GATE ECE 2009

The order of a differential equation is determined by the highest-order derivative present in the equation.

Let's examine the given equation:
$\frac{d^{2}y}{dt^{2}}+(\frac{dy}{dt})^{3}+y^{4}=e^{-t}$

The equation contains two derivative terms: $\frac{d^{2}y}{dt^{2}}$ (the second derivative) and $\frac{dy}{dt}$ (the first derivative). The highest order among these is 2.

The power of 3 on the term $(\frac{dy}{dt})^{3}$ refers to the degree of that term, not the order of the derivative. Since the highest derivative is the second derivative, the order of the entire equation is 2.

The symmetry of a periodic function dictates the terms present in its Fourier series. An even function, defined by $f(t) = f(-t)$, mirrors the symmetry of the cosine function itself. When calculating the sine coefficients ($b_n$), you integrate an even function $f(t)$ against an odd function $\sin(n\omega t)$. The result is an odd function, whose integral over a symmetric interval is always zero. This eliminates all sine terms, leaving only cosine terms (P).

Conversely, an odd function, defined by $f(t) = -f(-t)$, mirrors the symmetry of the sine function. Calculating the cosine coefficients ($a_n$) involves integrating an odd function $f(t)$ against an even function $\cos(n\omega t)$. This product is odd, making the coefficients zero. Consequently, the series consists only of sine terms (S).

To analyze the frequency components of the function, we should first simplify its expression using trigonometric identities. The power-reduction formula for sine squared is $\sin^2 t = \frac{1}{2}(1 - \cos 2t)$.

Substituting this into the given function, we get:
$f(t) = \frac{1}{2}(1 - \cos 2t) + \cos 2t = \frac{1}{2} - \frac{1}{2}\cos 2t + \cos 2t = \frac{1}{2} + \frac{1}{2}\cos 2t$.

This simplified form shows the function is a sum of two signals. The constant term, $\frac{1}{2}$, is a DC component, which has a frequency of 0 Hz. The other term, $\frac{1}{2}\cos 2t$, has an angular frequency of $\omega = 2$ rad/s.

To convert angular frequency $\omega$ to linear frequency $f$ in Hertz, we use the formula $f = \frac{\omega}{2\pi}$. This gives us the second frequency component: $f = \frac{2}{2\pi} = \frac{1}{\pi}$ Hz.

In a silicon crystal, the concentration of Si atoms is approximately $5 \times 10^{22} \text{cm}^{-3}$. Each Si atom has 4 valence electrons, so their concentration is about $2 \times 10^{23} \text{cm}^{-3}$. These values are much larger than the given concentration, ruling out options A and D.

An n-type semiconductor is created by adding donor dopant atoms. The value $4 \times 10^{19} \text{cm}^{-3}$ is a typical heavy doping concentration. At room temperature, these dopants are ionized, creating a free electron concentration of $n \approx 4 \times 10^{19} \text{cm}^{-3}$.

Using the mass-action law, $p = n_i^2 / n$, where the intrinsic carrier concentration $n_i \approx 1.5 \times 10^{10} \text{cm}^{-3}$ for silicon. This gives a hole concentration of $p \approx (1.5 \times 10^{10})^2 / (4 \times 10^{19}) \approx 5.6 \text{ cm}^{-3}$, which is extremely small, eliminating option B. Thus, the given value represents the concentration of dopant atoms.

This question tests your knowledge of common acronyms for digital logic families. Let's break down each term.

The abbreviation "TTL" stands for **T**ransistor-**T**ransistor **L**ogic. This name reflects the fact that logic gates in this family are built using transistors for both the logic operation and the amplification stages.

The abbreviation "CMOS" stands for **C**omplementary **M**etal-**O**xide-**S**emiconductor. This refers to the type of technology used, which employs pairs of complementary P-type and N-type metal-oxide-semiconductor field-effect transistors (MOSFETs) to create logic gates.

