GATE ECE 2018

To find the overall transconductance $g_m$, we consider the effect of the input voltage $V_{GS}$. This voltage controls the current in transistor $M_1$, given by $i_{d1} = g_{m1}V_{GS}$. Since $M_1$ and $M_2$ are connected in series, the drain current of $M_1$ must flow through $M_2$. Therefore, the output current $i_D$ is equal to $i_{d1}$, and the overall transconductance is simply $g_m = \frac{i_D}{V_{GS}} = g_{m1}$.

To find the output resistance $r_o$, we look into the drain of $M_2$ while grounding the input gate G. With its gate grounded, $M_1$ acts as a resistance of value $r_{o1}$ connected to the source of $M_2$. This circuit is a cascode configuration, where the resistance at the source of $M_2$ significantly boosts the output resistance. The total resistance looking into the drain of $M_2$ is $r_o = r_{o2} + r_{o1} + g_{m2}r_{o1}r_{o2}$. Since the term $g_{m2}r_{o1}r_{o2}$ is typically much larger than the other two terms, we can approximate the output resistance as $r_o \approx g_{m2}r_{o1}r_{o2}$.

For time $t > 0$, the circuit has a constant input current of $I = \frac{v_{IN}}{R} = \frac{1 \text{ V}}{1 \text{ k}\Omega} = 1 \text{ mA}$. This current flows through the feedback path, charging the capacitors. The output voltage, $v_{OUT}$, is the negative sum of the voltage across the top capacitor ($v_{C1}$) and the voltage across the parallel Zener/capacitor combination ($v_Z$).

The Zener diode breaks down when the voltage across it reaches $2.5 \text{ V}$, clamping the voltage $v_Z$ at this value. For the output to reach the target of $-10 \text{ V}$, the Zener must be in its breakdown region. Therefore, we can write the output voltage as $v_{OUT} = -(v_{C1} + v_Z)$.

Substituting the known values, we have $-10 \text{ V} = -(v_{C1} + 2.5 \text{ V})$. This implies the voltage across the top capacitor must be $v_{C1} = 7.5 \text{ V}$.

The time required for the constant $1 \text{ mA}$ current to charge the $1 \text{ }\mu\text{F}$ top capacitor to $7.5 \text{ V}$ is calculated from the capacitor's voltage-current relationship:
$v_{C1} = \frac{1}{C} \int_{0}^{t} I \,dt \implies 7.5 \text{ V} = \frac{1}{1 \text{ }\mu\text{F}} (1 \text{ mA} \times t)$.
This simplifies to $7.5 = 1000t$, which gives a time of $t = 7.5 \text{ ms}$.

A transimpedance amplifier is designed to convert an input current ($i_{\in}$) into an output voltage ($v_{out}$).

First, to accurately measure the incoming current, the amplifier's input impedance ($Z_{\in}$) must be as low as possible. This ensures that the amplifier acts like a virtual ground, drawing in the entire signal current from the source for a precise conversion.

Second, the amplifier's output must behave like an ideal voltage source for the next stage in the circuit. A low output impedance ($Z_{out}$) guarantees that the output voltage remains stable and is not diminished when driving a load.

A key requirement for a system to be linear is that a zero input must produce a zero output. This stems from the homogeneity property of linear systems. Let's test the equation $y = au + b$, where $b \neq 0$.

If we provide a zero input, $u=0$, the output becomes:
$y = a(0) + b = b$

Since we are given that $b \neq 0$, a zero input results in a non-zero output. This violates a fundamental condition for linearity. Therefore, the system is non-linear. Systems of this form are known as affine transformations.

Let's consider the relationship between the roots of a polynomial and its derivative, which is explained by Rolle's Theorem. This theorem states that between any two real roots of a differentiable function like $p(s)$, there must be at least one point where the derivative $p'(s)$ is zero.

The problem tells us that $p'(s)$ has no real roots. This implies that $p(s)$ cannot have two (or more) distinct real roots, because if it did, its derivative would have to have a root between them. Therefore, $p(s)$ can have at most one real root.

However, $p(s)$ is a cubic polynomial. Any odd-degree polynomial with real coefficients must have at least one real root. Since $p(s)$ must have at least one and at most one real root, it must have exactly one.

