DBATU SIGNALS & SYSTEMS

Course Code: 25AF1370MD406A / BTETC402

Branch Mapping: AI&DS, AI&ML, Computer Eng. (NEP 2020)

University Board: Dr. Babasaheb Ambedkar Technological University

Evaluation Metric: 60/60 End-Semester Target

Welcome to the full DBATU Signals & Systems revision guide. Navigate by unit to review solved questions, formulas, and exam-focused explanations. All supporting theory and solved examples are arranged for fast study and classroom revision.

Unit 1: Introduction to Signals and Systems

High Priority Topics for 10-CGPA

Even & Odd Decomposition: Guaranteed 4 to 6-mark question. Practice both trigonometric and exponential proofs.
Energy vs Power: Must know integration for continuous and summation for discrete. Decaying signals = Energy; Periodic signals = Power.
System Verification: 4-mark mandatory question. Focus heavily on testing Linearity (Superposition) and Causality (Time shifting).

Complete Formula Sheet

Even Component: $x_e(t) = \frac{x(t) + x(-t)}{2}$
Odd Component: $x_o(t) = \frac{x(t) - x(-t)}{2}$
Energy (Continuous): $E = \int_{-\infty}^{\infty} |x(t)|^2 dt$
Energy (Discrete): $E = \sum_{n=-\infty}^{\infty} |x(n)|^2$
Power (Continuous): $P = \lim_{T \to \infty} \frac{1}{2T} \int_{-T}^{T} |x(t)|^2 dt$
Linearity (Superposition): Output for $$ ax_1(t) + bx_2(t) $$ must equal $$ ay_1(t) + by_2(t) $$ .

Exam Tips & Examiner Traps

The Constant Trap: If an equation has an isolated constant, like $$ y(n) = 2x(n) - 3 $$ , it is always Non-Linear. A linear system MUST output 0 when the input is 0.
The Look-Ahead Trap: Any scaling multiplier inside the bracket, like $$ y(n) = x(2n) $$ , makes the system Non-Causal because finding the present requires a future value.
Periodic Test: For continuous sums like $\cos(\omega_1 t) + \sin(\omega_2 t)$ , find the periods $T_1 = \frac{2\pi}{\omega_1}$ and $T_2 = \frac{2\pi}{\omega_2}$ . If the ratio $$ T_1 / T_2 $$ is a rational fraction, it is periodic!

Objective Questions (Tap Option to Reveal Reason)

1 MarkSummer 2025

Which of the following is a periodic signal?

A) $ \sin(t) $

B) $ e^t $

C) $ t $

D) $ \ln(t) $

Detailed Reason: A periodic signal must satisfy the mathematical condition $ x(t) = x(t+T) $ for all $t$. The sine wave inherently oscillates and repeats its exact shape every $ 2\pi $ radians. Conversely, Exponential ($ e^t $), linear ($ t $), and logarithmic ($ \ln(t) $) functions grow infinitely over time and never repeat their values, classifying them as aperiodic.

1 MarkSummer 2025

Which operation is used to check linearity of a system?

A) Scaling

B) Additivity

C) Both a & b

D) None

Detailed Reason: Linearity requires a system to completely obey the Principle of Superposition. Superposition consists of two distinct mathematical properties: Additivity (if $ x_1 \rightarrow y_1 $ and $ x_2 \rightarrow y_2 $, then $ x_1 + x_2 \rightarrow y_1 + y_2 $) and Homogeneity/Scaling ($ a \cdot x \rightarrow a \cdot y $). Both must be verified to prove linearity.

1 MarkSummer 2025

The fundamental frequency of a periodic signal with period T is:

A) $ 2\pi/T $

B) $ 1/T $

C) $ T $

D) $ T/2 $

Detailed Reason: DBATU standardly asks for the fundamental angular frequency $ \omega_0 $ in formula contexts. The relationship is $ \omega_0 = 2\pi f $. Since the cyclic frequency is $ f = 1/T $ Hertz, substituting this gives the angular frequency as $ \omega_0 = \frac{2\pi}{T} $ radians/second.

1 MarkSummer 2025

A signal that does not repeat itself over time is:

A) Aperiodic

B) Even

C) Periodic

D) Odd

Detailed Reason: "Aperiodic" literally translates to "without period". If a signal fails to satisfy the mathematical condition $ x(t) = x(t+T) $ for all possible values of t, it means the signal does not possess a repeating cycle and is therefore classified as aperiodic.

1 MarkWinter 2024

The area under the curve $ \int_{-\infty}^{\infty} \delta(t) dt $ is:

A) $ \infty $

B) Unity

C) Zero

D) Undefined

Detailed Reason: The Dirac delta function $ \delta(t) $ is a mathematical construct representing a perfect impulse. While its amplitude (height) at exactly $ t=0 $ reaches towards infinity, its time-width approaches zero in such a specific way that the total area integrated underneath the impulse is strictly defined as exactly 1 (Unity). This property is crucial for convolution.

1 MarkWinter 2024

Periodic Signals are defined by:

A) $ x(t+T)=x(t) $

B) $ x(t-T)=x(t) $

C) $ x(t+kT)=x(t) $

D) All of the above

Detailed Reason: If a signal repeats every T seconds into the future (Option A: $x(t+T)$), logic dictates it must have also repeated T seconds in the past (Option B: $x(t-T)$). Furthermore, if it repeats once, it will logically continue to repeat at every integer multiple 'k' of the fundamental period T (Option C: $x(t+kT)$). Therefore, all three mathematical statements perfectly define periodicity.

Descriptive Questions (100% Comprehensive)

Step 1: State the Baseline Assumption

We begin by assuming the mathematical premise is true. Let an arbitrary continuous-time signal $ x(t) $ be written as the sum of an unknown even component $ x_e(t) $ and an unknown odd component $ x_o(t) $.

x(t) = x_e(t) + x_o(t) \quad \text{--- (Equation 1)}

Step 2: Apply Time Reversal Properties

To find the components, we need a second equation. We generate this by substituting $ t $ with $ -t $ in Equation 1 to find the time-reversed version of the signal.

$$x(-t) = x_e(-t) + x_o(-t)$$

By standard mathematical definition, an even signal remains unchanged when time-reversed: $ x_e(-t) = x_e(t) $. An odd signal becomes negative when time-reversed: $ x_o(-t) = -x_o(t) $. Substitute these fundamental properties back into the equation:

x(-t) = x_e(t) - x_o(t) \quad \text{--- (Equation 2)}

Step 3: Solve for the Even Component

We now have a system of two linear equations. To isolate the even part, we add Equation 1 and Equation 2 together. The positive and negative odd terms will perfectly cancel each other out.

$$x(t) + x(-t) = [x_e(t) + x_o(t)] + [x_e(t) - x_o(t)]$$

$$x(t) + x(-t) = 2x_e(t)$$

Divide both sides by 2 to yield the final expression for the even component:

x_e(t) = \frac{x(t) + x(-t)}{2}

Step 4: Solve for the Odd Component

To isolate the odd part, we subtract Equation 2 from Equation 1. This time, the even terms will cancel out, and the double negative will turn the odd term positive.

$$x(t) - x(-t) = [x_e(t) + x_o(t)] - [x_e(t) - x_o(t)]$$

$$x(t) - x(-t) = 2x_o(t)$$

Divide both sides by 2 to yield the final expression for the odd component:

x_o(t) = \frac{x(t) - x(-t)}{2}

Final Conclusion:
The proof is complete. Any real signal $ x(t) $ can be constructed exactly as $ x_e(t) + x_o(t) $ utilizing the derived expressions:
$ x_e(t) = \frac{x(t) + x(-t)}{2} $ and $ x_o(t) = \frac{x(t) - x(-t)}{2} $.

Part (i): Energy and Power Signals Detailed

Energy Signals: A signal is classified as an energy signal if its total calculated energy $ E $ evaluates to a finite, non-zero number ($ 0 < E < \infty $). Consequently, because the energy is finite over an infinite timeline, its average power $ P $ evaluates to strictly 0. These are typically transient signals that have a defined start and stop time, or decay to zero over time (non-periodic pulses).

Mathematical Example: An exponentially decaying pulse $ x(t) = e^{-at}u(t) $ for $ a > 0 $.

Power Signals: A signal is classified as a power signal if its average power $ P $ evaluates to a finite, non-zero number ($ 0 < P < \infty $). Because it constantly outputs power, its total accumulated energy $ E $ over infinite time approaches infinity ($ E = \infty $). These are typically signals that continue oscillating or existing forever without decaying (periodic signals or constant DC values).

Mathematical Example: A continuous sine wave $ x(t) = \sin(\omega_0 t) $ or a constant signal $ x(t) = 5 $.

Part (ii): Analog and Digital Signals Detailed

Analog (Continuous-Time, Continuous-Amplitude) Signals: These signals are completely continuous in both the time domain and the amplitude domain. The signal exists and has a mathematically defined value at every single micro-instant of time. Furthermore, its amplitude is not restricted; it can take on an infinite number of decimal values within a physical range.

