GATE ECE 2013

Truth table of XOR gate

So, from the XOR gate truth table it is clear that the bulb can be turned ON and also can be turned OFF by any one of the switches irrespective of the state of the other switch.

According to Stoke's theorem $\oint_{c} \vec{A} \cdot \vec{d} i=\iint_{s}(\nabla \times \vec{A}) \cdot \overrightarrow{d s}$

When two systems are connected in cascade, the output of the first system serves as the input to the second. To find the overall impulse response, we can trace an impulse, $\delta(t)$, through the chain.

The output of the first system is, by definition, its impulse response, $h_1(t)$. This signal, $h_1(t)$, then enters the second system. The final output is the result of convolving this input signal with the second system's impulse response, $h_2(t)$.

Therefore, the combined system's response to the initial impulse is the convolution of the two individual impulse responses:
$h_{overall}(t) = h_1(t) * h_2(t)$

When a pn-junction is forward biased, the potential barrier at the junction is lowered. This allows majority carriers-holes from the p-side and electrons from the n-side-to be "injected" across the junction into the opposite region. Once they cross, these carriers become minority carriers. A high concentration of these injected minority carriers near the junction causes them to diffuse deeper into the material. As they diffuse, they eventually recombine with the majority carriers present there, completing the current path. This entire sequence sustains the forward current flow.

In the fabrication of integrated circuits, dry oxidation involves using pure oxygen gas ($O_2$) to grow a silicon dioxide ($SiO_2$) layer. This process is significantly slower than wet oxidation because oxygen molecules diffuse through the existing $SiO_2$ layer less easily than the water molecules ($H_2O$) used in wet oxidation. This slower, more controlled growth results in a denser, more uniform oxide film with fewer structural defects and superior electrical insulating properties. Therefore, dry oxidation produces a superior quality oxide at a lower growth rate.

The small-angle approximation, $\sin \theta \approx \theta$, requires the angle $\theta$ to be expressed in radians. We are looking for the largest angle where the relative error is less than 10%. The error is calculated as $\frac{|\theta_{\text{rad}} - \sin\theta|}{\theta_{\text{rad}}}$.

Let's test the provided options, checking the boundary where the error might exceed 10%.
For $\theta = 18^{\circ} = \frac{18\pi}{180} \approx 0.314$ rad, we find that $\sin(18^{\circ}) \approx 0.309$.
The error is $\frac{0.314 - 0.309}{0.314} \approx 0.016$, or about 1.6%, which is well within the 10% tolerance.

Now, consider the next larger option, $\theta = 50^{\circ} \approx 0.873$ rad. In this case, $\sin(50^{\circ}) \approx 0.766$.
The error becomes $\frac{0.873 - 0.766}{0.873} \approx 0.122$, or 12.2%, which exceeds the 10% limit.
Thus, among the choices given, $18^{\circ}$ is the maximum angle for which the approximation holds to within 10% error.

To find the divergence of a vector field in Cartesian coordinates, we sum the partial derivatives of its components. The formula is $\nabla \cdot \overrightarrow{A} = \frac{\partial A_{x}}{\partial x}+\frac{\partial A_{y}}{\partial y}+\frac{\partial A_{z}}{\partial z}$.

The given vector field is $\overrightarrow{A}=x\hat{a_{x}}+y\hat{a_{y}}+z\hat{a_{z}}$. Its components are $A_x = x$, $A_y = y$, and $A_z = z$.

Now, we calculate the partial derivative of each component with respect to its corresponding variable:
$\nabla \cdot \overrightarrow{A} = \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial y}(y) + \frac{\partial}{\partial z}(z)$
Each of these derivatives evaluates to $1$.

Summing these results gives the final answer: $1 + 1 + 1 = 3$.

To find the system's output, we can analyze the problem in the Laplace domain, which turns time-domain convolution into simpler multiplication.

First, we find the Laplace transform of the impulse response, which gives us the system's transfer function: $H(s) = \mathcal{L}\{tu(t)\} = \frac{1}{s^2}$.

Next, we transform the input signal, which is a time-shifted unit step: $X(s) = \mathcal{L}\{u(t-1)\} = \frac{e^{-s}}{s}$.

The output in the Laplace domain, $Y(s)$, is the product of these two: $Y(s) = H(s)X(s) = \frac{1}{s^2} \cdot \frac{e^{-s}}{s} = \frac{e^{-s}}{s^3}$.