At equilibrium, two competing processes, diffusion and drift, balance each other. Diffusion occurs because of the concentration gradient: holes move from the p-side to the n-side, and electrons move from the n-side to the p-side. This movement creates a depletion region with a built-in electric field pointing from the n-side to the p-side. This field causes a drift of minority carriers. Holes drift in the direction of the field (n to p), while electrons, being negatively charged, drift opposite to the field (p to n). Therefore, an electron's diffusion motion (n to p) is in the exact opposite direction to its drift motion (p to n).

To determine the logic function, we can analyze the pull-down network (the NMOS transistors connected to ground). This network's logic defines when the output f(X,Y) is pulled to a low state (logic 0), giving us the expression for \overline{f(X,Y)}.

The parallel paths to ground are active for the input combinations \bar{X}\bar{Y} and XY. This gives the expression \overline{f(X, Y)} = \bar{X}\bar{Y} + XY. This is the definition of an XNOR gate, often written as X \odot Y.

The final output of the CMOS gate, f(X, Y), is the logical inverse of the pull-down network's function. Therefore, f(X, Y) = \overline{X \odot Y}. Inverting the XNOR function yields the XOR function, so f(X, Y) = X \oplus Y.

The provided function is expressed in the Sum of Products (SOP) form. Each product term, or minterm, represents an input combination for which the function's output is 1. By converting the variables to binary (e.g., A=1, $\bar{A}$=0), we can identify these minterms:

• $\bar{A}\bar{B}\bar{C}$ corresponds to binary 000, which is minterm $m_0$.
• $\bar{A}B\bar{C}$ corresponds to binary 010, which is minterm $m_2$.
• $A\bar{B}\bar{C}$ corresponds to binary 100, which is minterm $m_4$.
  Thus, the function is $F = \Sigma m(0, 2, 4)$.

The Product of Sums (POS) form is derived from the combinations where the function's output is 0. For a three-variable function, the total combinations are from 0 to 7. The maxterms are simply the combinations not present in the minterm list, so $F = \Pi M(1, 3, 5, 6, 7)$.

To get the final POS expression, we convert each maxterm index to a sum term. For maxterms, a variable is complemented if its corresponding bit is 1.

• $M_1 (001) \rightarrow (A+B+\bar{C})$
• $M_3 (011) \rightarrow (A+\bar{B}+\bar{C})$
• $M_5 (101) \rightarrow (\bar{A}+B+\bar{C})$
• $M_6 (110) \rightarrow (\bar{A}+\bar{B}+C)$
• $M_7 (111) \rightarrow (\bar{A}+\bar{B}+\bar{C})$
  Multiplying these terms gives the final POS form of the function.

On a Smith chart, the normalized impedance is represented as $z = r + jx$. We can identify the special cases as follows:

• Point P: This is the leftmost point on the horizontal axis. It corresponds to a reflection coefficient of $\Gamma = -1$, which occurs for a short circuit. For a short circuit, the impedance is zero, so the normalized impedance is $z = 0$, meaning $r=0$ and $x=0$.

• Point Q: This is the center of the chart. It corresponds to a reflection coefficient of $\Gamma = 0$, which signifies a perfectly matched load. The normalized impedance is $z = 1$, so $r=1$ and $x=0$.

• Point R: This is the rightmost point on the horizontal axis. It corresponds to a reflection coefficient of $\Gamma = +1$, which represents an open circuit. For an open circuit, the impedance is infinite, so the normalized impedance $z$ is also infinite.

Let's begin by assuming statement S2 is true. This means our $4 \times 4$ matrix $M$ has four distinct eigenvalues. A foundational theorem in linear algebra states that a set of eigenvectors, each corresponding to a different eigenvalue, is always linearly independent.

Since we have four distinct eigenvalues, we can find a corresponding eigenvector for each one. By the aforementioned theorem, this set of four eigenvectors must be linearly independent. This is exactly what statement S1 describes, proving that S2 implies S1.

First, let's simplify the function $f(x,y)$ by splitting the fraction:
$f(x,y) = \frac{ax^2}{xy} + \frac{by^2}{xy} = a\frac{x}{y} + b\frac{y}{x}$.