Real-World Example: Human speech captured by an analog microphone, or the continuous reading of a mercury thermometer rising.

Digital (Discrete-Time, Discrete-Amplitude) Signals: These signals are strictly discrete in both time and amplitude. The signal only exists at specific sampled time intervals (it is blank between samples), and its amplitude is rounded (quantized) to a specific set of predefined, finite binary values (e.g., 0s and 1s, or 0V and 5V).

Real-World Example: Audio data stored in an MP3 file on a hard drive, or binary packets transmitted over a computer network.

Step 1: The Fundamental Signals Defined

Unit Impulse (Delta) Signal $\delta(t)$: An infinitely tall, infinitely narrow spike occurring exactly at t=0. Its total area under the curve integrates exactly to 1. It is zero everywhere else. It represents a sudden, instantaneous shock to a system.

Unit Step Signal $u(t)$: A signal that is exactly 0 for all $t<0$ and abruptly jumps to an amplitude of 1 for all $t \ge 0$. It mathematically represents a switch turning on and staying on.

Unit Ramp Signal $r(t)$: A signal that is 0 for $t<0$ and grows linearly at a 45-degree angle (following the equation $r(t)=t$) for $t \ge 0$. It represents a constant, steady increase over time.

Step 2: Mathematical Integration Relationships

These three fundamental signals are directly linked sequentially through calculus. Moving "up" the complexity chain requires Integration.

1. Impulse to Step (Integration): Accumulating (integrating) an impulse spike from $-\infty$ up to time $t$ gives a solid step function of height 1.

u(t) = \int_{-\infty}^{t} \delta(\tau) d\tau

2. Step to Ramp (Integration): Integrating a constant step function of height 1 over time accumulates area, creating a linearly growing ramp function.

r(t) = \int_{-\infty}^{t} u(\tau) d\tau = t \cdot u(t)

Step 3: Mathematical Differentiation Relationships

Moving "down" the complexity chain requires Differentiation (finding the slope/rate of change).

3. Ramp to Step (Differentiation): The slope (derivative) of a linear ramp $ r(t)=t $ is a constant 1 (which is the step function).

\frac{d}{dt} r(t) = u(t)

4. Step to Impulse (Differentiation): The derivative of a step function is zero everywhere except at t=0, where it has an infinite vertical jump. This infinite rate of change is the impulse.

\frac{d}{dt} u(t) = \delta(t)

The 5 Core System Properties Explained

1. Linearity: A system is linear if it strictly obeys the principle of superposition. This requires two things: Additivity and Homogeneity (scaling). If input $x_1$ yields $y_1$, and $x_2$ yields $y_2$, then the response to a combined scaled input $ ax_1(t) + bx_2(t) $ must be exactly $ ay_1(t) + by_2(t) $.

2. Time-Invariance: A system is time-invariant if a delay in the input causes an identical, matching delay in the output, without changing the signal's core shape. Essentially, the system's behavior does not change depending on what time it is. If an input $ x(t) $ produces $ y(t) $, then a delayed input $ x(t-t_0) $ must produce exactly $ y(t-t_0) $.

3. Causality: A system is causal if the current output depends ONLY on present and past inputs. It cannot look into the future or anticipate data it hasn't received yet. For example, the system $ y(n) = x(n) + x(n-1) $ is causal, but the system $ y(n) = x(n+1) $ is non-causal because it requires tomorrow's input today.

4. Stability (BIBO): A system is Bounded-Input Bounded-Output (BIBO) stable if feeding it a finite input always produces a finite output. It does not blow up to infinity. Mathematically, for an LTI system, this requires the impulse response to be absolutely integrable: $ \int_{-\infty}^{\infty} |h(t)|dt < \infty $.

5. Memory (Static vs Dynamic): A static (memoryless) system's output depends ONLY on the current input at that exact instant (e.g., a simple resistor where $ V(t) = I(t) \cdot R $). A dynamic system requires memory of past or future states to calculate the present output (e.g., a capacitor or an integrator equation).

Signal Plotting Breakdown

Base Reference: $ u(t) $ is a standard unit step function jumping from 0 to 1 exactly at $ t=0 $ and heading right to positive infinity.

i) $ u(t-2) $ [Time Shifting]: This is delayed by 2 units. Set $ t-2 = 0 \implies t=2 $. It remains 0 until $ t=2 $, where it jumps to a height of 1 and continues right to infinity.

ii) $ u(-t) $ [Time Reversal]: The negative sign flips it. This is a mirror image across the Y-axis. It comes from $ -\infty $ at a steady height of 1, and drops to 0 exactly at $ t=0 $.

iii) $ u(-t+1) $ [Shift then Reverse]: To find the exact jump point, set the argument to zero: $ -t+1 = 0 \implies t=1 $. Because of the negative 't', it comes from the left. The signal comes from $ -\infty $ at a height of 1, and drops to 0 at $ t=1 $.

iv) $ u(-t-1) $ [Shift then Reverse]: Set the argument to zero: $ -t-1 = 0 \implies t=-1 $. Because of the negative 't', it comes from the left. The signal comes from $ -\infty $ at a height of 1, and drops to 0 early at $ t=-1 $.

v) $ u(2t) $ [Time Scaling]: Scaling time mathematically compresses the signal towards the y-axis. However, since a standard step function has infinite width from 0 to infinity, compressing infinity by a factor of 2 still leaves it as infinity. Thus, $ u(2t) $ visually looks exactly the same as $ u(t) $.

vi) $ 2 \cdot u(t) $ [Amplitude Scaling]: The multiplier is outside the function, so this affects the Y-axis, not the time axis. It is the standard step function starting at $ t=0 $, but instead of a height of 1, it jumps up to an amplitude of exactly 2.

Part (i): Trigonometric Signal Expansion

Given Formula: We will rely on the standard decomposition identities: $ x_e(t) = \frac{x(t) + x(-t)}{2} $ and $ x_o(t) = \frac{x(t) - x(-t)}{2} $.

Calculation of $ x(-t) $: First, we must find the time-reversed version by substituting $ t $ with $ -t $. Apply the fundamental trigonometric properties: $ \cos(-t) = \cos(t) $ [Even function] and $ \sin(-t) = -\sin(t) $ [Odd function].

x(-t) = \cos(-t) + \sin(-t) + \sin(-t)\cos(-t)

x(-t) = \cos(t) - \sin(t) - \sin(t)\cos(t)

Even Part Calculation: Add the original $ x(t) $ and the reversed $ x(-t) $. Observe how the sine terms containing negatives will completely cancel out.

x_e(t) = \frac{[\cos(t)+\sin(t)+\sin(t)\cos(t)] + [\cos(t)-\sin(t)-\sin(t)\cos(t)]}{2}

x_e(t) = \frac{2\cos(t)}{2} = \cos(t)

Odd Part Calculation: Subtract $ x(-t) $ from $ x(t) $. Notice the double negative makes the sine terms positive, while the cosine terms completely cancel out.

x_o(t) = \frac{[\cos(t)+\sin(t)+\sin(t)\cos(t)] - [\cos(t)-\sin(t)-\sin(t)\cos(t)]}{2}

x_o(t) = \frac{2\sin(t) + 2\sin(t)\cos(t)}{2} = \sin(t) + \sin(t)\cos(t)

Part (ii): Exponential Signal $ x(t)=e^{-2t}\cos(t) $

Calculation of $ x(-t) $: Replace $ t $ with $ -t $.

x(-t) = e^{-2(-t)}\cos(-t) = e^{2t}\cos(t)

Even Part Calculation: Plug into the even formula.

x_e(t) = \frac{e^{-2t}\cos(t) + e^{2t}\cos(t)}{2} = \cos(t) \left[ \frac{e^{-2t} + e^{2t}}{2} \right]

Recognize Euler's mathematical identity for hyperbolic cosine: $ \frac{e^{\theta} + e^{-\theta}}{2} = \cosh(\theta) $. Therefore:

x_e(t) = \cos(t) \cosh(2t)

Odd Part Calculation: Plug into the odd formula.

x_o(t) = \frac{e^{-2t}\cos(t) - e^{2t}\cos(t)}{2} = \cos(t) \left[ \frac{e^{-2t} - e^{2t}}{2} \right]

The identity for hyperbolic sine is $ \sinh(\theta) = \frac{e^{\theta} - e^{-\theta}}{2} $. Since our terms are reversed, we factor out a negative sign to match the identity, yielding $ -\sinh(2t) $.

x_o(t) = -\cos(t) \sinh(2t)

Step 1: Signal Definition and Index Mapping

The sequence $ y[n] = u[n] - u[n-4] $ evaluates to a rectangular window pulse. It turns on at index 0 and is cancelled out (turns off) at index 4.

It is exactly 1 at indices $ n=0, 1, 2, 3 $, and 0 everywhere else. Let's write the array: $ y[n] = \{1, 1, 1, 1\} $ starting at $n=0$.

