Discretization｜IO爱㛅

1. Formula
   Observation

   The top of the first chart on the left side of the screen is labeled $x(n)$, with n on the horizontal axis and $x(n)$ on the vertical axis.

$x(n)$

The original discrete-time signal sequence, whose independent variable is the integer n.

n is an integer.

1. Audio
   Observation

   The instructor says, "Let us interpret this n as n prime."

2. Formula
   Observation

   Handwritten on screen: $n = 2n$' and $n = n$'/2.

n'

An auxiliary variable used to solve for the independent variable of the new sequence; it represents the index of the transformed sequence.

n′ is an integer.

1. Formula
   Observation

   Inside the red box at the bottom of the screen: "i.e., $\delta$(an) = $\delta(n)$".

$\delta(n)$

The unit sample sequence (discrete-time impulse function).

n is an integer.

1. Formula
   Observation

   Label the horizontal axis n; the sequence notations $x(n)$, $x(2n)$, and $x(n/2)$ all contain n

n

discrete-time independent variable (integer index)

integer

1. Formula
   Observation

   The leftmost plot labels the vertical axis $x(n)$ and provides several sample values

$x(n)$

original discrete sequence

1. Formula
   Observation

   The middle plot labels the vertical axis $x(2n)$

$x(2n)$

compressed discrete sequence

1. Formula
   Observation

   The rightmost plot labels the vertical axis $x(n/2)$

$x(n/2)$

expanded discrete sequence

1. Formula
   Observation

   Note that the box states δ(an) = $δ(n)$

$δ(n)$

unit sample sequence (discrete impulse)

1. Audio
   Observation

   Here we change it to n prime

2. Formula
   Observation

   Handwritten: n' = -2, -1, 0, 1

n'

New variable used to index the original sequence in the extended sequence

1. Formula
   Observation

   Parameter a in δ(an) = $δ(n)$

a

Scaling factor (constant)

1. Formula
   Observation

   The horizontal axis is labeled n, representing the independent variable of a discrete-time sequence.

n

The independent variable of a discrete-time sequence (integer index).

The set of integers.

1. Formula
   Observation

   The vertical axis is labeled $x(n)$, representing the function values of a discrete-time sequence.

$x(n)$

The value of the discrete-time sequence at n.

The set of real numbers.

1. Audio
   Observation

   The instructor says, "Today we will analyze scaling transformations of discrete-time signals in Signals and Systems."

2. Formula
   Observation

   The screen title reads "(3) Scaling Transformation", and displays graphs of $x(2n)$ and $x(n/2)$."

Scaling of discrete-time signals includes compression (e.g., $x(2n)$) and expansion (e.g., $x(n/2)$). Unlike continuous-time signals, the independent variable of a discrete-time signal must be an integer; thus, sample loss or zero-valued samples may occur during scaling.

1. The independent variable n must be an integer

1. Formula
   Observation

   A red box at the bottom of the screen reads: "Note: For the unit sample sequence, scaling does not change the amplitude; that is, $\delta$(an) = $\delta(n)$."

When performing scaling, the amplitude of the unit sample sequence delta n remains unchanged, that is, delta a n equals delta n. This means that no matter how much compression or expansion is applied, the location and height of the impulse (at n equals0 is1) are unaffected.

1. a is a nonzero integer

1. Scaling of discrete-time signals

1. Formula
   Observation

   Title "(3) Scaling" and three sequence plots: $x(n)$, $x(2n)$, $x(n/2)$

2. Audio
   Observation

   Explanation focusing on the sampling rules for compression and expansion

Scaling the discrete sequence $x(n)$ yields x(an). Compression occurs when |a| > 1; expansion occurs when 0 < |a| < 1. In the discrete case, the independent variable must take integer values; non-integer points are discarded.

1. a is a nonzero constant

2. n is an integer

1. Audio
   Observation

   Moving it here corresponds to taking only even indices.

2. Formula
   Observation

   The graph of $x(2n)$ retains only the three points at $n = -1$, 0, and 1.

When computing $x(2n)$, the original sequence index corresponding to the new index n is 2n; thus, only samples at even positions of the original sequence are taken, equivalent to the 'even-index sampling' process.