To find the time-domain output $y(t)$, we take the inverse Laplace transform. We know that $\frac{1}{s^3}$ transforms to $\frac{t^2}{2}u(t)$. The $e^{-s}$ term acts as a time-shifting operator, delaying the signal by 1. This yields the final output: $y(t) = \frac{(t-1)^2}{2}u(t-1)$.

To determine the transfer function, we first find the slope of the Bode magnitude plot. The frequency increases by a factor of 10 (one decade) from 1 rad/s to 10 rad/s. Over this interval, the gain changes from 32 dB to -8 dB.

The slope is calculated as $\frac{\text{change \in gain}}{\text{change \in frequency}} = \frac{(-8 \text{ dB}) - (32 \text{ dB})}{1 \text{ decade}} = -40 \text{ dB/decade}$.

A slope of $-40$ dB/decade indicates the presence of two poles at the origin. This means the transfer function has the general form $G(s) = \frac{K}{s^2}$.

To find the constant $K$, we use the given gain at $\omega=1$ rad/s.
$20\log|G(j1)| = 20\log\left|\frac{K}{(j1)^2}\right| = 20\log(K) = 32 \text{ dB}$.

Solving for $K$ gives $K = 10^{(32/20)} = 10^{1.6} \approx 39.8$. Thus, the transfer function is $G(s) = \frac{39.8}{s^2}$.

This circuit uses an ideal op-amp in a negative feedback configuration. Because the non-inverting (+) input is grounded, the virtual ground principle dictates that the inverting (-) input is also at $0V$. The transistor's base is connected to this inverting input, so its voltage is $V_B = 0V$.

The op-amp will adjust its output voltage ($V_{out}$) to whatever value is necessary to maintain this virtual ground. The output is connected to the transistor's emitter, so $V_{out} = V_E$. For the silicon NPN transistor to be active and complete the feedback path, its base-emitter junction must be forward-biased with a voltage drop of approximately $0.7V$.

This gives us the relationship $V_{BE} = V_B - V_E = 0.7V$. Substituting the known base voltage, we get $0V - V_E = 0.7V$. Solving for the emitter voltage gives $V_E = -0.7V$, which means the output voltage is $V_{out} = -0.7V$.

To understand how the scaling works, let's start with the standard formula for converting from a delta to a star configuration. The resistance of one leg of the star network, let's say $R_A$, is given by:
$R_A = \frac{R_b R_c}{R_a + R_b + R_c}$
Now, if we scale every resistor in the delta network by a factor $k$, our new resistances are $kR_a$, $kR_b$, and $kR_c$. Let's calculate the new corresponding star resistor, which we'll call $R'_A$:
$R'_A = \frac{(kR_b)(kR_c)}{kR_a + kR_b + kR_c}$
We can factor out $k^2$ from the numerator and $k$ from the denominator:
$R'_A = \frac{k^2 (R_b R_c)}{k (R_a + R_b + R_c)} = k \left( \frac{R_b R_c}{R_a + R_b + R_c} \right)$
Since the term in the parenthesis is just the original $R_A$, we find that $R'_A = kR_A$. Thus, the elements of the star connection are scaled by the same factor $k$.

Initially, the accumulator A and register B are both loaded with the value $05H$. The program then enters a loop. In each pass, the current value of B is added to A, and then B is decremented. This loop executes for values of B from $05H$ down to $01H$.

The final value in A is the sum of its initial value and all the values added during the loop, plus a final addition after the loop.

This can be expressed as:
A_{final} = A_{initial} + (\text{\sum from loop}) + \text{final addition}
$A_{final} = 05H + (05H + 04H + 03H + 02H + 01H) + 03H$

Summing these values gives a decimal total of $23$, which is $17H$ in hexadecimal.

To achieve ISI-free transmission, we first need to determine the symbol rate from the given bit rate. For an M-ary Quadrature Amplitude Modulation (M-QAM) scheme, each symbol represents $k = \log_2(M)$ bits. In this case, with 32-QAM, we have $M=32$, so each symbol carries $k = \log_2(32) = 5$ bits.

The bit rate is given as $R$ kbits/s. The symbol rate, or baud rate ($R_s$), is the bit rate divided by the number of bits per symbol:
$R_s = \frac{\text{Bit Rate}}{\text{bits per symbol}} = \frac{R \text{ kbits/s}}{5} = \frac{R}{5} \text{ ksymbols/s}$

According to the Nyquist criterion for minimum bandwidth, the minimum required bandwidth for a passband signal to be free of inter-symbol interference is equal to its symbol rate. Therefore, the minimum bandwidth is $\frac{R}{5}$ kHz.