The given signal $x(n)$ is a combination of two separate sequences. Let's analyze the Region of Convergence (ROC) for each part.

The first term, $(\frac{1}{3})^{n}u(n)$, is a right-sided sequence. The ROC for a right-sided sequence is the region outside a circle with a radius equal to the magnitude of the pole, so we have $|z| > \frac{1}{3}$.

The second term, $-(\frac{1}{2})^{n}u(-n-1)$, is a left-sided sequence. Its ROC is the region inside a circle defined by its pole, which gives us $|z| < \frac{1}{2}$.

The overall ROC for $x(n)$ is the intersection of these two individual regions. The region that satisfies both $|z| > \frac{1}{3}$ and $|z| < \frac{1}{2}$ is the annular ring $\frac{1}{3} < |z| < \frac{1}{2}$.

The power of the output random process, $P_Y$, is determined by the input signal's power spectral density (PSD) and the characteristics of the filter. Specifically, the output PSD is the input PSD multiplied by the squared magnitude of the filter's frequency response, $S_Y(f) = S_X(f) |H(f)|^2$.

To find the total power, we integrate the output PSD over all frequencies. Since the input white noise has a constant PSD of $S_X(f) = 1 \times 10^{-10}$ W/Hz, the output power is this constant value times the area under the $|H(f)|^2$ graph.

$P_Y = \int_{-\infty }^{\infty } S_Y(f) df = (1 \times 10^{-10}) \times \text{Area under } |H(f)|^2$

The graph of $|H(f)|^2$ is a triangle with a base of $20$ kHz (from $-10$ kHz to $+10$ kHz) and a height of $1$. The area is therefore $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (20 \times 10^3 \text{ Hz}) \times 1 = 10^4 \text{ Hz}$.

Finally, multiplying the input PSD by this area gives the output power:
$P_Y = (1 \times 10^{-10} \text{ W/Hz}) \times (10^4 \text{ Hz}) = 1 \times 10^{-6} \text{ W}$.

Structure P is a coaxial line, a type of transmission line with two conductors. This two-conductor geometry allows for the propagation of a Transverse Electro-Magnetic (TEM) wave. A unique property of the TEM mode is that it has no lower frequency limit for propagation, meaning its cutoff frequency is zero ($f_c = 0$).

Conversely, structures Q and R are hollow waveguides. These single-conductor guides cannot support the TEM mode. Their fundamental modes will be either TE or TM, both of which have a non-zero cutoff frequency determined by the waveguide's physical dimensions.

Each toss of a fair coin is an independent event with a probability of $\frac{1}{2}$ for heads (H) and $\frac{1}{2}$ for tails (T). The problem specifies a single, unique sequence of outcomes for the 10 tosses: H, H, T, T, T, T, T, T, T, T. To find the probability of this specific sequence occurring, we multiply the probabilities of each individual outcome.

The probability is:
$P(\text{H, H, T, T, T, T, T, T, T, T}) = P(H) \times P(H) \times P(T) \times \dots \times P(T)$
This is the product of 10 individual probabilities, each of which is $\frac{1}{2}$.
$P(\text{sequence}) = (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) = (\frac{1}{2})^{10}$.

According to the Wiener-Khinchin theorem, a process's power spectral density (PSD) and its autocorrelation function, $R(\tau)$, form a Fourier transform pair. This means the autocorrelation function is the inverse Fourier transform of the PSD.

This problem uses a well-known Fourier transform pair: a triangular pulse in one domain corresponds to a sinc-squared function, written as $\text{sinc}^2(f)$, in the other.

The question implies the PSD has the shape of a $\text{sinc}^2$ function (often colloquially, if imprecisely, called a sine-squared function in this context). Therefore, its inverse Fourier transform-the autocorrelation function-must have the shape of a triangular pulse.

To solve this integral, we'll use the Residue Theorem. The contour is the unit circle, which encloses the only singularity at the origin, $z=0$.