Next, we'll find the partial derivatives with respect to $x$ and $y$.
For $x$: $\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left(a\frac{x}{y} + b\frac{y}{x}\right) = \frac{a}{y} - \frac{by}{x^2}$.
For $y$: $\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left(a\frac{x}{y} + b\frac{y}{x}\right) = -\frac{ax}{y^2} + \frac{b}{x}$.

Now, evaluate each partial derivative at the point $(x,y)=(1,2)$.
$\frac{\partial f}{\partial x}\bigg|_{(1,2)} = \frac{a}{2} - \frac{b(2)}{1^2} = \frac{a}{2} - 2b$.
$\frac{\partial f}{\partial y}\bigg|_{(1,2)} = -\frac{a(1)}{2^2} + \frac{b}{1} = -\frac{a}{4} + b$.

The problem states these are equal, so we set them equal and solve for the relationship between $a$ and $b$:
$\frac{a}{2} - 2b = -\frac{a}{4} + b \implies \frac{a}{2} + \frac{a}{4} = 3b \implies \frac{3a}{4} = 3b$, which simplifies to $a = 4b$.

For a discrete-time system to be stable, its Region of Convergence (ROC) must include the unit circle, $|z|=1$. The system has one pole inside the unit circle at $p_1=0.25$ and one pole outside at $p_2=2 \angle 30^{\circ}$.

To ensure the ROC contains the unit circle, it must be the region between these poles. This means the ROC must be the annular region defined by $0.25 < |z| < 2$. An ROC that is a ring like this corresponds to a system whose impulse response is two-sided, meaning it is non-zero for both $n > 0$ and $n < 0$. Therefore, stability is only possible if the impulse response is two-sided.

The function $x(t)$ has a fundamental period of $T=10$. If a function is periodic with period $T$, it is automatically periodic with any integer multiple of that period, such as $T' = 40 = 4 \times 10$.

When we expand $x(t)$ as a Fourier series using the new period $T'$, the basis functions change, and so do the indices of the coefficients. The new series will be sparser; coefficients $b_k$ will be non-zero only when their corresponding frequency matches one from the original series.

This means the set of non-zero coefficient values for $\{b_k\}$ is exactly the same as the set of coefficient values for $\{a_k\}$. We are just summing the same numbers. Therefore, the sum of their absolute values must also be the same.

$\sum_{k=-\infty }^{\infty } |b_k| = \sum_{k=-\infty }^{\infty } |a_k| = 16$

The provided signal is a standard amplitude modulated (AM) waveform. The central component with the largest amplitude, $4\cos(2400\pi t)$, is the carrier signal. The other two components, $\cos(2000\pi t)$ and $\cos(2800\pi t)$, are the sidebands which contain the message information.

The power of any sinusoidal signal $A\cos(\omega t)$ is given by $P = A^2/2$.

Using this formula, the power of the carrier signal with amplitude $A_c=4$ is $P_c = \frac{4^2}{2} = 8$.

The power of the message is the combined power of the two sidebands. Since each sideband has an amplitude of $A_s=1$, the total message power is $P_m = \frac{1^2}{2} + \frac{1^2}{2} = 1$.

Finally, the required ratio is $\frac{P_m}{P_c} = \frac{1}{8} = 0.125$.

We can determine an upper limit on the number of codewords, $M$, using standard coding theory bounds. The number of codewords is $M=2^k$, where $k$ is the number of message bits.

The Plotkin bound gives a relationship: $d_{min} \leq \frac{n \cdot 2^{k-1}}{2^k - 1}$.
Given $n=5$ and a minimum distance $d_{min}=2$, we have $2 \leq \frac{5 \cdot 2^{k-1}}{2^k - 1}$.
This inequality holds true for any reasonable value of $k$, so the Plotkin bound itself doesn't limit our specific choice.

The practical limitation is that for a code to have a minimum distance greater than 1, it must include redundancy. This means the number of message bits ($k$) must be strictly less than the codeword length ($n$). With $n=5$, the maximum possible integer value for $k$ is 4.

Therefore, the maximum number of codewords is $M = 2^k = 2^4 = 16$.