The given signal is a sum of three individual sinusoidal components. To find the fundamental frequency of the combined signal, we first identify the angular frequency of each component. By inspecting the terms, we find the frequencies are $\omega_1 = 100$, $\omega_2 = 300$, and $\omega_3 = 500$ rad/s.

The fundamental angular frequency, $\omega_0$, for the overall periodic signal is the largest frequency of which all component frequencies are integer multiples. This is mathematically equivalent to finding the greatest common divisor (GCD) of the individual frequencies.

$\omega_0 = \text{GCD}(100, 300, 500)$

The largest number that divides 100, 300, and 500 is 100. Therefore, the fundamental frequency is $100$ rad/s.

This circuit uses voltage-series feedback. The feedback network samples the output voltage and mixes a signal in series with the input voltage. This series mixing at the input increases the overall input impedance, governed by the equation $R_{\in,f} = R_{\in}(1 + A_v k)$. Conversely, the voltage sampling at the output decreases the overall output impedance, as described by $R_{out,f} = \frac{R_{out}}{1 + A_v k}$. The term $(1 + A_v k)$ represents the magnitude of the feedback effect. Therefore, increasing the feedback gain $k$ will cause the input impedance to increase and the output impedance to decrease.

To properly sample a signal without losing information, we must follow the Nyquist-Shannon sampling theorem. This principle states that the sampling frequency, $f_s$, must be at least twice the maximum frequency, $f_m$, present in the signal.

The signal's maximum frequency is given as $f_m = 5$ kHz.

Therefore, the minimum required sampling frequency (the Nyquist rate) is:
$f_{s, \text{min}} = 2 \times f_m = 2 \times 5 \text{ kHz} = 10 \text{ kHz}$

This means any valid sampling frequency must be greater than or equal to 10 kHz. Any frequency that does not satisfy the condition $f_s \geq 10$ kHz is considered invalid.

In an ideal MOSFET, the saturation drain current $I_D$ is independent of the drain-source voltage $V_{DS}$, which would imply an infinite output resistance. However, channel length modulation accounts for the fact that as $V_{DS}$ increases, the effective channel length decreases. Since $I_D$ in saturation is inversely proportional to the channel length, a shorter channel leads to a slightly larger drain current. This means $I_D$ now has a small, positive slope with respect to $V_{DS}$. The output resistance, $r_o$, is defined as the inverse of this slope, $r_o = (\frac{\partial I_D}{\partial V_{DS}})^{-1}$. Because the slope is no longer zero, the output resistance becomes a finite value, representing a decrease from the ideal infinite resistance.

For a continuous-time LTI system to be both causal and stable, all of its poles must lie strictly in the left-half of the s-plane, meaning their real parts must be negative (\text{Re}{s} < 0$). This ensures the region of convergence (ROC) includes the entire imaginary axis ($j\omega$-axis).

The statement that all poles must lie within the unit circle, $|s|<1$, is the stability condition for discrete-time systems, not continuous-time ones. In the s-plane, the unit circle $|s|<1$ contains areas in the right-half plane. For example, a pole at $s=0.5$ is inside the unit circle but has a positive real part. For a causal system, this pole would create an ROC of $\text{Re}\{s\} > 0.5$, which does not include the $j\omega$-axis, making the system unstable.

To find the eigenvalues ($\lambda$) of a matrix, we solve the characteristic equation $\det(A - \lambda I) = 0$. For the given matrix $A$, this requires calculating the determinant of:

$\begin{vmatrix} 3-\lambda & 5 & 2 \\ 5 & 12-\lambda & 7 \\ 2 & 7 & 5-\lambda \end{vmatrix} = 0$

Expanding this determinant results in a cubic polynomial in $\lambda$. After simplification, the characteristic equation is $-\lambda^3 + 20\lambda^2 - 33\lambda = 0$, or more simply, $\lambda^3 - 20\lambda^2 + 33\lambda = 0$.

A much quicker way to see this is to realize the determinant of the original matrix is $0$. This means the matrix is singular, which directly implies that one of its eigenvalues must be $0$. Since the other eigenvalues are positive, the minimum eigenvalue is $0$.