First, let's substitute the given expression for $f(z)$ into the integrand:
$\frac{1+f(z)}{z} = \frac{1 + (c_0 + c_1 z^{-1})}{z}$

Next, simplify the expression to find its Laurent series form around $z=0$:
$\frac{1 + c_0 + c_1/z}{z} = \frac{1+c_0}{z} + \frac{c_1}{z^2}$

According to the Residue Theorem, the value of the integral is $2\pi j$ times the sum of the residues inside the contour. The residue at $z=0$ is the coefficient of the $z^{-1}$ term in the Laurent series.

From our simplified expression, the coefficient of $z^{-1}$ is $(1+c_0)$.

Therefore, the integral is $2\pi j \times \text{Residue} = 2\pi j (1+c_0)$.

Since, the power is absorbed by 60 V source

The relationship between the diffusion coefficient ($D$) and carrier mobility ($\mu$) in a semiconductor is defined by the Einstein relation. This fundamental equation is expressed as $\frac{D}{\mu} = V_T$, where $V_T$ represents the thermal voltage.

The question asks for the units of the ratio of mobility to the diffusion coefficient, which is $\frac{\mu}{D}$. To find this, we can simply rearrange the Einstein relation:
$\frac{\mu}{D} = \frac{1}{V_T}$
Since the unit of thermal voltage ($V_T$) is Volts (V), the units of its reciprocal must be inverse volts, which is written as $V^{-1}$.

An interrupt is called 'vectored' when its service routine resides at a fixed memory address. This allows the processor to automatically jump to that specific location without needing external guidance. This matches the description "starts from a fixed location of memory."

An interrupt is called 'maskable' if the processor can choose to ignore or postpone it. The phrase "can be delayed or rejected" directly describes this ability to be masked. Therefore, the interrupt has both of these properties.

This circuit is a voltage divider where the output voltage is measured across the parallel combination of the load resistor $R_L$ and the capacitor $C$. First, let's find the impedance of this parallel section, which we'll call $Z_{out}$.
$Z_{out} = R_L || \frac{1}{sC} = \frac{R_L \cdot \frac{1}{sC}}{R_L + \frac{1}{sC}} = \frac{R_L}{1+sCR_L}$.

The overall transfer function is found using the voltage divider rule:
$\frac{V_o(s)}{V_i(s)} = \frac{Z_{out}}{R+Z_{out}} = \frac{\frac{R_L}{1+sCR_L}}{R+\frac{R_L}{1+sCR_L}}$.

To simplify, multiply the numerator and denominator by $(1+sCR_L)$, which gives:
$\frac{V_o(s)}{V_i(s)} = \frac{R_L}{R(1+sCR_L)+R_L} = \frac{R_L}{R+R_L+sCR R_L}$.

We are given that this transfer function equals $\frac{1}{2+sCR}$. By comparing the two expressions, we can see that if we set $R_L = R$, our derived function becomes $\frac{R}{R+R+sCR^2} = \frac{R}{2R+sCR^2}$. Dividing the numerator and denominator by $R$ yields $\frac{1}{2+sCR}$, which matches the given function. Therefore, the load resistance must be $R_L=R$.

To assess the system's controllability, we examine the rank of the controllability matrix, $Q_c = [B \quad AB]$. In this case, the matrix $A$ is the $2 \times 2$ identity matrix, $I$. Multiplying any matrix by the identity matrix leaves it unchanged, so $AB = IB = B$.

This means our controllability matrix is $Q_c = [B \quad B]$. Substituting the vector for $B$, we get:
$Q_c = \begin{bmatrix} p & p \\ q & q \end{bmatrix}$
The two columns of this matrix are identical. This makes them linearly dependent, regardless of the values of $p$ and $q$. A matrix with linearly dependent columns is singular, meaning its rank is always less than the dimension of the state space (which is 2). For a system to be controllable, this matrix must be full rank. Since it never is, the system is uncontrollable for all values of $p$ and $q$.