The time-reversed sequence $ y[-n] $ is the mirror image across the y-axis. It is exactly 1 at indices $ n=0, -1, -2, -3 $, and 0 everywhere else.

Step 2: Even Part $ y_e[n] = (y[n]+y[-n])/2 $

We calculate this formula point-by-point along the integer timeline:

At $ n=0 $: Both sequences evaluate to 1. Therefore, $ (1+1)/2 = 1 $.

At positive indices $ n=1, 2, 3 $: Only the right-sided $ y[n] $ equals 1. Therefore, $ (1+0)/2 = 0.5 $.

At negative indices $ n=-1, -2, -3 $: Only the left-sided $ y[-n] $ equals 1. Therefore, $ (0+1)/2 = 0.5 $.

Everywhere else outside this range: $ (0+0)/2 = 0 $.

Step 3: Odd Part $ y_o[n] = (y[n]-y[-n])/2 $

Calculate point-by-point:

At $ n=0 $: Both sequences equal 1. $ (1-1)/2 = 0 $ (Note: Odd signals mathematically MUST always evaluate to zero at the origin).

At positive indices $ n=1, 2, 3 $: Only $ y[n] $ equals 1. $ (1-0)/2 = 0.5 $.

At negative indices $ n=-1, -2, -3 $: Only $ y[-n] $ equals 1. $ (0-1)/2 = -0.5 $.

Everywhere else outside this range: 0.

Final Arrays:
$ y_e[n] $ has values $ \{0.5, 0.5, 0.5, \mathbf{1}, 0.5, 0.5, 0.5\} $ from $ n=-3 $ to $ 3 $.
$ y_o[n] $ has values $ \{-0.5, -0.5, -0.5, \mathbf{0}, 0.5, 0.5, 0.5\} $ from $ n=-3 $ to $ 3 $.

Step 1: Energy Calculation via Summation

The standard mathematical formula for the total energy of a discrete-time signal is:

E = \sum_{n=-\infty}^{\infty} |x(n)|^2

Because of the unit step function $ u(n) $, the signal is strictly zero for all $ n < 0 $. We adjust the summation limits to start at 0:

E = \sum_{n=0}^{\infty} (a^n)^2 = \sum_{n=0}^{\infty} (a^2)^n

This represents an infinite geometric progression sum. The standard formula is $ S_{\infty} = \frac{\text{initial\_term}}{1-r} $. Here, our initial term at n=0 is 1, and the common ratio is $ r = a^2 $. Since the problem states $ |a| < 1 $, the series converges safely to a finite value.

E = \frac{1}{1 - a^2} \text{ Joules}

Step 2: Power Calculation via Limits

By mathematical definition, the average power is the total energy divided by infinite time: $ P = \lim_{N \to \infty} \frac{E}{2N+1} $.

Since we proved the total Energy $ E $ evaluates to a finite, constant number, dividing a constant by infinity yields exactly zero.

Final Answer: Total Energy $ E = \frac{1}{1 - a^2} $. Average Power $ P = 0 $.

(i) Check Linearity for $ y(n) = n x^2(n) $

Concept: We must mathematically prove that the system obeys superposition: $ T\{a x_1(n) + b x_2(n)\} = a T\{x_1(n)\} + b T\{x_2(n)\} $.

Right Hand Side (RHS): Apply the inputs individually, square them as the system dictates, and multiply by constants.

\text{RHS} = a \cdot y_1(n) + b \cdot y_2(n) = a [n x_1^2(n)] + b [n x_2^2(n)]

Left Hand Side (LHS): Combine the inputs into a single mixed input first: $ x_3(n) = a x_1(n) + b x_2(n) $, then pass it through the system.

$$y_3(n) = n [x_3(n)]^2 = n [a x_1(n) + b x_2(n)]^2$$

Expanding using the algebraic identity $ (A+B)^2 = A^2 + B^2 + 2AB $:

\text{LHS} = n [a^2 x_1^2(n) + b^2 x_2^2(n) + 2ab x_1(n)x_2(n)]

Conclusion: LHS $\neq$ RHS due to the squared constants ($a^2$) and the cross-product term. The system is NON-LINEAR.

(ii) Check Causality for $ y(n) = x(2n) $

Concept: A causal system depends ONLY on present and past inputs. It cannot access future data points.

Calculation: Test a positive integer time step, let $ n = 1 $.

y(1) = x(2 \cdot 1) = x(2)

Conclusion: To calculate the output at time index 1, the system requires the input at time index 2. Because time index 2 is in the future relative to time index 1, the system relies on look-ahead data, making it NON-CAUSAL.

(iii) Check Linearity for $ y(n) = 2x(n) - 3 $

Concept: The fastest test for linearity is the "Zero-Input" test. A perfectly linear system must pass through the origin (output zero when the input is zero).

Calculation: Set the input $ x(n) = 0 $.

$$y(n) = 2(0) - 3 = -3$$

Conclusion: Because $ y(n) = -3 $ and not $ 0 $, the system generates an output without an input. It fails the baseline zero-input property. The system is NON-LINEAR.

Part (i): $ x(t) = \cos(4t)+2\sin(8t) $

First, find the fundamental period of each individual sinusoid using the formula $ T = 2\pi / \omega $.

For the first term: $ \omega_1 = 4 \implies T_1 = 2\pi/4 = \pi/2 $.

For the second term: $ \omega_2 = 8 \implies T_2 = 2\pi/8 = \pi/4 $.

To be periodic, the ratio of their periods $ T_1/T_2 $ must evaluate to a rational number (an integer fraction).

Ratio: $ \frac{\pi/2}{\pi/4} = 2 $. Since 2 is a rational number, the combined signal is Periodic.

The overall Fundamental Period is the Least Common Multiple: $ T = \text{LCM}(T_1, T_2) = \pi/2 $.

Part (ii): $ x(t) = \cos(3\pi t)+2\cos(4\pi t) $

For the first term: $ \omega_1 = 3\pi \implies T_1 = 2\pi/3\pi = 2/3 $.

For the second term: $ \omega_2 = 4\pi \implies T_2 = 2\pi/4\pi = 1/2 $.

Ratio: $ \frac{2/3}{1/2} = \frac{4}{3} $. Since 4/3 is a rational fraction, the combined signal is Periodic.

Fundamental Period $ T = \text{LCM}(2/3, 1/2) $. The Least Common Multiple of fractions is calculated as LCM(numerators) / GCD(denominators).

LCM(2,1) / GCD(3,2) = 2 / 1 = 2 seconds.

Step 1: Map the Original Signal Indices

The sequence has 5 elements. The problem states $ X[0]=2 $, meaning the element '2' resides at index 0. Therefore, the indices map as follows:

$ x[-1]=1, \quad x[0]=2, \quad x[1]=3, \quad x[2]=4, \quad x[3]=5 $.

Part (i): $ y[n] = x[3n-1] $ (Time Scaling/Decimation & Shift)

Evaluate the new array index by substituting integer values for $ n $:

For $ n=0 $: $ y[0] = x[3(0)-1] = x[-1] = 1 $.

For $ n=1 $: $ y[1] = x[3(1)-1] = x[2] = 4 $.

Any other integer $ n $ (e.g., n=-1 gives x[-4], n=2 gives x[5]) results in indices that are outside the original array bounds, so they evaluate to 0.

Resulting Array: A compressed sequence $ \{1, 4\} $ located exactly at indices $ n=0 $ and $ n=1 $.

Part (ii): $ y[n] = x[-n+1] $ (Time Reversal & Shift)

Evaluate index by index to find the new bounds:

For $ n=-2 $: $ y[-2] = x[-(-2)+1] = x[3] = 5 $.

For $ n=-1 $: $ y[-1] = x[-(-1)+1] = x[2] = 4 $.

For $ n=0 $: $ y[0] = x[-(0)+1] = x[1] = 3 $.

For $ n=1 $: $ y[1] = x[-(1)+1] = x[0] = 2 $.

For $ n=2 $: $ y[2] = x[-(2)+1] = x[-1] = 1 $.

Resulting Array: A reversed and shifted sequence $ \{5, 4, 3, 2, 1\} $ spanning from indices $ n=-2 $ to $ n=2 $.

Step 1: Understand the Base Signal

The signal $ x(t) $ is a right-sided right triangle. It is 0 for $ t < 0 $, rises linearly to an amplitude of 1 at $ t=1 $, and drops immediately to 0. Equation: $ x(t) = t $ for $ 0 \le t \le 1 $.

The time-reversed signal $ x(-t) $ is a left-sided right triangle. It rises linearly from 0 at $ t=-1 $ to an amplitude of 1 at $ t=0 $. Equation: $ x(-t) = -t $ for $ -1 \le t \le 0 $.

Step 2: Even Part $ x_e(t) = (x(t) + x(-t))/2 $

We add the two triangles and divide the amplitude by 2.

