Networks, Signal & Systems PYQ Set (2000 to 2025)
Solution =
Step 1: Understand Periodicity Condition
For a sampled signal to be periodic, there must exist integers \( N \) and \( k \) such that: \[ NT_s = kT \] This means the sampling period \( T_s \) must be a rational multiple of the original period \( T \).
Step 2: Analyze Each Option
Option A: \( T = \sqrt{2}T_s \)
\[ \frac{T}{T_s} = \sqrt{2} \text{ (irrational)} \] No integers \( N \), \( k \) satisfy \( NT_s = kT \). Not periodic.
Option B: \( T = 1.2T_s \)
\[ 1.2 = \frac{6}{5} \Rightarrow 5T = 6T_s \] Here \( N=6 \), \( k=5 \) satisfy the condition. Periodic with period \( 6T_s \) or \( 5T \).
Option C: Always
False, as shown in Option A where the ratio is irrational.
Option D: Never
False, as shown in Option B where the ratio is rational.
Step 3: Mathematical Verification
The general condition can be rewritten as: \[ \frac{T_s}{T} = \frac{k}{N} \text{ must be rational} \] This is satisfied only when \( T/T_s \) is rational.
Step 4: Conclusion
Only Option B satisfies the periodicity condition, as \( 1.2 = 6/5 \) is a rational number.
Final Answer: \(\boxed{B}\)
Solution =
To determine which function is an eigenfunction of the class of all continuous-time, linear, time-invariant (LTI) systems, let’s recall the definition of an eigenfunction in this context:
An eigenfunction of an LTI system is a signal that, when input to the system, produces an output that is a scaled version of the input signal, where the scaling factor is a complex constant (the eigenvalue). Mathematically, for an LTI system with impulse response \( h(t) \), the input \( x(t) \) is an eigenfunction if the output \( y(t) \) satisfies:
\[ y(t) = H(\omega) x(t) \] where \( H(\omega) \) is the system’s frequency response (the eigenvalue).
The eigenfunctions of all continuous-time LTI systems are the complex exponentials of the form:
\[ e^{j\omega_0 t} \] This is because:
\[ y(t) = \int_{-\infty}^{\infty} h(\tau) e^{j\omega_0 (t – \tau)} d\tau \]\[ = e^{j\omega_0 t} \int_{-\infty}^{\infty} h(\tau) e^{-j\omega_0 \tau} d\tau = H(\omega_0) e^{j\omega_0 t} \] where \( H(\omega_0) \) is the Fourier transform of \( h(t) \) evaluated at \( \omega_0 \).
Now let’s analyze the options:
Option A: \( e^{j\omega_0 t} u(t) \) This is a causal complex exponential (multiplied by the unit-step function \( u(t) \)). It is not an eigenfunction of all LTI systems because the presence of \( u(t) \) restricts the signal to \( t \geq 0 \), breaking the time-invariance property for \( t < 0 \).
Option B: \( \cos(\omega_0 t) \) While cosine can be expressed as a sum of complex exponentials (\( \cos(\omega_0 t) = \frac{e^{j\omega_0 t} + e^{-j\omega_0 t}}{2} \)), it is not itself an eigenfunction. It is the sum of two eigenfunctions.
Option C: \( e^{j\omega_0 t} \) This is the correct eigenfunction of all continuous-time LTI systems, as explained above.
Option D: \( \sin(\omega_0 t) \) Similar to cosine, sine can be written as a sum of complex exponentials (\( \sin(\omega_0 t) = \frac{e^{j\omega_0 t} – e^{-j\omega_0 t}}{2j} \)), but it is not itself an eigenfunction.
Final Answer: C. \( e^{j\omega_0 t} \)
Solution =
The Laplace equation \(\nabla^2 f = 0\) is a second-order partial differential equation that appears in many areas of physics, such as electrostatics, fluid dynamics, and heat conduction. Its solutions, called harmonic functions, have several important properties:
Option A: The solutions have neither maxima nor minima anywhere except at the boundaries. This is correct. Harmonic functions obey the maximum principle, which states that their maximum and minimum values must occur on the boundary of the domain. They cannot have local maxima or minima in the interior.