When the message signal $m(t)$ modulates the carrier, it produces what's known as a double-sideband (DSB) signal. For the given signals, this DSB signal is $m(t)\cos(2\pi f_c t) = \cos(2\pi f_m t)\cos(2\pi f_c t)$.

Using a product-to-sum identity, this expands into two distinct frequency components:
$\frac{1}{2}\cos[2\pi(f_c-f_m)t] + \frac{1}{2}\cos[2\pi(f_c+f_m)t]$
These represent the lower sideband (LSB) and upper sideband (USB), respectively. A single side-band (SSB) signal transmits only one of these. The expression $\cos[2\pi(f_c+f_m)t]$ is a pure sinusoid at the upper sideband frequency, making it a perfect example of an SSB signal.

$H_{dicection}$ due to any current wire = $I$ flow direction $\times$ (cross) radial vector from current to point of consideration.

We are asked to find the conditional probability $P(X+Y=2|X-Y=0)$. The formula for conditional probability tells us this is equal to $\frac{P(X+Y=2 \text{ and } X-Y=0)}{P(X-Y=0)}$.

First, let's calculate the probability of the condition, $P(X-Y=0)$. This event occurs when $X=Y$. The possible cases are $(0,0)$, $(1,1)$, and $(2,2)$. Since $X$ and $Y$ are independent, we can find the total probability by summing the probabilities of each case:
$P(X=Y) = (\frac{1}{2})(\frac{1}{2}) + (\frac{1}{4})(\frac{1}{4}) + (\frac{1}{4})(\frac{1}{4}) = \frac{1}{4} + \frac{1}{16} + \frac{1}{16} = \frac{6}{16} = \frac{3}{8}$.

Next, let's find the probability of the joint event in the numerator. For both $X+Y=2$ and $X-Y=0$ to be true, we must have $X=1$ and $Y=1$. The probability of this specific outcome is $P(X=1, Y=1) = P(X=1) \times P(Y=1) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$.

Finally, we divide the probability of the joint event by the probability of the condition:
$P(X+Y=2|X-Y=0) = \frac{1/16}{3/8} = \frac{1}{16} \times \frac{8}{3} = \frac{1}{6}$.

To find the Taylor series of a function centered at a point other than zero, it's often easiest to use a substitution. Let's set $u = x-\pi$, so that we are looking for a series in powers of $u$ centered at $u=0$.

Substituting $x = u+\pi$ into the original function gives $\frac{\sin(u+\pi)}{u}$. Using the trigonometric identity $\sin(\theta+\pi) = -\sin\theta$, this expression simplifies to $-\frac{\sin u}{u}$.

Now, we can use the standard Maclaurin series for $\sin u$, which is $\sin u = u - \frac{u^3}{3!} + \frac{u^5}{5!} - \dots$.

Plugging this series into our simplified function, we get:
$-\frac{1}{u}\left(u - \frac{u^3}{3!} + \dots\right) = -\left(1 - \frac{u^2}{3!} + \dots\right) = -1 + \frac{u^2}{3!} - \dots$

Finally, replacing $u$ with $x-\pi$ gives us the desired Taylor series:
$-1 + \frac{(x-\pi)^2}{3!} - \dots$

This question is a direct application of Stokes' Theorem from vector calculus. The theorem relates the line integral of a vector field around a closed loop to the surface integral of its curl over the surface enclosed by the loop.