Let's proceed by factoring the characteristic equation we found:

$\lambda(\lambda^2 - 20\lambda + 33) = 0$

This equation clearly shows that one of the eigenvalues is $\lambda = 0$. The other two eigenvalues are the roots of the quadratic factor $\lambda^2 - 20\lambda + 33 = 0$. Since these roots are positive, the smallest eigenvalue of the matrix is $0$.

We can determine the number of real roots and their signs using Descartes' Rule of Signs, which looks at sign changes in the polynomial's coefficients.

First, let's examine the polynomial $f(x) = a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x-a_{0}$. Since all coefficients $a_4, a_3, a_2, a_1, a_0$ are positive, the sequence of signs of the coefficients of $f(x)$ is $(+, +, +, +, -)$. There is exactly one change in sign. This means the polynomial has exactly one positive real root.

Next, to find the number of negative roots, we examine $f(-x)$:
$f(-x) = a_4(-x)^4 + a_3(-x)^3 + a_2(-x)^2 + a_1(-x) - a_0 = a_4x^4 - a_3x^3 + a_2x^2 - a_1x - a_0$.
The signs of the coefficients for $f(-x)$ are $(+, -, +, -, -)$. Counting the sign changes, we have three. This indicates that $f(x)$ has either three or one negative real roots.

In either case, the polynomial is guaranteed to have at least one positive real root and at least one negative real root.

To determine the system's response, we can work in the Laplace domain. The input is a unit step function, $x(t) = u(t)$, and its Laplace transform is $X(s) = \frac{1}{s}$. The system itself is an integrator, which has a transfer function of $H(s) = \frac{1}{s}$.

The output signal in the s-domain, $Y(s)$, is the product of the system's transfer function and the input signal's transform.

$Y(s) = H(s) X(s) = \left(\frac{1}{s}\right) \left(\frac{1}{s}\right) = \frac{1}{s^2}$

To find the response in the time domain, $y(t)$, we take the inverse Laplace transform of $Y(s)$. The inverse transform of $\frac{1}{s^2}$ is the standard ramp function, $t u(t)$.

To find the transfer function, we'll analyze the circuit as a voltage divider in the s-domain. The impedance of the resistor is $R = 10^4 \, \Omega$, and the impedance of each capacitor is $Z_C = \frac{1}{sC} = \frac{1}{s(100 \times 10^{-6})} = \frac{10^4}{s}$.

The output voltage $V_2(s)$ is taken across the resistor and one capacitor. The total impedance is across the resistor and both capacitors in series. The transfer function is therefore the ratio of these impedances:
$\frac{V_2(s)}{V_1(s)} = \frac{R + Z_C}{R + Z_C + Z_C} = \frac{10^4 + \frac{10^4}{s}}{10^4 + \frac{2 \cdot 10^4}{s}}$
To simplify this expression, multiply the numerator and denominator by $s$:
$\frac{s(10^4) + 10^4}{s(10^4) + 2 \cdot 10^4} = \frac{10^4(s+1)}{10^4(s+2)} = \frac{s+1}{s+2}$

This problem deals with a special case of the maximum power transfer theorem. While maximum power transfer generally occurs when the load impedance is the complex conjugate of the source impedance, the condition changes when the load is restricted to be purely resistive.

For a purely resistive load $R_L$ to draw maximum power from a source with a complex internal impedance $Z_s$, the load resistance must equal the magnitude of the source impedance.

Given the source impedance $Z_s = (4+j3) \Omega$, we find its magnitude:
$R_L = |Z_s| = |4+j3| = \sqrt{4^2 + 3^2}$
$R_L = \sqrt{16 + 9} = \sqrt{25} = 5 \Omega$.

Here is a clearer, more pedagogical explanation:

Return loss (RL) is related to the magnitude of the reflection coefficient, $|\Gamma|$, by the formula $RL = -20 \log_{10}(|\Gamma|)$. We are given that the return loss is 20 dB.

By setting up the equation $20 = -20 \log_{10}(|\Gamma|)$, we can solve for $|\Gamma|$. This simplifies to $\log_{10}(|\Gamma|) = -1$, which means the magnitude of the reflection coefficient is $|\Gamma| = 10^{-1} = 0.1$.

Next, we use this value to find the Voltage Standing Wave Ratio (VSWR) with the formula $VSWR = \frac{1+|\Gamma|}{1-|\Gamma|}$.

Substituting $|\Gamma| = 0.1$ gives us $VSWR = \frac{1+0.1}{1-0.1} = \frac{1.1}{0.9} \approx 1.22$.