The total probability of error, $P_e$, is the sum of the probabilities of the two ways an error can happen. Given the detection threshold is zero, an error occurs under two conditions:

1. If $X = -1$ is sent, the detector makes an error if the received signal $R = -1 + N > 0$, which means the noise $N$ must be greater than 1.
2. If $X = 1$ is sent, an error occurs if $R = 1 + N < 0$, which means the noise $N$ must be less than -1.

We find the probabilities of these noise events by calculating the area under the given noise PDF, $f_N(n)$. The probability $P(N>1)$ is the area of the triangle from $n=1$ to $n=2$. The height at $n=1$ is $0.25$, so the area is $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}(1)(0.25) = 0.125$. By symmetry of the PDF, $P(N<-1)$ is also $0.125$.

Since the symbols are equiprobable ($P(X=1) = P(X=-1) = 0.5$), the total error probability is:
$P_e = P(X=1)P(N<-1) + P(X=-1)P(N>1) = (0.5)(0.125) + (0.5)(0.125) = 0.125$.

Total number of unique states required $=\frac{70+5+75}{5}=30$ Minimum number of flip-flops required is, $n=\left\lceil\log _{2}(30)\right\rceil=\lceil 4.91\rceil=5$

In photolithography, the smallest feature you can create is fundamentally limited by the diffraction of the light source. A shorter wavelength of light allows for finer resolution. This means the minimum feature size, $L_{min}$, is directly proportional to the wavelength, $\lambda$.

Given this direct proportionality ($L_{min} \propto \lambda$), we can find the ratio of the minimum feature sizes for the two systems by simply taking the ratio of their corresponding wavelengths.

$\frac{L_{min1}}{L_{min2}} = \frac{\lambda_1}{\lambda_2}$

Substituting the given values for System 1 and System 2:

$\frac{156 \text{ nm}}{325 \text{ nm}} = 0.48$

The defining characteristic of a distortionless line is the condition $\frac{R}{L} = \frac{G}{C}$, which greatly simplifies its parameters. For such a line, the attenuation constant is $\alpha = \sqrt{RG}$ and the characteristic impedance is $Z_0 = \sqrt{R/G}$. We can cleverly combine these two equations to find $\alpha$ without needing to know the conductance, $G$. By rearranging the impedance equation as $\sqrt{G} = \sqrt{R}/Z_0$ and substituting it into the expression for $\alpha$, we get a direct formula:
$\alpha = \sqrt{R} \cdot \sqrt{G} = \sqrt{R} \cdot \frac{\sqrt{R}}{Z_0} = \frac{R}{Z_0}$
Plugging in the given values for resistance and impedance:
$\alpha = \frac{0.05 \text{ } \Omega/\text{m}}{50 \text{ } \Omega} = 0.001 \text{ Np/m}$

The homogeneous system of equations $Ax=0$ has infinitely many solutions precisely when the matrix $A$ is singular. A square matrix is singular if its determinant is zero.

Let's compute the determinant of $A$ and set it to zero:
$\text{det}(A) = k(k^2) - (2k)(k^{2-}k) = 0$
Expanding and simplifying the equation gives:
$k^3 - 2k^3 + 2k^2 = 0$
$-k^3 + 2k^2 = 0$
Factoring out $k^2$ from the expression, we get $k^2(2-k) = 0$. The distinct real values of $k$ that satisfy this equation are $k=0$ and $k=2$. Thus, there are 2 such values.

The four random variables $X_1, X_2, X_3,$ and $X_4$ are independent and follow the same distribution. This fundamental symmetry means that any one of them is equally likely to take on any rank (such as smallest, largest, etc.) compared to the others. Since there are four variables, each has an equal chance of being the smallest. Therefore, the probability that $X_4$ is the smallest is simply $\frac{1}{4}$.

To find the coefficient $a_2$ for the Taylor series of $f(x)$ centered at $x=0$, we use the formula $a_2 = \frac{f''(0)}{2!}$.

First, we need to find the second derivative of $f(x) = \int_{0}^{x} e^{-t^2/2} dt$.

