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Ct 1 pyq

Nakib Noor11 min read

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Here is the complete compilation of all Chapter 1: Signal Fundamentals & Operations past year university questions (PYQs) with compact, exam-scoring answers.

This sheet is categorized by core topics so you can use it directly to secure full marks on your upcoming Class Test (CT).

📂 Group 1: Core Definitions & Singularity Functions

*(Target: 2-Mark to 8-Mark Descriptive Questions)*

Q1: What are a signal and a system? Discuss their physical importance in communication engineering. [2025 - 4 Marks | 2024 - 6 Marks | 2023 - 4 Marks]

  • Signal: Any physical quantity that carries information and varies with one or more independent variables (such as time \(t\) or sample index \(n\)) [1.00, 1.11]. Continuous: \(x(t)\); Discrete: \(x[n]\) [1.00, 1.12].
  • System: A physical device, process, or mathematical algorithm that operates on an input signal \(x(t)\) (excitation) to produce a transformed output signal \(y(t)\) (response) [1.00, 1.13]: \[y(t) = \mathbb{T}[x(t)] \quad [1.13]\]
  • Communication Importance: Raw physical signals are rarely usable in their original form. Systems allow engineers to transmit data over long distances, filter out-of-band noise, and multiplex channels [1.06, 1.14].
  • Key Applications: ECG/EEG monitoring, radar/sonar echo range-finding, GPS satellite synchronization, seismic tracking, and voice/TV transmission [1.00, 1.14].

Q2: Define: (i) Unit step function, (ii) Unit impulse function, (iii) Ramp function, and (iv) Signum function. [2022 - 8 Marks]

  • (i) Unit Step Function \(u(t)\): Represents a DC switch closing instantaneously at \(t=0\) [1.00, 1.16]: \[u(t) = \begin{cases} 1, & t > 0 \ 0, & t < 0 \end{cases} \quad [1.00, 1.16]\]
  • (ii) Unit Impulse Function \(\delta(t)\): An infinitely narrow, infinitely tall spike centered at \(t=0\) containing an area of exactly 1 [1.00, 1.17]: \[\delta(t) = 0 \quad \text{for } t \neq 0, \quad \text{and} \quad \int_{-\infty}^{\infty} \delta(t) , dt = 1 \quad [1.00, 1.17]\]
  • (iii) Unit Ramp Function \(r(t)\): A linearly growing signal representing the time integral of the step function [1.00, 1.18]: \[r(t) = \begin{cases} t, & t > 0 \ 0, & t \le 0 \end{cases} = t \cdot u(t) \quad [1.00, 1.18]\]
  • (iv) Signum Function \(\text{sgn}(t)\): A polarity indicator extracting the algebraic sign of the time variable \(t\) [1.00, 1.19]: \[\text{sgn}(t) = \begin{cases} 1, & t > 0 \ 0, & t = 0 \ -1, & t < 0 \end{cases} = u(t) - u(-t) \quad [1.00, 1.19]\] (Sketch the simple 2D waveforms showing axes and vertical boundaries for full marks!) [1.00, 1.16, 1.17, 1.18, 1.19]

📂 Group 2: Signal Operations & Plotting

*(Target: 4-Mark to 8-Mark Graphical Sketching)*

Q3: Sketch the following signals: (i) \(x(t) = -4r(3t-1)\), (ii) \(x(t) = \Pi(2t+5)\). [2024 - 4 Marks | 2019 - 8 Marks]

  • (i) \(x(t) = -4r(3t-1)\):
    1. Precedence Rule: Shifting must always be performed before scaling [1.00, 1.22].
    2. Shift: Delay unit ramp \(r(t)\) right by 1 to get \(r(t-1)\) [1.23].
    3. Amplitude Scale: Multiply by \(-4\) to get \(-4r(t-1)\) (starts at \(t=1\), slope is \(-4\)) [1.23].
    4. Time Scale: Compress the time axis by 3 [1.23]. The new starting boundary is \(3t-1=0 \implies t = 1/3\).
    5. The slope scales proportionally: \(\text{New Slope} = -4 \times 3 = -12\) [1.23].
    6. Final piecewise form: \(x(t) = -12t + 4\) for \(t \ge 1/3\), and \(x(t) = 0\) for \(t < 1/3\) [1.23].
  • (ii) \(x(t) = \Pi(2t+5)\):
    1. Factor the scaling constant inside: \(x(t) = \Pi(2(t+2.5))\) [1.22].
    2. The standard gate pulse \(\Pi(t)\) is centered at 0 with width 1 [1.01].
    3. Shift left by \(2.5\) units to center the gate at \(t = -2.5\) [1.22, 1.24].
    4. Compress the pulse width by a factor of 2 (new width \(\tau = 0.5\)) [1.22, 2.05].
    5. Final pulse boundaries: Active with amplitude 1 strictly between \(t = -2.75\) and \(t = -2.25\) [1.24, 2.05].