Let's find the solution in the frequency domain. The Fourier transform of the output, let's call it $Y(f)$, is the product of the input's Fourier transform, $G(f)$, and the filter's frequency response, $H(f)$.

First, the input signal $g(t) = e^{-\pi t^2}$ is a special Gaussian function that is its own Fourier transform, meaning $G(f) = e^{-\pi f^2}$.

A matched filter has a frequency response $H(f)$ that is the complex conjugate of the input's spectrum, so $H(f) = G^*(f)$. Since our $G(f)$ is purely real, this simplifies to $H(f) = G(f) = e^{-\pi f^2}$.

Therefore, the output's Fourier transform is $Y(f) = G(f)H(f) = (e^{-\pi f^2})(e^{-\pi f^2})$, which simplifies to $Y(f) = e^{-2\pi f^2}$.

We want to find the probability $P(3V \geq 2U)$, which is equivalent to finding $P(W \geq 0)$ for the random variable $W = 3V - 2U$.

Since $W$ is a linear combination of two independent Gaussian variables, it is also a Gaussian variable. Its mean is $E[W] = 3E[V] - 2E[U] = 3(0) - 2(0) = 0$. Because $U$ and $V$ are independent, the variance of $W$ is $Var(W) = 3^2Var(V) + (-2)^2Var(U)$.

Substituting the given variances, we get $Var(W) = 9(\frac{1}{9}) + 4(\frac{1}{4}) = 1 + 1 = 2$. So, $W$ is a Gaussian random variable with a mean of 0. The probability density function of any zero-mean Gaussian is symmetric about the origin. Therefore, the probability that $W$ is greater than or equal to its mean is precisely $\frac{1}{2}$.

Instead of a brute-force calculation, we can solve this elegantly by using the hint provided in the problem. First, notice the structure of the matrix: it's the $4\times4$ identity matrix, $I_4$, plus a matrix where every entry is 1.

Let's call this all-ones matrix $J$. So we want to find the determinant of $M = I_4 + J$. The matrix $J$ can be written as the product of a column vector and a row vector. Let $A$ be a $4 \times 1$ column of ones, and $B$ be a $1 \times 4$ row of ones. Then $AB = J$.

The determinant we want is $\det(I_4 + AB)$. The problem states that $\det(I_m + AB) = \det(I_n + BA)$. Applying this property, we get $\det(I_4 + AB) = \det(I_1 + BA)$.

Now we just need to compute the $1 \times 1$ matrix $BA$:

.

Finally, we calculate the determinant: $\det(I_1 + BA) = \det([1] + [4]) = \det([5]) = 5$.

To find the Thevenin voltage ($V_{Th}$), we begin by open-circuiting the terminals across the load resistance $R_L$. This action forces the current in the right-hand loop, let's call it $I_2$, to be zero.

As a result, the dependent voltage source $j40I_2$ in the left loop becomes a short circuit ($0V$). This allows us to find the controlling voltage $V_{x}$ across the $j4\Omega$ inductor using a simple voltage divider:
$V_{x} = V_s \frac{j4}{3+j4} = (100\angle 53.13^{\circ}) \frac{4\angle 90^{\circ}}{5\angle 53.13^{\circ}} = 80\angle 90^{\circ}V$.

Since $I_2=0$, there is no voltage drop across the $3\Omega$ resistor or the $j6\Omega$ inductor. Therefore, the Thevenin voltage seen at the open terminals is equal to the dependent source $10V_{x}$:
$V_{Th} = 10V_{x} = 10 \times (80\angle 90^{\circ}) = 800\angle 90^{\circ}V$.

First, we identify the open-loop time constant from the given transfer function, $\frac{10}{1+10s}$. By comparing this to the standard first-order form $\frac{K}{1+\tau s}$, we see the time constant $\tau_{ol}$ is $10$.

Our goal is to reduce this time constant by a factor of 100, so the new, desired closed-loop time constant is $\tau_{cl} = \frac{\tau_{ol}}{100} = \frac{10}{100} = 0.1$.

Now, let's find the closed-loop transfer function using the feedback formula $T(s) = \frac{G_{forward}}{1+G_{loop}}$:
$T(s) = \frac{K_a \frac{10}{1+10s}}{1+K_a \frac{10}{1+10s}} = \frac{10K_a}{1+10s+10K_a}$
To find the closed-loop time constant, we rearrange the denominator to match the standard form $1+\tau_{cl}s$:
$T(s) = \frac{\frac{10K_a}{1+10K_a}}{1+\frac{10s}{1+10K_a}}$
From this, we can see that the closed-loop time constant is $\tau_{cl} = \frac{10}{1+10K_a}$.