Using the Fundamental Theorem of Calculus, the first derivative is simply the integrand with $t$ replaced by $x$:
$f'(x) = e^{-x^2/2}$

Next, we differentiate $f'(x)$ using the chain rule to find the second derivative:
$f''(x) = e^{-x^2/2} \cdot \frac{d}{dx}(-\frac{x^2}{2}) = e^{-x^2/2} \cdot (-x)$

Now, we evaluate the second derivative at $x=0$:
$f''(0) = e^0 \cdot (-0) = 1 \cdot 0 = 0$

Finally, we calculate the coefficient $a_2$:
$a_2 = \frac{f''(0)}{2!} = \frac{0}{2} = 0$

To find the ABCD parameter $B$, we use its definition: $B = -\frac{V_1}{I_2}$ when the output port is short-circuited ($V_2 = 0$).

When port 2 is shorted, the input impedance seen from port 1 is the $2 \Omega$ resistor in series with the parallel combination of the $5 \Omega$ and the other $2 \Omega$ resistor.
$Z_{\in} = 2 + (5 \parallel 2) = 2 + \frac{5 \times 2}{5 + 2} = 2 + \frac{10}{7} = \frac{24}{7} \Omega$.
The input current is then $I_1 = \frac{V_1}{Z_{\in}} = \frac{7V_1}{24}$.

Next, we use the current divider rule to relate $I_2$ to $I_1$. Note the direction of $I_2$ is into the port, so it gets a negative sign in the standard current division formula.
$I_2 = -I_1 \times \left( \frac{5}{5+2} \right) = -I_1 \frac{5}{7}$.

Substituting the expression for $I_1$, we get $I_2 = -\left(\frac{7V_1}{24}\right) \frac{5}{7} = -\frac{5V_1}{24}$.
Finally, we can calculate $B$:
$B = -\frac{V_1}{I_2} = -\frac{V_1}{-5V_1/24} = \frac{24}{5} = 4.80 \Omega$.

First, let's establish the operating conditions. The input voltage $V_I$ varies from a nominal 6 V by $\pm 5\%$, so its range is $5.7V \leq V_I \leq 6.3V$. The Zener diode maintains a constant $5V$ across the $1k\Omega$ load, meaning the load current $I_L$ is fixed at $I_L = \frac{5V}{1k\Omega} = 5mA$.

To ensure the diode remains in breakdown, we must design for the worst-case scenario, which is when the input voltage is at its minimum ($V_{I_{min}} = 5.7V$). At this point, the current from the source, $I_S$, must be enough to supply both the load ($5mA$) and the minimum required Zener current ($I_{Z_{min}} = 2mA$). Therefore, the minimum source current is $I_{S_{min}} = 5mA + 2mA = 7mA$. We can now calculate the series resistor $R$ from the voltage drop across it:
$R = \frac{V_{I_{min}} - V_Z}{I_{S_{min}}} = \frac{5.7V - 5V}{7mA} = 100\Omega$.

Next, to find the required power rating for the diode, we must calculate the maximum power it will dissipate. This occurs when the current through the Zener is highest, which happens when the input voltage is at its maximum ($V_{I_{max}} = 6.3V$). The maximum source current is $I_{S_{max}} = \frac{V_{I_{max}} - V_Z}{R} = \frac{6.3V - 5V}{100\Omega} = 13mA$. The maximum Zener current is this source current minus the constant load current: $I_{Z_{max}} = I_{S_{max}} - I_L = 13mA - 5mA = 8mA$.

The minimum required power rating is therefore the maximum power the diode will experience: $P_{Z_{max}} = V_Z \times I_{Z_{max}} = 5V \times 8mA = 40mW$.

The input signal to the non-linear device is $v_{i}(t) = A_{c} \cos(2\pi f_{c}t) + \cos(2\pi f_{m}t)$. The device output is $v_{o}(t) = a v_{i}(t) + b v_{i}^{2}(t)$. When this signal is passed through the band-pass filter centered at $f_c$, only frequency components near $f_c$ are retained. These are the original carrier and the sidebands generated by the product of the carrier and message signals.