Option B: The solutions are not separable in the coordinates. This is incorrect. Many solutions to the Laplace equation are separable in various coordinate systems (e.g., Cartesian, spherical, cylindrical). Separation of variables is a common technique for solving the Laplace equation.
Option C: The solutions are not continuous. This is incorrect. Solutions to the Laplace equation are infinitely differentiable (\(C^\infty\)) and thus highly continuous within their domain of definition.
Option D: The solutions are not dependent on the boundary conditions. This is incorrect. The solutions to the Laplace equation are uniquely determined by the boundary conditions (for well-posed problems). This is a consequence of the uniqueness theorems for harmonic functions.
Final Answer:
A. The solutions have neither maxima nor minima anywhere except at the boundaries.
This property reflects the maximum principle of harmonic functions, which is a fundamental characteristic of solutions to the Laplace equation.
Solution =
In Linear Time-Invariant (LTI) systems, the impulse response \( h(t) \) is the derivative of the unit step response \( s(t) \). This relationship arises because the unit step function \( u(t) \) is the integral of the impulse function \( \delta(t) \), and differentiation reverses this operation.
Mathematically:
\[ h(t) = \frac{d}{dt} s(t) \]
Why not the other options?
A: Differentiating the unit ramp response yields the unit step response, not the impulse response.
C: Integrating the unit ramp response does not produce the impulse response.
D: Integrating the unit step response results in the unit ramp response, not the impulse response.
Thus, the correct method to obtain the impulse response is by differentiating the unit step response.
Answer: B – differentiating the unit step response
(A) \( x(t) \in R \)
(B) \( x(t) \in P \)
(C) \( x(t) \in (C – R) \)
(D) the information given is not sufficient to draw any conclusion about \( x(t) \)
Solution =
Reason:
From the magnitude plot: \[ |a_{3}|=3,\qquad |a_{2}|=2,\qquad |a_{0}|=1, \] and the plot shows symmetry \( |a_{-k}| = |a_{k}| \).
From the phase plot the phases of nonzero coefficients are either \(0\) (for \(k=0\)) or \(-\pi\) (for \(k=\pm2,\pm3\)).
Thus the nonzero Fourier coefficients are \[ a_0 = 1,\qquad a_{\pm2} = 2e^{-j\pi} = -2,\]\[ a_{\pm3} = 3e^{-j\pi} = -3, \] i.e. all nonzero \(a_k\) are purely real numbers.
Since every \(a_k\) is real and the coefficients satisfy the conjugate symmetry relation \(a_{-k}=a_k^*\) (here they are equal and real), the time-domain signal \(x(t)\) is real-valued.
Therefore: \(x(t)\in\mathbb{R}.\)
(A) \( 1.25 \sqrt{2} \sin(5t – 0.25\pi) \)
(B) \( 1.25 \sqrt{2} \sin(5t – 0.125\pi) \)
(C) \( 2.5 \sqrt{2} \sin(5t – 0.25\pi) \)
(D) \( 2.5 \sqrt{2} \sin(5t – 0.125\pi) \)
Solution =
Given: \(v_s(t)=5\sin(5t)\), \(R_1=R_2=100\ \text{k}\Omega\), \(C=1\ \mu\text{F}\), \(\omega=5\ \text{rad/s}\).
