1.02 Elementary Operations on Signals

Focus & Objective

Continuous and discrete signals undergo mathematical transformations along their horizontal (independent variable) axis. Shifting, scaling, and reversal are heavily tested, both individually and in combination.

1. Core Operations on the Independent Variable (Time)

Manipulating the independent variable (time $t$ for continuous, or index $n$ for discrete) alters when and how fast a signal occurs without changing its amplitude.

1.1 Time Shifting

Time shifting delays or advances the signal along the horizontal axis. It does not change the shape or width of the signal.

Time Delay (Right Shift): Mathematically represented as $x (t - t_{0})$ where $t_{0} > 0$ . The signal occurs later.
Time Advance (Left Shift): Mathematically represented as $x (t + t_{0})$ where $t_{0} > 0$ . The signal occurs earlier.

left = -6; right = 6
top = 2; bottom = -0.5
---
y = 1 | -1 < x < 1 | BLUE | SOLID
y = 1 | 2 < x < 4 | RED | DASHED
y = 1 | -4 < x < -2 | GREEN | DASHED
(0, 1) | label:x(t) | BLUE
(3, 1) | label:x(t-3) [Delay] | RED
(-3, 1) | label:x(t+3) [Advance] | GREEN

self-question why right shift is called delay when its like in advance in x axis?

1.2 Time Scaling

Time scaling speeds up or slows down the signal by multiplying the time variable.

Compression: Mathematically represented as $x (a t)$ where $a > 1$ . The signal occurs faster, shrinking horizontally. (e.g., $x (2 t)$ is twice as fast, taking half the time).
Expansion: Mathematically represented as $x (a t)$ where $0 < a < 1$ . The signal occurs slower, stretching horizontally. (e.g., $x (0.5 t)$ takes twice as long).

Below is the original triangle $x (t)$ (blue), compressed $x (2 t)$ (red dashed), and expanded $x (t /2)$ (green dotted):

left = -5; right = 5
top = 1.5; bottom = -0.5
---
y = 1 - 0.5*abs(x) | -2 < x < 2 | BLUE | SOLID
y = 1 - abs(x) | -1 < x < 1 | RED | DASHED
y = 1 - 0.25*abs(x) | -4 < x < 4 | GREEN | DOTTED
(0, 1.1) | label:x(t) | BLUE
(0, 0.7) | label:x(2t) | RED
(0, 0.3) | label:x(t/2) | GREEN

1.3 Time Reversal (Folding)

Time reversal flips the signal across the vertical axis ( $t = 0$ ).

Mathematical Representation: $x (- t)$ .
Whatever happened at $t = t_{0}$ now happens at $t = - t_{0}$ . If a signal is symmetric about $t = 0$ , then $x (t) = x (- t)$ (an even signal), and reversal has no visible effect.

Below is an asymmetrical original signal $x (t)$ (blue) and its reversed version $x (- t)$ (red):

left = -3; right = 3
top = 1.5; bottom = -0.5
---
y = 0.5*x | 0 < x < 2 | BLUE | SOLID
y = -0.5*x | -2 < x < 0 | RED | SOLID
(1.5, 0.8) | label:x(t) | BLUE
(-1.5, 0.8) | label:x(-t) | RED

2. Processing of Discrete Signals

High-Yield PYQ

[PYQ 2019, 2018, 2017, 2015]: What is signal? Explain the operations performed on the independent variables in the processing of discrete signals mathematically and graphically. (12 Marks)

When answering for discrete signals $x [n]$ , the operations must use integer indices $n$ and sample-by-sample representations.

2.1 Discrete Time Shifting

Mathematical: $y [n] = x [n - k]$ . If $k > 0$ , the sequence is shifted right (delayed) by $k$ samples. If $k < 0$ , it is shifted left (advanced) by $∣ k ∣$ samples.
Graphical (ASCII Representation):

x[n]:
     3 |     o
     2 |   o |
     1 | o | |
-------+-+-+-+-+--> n
       0 1 2 3
 
y[n] = x[n-2] (Delayed by 2):
     3 |         o
     2 |       o |
     1 |     o | |
-------+-+-+-+-+-+--> n
       0 1 2 3 4

2.2 Discrete Time Reversal (Folding)

Mathematical: $y [n] = x [- n]$ . The sequence is mirrored around the origin $n = 0$ .
Graphical (ASCII Representation):

x[n]:
     2 |   o
     1 | o |
-------+-+-+-+--> n
       0 1 2
 
y[n] = x[-n] (Folded):
     2 | o
     1 | | o
-------+-+-+-+--> n
      -2 -1 0

2.3 Discrete Time Scaling

Unlike continuous time scaling, discrete time scaling is non-trivial because the independent variable must remain an integer.

