Here, the currents $I$ and $0.4I$ flow into the node, while 14 A flows out. This relationship gives us the equation $I + 0.4I = 14$.

Solving this equation, we get $1.4I = 14$, which yields $I = 10$ A.

Finally, the power supplied by the 25 V source is the product of its voltage and the current it delivers:
$P = V \times I = 25 \text{ V} \times 10 \text{ A} = 250$ W.

This circuit is a single-phase full-bridge inverter, which creates a square wave output $v_o(t)$. The amplitude of this square wave is equal to the DC input voltage, $V_s = 220$ V.

From the Fourier series of a square wave, the peak amplitude of the fundamental component is $\frac{4V_s}{\pi}$. The RMS value of a sinusoidal signal is its peak value divided by $\sqrt{2}$.

Therefore, the RMS value of the fundamental component of the output voltage is:
$V_{o1, \text{rms}} = \frac{\text{Peak Value}}{\sqrt{2}} = \frac{4V_s}{\sqrt{2}\pi}$
Plugging in the given source voltage:
$V_{o1, \text{rms}} = \frac{4 \times 220}{\sqrt{2}\pi} \approx 198.06 \text{ V}$

To find the simplified expression, we group the 1s in the Karnaugh map into the largest possible rectangular blocks.

First, we can form a large group of four 1s by combining the cells where $B=1$ (the bottom two rows) and $D=0$ (the first and last columns). In this group, the variables $A$ and $C$ change, so they are eliminated, giving us the term $B\bar{D}$.

Next, there is one '1' left uncovered in the cell corresponding to $ABCD = 1111$. To cover it, we can create a pair with the adjacent '1' at $ABCD = 1110$. For this group, the variables $A$, $B$, and $C$ are constant, while $D$ changes. This yields the term $ABC$.

The complete minimal expression is the sum of the terms from these two essential groups: $B\bar{D} + ABC$.

Here's a clearer way to understand the solution.

First, let's establish the relationship between speed and current. We are given that load torque is proportional to the square of the speed ($T \propto N^2$). For a DC series motor, flux is proportional to armature current ($\phi \propto I_a$), which means torque is proportional to the square of the armature current ($T \propto \phi I_a \propto I_a^2$). Combining these two facts, we see that $N^2 \propto I_a^2$, or simply $N \propto I_a$.

Since the speed is reduced by 50% ($N_2 = 0.5 N_1$), the armature current must also decrease by the same factor: $I_{a2} = 0.5 I_{a1} = 0.5 \times 30 \text{ A} = 15 \text{ A}$.

Now, let's calculate the back EMF in both cases. The initial back EMF is $E_{b1} = V - I_{a1}(R_a + R_{se}) = 220 - 30(0.4 + 0.1) = 205 \text{ V}$. The speed of a series motor is also governed by $N \propto E_b/I_a$. We can use this to find the final back EMF: $\frac{N_2}{N_1} = \frac{E_{b2}}{E_{b1}} \frac{I_{a1}}{I_{a2}}$, which gives $0.5 = \frac{E_{b2}}{205} \times \frac{30}{15}$. Solving this yields $E_{b2} = 51.25 \text{ V}$.

Finally, we find the required external resistance, $R_{ext}$. The final back EMF is given by $E_{b2} = V - I_{a2}(R_a + R_{se} + R_{ext})$. Plugging in the values, we get $51.25 = 220 - 15(0.4 + 0.1 + R_{ext})$. Solving this equation for $R_{ext}$ gives a value of $10.75 \, \Omega$.

To determine the system's transfer function, we convert the given state-space representation using the standard formula $G(s) = C(sI-A)^{-1}B+D$. The state-space matrices are

,

, $C=[1 \ 0]$, and $D=0$.

First, we construct the matrix $(sI-A)$:

Next, we find its inverse, $(sI-A)^{-1}$. The determinant is $s(s-1) - (-2)(-2) = s^{2-}s-4$.

Finally, we substitute all matrices into the transfer function formula and multiply:

Excellent! Let's break down this problem step-by-step for better clarity.

Initially, with the switch on N, the three wattmeters measure the power per phase. The total power is simply the sum of these readings: $P_{total} = 3 \times 577.35 W = 1732 W$. We can determine the load's power factor angle, $\phi$, by comparing this real power to the given apparent power of $3464 VA$. The power factor is $\cos(\phi) = P/S = 1732/3464 = 0.5$, which gives us a lagging angle of $\phi = 60^\circ$.

When the switch is moved to position Y, the setup becomes a variation of the two-wattmeter method, with the Y-phase line serving as the common point for the pressure coils.

The reading for $W_1$ is given by the formula $V_{RY} I_R \cos(\angle V_{RY}, I_R)$, where the angle is $(30^\circ + \phi)$. Substituting $\phi = 60^\circ$, the angle becomes $90^\circ$, and since $\cos(90^\circ)=0$, the reading $W_1=0$. The reading for $W_2$ is also zero because its pressure coil is connected between the Y-phase and itself, meaning there is no voltage difference.

The total power drawn by the load doesn't change, so it remains $1732 W$. In this new configuration, the total power is the sum of the readings $W_1$ and $W_3$. Since we've found that $W_1=0$, all the power must be measured by $W_3$. Therefore, $W_3 = 1732 W$.

To solve this problem, we first need to find the overall transmission parameters for the two networks connected in cascade. When using ABCD parameters, the overall matrix for a cascaded system is the product of the individual matrices. This gives us a single equivalent network relating port 1 to port 3.

The overall parameters $A$ and $B$ can be found by multiplying the matrices:

This gives $A = A_1A_2+B_1C_2$ and $B = A_1B_2+B_1D_2$. The resulting equation for the entire system is $V_1 = AV_3 + BI_3$.

The Thevenin voltage, $V_T$, is the open-circuit voltage at port 3. We find this by setting the output current $I_3 = 0$. The equation becomes $V_1 = AV_3$. Thus, $V_T = V_3 = \frac{V_1}{A}$, which means $V_T = \frac{V_1}{A_1A_2+B_1C_2}$.