1. a equals 2

2. n is an integer

1. Scaling of discrete sequences

1. Audio
   Observation

   Here division is applied, so expansion must involve multiplication.

2. Formula
   Observation

   Handwritten: n' = -2, -1, 0, 1 and corresponding 2n' = -4, -2, 0, 2

When computing $x(n/2)$, let n' = $n/2$, so $n = 2n$'. Each integer n in the new sequence corresponds to the original sequence index n' = $n/2$; n' is an integer only when n is even, so nonzero samples occur only at even positions.

1. a equals one half

2. n' is an integer when n is even

1. Scaling of discrete sequences

1. Formula
   Observation

   Title (3): Scaling Transformation; Handwritten Notes: $x(2n)$, $x(n/2)$

2. Audio
   Observation

   The instructor says, 'After expansion, the signal takes values at points negative four, negative two, zero, two, four, and six.'

3. Diagram
   Observation

   The right-hand graph shows the waveform of $x(n/2)$, where nonzero values occur at even integers and odd integers yield zero.

Scaling transformation of discrete-time signals includes compression (e.g., $x(2n)$) and expansion (e.g., $x(n/2)$). For the expansion operation $x(n/2)$, when n is odd, $n/2$ is not an integer, so the corresponding sequence value is zero; graphically, this appears as zero-valued points inserted between the original sequence's nonzero values.

1. a is a positive integer

2. n is an integer

1. Formula
   Observation

   Bottom text box: Note: Under scaling transformation, the unit sample sequence retains unchanged amplitude; that is, $\delta$(an) = $\delta(n)$

When performing scaling of a discrete-time signal, if the original signal is the unit sample sequence (that is, it equals 0 only at $n = 1$ and equals 0 elsewhere), then no matter what integer scaling factor is applied (for example, x of a times n), its amplitude remains unchanged, that is, delta of a times n equals delta of n.

1. a is a nonzero integer

1. Scaling transformation of discrete-time signals

1. Formula
   Observation

   Bottom formula: (4) Sequence difference A: Forward difference $\Delta f(n) = f(n+1) - f(n)$

The forward difference is a fundamental operation in discrete signal processing, defined as the value of the sequence at $n + 1$ minus its value at n. It is commonly used to convert differential equations into difference equations.

1. $f(n)$ is a discrete sequence

1. Formula
   Observation

   Note the box: 'The amplitude remains unchanged under scaling of the unit sample sequence, i.e., δ(an) = $δ(n)$'

When scaling the unit sample sequence $δ(n)$, its amplitude remains unchanged, satisfying δ(an) = $δ(n)$.

1. $δ(n)$ is the unit sample sequence

2. a is a nonzero constant

for all nonzero constants a and integers n

1. Audio
   Observation

   The instructor says, "For compression, we examine 2n... Let us interpret this n as n'... Here, the original n equals 2n'."

2. Formula
   Observation

   The screen shows handwritten $n = 2n$', followed below by the calculation process of dividing -2, -1, 0, 1, 2, and 3 by 2

1. Expression

   $$n = 2n'$$
   Explanation

   Define the relationship between the original sequence's independent variable n and the new sequence's independent variable n'.

   Justification

   To compute $x(2n)$, we need to determine the values of 2n' for which the original sequence is defined.

   Shown in the video
2. Expression

   $$n' = n / 2$$
   Explanation

   Rearrange the relation to express the new index n' in terms of the original index n.

   Justification

   Algebraic manipulation.

   Shown in the video
3. Expression

   $$-2/2, -1/2, 0/2, 1/2, 2/2, 3/2$$
   Explanation

   Substitute the known indices of the original sequence to compute the new index values.

   Justification

   Compute using n' = $n / 2$.

   Shown in the video
4. ExpressionKeep the integer indices
   Explanation

   In discrete-time signals, a sample exists only when the computed n' is an integer; otherwise, no sample exists at that point (or it is treated as 0, depending on the specific definition—but here we emphasize finding integer points).

   Justification

   The independent variable of a discrete-time signal must be an integer.