Finally, we set this expression equal to our target value of $0.1$ and solve for $K_a$:
$\frac{10}{1+10K_a} = 0.1 \implies 100 = 1+10K_a \implies K_a = 9.9$
The closest value to $9.9$ is $10$.

This circuit uses a Zener diode to maintain a constant 5 V across the load resistor, $R_L$. First, let's find the total current $I$ supplied by the source through the $100 \, \Omega$ resistor. This current is constant as long as the Zener is regulating:
$I = \frac{V_{\in} - V_z}{R} = \frac{10 \text{ V} - 5 \text{ V}}{100 \, \Omega} = 50 \text{ mA}$.

This total current $I$ splits between the Zener diode ($I_z$) and the load ($I_L$). The minimum value of $R_L$ corresponds to the maximum load current $I_{L,max}$. This occurs when the Zener current is at its minimum, the knee current $I_{z,min} = 10 \text{ mA}$.
$I_{L,max} = I - I_{z,min} = 50 \text{ mA} - 10 \text{ mA} = 40 \text{ mA}$.
Therefore, $R_{L,min} = \frac{V_z}{I_{L,max}} = \frac{5 \text{ V}}{40 \text{ mA}} = 125 \, \Omega$.

For the power rating, we must consider the worst-case scenario for the Zener, which is maximum power dissipation. This happens when the Zener current $I_z$ is at its maximum. This occurs when the load is disconnected ($R_L \to \infty$, so $I_L = 0$), forcing the entire 50 mA source current through the Zener.
$P_{z,max} = V_z \times I_{z,max} = 5 \text{ V} \times 50 \text{ mA} = 250 \text{ mW}$.

Let's analyze the circuit for each case.

In the first scenario, the input voltage $V_{WX1} = 100V$ is applied to the primary coil. The transformer steps up this voltage by a factor of 1.25, producing $100V \times 1.25 = 125V$ across the secondary. This voltage is then reduced by the attenuator's factor of 0.8, resulting in $V_{YZ1} = 125V \times 0.8 = 100V$.

In the second scenario, the input $V_{YZ2} = 100V$ is applied. This voltage appears across the secondary winding. The transformer now works in reverse, acting as a step-down transformer. The output voltage across WX is $V_{WX2} = \frac{100V}{1.25} = 80V$.

Therefore, the respective ratios are $\frac{V_{YZ1}}{V_{WX1}} = \frac{100}{100}$ and $\frac{V_{WX2}}{V_{YZ2}} = \frac{80}{100}$.

The Q factor of an inductor is defined as the ratio of its reactance to its resistance, so $q = \frac{\omega L}{R}$. For the two coils in series, the effective Q factor will be the total reactance divided by the total resistance.

When connected in series, the total resistance is $R_{eff} = R_1 + R_2$, and the total reactance is $X_{eff} = \omega L_1 + \omega L_2$. Therefore, the effective Q factor is:
$q_{eff} = \frac{X_{eff}}{R_{eff}} = \frac{\omega L_1 + \omega L_2}{R_1 + R_2}$

We can express the individual reactances in terms of their Q factors: $\omega L_1 = q_1 R_1$ and $\omega L_2 = q_2 R_2$. Substituting these into the equation for $q_{eff}$ gives the final result:
$q_{eff} = \frac{q_1 R_1 + q_2 R_2}{R_1 + R_2}$

The step response of a system, let's call it $s(t)$, is the time integral of its impulse response, $h(t)$.
Given the impulse response $h(t) = \delta(t-1) + \delta(t-3)$, we can find the step response by integrating:
$s(t) = \int_{-\infty }^{t} h(\tau)d\tau = \int_{-\infty }^{t} [\delta(\tau-1) + \delta(\tau-3)]d\tau$

The integral of a time-shifted delta function $\delta(t-a)$ is a time-shifted unit step function $u(t-a)$. Applying this rule, we get:
$s(t) = u(t-1) + u(t-3)$

Now, we evaluate the step response at $t=2$:
$s(2) = u(2-1) + u(2-3) = u(1) + u(-1)$

By the definition of the unit step function, $u(1)=1$ and $u(-1)=0$. Therefore, the value is $1 + 0 = 1$.