Q4: Sketch the following signal: \(x(t) = r(-0.5t+2)\). [2015 - 8 Marks]

  • Step 1 (Factoring): Rewrite as \(x(t) = r(-0.5(t-4))\) to isolate the true shift [2.11].
  • Step 2 (Time Reversal & Scaling): Fold unit ramp \(r(t)\) to get \(r(-t)\) (points to the left) [2.11]. Expand the time axis by 2 to shallow out the slope to \(0.5\) (\(r(-0.5t)\)) [2.11].
  • Step 3 (Shifting): Shift the entire expanded waveform right by 4 units [2.11].
  • Key coordinates to plot: Starts at \(t = 4\) (where \(-0.5(4)+2=0\)) and exists only for \(t \le 4\) [2.11]. At \(t = 0 \implies x(0) = 2\) [2.11]; at \(t = -2 \implies x(-2) = 3\) [2.11].

Q5: Sketch the signals: (i) \(u(t-5)-u(t-7)\), (ii) \(t^2[u(t-1)-u(t-2)]\), (iii) \(\sin(50t-\pi/4)\). [2021 - 5 Marks]

  • (i) \(u(t-5)-u(t-7)\): Step \(u(t-5)\) turns on (+1) at \(t=5\), step \(-u(t-7)\) subtracts 1 at \(t=7\). This sum yields a flat rectangular pulse of amplitude 1 from \(t=5\) to \(t=7\) [1.00, 2.12].
  • (ii) \(t^2[u(t-1)-u(t-2)]\): The step difference acts as a gating window between \(t=1\) and \(t=2\) [1.00, 2.12]. Plot a curved parabolic segment of \(t^2\) that turns on at \(t=1\) (height 1) and turns off at \(t=2\) (height 4).
  • (iii) \(\sin(50t-\pi/4)\): A continuous sinusoid of angular frequency \(\Omega = 50\text{ rad/s}\) shifted right by a phase delay of \(\pi/4\) rad (delay time \(t_0 = \pi/200\text{ s}\)) [1.25].

Q6: Sketch the following signals: (i) \(x(t) = \Pi(2t+3)\), (ii) \(x(t) = \cos(20\pi t - 5\pi)\). [2016 - 8 Marks]

  • (i) \(x(t) = \Pi(2t+3)\): Factor to \(\Pi(2(t+1.5))\) [1.22]. Centered at \(t = -1.5\) with a compressed total width of 0.5 (active from \(t = -1.75\) to \(t = -1.25\)) [1.24].
  • (ii) \(x(t) = \cos(20\pi t - 5\pi)\): Shift of an odd integer multiple of \(\pi\) inverts the waveform: \(\cos(\theta - 5\pi) = -\cos(\theta)\) [1.138]. Sketch a standard inverted cosine wave starting at a valley of \(-1\) at \(t=0\), with a fundamental period of \(T_0 = 2\pi/20\pi = 0.1\) seconds.

Q7: Write down the equations for a stair step signal \(x(t)\) (levels: 2 from \(0 \le t < 1\), and 3 from \(1 \le t < 3\)) via: i) addition, ii) multiplication of unit step functions. [2015 - 8 Marks]

  • i) Addition Method (Step Accumulation): \[x(t) = 2u(t) + 1u(t-1) - 3u(t-3) \quad [1.00]\]
  • ii) Multiplication Method (Gate Windowing): \[x(t) = 2[u(t) - u(t-1)] + 3[u(t-1) - u(t-3)] \quad [1.00]\]

📂 Group 3: Symmetry & Decompositions

*(Target: 7-Mark to 10-Mark Mathematical Proofs)*

Q8: If \(x_e(t)\) and \(x_o(t)\) are even and odd parts of \(x(t)\), prove that: \(\int_{-\infty}^{\infty} x^2(t) dt = \int_{-\infty}^{\infty} x_e^2(t) dt + \int_{-\infty}^{\infty} x_o^2(t) dt\) [2024, 2023, 2021 - 7 to 10 Marks]