   Derived from the video

By setting $n = 2n$' and solving for integer n', we obtain the positions of nonzero samples of the compressed sequence $x(2n)$.

1. Audio
   Observation

   The instructor says, 'Also equals n' over two for this expansion.'

2. Formula
   Observation

   The handwritten note on screen reads $n = n$'/2.

1. The instructor does not elaborate on the specific computational steps for expansion, only giving the relation.

1. Expression

   $$n = n' / 2$$
   Explanation

   Define the relationship between the original sequence's independent variable n and the new sequence's independent variable n'.

   Justification

   To compute $x(n/2)$, we need to know the values of n'/2 for which the original sequence is defined.

   Shown in the video
2. Expression

   $$n' = 2n$$
   Explanation

   Rearrange the relation to express the new index n' in terms of the original index n.

   Justification

   Algebraic manipulation.

   Derived from the video
3. ExpressionThe new indices are even
   Explanation

   Since n is an integer, n' = 2n must be even. This means the expanded sequence retains the original sequence’s values only at even positions, while odd positions are typically zero-padded.

   Justification

   Property of discrete-signal expansion.

   Derived from the video

By setting $n = n$'/2 (i.e., n' = 2n), we see that the nonzero samples of the expanded sequence occur at indices twice those of the original samples.

1. Audio
   Observation

   We discard all non-integer points… only the integer points negative one, zero, and one are valid.

2. Formula
   Observation

   x of 2n. In the graph, n equals negative 1,0,1 corresponding values 2,0,2

1. Expression

   $$n' = 2n$$
   Explanation

   Map the compressed new index n back to the original sequence index n'

   Justification

   Definition of scaling: $x(2n)$

   Shown in the video
2. Expression

   $$n \in \{-1, 0, 1\}$$
   Explanation

   Within the original sequence support [-2, 3], the only integers n for which 2n remains an integer within that range are -1, 0, 1

   Justification

   Discrete sequences take values only at integer points

   Derived from the video
3. Expression

   $$x(2\cdot(-1)) = x(-2) = 2$$
   Explanation

   At $n = -1$, the corresponding original sequence index is n' = -2, whose value is 2

   Justification

   Original sequence diagram data

   Shown in the video
4. Expression

   $$x(2\cdot 0) = x(0) = 0$$
   Explanation

   When $n = 0$, the corresponding original sequence index is n' = 0, and the value found is 0

   Justification

   Original sequence diagram data

   Shown in the video
5. Expression

   $$x(2\cdot 1) = x(2) = 2$$
   Explanation

   When $n = 1$, the corresponding original sequence index is n' = 2, and the value found is 2

   Justification

   Original sequence diagram data

   Shown in the video

$x(2n)$ is nonzero only at $n = -1$, 0, 1, with respective values 2, 0, 2.

1. Audio
   Observation

   “Here we divide, so over here expansion must involve multiplication.”

2. Formula
   Observation

   Handwritten: n' = -2, -1, 0, 1 and 2n' = -4, -2, 0, 2

1. The segment ends after listing the mappings, without fully showing the evaluation process of $x(n/2)$ at each point

1. Expression

   $$n' = n/2 \Rightarrow n = 2n'$$
   Explanation

   Introduce a new variable n' to represent the original sequence index and establish the relationship between the new and old indices

   Justification

   Definition of scaling transformation: $x(n/2)$

   Shown in the video
2. Expression

   $$n' \in \{-2, -1, 0, 1\}$$
   Explanation

   To ensure that $n = 2n$' falls within the visible range of the expanded sequence, select these values of n'.

   Justification

   Original sequence support and expansion factor

   Derived from the video
3. Expression

   $$n = 2n' \in \{-4, -2, 0, 2\}$$
   Explanation

   Compute the corresponding new sequence integer positions

   Justification

   Substitute $n = 2n$'

   Shown in the video

The nonzero samples of the expanded sequence $x(n/2)$ can only occur at even positions such as $n = -4$, -2, 0, 2.