Because the triangles do not overlap (one is strictly left of the Y-axis, one is strictly right), adding them creates a large, symmetric triangle centered at 0.

The resulting even signal is a symmetric triangle with a base from -1 to 1, with a peak amplitude of 0.5 exactly at $ t=\pm 1 $.

Step 3: Odd Part $ x_o(t) = (x(t) - x(-t))/2 $

We subtract the left triangle from the right triangle and divide by 2.

The right side ($ 0 < t \le 1 $) remains positive, peaking at an amplitude of 0.5.

The left side ($ -1 \le t < 0 $) is subtracted, meaning it flips upside down into the negative quadrant, peaking at an amplitude of -0.5.

The resulting odd signal looks mathematically like a positive slope crossing the origin, similar to an 'N' shape centered at the origin.

Unit 2: LTI Systems (Convolution)

High Priority Topics for 10-CGPA

Tabular Convolution: Mandatory 6-mark numerical. Never skip writing the bounds explicitly ($n_{start} = n_{x} + n_{h}$).
Continuous Convolution Integral: Highly likely to be asked with $ e^{-at}u(t) $. Master the integration bounds change to (0 to t).
Properties Proofs: Frequently asked to prove that Convolution of Odd and Even is Odd. Follow the dummy variable $ k $ substitution closely.

Complete Formula Sheet

Continuous Convolution Integral: $y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau$
Discrete Convolution Sum: $y[n] = \sum_{k=-\infty}^{\infty} x[k]h[n-k]$
Array Output Length: For discrete arrays, length $$ L = L_x + L_h - 1 $$ .
Cascaded Systems: Equivalent impulse response is the convolution: $h_{eq} = h_1 * h_2$ .

Exam Tips & Examiner Traps

The Integral Bounds Shift: The standard formula has limits from $-\infty$ to $\infty$. BUT, if the signals contain the unit step function $ u(t) $ (meaning they are causal and start at 0), the integration bounds immediately restrict to 0 to t. If you don't physically change the bounds on your paper, you will lose marks!
Tabular Matrix Check: Always mathematically verify your final array length using $ L_x + L_h - 1 $ before submitting your paper to ensure you didn't accidentally skip adding a diagonal.

Objective Questions

1 MarkSummer 2025

If $ h_1, h_2 $ and $ h_3 $ are cascaded, find the overall impulse response:

A) $ h_1 * h_2 * h_3 $

B) $ h_1 + h_2 + h_3 $

C) $ h_1 - h_2 \cdot h_3 $

D) $ h_1 - h_2 + h_3 $

Detailed Reason: In Linear Time-Invariant (LTI) systems, when subsystem blocks are placed in series (cascaded one after another), their individual impulse responses are mathematically convolved ($ * $) with each other. If they were placed in parallel, their outputs would simply be added (+).

1 MarkWinter 2024

The convolution of a rectangular pulse with itself is:

A) another rectangular pulse

B) square pulse

C) triangular pulse

D) sinc pulse

Detailed Reason: Visualizing the integral: As one rectangle slides horizontally over another identical stationary rectangle, the area of intersection (overlap) increases linearly from zero to a peak (when perfectly aligned). Then, as it continues sliding past, the overlap decreases linearly back to zero, creating a perfect triangular shape.

1 MarkWinter 2024

Unit step response of an LTI system with impulse response $ h(n) = \delta(n) - \delta(n-1) $ is:

A) $ \delta(n-1) $

B) $ \delta(n) $

C) $ u(n-1) $

D) $ u(n) $

Detailed Reason: The unit step response $ s(n) $ is mathematically defined as the running sum of the impulse response: $ s(n) = \sum_{k=-\infty}^n h(k) $. Summing the sequence $ \delta(n) - \delta(n-1) $ from 0 to n results in: at n=0, sum = 1; at n=1, sum = 1 - 1 = 0; for all $ n \ge 1 $, sum = 0. An array that is 1 at n=0 and 0 everywhere else is exactly the definition of $ \delta(n) $.

Descriptive Questions (100% Comprehensive)

Step 1: Determine Bounds and Output Length

Input $ x(n) $: Length $ L_x = 3 $. (Assuming it starts at index 0 based on standard DBATU notation).

Impulse $ h(n) $: Length $ L_h = 4 $. (Starts at index 0).

Output $ y(n) $: The total length will be $ L = L_x + L_h - 1 = 3 + 4 - 1 = 6 $.

The output array will map from index 0 to index 5.

Step 2: Construct the Multiplication Matrix

Place $ x(n) $ horizontally and $ h(n) $ vertically. Multiply the row headers by the column headers to fill the interior grid.

[ x(n) ] \to 123 ------------------------- 1 [h0] | 1 2 3 2 [h1] | 2 4 6 1 [h2] | 1 2 3 -1 [h3] | -1 -2 -3

Step 3: Add the Diagonals

Sum the numbers diagonally from the top-right to the bottom-left to generate each specific sequence value.

$ y(0) = \mathbf{1} $

$ y(1) = 2 + 2 = \mathbf{4} $

$ y(2) = 1 + 4 + 3 = \mathbf{8} $

$ y(3) = -1 + 2 + 6 = \mathbf{7} $

$ y(4) = -2 + 3 = \mathbf{1} $

$ y(5) = \mathbf{-3} $

Final Output Array: $ y(n) = \{1, 4, 8, 7, 1, -3\} $

Step 1: Bounds and Length

Input Length $ L_x = 4 $, Impulse Length $ L_h = 3 $. Output length = 4 + 3 - 1 = 6.

Step 2: Multiplication Matrix

Construct the grid and multiply terms:

* | 1 1 2 3 --|--------------- 1 | 1 1 2 3 1 | 1 1 2 3 1 | 1 1 2 3

Step 3: Add the Diagonals

Group the numbers diagonally (top-right to bottom-left):

$ y(0) = 1 $

$ y(1) = 1 + 1 = 2 $

$ y(2) = 2 + 1 + 1 = 4 $

$ y(3) = 3 + 2 + 1 = 6 $

$ y(4) = 3 + 2 = 5 $

$ y(5) = 3 $

Final Output Array: $ y(n) = \{1, 2, 4, 6, 5, 3\} $

Step 1: Setup the Convolution Integral

Formula:

y(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau

Applying the Limits: Because both functions contain the unit step $ u(t) $, they evaluate to zero for $ t < 0 $. Because of this causality, the integration limits restrict exactly to the overlapping active region from $ 0 $ to $ t $. If you integrate to infinity, you fail the question.

y(t) = \int_0^t (1) \cdot e^{-(t-\tau)} d\tau

Step 2: Solve the Integration

Since the integral is with respect to the dummy variable $ \tau $, we can pull the constant $ e^{-t} $ completely outside the integral to make it easier.

y(t) = e^{-t} \int_0^t e^{\tau} d\tau

The integral of $ e^\tau $ is simply $ e^\tau $. Applying the limits from 0 to t:

y(t) = e^{-t} [e^t - e^0]

Substitute $ e^0 = 1 $:

y(t) = e^{-t}(e^t - 1)

Step 3: Simplify

Multiply the term inside the bracket: $ (e^{-t} \cdot e^t) - (e^{-t} \cdot 1) = 1 - e^{-t} $.

Multiply the entire result by $ u(t) $ to mathematically indicate it remains a causal signal starting at t=0.

Final Output: $ y(t) = (1 - e^{-t})u(t) $

Visual Explanation Method

Let the rectangle signal $ x(t) $ exist from time $ -T $ to $ T $ with a constant amplitude of 1.

When a signal is convolved with itself, we mathematically flip one copy (time reversal) and slide it across the stationary copy from $ -\infty $ to $ \infty $. Since the rectangle is symmetric, the flip doesn't visually change it.

Phase 1: Approaching: The front edge of the sliding rectangle hits the back edge of the stationary rectangle at exactly $ t = -T - T = -2T $. The overlap begins, increasing linearly as it slides further in.

Phase 2: Full Overlap: The two rectangles perfectly align at exactly $ t = 0 $. The overlap area reaches its maximum possible value: Area = width × height = $ 2T \times 1 = 2T $.

Phase 3: Exiting: The trailing edge of the sliding rectangle leaves the front edge of the stationary rectangle at $ t = T + T = 2T $. The overlap area decreases linearly back down to zero.

Resulting Shape: The linear increase to a peak followed by a linear decrease creates a perfect Triangular pulse. The triangle starts at -2T, peaks at an amplitude of 2T (at t=0), and ends exactly at +2T.

Step 1: Formula Setup

y(n) = \sum_{k=-\infty}^{\infty} x(k)h(n-k)

Because both the input sequence and the impulse response sequence contain the step function $ u(n) $, the system and the signal are strictly causal. Therefore, the convolution summation limits mathematically collapse from $ -\infty \rightarrow \infty $ down to exactly $ 0 \rightarrow n $.

y(n) = \sum_{k=0}^{n} \beta^k \alpha^{n-k}

Step 2: Algebraic Manipulation

Separate the $\alpha$ terms using basic exponent rules: $ \alpha^{n-k} = \alpha^n \cdot \alpha^{-k} $.