1. Phasor form of the source
Using a sine reference, the source phasor is
\[ \underline{V}_s = 5\angle 0. \] 2. Impedance of the capacitor
\[ Z_C=\frac{1}{j\omega C}=\frac{1}{j(5)(1\times10^{-6})} \]\[= -j\,200{,}000\ \Omega. \] 3. Total series impedance
\[ Z_{\text{tot}} = R_1 + R_2 + Z_C \]\[= 100{,}000 + 100{,}000 – j\,200{,}000 \]\[= 200{,}000 – j\,200{,}000. \] 4. Voltage divider (phasor) for the capacitor
\[ \underline{V}_C = \underline{V}_s \cdot \frac{Z_C}{Z_{\text{tot}}}. \] Compute the magnitude and angle of the ratio: \[ |Z_C| = 200{,}000,\]\[ |Z_{\text{tot}}| = \sqrt{(200{,}000)^2+(200{,}000)^2}\]\[=200{,}000\sqrt{2}, \] so \[ \left|\frac{Z_C}{Z_{\text{tot}}}\right| = \frac{200{,}000}{200{,}000\sqrt{2}}=\frac{1}{\sqrt{2}}. \] Angles: \[ \angle Z_C = -\frac{\pi}{2},\qquad \angle Z_{\text{tot}} = -\frac{\pi}{4}, \] thus \[ \angle\frac{Z_C}{Z_{\text{tot}}} = -\frac{\pi}{2} -\left(-\frac{\pi}{4}\right) = -\frac{\pi}{4}. \] 5. Phasor magnitude and time-domain result
\[ |\underline{V}_C| = 5\cdot\frac{1}{\sqrt{2}} = \frac{5}{\sqrt{2}} = 2.5\sqrt{2}, \]\[ \angle\underline{V}_C = -\frac{\pi}{4}. \] Since we used the sine reference, the time-domain voltage across the capacitor is \[ \boxed{\,v_C(t)=2.5\sqrt{2}\,\sin\big(5t-\tfrac{\pi}{4}\big)\, }. \] (This matches option C.)
Solution =
To determine the condition for a reciprocal two-port network with the transmission matrix \( T = \begin{pmatrix} A & B \\ C & D \end{pmatrix} \), let’s analyze the properties of reciprocal networks.
Reciprocity in Two-Port Networks:
A two-port network is reciprocal if the interchange of an ideal voltage source at one port and an ideal ammeter at the other port gives the same reading. For the transmission (ABCD) parameters, reciprocity requires that the determinant of the transmission matrix equals 1.
Specifically, for reciprocal networks:
\[ AD – BC = 1 \] This is equivalent to saying:
\[ \text{Determinant}(T) = 1 \]
Now, let’s evaluate the options:
A \( T^{-1} = T \): This would imply \( T^2 = I \), which is not generally true for reciprocal networks.
B \( T^2 = T \): This is not a requirement for reciprocity.
C Determinant (T) = 0: This would imply the network is singular (not reciprocal).
D Determinant (T) = 1: This is the condition for reciprocity.
Thus, the correct answer is D.
Final Answer:
\[ \boxed{\text{D}} \]
Solution =
Norton’s theorem says:
A linear two-terminal network containing independent/dependent sources and impedances can be replaced by an equivalent current source in parallel with an equivalent impedance.
So, the correct answer is:
A complex network connected to a load can be replaced with an equivalent impedance in parallel with a current source.
Correct Answer: B
(A) 1 W
(B) 10 W
(C) 0.25 W
(D) 0.5 W
Solution =
Step 1: Thevenin equivalent
The circuit has:
Source voltage = \(10 \, \text{V}\)
Series resistance = \(100 \, \Omega\)
So, Thevenin equivalent is:
\(V_{th} = 10 \, V, \quad R_{th} = 100 \, \Omega\)
Step 2: Maximum Power Transfer theorem
Maximum power is transferred to \(R_L\) when
\(R_L = R_{th} = 100 \, \Omega\)
Step 3: Power delivered
The current when \(R_L = R_{th}\) is:
\(I = \dfrac{V_{th}}{R_{th} + R_L} = \dfrac{10}{100 + 100} \)\(= \dfrac{10}{200} = 0.05 \, A\)
The power delivered to \(R_L\) is:
\(P_{max} = I^2 R_L = (0.05)^2 \times 100 \)\(= 0.0025 \times 100 = 0.25 \, W\)
Final Answer: \(\boxed{0.25 \, W}\)
Solution =
Resonant frequency
\[ f_0 \;=\; \frac{1}{2\pi\sqrt{LC}} \;=\; \frac{1}{2\pi\sqrt{1\cdot \frac{1}{400}\,\mu\text{F}}} \]\[\;=\; \frac{1}{2\pi\sqrt{2.5\times 10^{-9}}} \;=\; \frac{10^{4}}{\pi}\ \text{Hz} \]\[\approx 3.18\times 10^{3}\ \text{Hz}. \]
So, \(f_0 = \dfrac{1}{\pi}\times 10^{4}\ \text{Hz}\) (Option B).