1. Audio
   Observation

   The instructor step by step computes the values of $x(n/2)$ at $n = -4$, -2, 0, 2, 4, and 6, and points out that the values at odd indices are zero

2. Diagram
   Observation

   The right-hand graph dynamically illustrates the plotting process of $x(n/2)$, with yellow circles highlighting the newly inserted zeros

1. Expression

   $$n = -4 \Rightarrow x(-4/2) = x(-2) = 2$$
   Explanation

   When $n = -4$, substituting into $x(n/2)$ yields $x(-2)$, whose value read from the original graph is 2

   Justification

   definition of scaling transformation

   Shown in the video
2. Expression

   $$n = -2 \Rightarrow x(-2/2) = x(-1) = 1$$
   Explanation

   When $n = -2$, substitution yields $x(-1)$, whose value from the original graph is 1

   Justification

   definition of scaling transformation

   Shown in the video
3. Expression

   $$n = 0 \Rightarrow x(0/2) = x(0) = 0$$
   Explanation

   When $n = 0$, substitution yields $x(0)$, whose value from the original graph is 0

   Justification

   definition of scaling transformation

   Shown in the video
4. Expression

   $$n = 2 \Rightarrow x(2/2) = x(1) = 1$$
   Explanation

   When n equals 2, substituting yields $x(1)$, and the original graph value is 1

   Justification

   Definition of scaling transformation

   Shown in the video
5. Expression

   $$n = 4 \Rightarrow x(4/2) = x(2) = 2$$
   Explanation

   When n equals 4, substituting yields $x(2)$, and the original graph value is 2

   Justification

   Definition of scaling transformation

   Shown in the video
6. Expression

   $$n = 6 \Rightarrow x(6/2) = x(3) = 3$$
   Explanation

   When n equals 6, substituting yields $x(3)$, and the original graph value is 3

   Justification

   Definition of scaling transformation

   Shown in the video
7. Expression

   $$n \text{is odd} \Rightarrow x(n/2) = 0$$
   Explanation

   Because a discrete sequence is defined only at integer points, when n is odd, n over 2 is not an integer, so its value is 0

   Justification

   Domain restriction of discrete signals

   Shown in the video

The expanded sequence x(n over 2) retains the original sequence’s values at even indices and inserts zero values at odd indices.

1. Formula
   Observation

   Three side-by-side sequence plots and handwritten calculations

Given a discrete sequence $x(n)$, find the effective sample points and their values for the compressed sequence $x(2n)$ and the expanded sequence $x(n/2)$.

1. The original sequence $x(n)$ is defined at $n = -2$, -1, 0, 1, 2, 3.

2. Illustrated sample values: $x(-2) = 2$, $x(-1) = 1$, $x(0) = 0$, $x(1) = 1$, $x(2) = 2$, $x(3) = 1$

3. Scaling rule: values are taken only at integer points

Determine the valid integer indices and corresponding amplitudes for $x(2n)$ and $x(n/2)$

1. Expression

   $$x(2n): n \in \{-1,0,1\}$$
   Explanation

   During compression, retain only those n for which 2n is an integer index of the original sequence.

   Justification

   Integer-point sampling principle for discrete sequences.

   Derived from the video
2. Expression

   $$x(2n)|_{n=-1}=2,\; x(2n)|_{n=0}=0,\; x(2n)|_{n=1}=2$$
   Explanation

   Obtain the compressed sequence's amplitudes by looking up the values of the original sequence at even-numbered points.

   Justification

   Graphical representation of the original sequence's data.

   Shown in the video
3. Expression

   $$x(n/2): n=2n',\; n' \in \{-2,-1,0,1\}$$
   Explanation

   During expansion, map the new indices back to the original indices via $n = 2n$'.

   Justification

   Definition of scaling transformation.

   Shown in the video
4. Expression

   $$n \in \{-4,-2,0,2\}$$
   Explanation

   Compute the integer positions that may exist in the expanded sequence.

   Justification

   Substitute $n = 2n$'.

   Derived from the video

The valid points of $x(2n)$ occur at $n = -1$, 0, 1, with corresponding values 2, 0, 2; the nonzero samples of $x(n/2)$ appear only at even positions $n = -4$, -2, 0, 2 (exact values require further lookup in the original sequence; the segment shown is incomplete).