Down-sampling (Decimation / Compression): $y [n] = x [M n]$ where $M$ is an integer $> 1$ . The system keeps only every $M$ -th sample and discards the rest. This results in data loss.
Up-sampling (Interpolation / Expansion): $y [n] = x [n / M]$ only for indices where $n / M$ is an integer. For all other values, $y [n] = 0$ . This inserts $M - 1$ zeros between every sample.

x[n] = {0, 1, 2, 3, 4} for n=0..4:
     4 |         o
     3 |       o |
     2 |     o | |
     1 |   o | | |
-------+-+-+-+-+-+--> n
       0 1 2 3 4
 
y[n] = x[2n] (Decimated by M=2; sample 1 and 3 are lost):
     4 |     o
     2 |   o |
-------+-+-+-+--> n
       0 1 2
 
y[n] = x[n/2] (Interpolated by M=2; zeros inserted):
     4 |                 o
     3 |             o   |
     2 |         o   |   |
     1 |     o   |   |   |
-------+-+-+-+-+-+-+-+-+-+--> n
       0 1 2 3 4 5 6 7 8

3. Combined Operations & Sketching

Order of Operations

To avoid boundary errors when sketching $x (a t - b)$ , always follow this sequence:

Shift First: Plot the delayed or advanced signal $x (t - b)$ .

Scale Second: Replace $t$ with $a t$ to compress or expand.

Reverse Last (if $a < 0$ ): Flip the entire scaled result across the vertical $y$ -axis.

Sketching PYQ 1

Question: Sketch the following signal: $x (t) = 4 r (3 t - 1)$ (2024, 2019)

Step-by-Step Solution:

Base Signal: The ramp function $r (t) = t \cdot u (t)$ , starting at origin with slope = 1.

Amplitude Scaling: $4 r (t)$ . Slope is now 4.

Time Shifting: $4 r (t - 1)$ . Shift the starting point right to $t = 1$ . The formula is $4 (t - 1)$ for $t \geq 1$ .

Time Scaling (Compression): $4 r (3 t - 1)$ .

Starting point: $3 t - 1 = 0 ⟹ t = 1/3$ .

The new slope: Amplitude factor $\times$ Compression factor $= 4 \times 3 = 12$ .

Mathematical formula: $x (t) = 12 t - 4$ for $t \geq 1/3$ .

left = -1; right = 2
top = 10; bottom = -2
---
y = 12*x - 4 | x > 1/3 | RED | SOLID
y = 0 | x <= 1/3 | RED | SOLID
(0.333, 0) | label:t=1/3 | RED
(1, 8) | label:Slope = 12 | RED

Sketching PYQ 2

Question: Sketch the following signal: $x (t) = r (- 0.5 t + 2)$ (2015)

Step-by-Step Solution:

Factor terms inside: $r (- 0.5 (t - 4))$ to prevent scaling-shifting confusion.

Base Signal: $r (t)$ , starting at 0, slope 1.

Time Reversal: $r (- t)$ . The ramp now goes to the left (exists for $t \leq 0$ ).

Time Scaling (Expansion): $r (- 0.5 t)$ . The ramp expands, slope becomes shallower ( $0.5$ ).

Time Shifting: Shift right by 4 units $⟹ r (- 0.5 (t - 4))$ .

Boundary: $- 0.5 t + 2 = 0 ⟹ t = 4$ .

Due to reversal, signal exists for $t \leq 4$ .

Value at $t = 4$ is $0$ .

Value at $t = 0$ is $r (2) = 2$ .

Value at $t = - 2$ is $r (3) = 3$ .

left = -2; right = 6
top = 4; bottom = -1
---
y = -0.5*x + 2 | x < 4 | GREEN | SOLID
y = 0 | x >= 4 | GREEN | SOLID
(4, 0) | label:t=4 | GREEN
(0, 2) | label:(0, 2) | GREEN

Sketching PYQ 3

Question: Sketch the signal: $x (t) = u (t - 5) - u (t - 7)$ (2021)

Step-by-Step Solution:

$u (t - 5)$ is a step function turning ON (+1) at $t = 5$ .

$- u (t - 7)$ is a negative step turning OFF (-1 subtraction) at $t = 7$ .

Summing them creates a rectangular window pulse of width 2 between $t = 5$ and $t = 7$ .

left = 3; right = 9
top = 2; bottom = -1
---
y = 1 | 5 < x < 7 | PURPLE | SOLID
y = 0 | x < 5 | PURPLE | SOLID
y = 0 | x > 7 | PURPLE | SOLID
(5, 0) | open | PURPLE
(5, 1) | open | PURPLE
(7, 0) | open | PURPLE
(7, 1) | open | PURPLE

007_index

1.02 Elementary Operations on Signals

1.02 Elementary Operations on Signals

1. Core Operations on the Independent Variable (Time)

1.1 Time Shifting

1.2 Time Scaling

1.3 Time Reversal (Folding)

2. Processing of Discrete Signals

2.1 Discrete Time Shifting

2.2 Discrete Time Reversal (Folding)

2.3 Discrete Time Scaling

3. Combined Operations & Sketching

Graph View

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