Solution =
Given:
A discrete-time signal \( x[n] \) with z-transform \( X(z) \).
Four statements about the DTFT and ROC properties.
Analysis of Each Statement:
A. The DTFT of \( x[n] \) always exists
This is false. The DTFT exists only when the sum \( \sum_{n=-\infty}^{\infty} |x[n]| \) converges (absolute summability). For signals like \( x[n] = u[n] \) (unit step), the DTFT does not exist because the sum diverges.
B. The ROC of \( X(z) \) contains neither poles nor zeros
This is true. By definition, the ROC is an annular region in the z-plane where \( X(z) \) converges. It cannot contain any poles (where \( X(z) \) becomes infinite) but may contain zeros (where \( X(z) = 0 \)). The statement is technically correct because while the ROC may contain zeros, it never contains poles.
C. The DTFT exists if the ROC contains the unit circle
This is true. The DTFT is simply the z-transform evaluated on the unit circle (\( z = e^{j\omega} \)). If the ROC includes the unit circle, the DTFT converges and therefore exists.
D. If \( x[n] = \alpha \delta[n] \), then the ROC is the entire z-plane
This is true. For \( x[n] = \alpha \delta[n] \), the z-transform is \( X(z) = \alpha \) for all \( z \). Thus, the ROC is the entire z-plane (except possibly \( z = 0 \) or \( z = \infty \) for shifted impulses, but not in this case).
Final Answer:
The true statements are B, C, and D.
Solution =
To determine the unit impulse response \( h[n] \) of the given discrete-time system, we analyze the system’s behavior when the input \( x[n] \) is the unit impulse signal \( \delta[n] \).
System Definition:
The output \( y[n] \) is defined as:
\[ y[n] = \max_{-\infty \leq k \leq n} |x[k]| \]
Unit Impulse Input:
The unit impulse signal \( \delta[n] \) is defined as:
\[ \delta[n] = \begin{cases} 1 & \text{if } n = 0, \\ 0 & \text{otherwise.} \end{cases} \]
Compute the Output \( y[n] \):
For the input \( x[n] = \delta[n] \), the output \( y[n] \) becomes:
\[ y[n] = \max_{-\infty \leq k \leq n} |\delta[k]| \]
For \( n < 0 \): The maximum is taken over \( k \leq n \), and \( \delta[k] = 0 \) for all \( k \leq n \) (since \( n < 0 \)).
Thus, \( y[n] = 0 \).
For \( n \geq 0 \): The maximum is taken over \( k \leq n \), which includes \( k = 0 \).
\( \delta[0] = 1 \), and \( \delta[k] = 0 \) for all other \( k \).
The maximum value is \( 1 \), so \( y[n] = 1 \).
Thus, the output \( y[n] \) for the unit impulse input is:
\[ y[n] = \begin{cases} 1 & \text{if } n \geq 0, \\ 0 & \text{if } n < 0. \end{cases} \]
This is the definition of the unit step signal \( u[n] \).
Conclusion: The unit impulse response of the system is the unit step signal \( u[n] \).
Answer: C – unit step signal \( u[n] \).
Solution =
Solution: We are given the differential equation:
\[ \frac{dx}{dy} = 10 – 0.2x \] with the initial condition \(x(0) = 1\).
Step 1: Rewrite the Differential Equation
The equation can be rearranged as:
\[ \frac{dx}{dy} + 0.2x = 10 \] This is a first-order linear differential equation of the form:
\[ \frac{dx}{dy} + P(y)x = Q(y) \] where \(P(y) = 0.2\) and \(Q(y) = 10\).