Because $ \alpha^n $ does not depend on the summation variable $ k $, we can treat it as a constant and pull it outside the sum entirely.

y(n) = \alpha^n \sum_{k=0}^{n} \beta^k \alpha^{-k} = \alpha^n \sum_{k=0}^{n} \left(\frac{\beta}{\alpha}\right)^k

Step 3: Apply Geometric Series Formula

The standard mathematical formula for the sum of a finite geometric series is $ S = \frac{1 - r^{n+1}}{1 - r} $ where our common ratio is $ r = \frac{\beta}{\alpha} $.

y(n) = \alpha^n \left[ \frac{1 - (\beta/\alpha)^{n+1}}{1 - (\beta/\alpha)} \right]

Multiply the numerator and denominator by $ \alpha $ to simplify the complex fraction:

y(n) = \alpha^n \left[ \frac{\alpha - \beta(\beta/\alpha)^n}{\alpha - \beta} \right]

Final Output: $ y(n) = \frac{\alpha^{n+1} - \beta^{n+1}}{\alpha - \beta} u(n) $ (assuming $ \alpha \neq \beta $)

Proof Foundation

Let the convolution output be defined as $ y(t) = x(t) * h(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau $. We want to analyze $ y(-t) $ to check for symmetry.

y(-t) = \int_{-\infty}^{\infty} x(\tau)h(-t-\tau)d\tau

Let's perform a substitution. Let dummy variable $ k = -\tau $, which means $ d\tau = -dk $. As $ \tau $ goes from $ -\infty $ to $ \infty $, $ k $ goes from $ \infty $ to $ -\infty $. The negative sign from $ dk $ flips the integral limits back to normal.

y(-t) = \int_{-\infty}^{\infty} x(-k)h(-t+k)dk \quad \text{--- (Master Equation)}

Case 1: Odd * Even

Assume $ x $ is an Odd signal ($ x(-k) = -x(k) $) and $ h $ is an Even signal ($ h(-t+k) = h(t-k) $).

Substitute these specific properties into our Master Equation:

y(-t) = \int -x(k)h(t-k)dk = -\int x(k)h(t-k)dk = -y(t)

Since $ y(-t) = -y(t) $, the convolution of an Odd and Even signal is ODD.

Case 2: Even * Even

Assume $ x $ is an Even signal ($ x(-k) = x(k) $) and $ h $ is an Even signal ($ h(-t+k) = h(t-k) $).

Substitute these specific properties into our Master Equation:

y(-t) = \int x(k)h(t-k)dk = y(t)

Since $ y(-t) = y(t) $, the convolution of an Even and Even signal is EVEN.

Case 3: Odd * Odd (The Exam Anomaly)

Assume $ x $ is an Odd signal ($ x(-k) = -x(k) $) and $ h $ is an Odd signal ($ h(-t+k) = -h(t-k) $).

Substitute these specific properties into our Master Equation:

y(-t) = \int (-x(k))(-h(t-k))dk = \int x(k)h(t-k)dk = y(t)

Note for the Exam: The DBATU question paper specifically asked to prove Odd*Odd is Odd, which is mathematically false. Two negative signs multiply together to create a positive. State the mathematical proof clearly to show the examiner that Odd*Odd is actually EVEN.

Unit 3: Fourier Series

High Priority Topics for 10-CGPA

Dirichlet Conditions: Extremely common 6-mark theory question. Must know all 3 points perfectly.
Symmetry Rules Derivation: Deriving $ a_0, a_n, b_n $ for Even and Odd signals is a staple long-form question. The trick is to split the bounds.
Mathematical FS Calculation: Practice calculating the Trig FS for a Sawtooth or Square wave using Integration by parts.

Complete Formula Sheet

Fourier Series: $x(t) = a_0 + \sum_{n=1}^{\infty} [a_n \cos(n\omega_0 t) + b_n \sin(n\omega_0 t)]$
Coefficient $a_0$: DC offset. $a_0 = \frac{1}{T}\int x(t) dt$
Coefficient $a_n$: Cosine amplitudes. $a_n = \frac{2}{T}\int x(t)\cos(n\omega_0 t) dt$
Coefficient $b_n$: Sine amplitudes. $b_n = \frac{2}{T}\int x(t)\sin(n\omega_0 t) dt$

Exam Tips & Examiner Traps

The Symmetry Hack: Before calculating massive integrals, look at the graph. If the left side mirrors the right side exactly (Even symmetry), write down $ b_n = 0 $ immediately. If the right side is the inverted mirror of the left side (Odd symmetry), write $ a_0 = 0, a_n = 0 $ immediately. You will save 15 minutes of integration time.

Objective Questions

1 MarkWinter 2024

The Fourier series of a real, even periodic signal will contain only:

A) cosine terms

B) sine terms

C) even terms

D) odd harmonics

Detailed Reason: A cosine wave is a perfectly "even" mathematical function (it reflects perfectly symmetrically across the y-axis). A sine wave is mathematically "odd". Therefore, an even periodic signal can only be constructed using even building blocks (cosines). All sine coefficients ($ b_n $) evaluate to exactly zero.

1 MarkWinter 2024

If the Fourier series coefficients of a signal are periodic then the signal must be:

A) continuous-time, periodic

B) discrete-time, periodic

C) continuous-time, non-periodic

D) discrete-time, non-periodic

Detailed Reason: The Discrete Time Fourier Series (DTFS) has a unique mathematical property where its frequency spectrum coefficients are always periodic (repeating infinitely every $ 2\pi $ radians). Continuous-time signals, on the other hand, have aperiodic, infinite spectrums that do not loop.

Descriptive Questions (100% Comprehensive)

Introduction

The Dirichlet conditions are the three fundamental mathematical rules that a periodic signal $ x(t) $ must obey for its Fourier Series to converge (exist) successfully. If a signal fails these rules, it cannot be reliably represented as a sum of sines and cosines.

Condition 1: Absolute Integrability

The signal must be absolutely integrable over one complete time period $ T_0 $. This ensures that the total area under the curve is finite, which keeps the $ a_0 $ (DC component) from evaluating to infinity.

\int_{0}^{T_0} |x(t)| dt < \infty

Condition 2: Finite Number of Maxima and Minima

Within any single time period $ T_0 $, the signal must have a finite (countable) number of peaks (maxima) and valleys (minima). It cannot oscillate infinitely fast (like the theoretical mathematical function $ \sin(1/t) $ near zero).

Condition 3: Finite Number of Discontinuities

Within any single time period $ T_0 $, the signal can have jumps or breaks (discontinuities), but the total number of these breaks must be finite, and the size (amplitude) of each individual jump must also be a finite, countable number.

Step 1: Define the Base Property

An Even signal mathematically satisfies the condition: $ x(t) = x(-t) $.

When calculating integrals for symmetric bounds ($-T/2$ to $T/2$), the integral of an even function doubles (because left area = right area), while the integral of an odd function perfectly cancels out to exactly zero.

Step 2: Derive DC Component ($ a_0 $)

The base definition formula is:

a_0 = \frac{1}{T} \int_{-T/2}^{T/2} x(t) dt

Since $ x(t) $ is even, the area on the left of the y-axis equals the area on the right. We change the integration bounds to evaluate only the right side ($ 0 $ to $ T/2 $) and multiply the result by 2:

a_0 = \frac{2}{T} \int_{0}^{T/2} x(t) dt

Step 3: Derive Cosine Coefficients ($ a_n $)

The base definition formula is:

a_n = \frac{2}{T} \int_{-T/2}^{T/2} x(t)\cos(n\omega_0 t) dt

We know $ x(t) $ is Even. We know the cosine function is Even.

The mathematical product of [Even × Even] results in an Even function. Thus, we can apply the same boundary trick and double the integral:

a_n = \frac{4}{T} \int_{0}^{T/2} x(t)\cos(n\omega_0 t) dt

Step 4: Derive Sine Coefficients ($ b_n $)

The base definition formula is:

b_n = \frac{2}{T} \int_{-T/2}^{T/2} x(t)\sin(n\omega_0 t) dt

We know $ x(t) $ is Even. We know the sine function is Odd.

The mathematical product of [Even × Odd] results in an ODD function. The integral of an odd function over symmetric bounds is exactly zero.

$$b_n = 0$$

Final Result: The Fourier Series for an Even signal only contains $ a_0 $ and cosine harmonics ($ a_n $). All sine components $ b_n = 0 $.

Step 1: Identify Parameters

This describes a repeating linear sawtooth wave. The fundamental period is given as the full interval $ T = 2\pi $.

Therefore, the fundamental angular frequency is $ \omega_0 = \frac{2\pi}{T} = \frac{2\pi}{2\pi} = 1 $.