For any vector field, let's call it $\vec{F}$, Stokes' Theorem is written as:
$\oint_{C} \vec{F} \cdot d\vec{l} = \iint_{S_{C}} (\nabla \times \vec{F}) \cdot d\vec{S}$
Let's apply this theorem to the vector field $\vec{A}$ given in the problem. We substitute $\vec{F}$ with $\vec{A}$:
$\oint_{C} \vec{A} \cdot d\vec{l} = \iint_{S_{C}} (\nabla \times \vec{A}) \cdot d\vec{S}$
The question states the relationship $\vec{V} = \nabla \times \vec{A}$. We can substitute $\vec{V}$ for the term $\nabla \times \vec{A}$ in the equation above. This gives us the final, correct relationship:
$\oint_{C} \vec{A} \cdot d\vec{l} = \iint_{S_{C}} \vec{V} \cdot d\vec{S}$

This question asks for the Laplace transform of an integral, which is a standard property. The operation of integration in the time domain, $\int_{0}^{t} f(\tau) d\tau$, has a direct counterpart in the s-domain.

The property connects the result to the transform of the original function, $F(s)$, and its initial value, $f(0)$. The rule states that you first adjust the transform $F(s)$ by subtracting the initial condition, yielding the term $F(s) - f(0)$. Then, to account for the integration, this entire expression is divided by $s$. This two-step process gives the final expression for the transform of the integral.

To find the family of solution curves for each differential equation, we'll use the method of separation of variables and integration.

For equation P, separating variables in $\frac{dy}{dx} = \frac{y}{x}$ gives $\frac{dy}{y} = \frac{dx}{x}$. Integrating both sides yields $\ln|y| = \ln|x| + c$, which simplifies to $y = kx$, the equation for a family of straight lines.

For equation Q, $\frac{dy}{dx} = -\frac{y}{x}$ separates to $\frac{dy}{y} = -\frac{dx}{x}$. Integration gives $\ln|y| = -\ln|x| + c$, which can be rewritten as $xy = k$. This is the equation for a family of hyperbolas.

For equation R, separating variables in $\frac{dy}{dx} = \frac{x}{y}$ gives $y\,dy = x\,dx$. Integrating results in $\frac{y^2}{2} = \frac{x^2}{2} + c$, or $y^2 - x^2 = k$, which also represents a family of hyperbolas.

Finally, for equation S, $\frac{dy}{dx} = -\frac{x}{y}$ becomes $y\,dy = -x\,dx$. Integrating this gives $\frac{y^2}{2} = -\frac{x^2}{2} + c$, or $x^2 + y^2 = k$, the equation for a family of circles.

Therefore, the matching is: P→2 (Straight lines), Q→3 (Hyperbolas), R→3 (Hyperbolas), and S→1 (Circles).

A fundamental property of matrices is that the sum of the eigenvalues equals the trace of the matrix (the sum of the elements on the main diagonal). Let's calculate the trace of the given matrix:
$\text{Trace} = -1 + (-1) + 3 = 1$
Now, we check which set of options sums to 1. For option D, the sum of the eigenvalues is:
$3 + (-1 + 3j) + (-1 - 3j)$
The imaginary parts cancel each other out ($3j - 3j = 0$), simplifying the sum to $3 - 1 - 1 = 1$. This matches our calculated trace. Additionally, for a matrix with all real entries, any complex eigenvalues must appear as conjugate pairs. The values $-1 + 3j$ and $-1 - 3j$ are a conjugate pair, satisfying this condition.

To find the reactive power consumed by the load, we first need the total current flowing in the circuit. The total impedance is the sum of the source and load impedances: $Z_{total} = Z_s + Z_L = (1+2j)\Omega + (7+4j)\Omega = (8+6j)\Omega$.

Next, we calculate the magnitude of the RMS current using Ohm's Law.
$|I_{rms}| = \frac{|V_{rms}|}{|Z_{total}|} = \frac{20 \text{ V}}{|(8+6j)\Omega|} = \frac{20}{\sqrt{8^2+6^2}} = \frac{20}{10} = 2$ A.

The reactive power ($Q_L$) absorbed by the load is calculated as the squared current magnitude times the load's reactance ($X_L$). From the load impedance $Z_L = (7+4j)\Omega$, the reactance is $X_L = 4\Omega$.