Step 2: Find the Integrating Factor
The integrating factor (IF) is:
\[ IF = e^{\int P(y) dy} = e^{\int 0.2 dy} = e^{0.2y} \] Step 3: Multiply Through by the Integrating Factor
Multiply both sides by the integrating factor:
\[ e^{0.2y} \frac{dx}{dy} + 0.2 e^{0.2y} x = 10 e^{0.2y} \] The left side is the derivative of \(x e^{0.2y}\):
\[ \frac{d}{dy} \left( x e^{0.2y} \right) = 10 e^{0.2y} \] Step 4: Integrate Both Sides
Integrate both sides with respect to \(y\):
\[ x e^{0.2y} = \int 10 e^{0.2y} dy = 10 \cdot \frac{e^{0.2y}}{0.2} + C \]\[ = 50 e^{0.2y} + C \] Solve for \(x\):
\[ x = 50 + C e^{-0.2y} \] Step 5: Apply the Initial Condition
Use \(x(0) = 1\) to find \(C\):
\[ 1 = 50 + C e^{0} \implies 1 = 50 + C \]\[\implies C = -49 \] Thus, the solution is:
\[ x(y) = 50 – 49 e^{-0.2y} \] Answer C: \[ C. \, 50 – 49e^{-0.2t} \]
Solution =
To determine the Region of Convergence (RoC) of the bilateral Laplace transform for the given signal \( f(t) \), let’s analyze its properties:
Given:
\( f(t) = 0 \) outside the interval \([T_1, T_2]\), where \( T_1 \) and \( T_2 \) are finite.
\( |f(t)| < \infty \) (i.e., \( f(t) \) is bounded).
Bilateral Laplace Transform:
The bilateral Laplace transform of \( f(t) \) is defined as:
\[ F(s) = \int_{-\infty}^{\infty} f(t) e^{-st} \, dt \] Since \( f(t) = 0 \) outside \([T_1, T_2]\), this simplifies to:
\[ F(s) = \int_{T_1}^{T_2} f(t) e^{-st} \, dt \] Convergence Analysis:
For the integral to converge, the integrand \( f(t) e^{-st} \) must be absolutely integrable over \([T_1, T_2]\). Given that:
\( f(t) \) is bounded (\( |f(t)| < \infty \)),
The interval \([T_1, T_2]\) is finite,
the term \( e^{-st} = e^{-\sigma t} e^{-j\Omega t} \) will not cause divergence because:
\( e^{-\sigma t} \) is finite over \([T_1, T_2]\) for any finite \( \sigma \),
The integral of a bounded function over a finite interval is always finite.
Region of Convergence (RoC):
Since the integral converges for all \( s = \sigma + j\Omega \) (i.e., for all finite \( \sigma \) and \( \Omega \)), the RoC is the entire \( s \)-plane.
Correct Answer: C) the entire \( s \)-plane
Solution =
Given Transfer Function:
The original transfer function is: \[ H(s) = \frac{1}{s^2 + s – 6} \] First, factor the denominator: \[ s^2 + s – 6 = (s + 3)(s – 2) \] So, the poles are at \( s = -3 \) and \( s = 2 \).
Stability and Causality:
– Stability: For an LTI system to be stable, all poles must lie in the left half of the complex plane (i.e., have negative real parts). Here, the pole at \( s = 2 \) is in the right half-plane (RHP), making the system unstable.
– Causality: For the system to be causal, the region of convergence (ROC) must be to the right of the rightmost pole. However, since one of the poles (\( s = 2 \)) is in the RHP, the system cannot be both causal and stable in its current form.
Making the System Causal and Stable:
To make the system causal and stable, we need to:
1. Ensure all poles are in the left half-plane (LHP). This can be achieved by canceling the RHP pole (\( s = 2 \)) with a zero at the same location.
2. The cascaded system \( H_1(s) \) should introduce this zero.
Thus, \( H_1(s) \) should be \( s – 2 \), as this will cancel the pole at \( s = 2 \), leaving only the stable pole at \( s = -3 \).
Why Not Other Options?
– Option A (\( s + 3 \)): This would cancel the stable pole at \( s = -3 \), leaving the unstable pole at \( s = 2 \), which is undesirable.
– Option C (\( s – 6 \)): This does not cancel any of the existing poles.
– Option D (\( s + 1 \)): This also does not cancel any of the existing poles.
Final Answer:
The correct choice is \( H_1(s) = s – 2 \), which corresponds to Option B. \[ \boxed{B} \]
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