Step 2: Find DC Component ($ a_0 $)

a_0 = \frac{1}{2\pi} \int_0^{2\pi} \frac{10t}{2\pi} dt = \frac{10}{4\pi^2} \int_0^{2\pi} t dt

a_0 = \frac{10}{4\pi^2} \left[ \frac{t^2}{2} \right]_0^{2\pi} = \frac{10}{4\pi^2} \cdot \frac{4\pi^2}{2} = 5

Step 3: Find Cosine Coefficients ($ a_n $)

a_n = \frac{2}{2\pi} \int_0^{2\pi} \frac{10t}{2\pi} \cos(nt) dt = \frac{10}{2\pi^2} \int_0^{2\pi} t \cos(nt) dt

Using integration by parts: $ \int u dv = uv - \int v du $, where $ u = t, dv = \cos(nt)dt $. Because of the symmetry of the wave and the bounds, this entire integral evaluates to exactly 0 over a full period. $ a_n = 0 $.

Step 4: Find Sine Coefficients ($ b_n $)

b_n = \frac{2}{2\pi} \int_0^{2\pi} \frac{10t}{2\pi} \sin(nt) dt = \frac{10}{2\pi^2} \int_0^{2\pi} t \sin(nt) dt

Using integration by parts (DI method):

b_n = \frac{10}{2\pi^2} \left[ \frac{-t\cos(nt)}{n} + \frac{\sin(nt)}{n^2} \right]_0^{2\pi}

Plugging in the upper and lower boundaries (remembering $ \sin(2\pi n) = 0 $ and $ \cos(2\pi n) = 1 $):

b_n = \frac{10}{2\pi^2} \left( \frac{-2\pi(1)}{n} - 0 \right) = -\frac{10}{n\pi}

Final Series: $ x(t) = 5 - \sum_{n=1}^{\infty} \frac{10}{n\pi} \sin(nt) $

Step 1: Symmetry Check

The function is $ f(x) = x^3 $. Testing for mathematical symmetry: $ f(-x) = (-x)^3 = -x^3 = -f(x) $.

Because the negative sign pulls out perfectly, the function is perfectly ODD. Therefore, $ a_0 = 0 $ and all $ a_n = 0 $. The series will only contain sine terms ($ b_n $).

Step 2: Set up the integral for $ b_n $

The period is $ T = 2\pi $. The specific formula for $ b_n $ for an odd function bounded from $-\pi$ to $\pi$ is:

b_n = \frac{2}{\pi} \int_{0}^{\pi} x^3 \sin(nx) dx

Step 3: Integration by Parts

Applying tabular integration by parts on the product $ x^3 \sin(nx) $:

b_n = \frac{2}{\pi} \left[ x^3\left(\frac{-\cos nx}{n}\right) - 3x^2\left(\frac{-\sin nx}{n^2}\right) + 6x\left(\frac{\cos nx}{n^3}\right) - 6\left(\frac{\sin nx}{n^4}\right) \right]_0^\pi

Evaluate at bounds (noting that $ \sin(n\pi) = 0 $ and $ \cos(n\pi) = (-1)^n $):

b_n = \frac{2}{\pi} \left[ \frac{-\pi^3 (-1)^n}{n} + \frac{6\pi (-1)^n}{n^3} \right]

Factor out the common terms:

b_n = 2(-1)^n \left[ \frac{6}{n^3} - \frac{\pi^2}{n} \right]

Final Series: $ f(x) = \sum_{n=1}^{\infty} 2(-1)^n \left[ \frac{6}{n^3} - \frac{\pi^2}{n} \right] \sin(nx) $

Unit 4: Fourier Transforms & Sampling

High Priority Topics for 10-CGPA

Nyquist Rate Calculations: The most heavily tested concept. You MUST know how to mathematically expand squared sines and multiplied sines using trig identities to find $ f_{max} $.
FT Integrations: Practice the integration for $ e^{-at}u(t) $ and $ e^{at}u(-t) $.
DTFT Infinite Sums: Master the geometric progression formula $ S = \frac{1}{1-r} $ for discrete sequences like $ a^n u(n) $.

Complete Formula Sheet

Continuous-Time FT: $X(j\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t} dt$
Discrete-Time FT (DTFT): $X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x(n) e^{-j\omega n}$
Nyquist Rate: Sampling frequency must be strictly greater than or equal to twice the maximum frequency. $f_s \ge 2 f_{max}$
Nyquist Interval: The maximum time between samples. $T_s = \frac{1}{f_s} = \frac{1}{2 f_{max}}$

Exam Tips & Examiner Traps

The Non-Linear Nyquist Trap: If a signal is squared, like $\sin^2(100\pi t)$ , you CANNOT just take 100 and divide by 2. Squaring a signal doubles its frequency in the real world. You must mathematically expand it using trigonometric identities: $\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}$ before finding the true $f_{max}$ .
Multiplication Nyquist Trap: Multiplying two sines/cosines adds their frequencies together! $ \sin(\omega_1 t)\cos(\omega_2 t) $ creates a new maximum frequency bound of $ \omega_1 + \omega_2 $.

Objective Questions

1 MarkSummer 2025

Convolution in the time domain is equivalent to _______ in the frequency domain.

A) Division

B) Addition

C) Convolution

D) Multiplication

Detailed Reason: This is the fundamental Convolution Theorem. $ x(t) * h(t) \leftrightarrow X(j\omega) \cdot H(j\omega) $. This crucial property makes complex, time-consuming time-domain integral math easy by turning it into simple algebraic multiplication in the frequency domain.

1 MarkSummer 2025Supp 2022

The Nyquist rate for a signal with maximum frequency $ f_{max} = 10 \text{ kHz} $ is:

A) 5 kHz

B) 10 kHz

C) 15 kHz

D) 20 kHz

Detailed Reason: The Nyquist-Shannon sampling criterion states that to avoid aliasing (signal distortion during sampling), the sampling frequency $ f_s $ must be at least twice the highest frequency component present in the signal. $ 2 \times 10\text{kHz} = 20\text{kHz} $.

1 MarkSummer 2025

The DTFT of $ x[n]=\delta[n] $ is:

A) 1

B) 0

C) $ e^{-j\omega} $

D) None

Detailed Reason: The Discrete Time Fourier Transform of a unit impulse $ \delta[n] $ evaluates to exactly 1 across all frequencies. A sharp impulse in time contains all frequencies equally across the spectrum (which is why impulse testing is so useful for finding a system's full frequency response).

1 MarkWinter 2024

The Fourier transform of a signal $ x(t)=e^t u(-t) $ is given by:

A) $ \frac{1}{2-j\omega} $

B) $ \frac{1}{1-j\omega} $

C) $ \frac{1}{2j-\omega} $

D) $ \frac{2}{2j-\omega} $

Detailed Reason: Using the base FT integral $ \int_{-\infty}^{0} e^t e^{-j\omega t} dt = \int_{-\infty}^{0} e^{(1-j\omega)t} dt $. Evaluating this integral at the upper bound of 0 gives $ e^0 = 1 $, yielding $ \frac{1}{1-j\omega} $.

1 MarkWinter 2024

Fourier transform of a rectangular pulse is a:

A) rectangular pulse

B) triangular pulse

C) sinc pulse

D) impulse function

Detailed Reason: A rectangular window block in the time domain mathematically transforms into the $ \text{sinc}(x) = \frac{\sin(x)}{x} $ function in the frequency domain. This is one of the most fundamental pairs in signal processing.

1 MarkWinter 2024

Fourier transform of a real and even signal $ x(n) $ is:

A) real & even

B) real & odd

C) complex

D) imaginary

Detailed Reason: By the symmetry properties of the Fourier Transform, if a signal is purely real and perfectly even in the time domain, its spectrum will also be purely real and perfectly even in the frequency domain (the imaginary sine components completely cancel out).

Descriptive Questions (100% Comprehensive)

Part (a): Linear Sine Wave

Given: $ x(t) = \sin(200\pi t) $.

Calculation: Compare this to the standard format $ \sin(2\pi f_{max} t) $.

Equate the arguments: $ 2\pi f_{max} = 200\pi $.

Divide by $ 2\pi $: $ f_{max} = 100 $ Hz.

Conclusion: Using the Nyquist formula, $ f_s \ge 2 \cdot f_{max} $. Therefore, $ f_s = 2 \cdot 100 = \mathbf{200 \text{ Hz}} $.

Part (b): Squared Sine Wave (The Trap)

Given: $ x(t) = \sin^2(200\pi t) $.

Calculation: You cannot find the frequency of a squared waveform directly. Expand it using the trigonometric identity: $ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} $.

x(t) = \frac{1 - \cos(2 \cdot 200\pi t)}{2} = \frac{1}{2} - \frac{1}{2}\cos(400\pi t)

Now, compare the new oscillating term $ \cos(400\pi t) $ to $ \cos(2\pi f_{max} t) $.

Equate the arguments: $ 2\pi f_{max} = 400\pi \implies f_{max} = 200 $ Hz.