The three labeled points on the graph of $x(2n)$ in the video match the handwritten calculation; the horizontal axis of the $x(n/2)$ graph uses even-numbered spacing, consistent with the derivation.

1. Diagram
   Observation

   The left plot shows the values of the original sequence $x(n)$: x of negative 2 equals 2, x of negative 1 equals 1, x of 0 equals 0, x of 1 equals 1, x of 2 equals 2, and x of 3 equals 3

2. Diagram
   Observation

   The right plot shows the transformed sequence x(n over 2)

Given the waveform of the discrete sequence $x(n)$ shown in the figure, find the waveform of $x(n/2)$.

1. The value of x at negative two equals two.

2. The value of x at negative one equals one.

3. The value of x at zero equals zero.

4. The value of x at one equals one.

5. The value of x at two equals two.

6. The value of x at three equals three.

Plot and determine the values of $x(n/2)$ at each integer point.

1. Expression

   $$x(-4/2) = x(-2) = 2$$
   Explanation

   Compute the value when n equals negative 4.

   Justification

   Substitution method.

   Shown in the video
2. Expression

   $$x(-3/2) = 0$$
   Explanation

   n equals negative 3 is odd, yielding a non-integer index, so the value is 0.

   Justification

   Definition of a discrete signal.

   Shown in the video
3. Expression

   $$x(-2/2) = x(-1) = 1$$
   Explanation

   Compute the value when $n = -2$

   Justification

   Substitution method

   Shown in the video
4. Expression

   $$x(-1/2) = 0$$
   Explanation

   $n = -1$ is odd, and the value is 0

   Justification

   Definition of a discrete signal

   Shown in the video
5. Expression

   $$x(0/2) = x(0) = 0$$
   Explanation

   Compute the value when $n = 0$

   Justification

   Substitution method

   Shown in the video
6. Expression

   $$x(1/2) = 0$$
   Explanation

   $n = 1$ is odd, and the value is 0

   Justification

   Definition of a discrete signal

   Shown in the video
7. Expression

   $$x(2/2) = x(1) = 1$$
   Explanation

   Compute the value when $n = 2$

   Justification

   Substitution method

   Shown in the video
8. Expression

   $$x(3/2) = 0$$
   Explanation

   $n = 3$ is odd, and the value is 0

   Justification

   Definition of a discrete signal

   Shown in the video
9. Expression

   $$x(4/2) = x(2) = 2$$
   Explanation

   Compute the value at $n = 4$

   Justification

   Substitution method

   Shown in the video
10. Expression

    $$x(5/2) = 0$$
    Explanation

    $n = 5$ is odd, and the value is 0

    Justification

    Definition of a discrete-time signal

    Shown in the video
11. Expression

    $$x(6/2) = x(3) = 3$$
    Explanation

    Compute the value at $n = 6$

    Justification

    Substitution method

    Shown in the video

The sequence values of $x(n/2)$ are: ..., 0, 2, 0, 1, 0, 0, 0, 1, 0, 2, 0, 3, ... (original values retained at even indices; values at odd indices are 0)

Exactly matches the waveform of $x(n/2)$ plotted in the right-hand graph

1. Diagram
   Observation

   The screen shows three discrete-signal waveforms side by side: $x(n)$, $x(2n)$, and $x(n/2)$.

1. Original signal $x(n)$.

2. Compressed signal $x(2n)$.

3. Expanded signal $x(n/2)$.

1. In the $x(2n)$ plot, sample points become sparser, retaining only those from even positions of the original plot.

2. The sample points in the graph of $x(n/2)$ become denser; the original sample positions are stretched, and zero-valued samples are inserted in between.

1. The vertical-axis amplitude remains unchanged at the corresponding sample points.

2. The unit sample sequence $\delta(n)$ remains $\delta(n)$ after transformation.

It intuitively illustrates how downsampling (compression) and upsampling (zero insertion) of discrete sequences affect sample positions and sample count.

1. Diagram
   Observation

   The video displays side-by-side stem plots of three discrete sequences: $x(n)$, $x(2n)$, and $x(n/2)$.