  • Proof:
    1. Substitute the even-odd decomposition \(x(t) = x_e(t) + x_o(t)\) [1.00, 1.132]: \[\int_{-\infty}^{\infty} x^2(t) , dt = \int_{-\infty}^{\infty} [x_e(t) + x_o(t)]^2 , dt \quad [1.132]\]
    2. Expand the squared integrand [1.132]: \[\int_{-\infty}^{\infty} x^2(t) , dt = \int_{-\infty}^{\infty} x_e^2(t) , dt + \int_{-\infty}^{\infty} x_o^2(t) , dt + 2\int_{-\infty}^{\infty} x_e(t)x_o(t) , dt \quad [1.132]\]
    3. Analyze the cross-product term: \(g(t) = x_e(t)x_o(t)\). Since even \(\times\) odd produces an odd function (\(g(-t) = x_e(-t)x_o(-t) = x_e(t)[-x_o(t)] = -g(t)\)) [1.132, 2.29]:
    4. The definite integral of any odd function over symmetrical limits is strictly zero [1.00, 1.132]: \[2\int_{-\infty}^{\infty} x_e(t)x_o(t) , dt = 0 \quad [1.00, 1.132]\]
    5. Therefore, the cross-product vanishes, proving the power identity [1.00, 1.132]: \[\int_{-\infty}^{\infty} x^2(t) , dt = \int_{-\infty}^{\infty} x_e^2(t) , dt + \int_{-\infty}^{\infty} x_o^2(t) , dt \quad \blacksquare \quad [1.00, 1.132]\]

Q9: Find the even and odd components for: (i) \(x(t) = u(t)\); (ii) \(x(t) = e^{-\alpha t} u(t)\). [2020 - 8 Marks]

  • (i) Unit Step Components: \[x_e(t) = \frac{u(t) + u(-t)}{2} = \mathbf{\frac{1}{2}} \quad \text{and} \quad x_o(t) = \frac{u(t) - u(-t)}{2} = \mathbf{\frac{1}{2}\text{sgn}(t)} \quad [1.138]\]
  • (ii) Causal Decay Components: \[x_e(t) = \mathbf{\frac{e^{-\alpha t}u(t) + e^{\alpha t}u(-t)}{2}} \quad \text{and} \quad x_o(t) = \mathbf{\frac{e^{-\alpha t}u(t) - e^{\alpha t}u(-t)}{2}} \quad [1.138]\]

Q10: Find the Even and Odd components of \(x(t) = \cos(20\pi t - 5\pi)\) and sketch. [2018, 2015 - 10 Marks]

  • Step 1: Simplify using cosine symmetries: \(\cos(20\pi t - 5\pi) = -\cos(20\pi t)\) [1.138].
  • Step 2: Since \(\cos(-\theta) = \cos(\theta)\), the signal is purely even [1.138]:
    • Even component: \(x_e(t) = \mathbf{-\cos(20\pi t)}\) [1.138].
    • Odd component: \(x_o(t) = \mathbf{0}\) [1.138].

📂 Group 4: Composite Periodicity Tests

*(Target: 5-Mark to 9-Mark Rationality Calculations)*

Q11: Determine the periodicity and period of \(x(t) = 2\cos(4\pi t) + 3\sin(3\pi t)\). [2025 - 5 Marks]

  • Step 1 (Find individual periods):
    • \(T_1 = 2\pi/\omega_1 = 2\pi/4\pi = \mathbf{1/2\text{ s}}\) [1.128].
    • \(T_2 = 2\pi/\omega_2 = 2\pi/3\pi = \mathbf{2/3\text{ s}}\) [1.128].
  • Step 2 (Apply the Rationality Rule): \[\frac{T_1}{T_2} = \frac{1/2}{2/3} = \frac{3}{4} \in \mathbb{Q} \quad (\text{Ratio is rational} \implies \mathbf{Periodic}) \quad [1.128]\]
  • Step 3 (Calculate Fundamental Period \(T_0\) via fractional LCM): \[T_0 = \text{LCM}\left(\frac{1}{2}, \frac{2}{3}\right) = \frac{\text{LCM}(1, 2)}{\text{HCF}(2, 3)} = \frac{2}{1} = \mathbf{2\text{ seconds}} \quad [1.128]\]

Q12: Determine whether \(x(t) = \sin(20\pi t) + \sin(5\pi t)\) is periodic or not. If periodic, find its fundamental period. [2017 - 6 Marks]

  • Step 1: \(T_1 = 2\pi/20\pi = 1/10\text{ s}\) and \(T_2 = 2\pi/5\pi = 2/5\text{ s}\) [2.19].
  • Step 2: Ratio \(\frac{T_1}{T_2} = \frac{1/10}{2/5} = \frac{1}{4} \in \mathbb{Q} \implies\) Periodic [2.19].
  • Step 3: \(T_0 = \text{LCM}\left(\frac{1}{10}, \frac{2}{5}\right) = \frac{\text{LCM}(1, 2)}{\text{HCF}(10, 5)} = \frac{2}{5} = \mathbf{0.4\text{ seconds}}\) [2.19].