Therefore, the reactive power is $Q_L = |I_{rms}|^2 X_L = (2\text{ A})^2 \times 4\Omega = 16$ VAR.

For a long time before $t=0$, the switch at position 'a' fully charged the capacitor to the source voltage. Because capacitor voltage cannot change instantaneously, the initial voltage at $t=0^+$ is $v_C(0^+) = 100 \text{ V}$.

After the switch moves to position 'b', the capacitor discharges through the resistor. The current for this first-order RC circuit is given by $i(t) = i(0^+) e^{-t/\tau}$, where $\tau$ is the time constant.

The initial current is found using Ohm's law: $i(0^+) = v_C(0^+) / R = 100 \text{ V} / 5 \text{ k}\Omega = 20 \text{ mA}$.
The time constant is $\tau = RC = (5 \times 10^3 \Omega)(0.16 \times 10^{-6} \text{ F}) = 0.8 \text{ ms}$, so the decay rate is $1/\tau = 1250 \text{ s}^{-1}$.

Combining these results, the current for $t>0$ is $i(t) = 20e^{-1250t}u(t) \text{ mA}$.

For maximum power to be delivered to the resistor $R_L$, its value must be equal to the Thevenin equivalent resistance ($R_{Th}$) of the source circuit.

To find $R_{Th}$, we deactivate the independent 10V source (replacing it with a short) and apply a test voltage $V_T$ across the output terminals. Let's conveniently choose $V_T=1$ V. By definition, the controlling voltage $V_x$ across the output terminals is now 1 V, which means the dependent voltage source also has a value of 1 V.

Notice that the voltage across the top $4\,\Omega$ resistor is driven by the dependent source and opposed by the test source. An outer-loop KVL shows these two 1 V sources cancel each other out, leaving 0 V across the top resistor and thus zero current in that branch. All the test current, $I_T$, must therefore flow through the right-hand $4\,\Omega$ resistor.

This current is $I_T = V_x / 4\,\Omega = 1\,\text{V} / 4\,\Omega = 0.25\,\text{A}$. Finally, the Thevenin resistance is calculated as $R_{Th} = V_T / I_T = 1\,\text{V} / 0.25\,\text{A} = 4\,\Omega$. Therefore, $R_L$ should be $4\,\Omega$.

To determine the steady state value of the current, we analyze the circuit's behavior as time approaches infinity ($t \rightarrow \infty$). The governing equation is $L\frac{di}{dt}+Ri=V_{0}(1+Be^{-Rt/L}\sin t)u(t)$.

As $t \rightarrow \infty$, the exponential term $e^{-Rt/L}$ decays to zero. Consequently, the input voltage source term simplifies:
$V_{\in}(t \rightarrow \infty ) = \lim_{t \to \infty } V_{0}(1+Be^{-Rt/L}\sin t) = V_{0}(1+0) = V_{0}$
The circuit is now driven by a constant DC voltage $V_{0}$. In a DC steady state, the current becomes constant, so its derivative $\frac{di}{dt}$ is zero. The inductor acts as a short circuit.

The original differential equation simplifies to:
$L(0) + Ri(\infty ) = V_{0}$
Solving for the steady state current $i(\infty )$ gives:
$i(\infty ) = \frac{V_{0}}{R}$
The initial condition affects the transient response but not the final steady state value for this stable system.

The behavior of this circuit is determined by the state of the ideal diode. The diode's cathode is connected to a 1V source, setting its potential to 1V.

When the input voltage $V_i \le 1V$, the potential at the diode's anode (node $V$) tends to be less than or equal to 1V. This reverse-biases the diode, causing it to act as an open circuit. With no current flowing through the resistor $R$, there is no voltage drop across it, so the output voltage $V$ equals the input voltage $V_i$.

When the input voltage $V_i > 1V$, the anode potential tends to exceed the 1V cathode potential. This forward-biases the ideal diode, making it act like a short circuit. As a result, the output node is connected directly to the 1V source, clamping the output voltage at $V = 1V$.