The filtered output, which is our AM wave, is thus:
$y(t) = a A_{c} \cos(2\pi f_{c}t) + 2b A_{c} \cos(2\pi f_{c}t)\cos(2\pi f_{m}t)$

We can factor this into the standard AM signal format:
$y(t) = a A_{c} [1 + \frac{2b}{a} \cos(2\pi f_{m}t)] \cos(2\pi f_{c}t)$

From this form, we can identify the modulation index as $\mu = \frac{2b}{a}$.

The relationship between total sideband power ($P_{SB}$) and carrier power ($P_C$) is $P_{SB} = P_C \frac{\mu^2}{2}$. We are given that $P_{SB} = \frac{1}{2}P_C$, which implies $\frac{\mu^2}{2} = \frac{1}{2}$, so $\mu^2=1$ and $\mu=1$.

Finally, setting our expression for the modulation index equal to 1 gives $\frac{2b}{a} = 1$. Rearranging this for the desired ratio yields $\frac{a}{b} = 2$.

The power in the output signal, $P_Y$, is found by integrating the output's power spectral density (PSD), which is the input PSD, $S_N(f)$, multiplied by the filter's squared frequency response, $|H(f)|^2$. Since the input noise PSD is a constant $S_N(f) = 0.5$, the output power is $P_Y = 0.5 \int_{-\infty }^{\infty } |H(f)|^2 df$. By Parseval's theorem, we can equate the energy integral in the frequency domain to one in the time domain: $\int_{-\infty }^{\infty } |H(f)|^2 df = \int_{-\infty }^{\infty } |h(t)|^2 dt$. This simplifies our calculation to $P_Y = 0.5 \int_{-\infty }^{\infty } |h(t)|^2 dt$. Substituting the given impulse response $h(t) = 0.5e^{-t^2/2}$, we get:
$P_Y = 0.5 \int_{-\infty }^{\infty } \left(0.5e^{-t^2/2}\right)^2 dt = 0.125 \int_{-\infty }^{\infty } e^{-t^2} dt$.
Using the standard Gaussian integral identity $\int_{-\infty }^{\infty } e^{-x^2} dx = \sqrt{\pi}$, the power is $P_Y = 0.125\sqrt{\pi} \approx 0.22$ W.

To find the transfer function from a state-space representation, we use the standard formula $T(s) = C(sI - A)^{-1}B$. Let's start by calculating the $(sI - A)$ term.

Next, we find the inverse of this matrix. The determinant is $s(s+4) - (1.5)(-4) = s^{2+}4s+6$. The inverse is the adjugate matrix divided by the determinant:

Now, we substitute everything back into the transfer function formula and perform matrix multiplication:

Multiplying the last two matrices first gives

. Then, multiplying by the $C$ matrix results in $[(1.5)(2s) + (0.625)(8)] = [3s+5]$. Therefore, the final transfer function is:
$T(s) = \frac{3s+5}{s^{2+}4s+6}$

The energy of light is inversely proportional to its wavelength ($\lambda$), with blue light having a shorter wavelength than green, and green shorter than red. This means the photon energies are ordered as $E_{Blue} > E_{Green} > E_{Red}$.

For an LED, the energy of the emitted photon is approximately equal to the semiconductor's band gap energy, $E_g$. Therefore, the band gaps of the three diodes are related as $E_{g,B} > E_{g,G} > E_{g,R}$.

The built-in voltage ($V_{bi}$) of a p-n junction is directly related to the material's band gap. For constant doping levels as stated, a larger band gap corresponds to a larger built-in voltage.

Thus, the built-in voltages follow the same order as the band gaps: $V_{B} > V_{G} > V_{R}$, or written another way, $V_{R} < V_{G} < V_{B}$.

Let's break down the circuit, starting with the multiplexer on the left. This MUX has select lines $U$ and $V$. Given its inputs ($I_0=0, I_1=1, I_2=1, I_3=0$), its output, which we'll call $F_1$, is $F_1 = \bar{U}V(1) + U\bar{V}(1)$. This is the XOR function: $F_1 = U \oplus V$.

Now, let's look at the second multiplexer. Its select lines are $W$ and $X$. The output of the first MUX, $F_1$, is connected to inputs $I_0$ and $I_1$, while inputs $I_2$ and $I_3$ are grounded (0). The expression for the final output $F$ is $F = \bar{W}\bar{X}(F_1) + \bar{W}X(F_1)$.