Q13: Determine whether \(x(t) = \sin\sqrt{2}\pi t + \sin 15\pi t\) is periodic or not, find period, and compute its total energy. [2019 - 9 Marks | 2015 - 5 Marks]

  • Periodicity: \(T_1 = \sqrt{2}\text{ s}\) and \(T_2 = 2/15\text{ s}\) [2.18]. Ratio \(\frac{T_1}{T_2} = \frac{15\sqrt{2}}{2}\) is irrational (\(\notin \mathbb{Q}\)), so the composite signal is Aperiodic [2.18].
  • Energy: Because sines are periodic and oscillate infinitely, the total integrated energy diverges to infinity (\(\mathbf{E = \infty}\)) [1.138].

📂 Group 5: Energy vs. Power Classifications

*(Target: 2-Mark to 6-Mark Definite Integrals)*

Q14: Determine the power and energy of the step signal \(x(t) = u(t)\). [2022 - 2 Marks]

  • Energy (\(E\)): \(E = \int_{0}^{\infty} (1)^2 , dt = [t]_0^{\infty} = \mathbf{\infty}\) [1.136].
  • Power (\(P\)): \(P = \lim_{T \to \infty} \frac{1}{T} \int_{0}^{T/2} (1)^2 , dt = \lim_{T \to \infty} \frac{T/2}{T} = \mathbf{0.5\text{ Watts}}\) [1.136].
  • Classification: Since \(E=\infty\) and \(P=0.5\text{ W}\), it is a Power Signal [1.136].

Q15: Determine whether \(x[n] = \cos(\pi n) u[n]\) is energy or power. [2023 - 5 Marks | 2019 - 6 Marks]

  • Note: For integer indices \(n\), \(\cos(\pi n) = (-1)^n\) [1.135].
  • Energy (\(E\)): \(E = \sum_{n=0}^{\infty} |(-1)^n|^2 = \sum_{n=0}^{\infty} 1 = \mathbf{\infty}\) [1.135].
  • Power (\(P\)): \(P = \lim_{N \to \infty} \frac{1}{2N+1} \sum_{n=0}^{N} (1) = \lim_{N \to \infty} \frac{N+1}{2N+1} = \mathbf{0.5\text{ Watts}}\) [1.135].
  • Classification: Power Signal [1.135].

Q16: Determine whether \(x[n] = \sin(\pi n) u[n]\) is energy or power. [2025 - 5 Marks]

  • Important Check: For any integer \(n\), \(\sin(\pi n) = 0\) [1.134].
  • Metrics: \(\mathbf{E = 0}\) and \(\mathbf{P = 0}\) [1.134].
  • Classification: Neither (Trivial Zero Signal) [1.134].

Q17: Determine whether \(x[n] = \cos(\pi n)\) for \(-4 \le n \le 4\) is energy or power. [2018 - 6 Marks]

  • Rule: Any bounded, finite-duration signal is always an energy signal [1.137].
  • Energy (\(E\)): \(E = \sum_{n=-4}^{4} |(-1)^n|^2 = \sum_{n=-4}^{4} 1 = \mathbf{9\text{ Joules}}\) [1.137].
  • Power (\(P\)): \(\mathbf{P = 0}\) [1.137].
  • Classification: Energy Signal [1.137].

📂 Group 6: Signal Causality

*(Target: 4-Mark to 5-Mark Boundary Identifications)*

Q18: State whether the following signals are causal, anticausal, or noncausal: (i) \(x(t) = e^{-2t}u(t)\), (ii) \(x(t) = tu(t)\). [2025 - 5 Marks | 2020 - 5 Marks]

  • (i) \(x(t) = e^{-2t}u(t)\): The multiplier step function \(u(t)\) forces the signal to be 0 for all negative time (\(t < 0\)) [1.130]. This satisfies the causal boundary condition, making it strictly Causal [1.130].
  • (ii) \(x(t) = tu(t)\): Similarly, \(x(t) = 0\) for \(t < 0\) due to \(u(t)\) [1.130]. It is strictly Causal [1.130].

💡 CT Exam Hacks:

  1. Avoid the “Ramp Slope” Mistake: When compressing a ramp like \(4r(3t-1)\), remember the slope is multiplied by the compression factor (\(\text{Slope} = 4 \times 3 = 12\)) [1.23].
  2. Verify your Fractions on Periodicity: Make sure you do not invert the fractional LCM formula. It is always: \(\frac{\text{LCM of Numerators}}{\text{HCF of Denominators}}\) [1.06, 1.126].
  3. Orthogonality Proof: When writing out the Even-Odd energy proof, explicitly write a sentence stating that the integral of \(2x_e(t)x_o(t)\) is zero because the product of an even and odd function is odd, and integrating an odd function over symmetric limits always yields zero [1.06, 1.132].

📊 All Chapter 1 CT PYQs are now thoroughly prepared! Would you like me to generate a quick, 3-question practice quiz to test your step-by-step sketching or fractional LCM skills before tomorrow?

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