Conclusion: Using Nyquist, $ f_s = 2 \cdot f_{max} = 2 \cdot 200 = \mathbf{400 \text{ Hz}} $.

Part (c): Sine with DC Offset

Given: $ x(t) = 1+\sin(200\pi t) $.

The '1' is a constant DC component, which has a frequency of exactly 0 Hz. The sine component has a frequency of 100 Hz. The maximum frequency present in the overall signal is still 100 Hz.

Conclusion: $ f_s = 2 \cdot 100 = \mathbf{200 \text{ Hz}} $.

Part (d): Multiplied Trig Functions

Given: $ x(t) = \cos(150\pi t)\sin(100\pi t) $.

Use the standard identity $ \cos(A)\sin(B) = \frac{1}{2}[\sin(A+B) - \sin(A-B)] $.

x(t) = \frac{1}{2}[\sin(250\pi t) - \sin(50\pi t)]

The highest frequency component belongs to the $ 250\pi t $ term.

$ 2\pi f_{max} = 250\pi \implies f_{max} = 125 $ Hz.

Conclusion: $ f_s = 2 \cdot 125 = \mathbf{250 \text{ Hz}} $.

Step 1: Setup the Fourier Integral

Formula: $ X(j\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t} dt $

Applying Limits: The term $ u(-t) $ is a time-reversed step function. It is only active (equals 1) for negative time $ t < 0 $ and is completely zero for positive time. Therefore, the integration limits mathematically collapse to $ -\infty $ to $ 0 $.

X(j\omega) = \int_{-\infty}^{0} e^{at} e^{-j\omega t} dt

Step 2: Combine Exponents and Integrate

Combine the mathematical bases:

X(j\omega) = \int_{-\infty}^{0} e^{(a-j\omega)t} dt

Perform the integration with respect to $ t $:

X(j\omega) = \left[ \frac{e^{(a-j\omega)t}}{a-j\omega} \right]_{-\infty}^{0}

Step 3: Evaluate at Boundaries

Upper limit (t=0): $ e^{(a-j\omega)(0)} = e^0 = 1 $.

Lower limit (t=-$\infty$): Assuming $ a > 0 $, the term $ e^{a(-\infty)} $ drops completely to 0.

X(j\omega) = \frac{1 - 0}{a - j\omega} = \frac{1}{a - j\omega}

Step 4: Magnitude and Phase Extraction

Convert the complex fraction to standard magnitude and phase format.

Magnitude Spectrum: $ |X(j\omega)| = \frac{1}{\sqrt{a^2 + (-\omega)^2}} = \frac{1}{\sqrt{a^2 + \omega^2}} $

Phase Spectrum: $ \angle X(j\omega) = 0 - \tan^{-1}\left(\frac{-\omega}{a}\right) = \tan^{-1}\left(\frac{\omega}{a}\right) $

Final Transform: $ X(j\omega) = \frac{1}{a - j\omega} $

Part (i): $ x(n)=a^n u(n) $

The Discrete Time Fourier Transform (DTFT) standard formula is $ X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x(n) e^{-j\omega n} $.

Because of $ u(n) $, limits change from 0 to $ \infty $.

X(e^{j\omega}) = \sum_{n=0}^\infty a^n e^{-j\omega n} = \sum_{n=0}^\infty (ae^{-j\omega})^n

This is an infinite geometric series with ratio $ r = ae^{-j\omega} $. Since $ |a|<1 $, it converges. Using $ S = \frac{1}{1-r} $:

X(e^{j\omega}) = \frac{1}{1 - ae^{-j\omega}}

Part (ii): $ x(n)=a^n u(-n-1) $

Because of the term $ u(-n-1) $, the signal exists only for $ n \le -1 $.

X(e^{j\omega}) = \sum_{n=-\infty}^{-1} a^n e^{-j\omega n}

Substitute a dummy variable $ m = -n $ to change the limits to positive $ 1 $ to $ \infty $:

X(e^{j\omega}) = \sum_{m=1}^{\infty} a^{-m} e^{j\omega m} = \sum_{m=1}^{\infty} (a^{-1}e^{j\omega})^m

The geometric series starts at index 1 (not 0), so the sum formula is $ \frac{r}{1-r} $ where $ r = a^{-1}e^{j\omega} $.

X(e^{j\omega}) = \frac{a^{-1}e^{j\omega}}{1 - a^{-1}e^{j\omega}}

Multiply the numerator and denominator by $ a e^{-j\omega} $ to clean up the fraction:

X(e^{j\omega}) = \frac{1}{a e^{-j\omega} - 1} = \frac{-1}{1 - a e^{-j\omega}}

Step 1: Find Individual Transforms

Using the standard Fourier transform pair $ e^{-at}u(t) \leftrightarrow \frac{1}{a+j\omega} $:

Input FT: $ \mathcal{F}\{e^{-2t}u(t)\} \implies X(j\omega) = \frac{1}{2+j\omega} $.

Output FT: $ \mathcal{F}\{e^{-t}u(t)\} \implies Y(j\omega) = \frac{1}{1+j\omega} $.

Step 2: Find Frequency Response $ H(\omega) $

In the frequency domain, output = input × transfer function. So, $ H(j\omega) = \frac{Y(j\omega)}{X(j\omega)} $.

H(j\omega) = \frac{\frac{1}{1+j\omega}}{\frac{1}{2+j\omega}} = \frac{2+j\omega}{1+j\omega}

Step 3: Find Impulse Response $ h(t) $

To find the inverse Fourier transform, we must rewrite the improper fraction using algebraic manipulation. Add and subtract 1 in the numerator:

H(j\omega) = \frac{1 + 1 + j\omega}{1+j\omega} = \frac{1+j\omega}{1+j\omega} + \frac{1}{1+j\omega} = 1 + \frac{1}{1+j\omega}

Now take the inverse Fourier transform of each part separately. The inverse of a constant 1 is the delta function $ \delta(t) $.

$ h(t) = \delta(t) + e^{-t}u(t) $

Step 1: Signal Bounds

If $ x(0) = 4 $, the sequence indices are:

$ x(-2)=1, x(-1)=2, x(0)=4, x(1)=0, x(2)=1, x(3)=-2, x(4)=2 $.

Step 2: Evaluate $ X(e^{j0}) $

By definition, $ X(e^{j\omega}) = \sum x(n)e^{-j\omega n} $.

If $ \omega = 0 $, the exponential term becomes $ e^0 = 1 $. Therefore, $ X(e^{j0}) $ is simply the sum of all elements in the sequence.

X(e^{j0}) = \sum x(n) = 1 + 2 + 4 + 0 + 1 - 2 + 2 = 8

Step 3: Evaluate Integral

The inverse DTFT formula states that $ x(n) = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{j\omega})e^{j\omega n} d\omega $.

If we evaluate this at $ n=0 $, the exponential term becomes $ e^0 = 1 $:

x(0) = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{j\omega}) d\omega

Rearranging to solve for the integral:

\int_{-\pi}^{\pi} X(e^{j\omega}) d\omega = 2\pi \cdot x(0)

Since we know $ x(0) = 4 $:

\int_{-\pi}^{\pi} X(e^{j\omega}) d\omega = 2\pi \cdot 4 = 8\pi

Unit 5: Laplace & Z-Transforms

High Priority Topics for 10-CGPA

ROC Analysis (Partial Fractions): Absolutely mandatory 9-mark question. Must know how to build causal, anticausal, and non-causal signals from the exact same transfer function by checking the poles.
Inverse Z-Transform: Focus on utilizing shifting properties $ z^{-1} \implies u[n-1] $.

Complete Formula Sheet

Laplace Transform Eq: $X(s) = \int_{-\infty}^{\infty} x(t)e^{-st} dt$
Z-Transform Eq: $X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n}$
Time Multiplication (Laplace): $t x(t) \leftrightarrow -\frac{d}{ds}X(s)$ . If a signal is multiplied by 't', take the negative derivative of its s-domain transform.
Common Pair: $e^{-at}u(t) \leftrightarrow \frac{1}{s+a}$

Exam Tips & Examiner Traps

The ROC Exclusion Rule: The Region of Convergence (ROC) can NEVER contain a pole. When you find poles (e.g., -2 and -3), the ROC must either be greater than both (Causal), less than both (Anticausal), or between them (Non-causal). The exact same transfer function equation can produce three completely different time signals depending solely on the ROC bounds given.

Objective Questions

1 MarkWinter 2024

ROC of $ X(s) $ contains:

A) zeros

B) poles

C) no zero

D) no pole

Detailed Reason: A "pole" is a root value of the denominator that causes the mathematical transfer function to evaluate to infinity (meaning it fails to converge). Because the ROC (Region of Convergence) strictly defines the area where the function is finite, it can mathematically never encompass a pole.