Combining these two conditions, the output voltage $V$ is equal to $V_i$ when $V_i \le 1V$ and is 1V when $V_i > 1V$. This describes the function $V = min(V_i, 1)$.

In an n-channel MOSFET operating in the active region, a positive drain-source voltage ($V_{DS}$) causes a drain current to flow. Due to the resistance of the inversion layer, this current creates a voltage drop along the channel. As a result, the channel potential, $V(x)$, increases steadily from $0$ V at the source to $V_{DS}$ at the drain. Therefore, statement S2 is true.

The amount of inversion charge at any point is determined by the local gate-to-channel voltage, which is $V_{GS} - V(x)$. Since the channel potential $V(x)$ increases from source to drain (as per S2), the effective gate-to-channel voltage decreases. A lower effective voltage induces less charge, causing the inversion charge density to decrease from source to drain. This makes the channel tapered and confirms that statement S1 is also true. Thus, S2 is the reason for S1.

This circuit is an astable multivibrator, which generates a square wave output, $v_0(t)$. The amplitude of this wave is determined by the op-amp's positive and negative saturation voltages ($\pm V_{sat}$), which are independent of the external resistor $R_2$.

The frequency of oscillation depends on the time it takes the capacitor $C$ to charge and discharge through resistor $R_1$. The switching thresholds for the op-amp are set by the voltage divider formed by $R_2$ and $R_3$. The threshold voltage is $\pm \beta V_{sat}$, where the feedback factor is $\beta = \frac{R_3}{R_2+R_3}$. The oscillation period is $T = 2R_1C \ln\left(\frac{1+\beta}{1-\beta}\right)$. Since $\beta$ depends on $R_2$, the period $T$ and the frequency $f=1/T$ also depend on $R_2$. Thus, only the frequency of $v_0(t)$ depends on $R_2$.

This circuit uses negative feedback. The op-amp adjusts its output to ensure the transistor's emitter provides the necessary feedback voltage to the non-inverting input.

First, let's find the transistor's emitter current, $I_E$. This current is set by the 10 V source, the 5 V Zener diode, and the 5 kΩ resistor.
$I_E = \frac{10 \text{ V} - 5 \text{ V}}{5 \text{ k}\Omega} = 1 \text{ mA}$

This emitter current flows through the 1.4 kΩ resistor, establishing the voltage at the non-inverting input, which we'll call $V'$.
$V' = 1 \text{ mA} \times 1.4 \text{ k}\Omega = 1.4 \text{ V}$

The op-amp's output voltage, $V$, is connected to the base of the transistor. For the emitter to be at voltage $V'$, the base must be higher by the base-emitter voltage drop, $V_{BE}$.
$V = V' + V_{BE} = 1.4 \text{ V} + 0.6 \text{ V} = 2 \text{ V}$

This amplifier's response is frequency-dependent due to the capacitors, which act as a high-pass filter. First, let's determine the maximum possible voltage gain ($A_v$), which occurs in the amplifier's mid-band frequency range. This gain is given by $A_v \approx -h_{fe} \frac{R_C}{h_{ie}}$. Using the circuit values, we get $A_v \approx -150 \frac{3k\Omega}{3k\Omega} = -150$.

The input signal $v_i(t)$ contains two frequencies: a very low frequency ($\omega = 20$ rad/s) and a high frequency ($\omega = 10^6$ rad/s). The low-frequency component will be almost completely blocked by the coupling capacitors, as its frequency is far below the amplifier's lower cutoff frequency. Conversely, the high-frequency component lies well within the amplifier's passband and will receive the full mid-band gain.

Therefore, we only consider the amplification of the high-frequency term. Applying the gain of -150 to this part of the signal gives the approximate output:
$v_o(t) \approx A_v \times (B\sin{10^6t}) = -150 B\sin{10^6t}$.