1 MarkSummer 2025

The Laplace Transform is a generalization of:

A) Fourier Transform

B) Z-Transform

C) Convolution

D) None

Detailed Reason: The Laplace transform variable is $ s = \sigma + j\omega $. The Fourier transform variable is simply $ j\omega $. The Laplace transform is literally the Fourier transform but with an added real exponential convergence factor ($ \sigma $) to handle unstable signals that explode to infinity. If you set $ \sigma = 0 $, Laplace perfectly collapses into the Fourier transform.

1 MarkSummer 2025

What is the Laplace Transform of $ \delta(t) $?

A) 0

B) 1

C) $ s $

D) -1

Detailed Reason: By the integral definition $ \int \delta(t)e^{-st} dt $. Since the impulse $ \delta(t) $ only exists at exactly t=0, the integral simply evaluates the rest of the function at t=0. Thus, $ e^{-s(0)} = e^0 = 1 $.

1 MarkWinter 2024

Laplace transform converts convolution of time signals to:

A) addition

B) subtraction

C) multiplication

D) division

Detailed Reason: Just like the Fourier Transform, the Laplace Transform obeys the Convolution Theorem: $ x(t) * h(t) \leftrightarrow X(s) \cdot H(s) $. Complex time-domain integration becomes simple s-domain algebra.

1 MarkWinter 2024

The z-transform of $ \delta(n-m) $ is:

A) $ z $

B) $ z^{-m} $

C) $ z^m $

D) $ \frac{1}{z-m} $

Detailed Reason: The time-shifting property of the Z-transform states that delaying a signal by 'm' samples corresponds to multiplying its Z-transform by $ z^{-m} $. Since the ZT of $ \delta(n) = 1 $, the ZT of $ \delta(n-m) = 1 \cdot z^{-m} $.

1 MarkSummer 2025

The convolution of $ x(t) $ with $ \delta(t) $ gives:

A) $ \delta(t) $

B) $ x(t) $

C) $ x(t-1) $

D) $ x(0) $

Detailed Reason: The impulse function acts as the mathematical identity element under the operation of convolution. Convolving any signal with a delta function at the origin simply returns the original signal unchanged.

Descriptive Questions (100% Comprehensive)

Simplified Concept: The exact same $ H(s) $ equation can create three completely different time signals depending on the ROC. If ROC is > a pole, that part is a standard positive decaying signal. If ROC is < a pole, that part is a negative anti-causal signal.

Step 1: Partial Fraction Expansion

Given: $ H(s) = \frac{1}{s^2+5s+6} $.

Factor the denominator into its roots: $ s^2 + 5s + 6 = (s+2)(s+3) $.

H(s) = \frac{1}{(s+2)(s+3)} = \frac{A}{s+2} + \frac{B}{s+3}

Solve for A using the Cover-Up Method (cover $ s+2 $, set $ s = -2 $): $ A = \frac{1}{-2+3} = 1 $.

Solve for B using the Cover-Up Method (cover $ s+3 $, set $ s = -3 $): $ B = \frac{1}{-3+2} = -1 $.

H(s) = \frac{1}{s+2} - \frac{1}{s+3}

The system has two poles: $ p_1 = -2 $ and $ p_2 = -3 $.

Case 1: ROC is $ \text{Re}\{s\} > -2 $

The region of convergence is strictly to the right of the right-most pole (-2). Because it is greater than all poles, every component of the signal is right-sided (causal) and requires $ u(t) $.

h(t) = e^{-2t}u(t) - e^{-3t}u(t)

Case 2: ROC is $ \text{Re}\{s\} < -3 $

The region of convergence is strictly to the left of the left-most pole (-3). Because it is less than all poles, every component of the signal is left-sided (anticausal) and requires $ -u(-t) $.

h(t) = -(e^{-2t}u(-t)) - (-e^{-3t}u(-t))

h(t) = -e^{-2t}u(-t) + e^{-3t}u(-t)

Case 3: ROC is $ -3 < \text{Re}\{s\} < -2 $

The region of convergence lies exactly between the two poles.

- It is to the left of -2 (so the $ \frac{1}{s+2} $ component is anticausal: $ -u(-t) $).

- It is to the right of -3 (so the $ \frac{1}{s+3} $ component is causal: $ u(t) $).

h(t) = -e^{-2t}u(-t) - e^{-3t}u(t)

Step 1: Identify Base Transform

Ignore the 't' multiplier first. Focus on the standard base Laplace transform for the exponential function $ e^{-at}u(t) $, which is $ \frac{1}{s+a} $.

\mathcal{L}\{e^{-2t}u(t)\} = \frac{1}{s+2}

Step 2: Apply the Time Multiplication Property

The specific Laplace property states that multiplying a time-domain signal by $ t $ corresponds mathematically to taking the negative derivative of its Laplace transform with respect to $ s $.

\mathcal{L}\{t \cdot x(t)\} = -\frac{d}{ds}X(s)

Substitute our base transform into the property equation:

X(s) = -\frac{d}{ds} \left[ \frac{1}{s+2} \right]

Step 3: Perform Differentiation

Use the chain rule or power rule. The derivative of $ (s+2)^{-1} $ is $ -1 \cdot (s+2)^{-2} $.

X(s) = -\left( \frac{-1}{(s+2)^2} \right) = \frac{1}{(s+2)^2}

Final Answer: $ X(s) = \frac{1}{(s+2)^2} $, with ROC $ \text{Re}\{s\} > -2 $

Step 1: Separate the Transfer Function

Split the numerator to handle the terms individually using linearity:

X(z) = \frac{1}{1 - \frac{1}{3}z^{-1}} + \frac{\frac{1}{3}z^{-1}}{1 - \frac{1}{3}z^{-1}}

Step 2: Determine System Causality

The single pole is located at $ z = 1/3 $. Since the given ROC is $ |z| > 1/3 $, it is strictly exterior to the pole circle. This mathematically guarantees the system is completely right-sided (Causal), so all inverse transforms will be multiplied by $ u[n] $.

Step 3: Apply Inverse Transforms

First Term: The standard pair format $ \frac{1}{1-az^{-1}} \leftrightarrow a^n u(n) $.

\mathcal{Z}^{-1} \left\{ \frac{1}{1 - \frac{1}{3}z^{-1}} \right\} = \left(\frac{1}{3}\right)^n u[n]

Second Term: This is exactly the first term multiplied by $ \frac{1}{3}z^{-1} $. Multiplying by $ z^{-1} $ simply delays the signal by 1 unit in the time domain (Time Shifting property).

\mathcal{Z}^{-1} \left\{ \frac{1}{3}z^{-1} \cdot \frac{1}{1 - \frac{1}{3}z^{-1}} \right\} = \frac{1}{3} \cdot \left(\frac{1}{3}\right)^{n-1} u[n-1]

Final Output: $ x[n] = \left(\frac{1}{3}\right)^n u[n] + \left(\frac{1}{3}\right)^n u[n-1] $

Step 1: Identify Signal Boundaries

The sequence $ u(n+4) - u(n-5) $ creates a rectangular window. The signal "turns on" at $ n = -4 $ and "turns off" just before $ n = 5 $ (meaning the last active index is 4). Thus, the signal only exists for the finite interval: $ n = -4, -3, \dots, 3, 4 $.

Step 2: Setup Z-Transform Summation

Substitute the finite limits into the general Z-Transform definition:

X(z) = \sum_{n=-4}^{4} (0.5)^n z^{-n} = \sum_{n=-4}^{4} (0.5 z^{-1})^n

Step 3: Analyze the ROC

Because the summation is finite, we do not have to worry about infinite series convergence. However, we must mathematically check where the individual terms blow up to infinity.

- The summation contains positive powers of z (e.g., $ z^4 $ when n=-4). This blows up if $ z = \infty $.

- The summation contains negative powers of z (e.g., $ z^{-4} = \frac{1}{z^4} $ when n=4). This blows up if $ z = 0 $.

Therefore, the function converges everywhere on the complex plane except at exactly zero and exactly infinity.

Final Answer: The Z-Transform is the finite sum $ \sum_{n=-4}^{4} (0.5 z^{-1})^n $. The ROC is the entire Z-plane except $ z = 0 $ and $ z = \infty $.

Standard Transforms Breakdown

1) Unit Impulse $ \delta(n) $:

Formula: $ \sum \delta(n)z^{-n} $. Since it only exists at n=0, $ z^0 = 1 $.

ROC: Entire z-plane.

2) Delayed Impulse $ \delta(n-k) $:

Only exists at n=k. Substitute n=k into summation to get $ z^{-k} $.

ROC: Entire z-plane except $ z=0 $ (if k > 0).

3) Exponential $ a^n u(n) $:

Formula: $ \sum_{n=0}^\infty (az^{-1})^n = \frac{1}{1-az^{-1}} $.

ROC: $ |z| > |a| $.

4) Finite Window $ u(n)-u(n-10) $:

Finite sequence of 1s from n=0 to 9. Sum is $ \sum_{n=0}^9 z^{-n} = \frac{1 - z^{-10}}{1 - z^{-1}} $.

ROC: Entire z-plane except $ z=0 $.