1. Original sequence plot $x(n)$

2. Compressed sequence plot $x(2n)$

3. Expanded sequence plot $x(n/2)$

4. Handwritten annotations and calculations

1. The instructor highlights the three valid points of $x(2n)$ with a yellow circle

2. Label the corresponding original index values on the $x(n)$ plot

3. Blue arrows and the text "take even indices" added in post-production

4. Write the mapping table between n' and 2n'

1. All three plots share the same original sequence data

2. The horizontal axis represents integer indices n

3. The vertical axis represents sequence magnitude

Side-by-side diagrams intuitively illustrate how index compression and expansion affect sample positions and counts in scaling transformations.

1. Animation
   Observation

   Handwritten red text "Discrete signals take values only at integer points", appearing character by character with an underline

1. Red handwritten text

2. Underline

1. Text appears character by character from nothing

2. Underline added last for emphasis

1. Text content always states the discrete signal value rule

Highlighting the essential difference between discrete sequences and continuous signals: defined only at integer times; non-integer points are discarded outright.

1. Animation
   Observation

   When explaining $x(n/2)$, yellow circles highlight the zero values newly inserted at odd indices

2. Diagram
   Observation

   The label 'zero insertion' appears below the right-hand graph, with blue arrows indicating it

1. Waveform plot of $x(n/2)$

2. yellow circle

3. blue arrow

1. odd-indexed points are highlighted as zero values

2. the waveform is stretched horizontally

1. the values at even-indexed points match those of the original sequence at corresponding points

intuitively illustrates the mathematical process of inserting zero values at non-integer index positions during discrete signal expansion transformation ($x(n/a)$).

1. Audio
   Observation

   The instructor, while computing $n/2$, remarks, 'We find there are both integers and fractions, right?', then writes 'discrete signal' to emphasize this point.

When handling $x(2n)$ or $x(n/2)$, directly drawing smooth curves or retaining points with fractional indices, as one would for continuous signals.

The independent variable n of a discrete signal must be an integer. After scale transformation yields new indices, non-integer indices must be discarded, and only samples corresponding to integer indices retained.

1. Audio
   Observation

   We discard all points that are not at integer points.

When performing scaling transformations on discrete sequences, attempting to retain or interpolate samples at non-integer indices.

A discrete sequence is defined only at integer points; after scaling, if a new index corresponds to a non-integer original index, that point does not exist and must be discarded.

1. Formula
   Observation

   Note that the box explicitly states the amplitude remains unchanged.

By analogy with the factor 1/|a| in continuous-time signal scaling, one assumes the amplitude of the discrete-time unit sample sequence δ(an) also scales.

The amplitude of the discrete-time unit sample sequence remains unchanged under scaling; it always holds that δ(an) = $δ(n)$.

1. Formula
   Observation

   the bottom text box emphasizes: "Note: the amplitude of the unit sample sequence remains unchanged under scaling transformation."

students may analogize from the scaling transformation of continuous signals (e.g., $\delta$(at) = $\frac{1}{|a|}\delta(t)$) and assume that for discrete signals, the amplitude also scales proportionally under the x(an) transformation.

for the discrete unit sample sequence $\delta(n)$, regardless of the nonzero integer value of a, $\delta$(an) always equals $\delta(n)$, with amplitude constantly equal to 1 and never scaled.

1. Formula
   Observation

   The caution inside the red box immediately follows the scale transformation heading and its accompanying diagram.

The scaling property of the unit sample sequence is a special case and important supplement to the general scaling rules for discrete signals.

1. Formula
   Observation

   $x(2n)$ is the specific form of x(an) when $a = 2$.

The compressed sequence $x(2n)$ is a special case of the discrete-time scaling x(an) with scaling factor $a = 2$.

1. Formula
   Observation

   $x(n/2)$ is the specific form of x(an) when $a = 1/2$

The expanded sequence $x(n/2)$ is a special case of the discrete scaling transformation x(an) with scaling factor $a = 1/2$.

1. Formula
   Observation

   Note the box treats δ(an) = $δ(n)$ as a special remark on scaling transformation.

The unit sample sequence’s amplitude-invariance property is a result of applying discrete scaling transformation theory to a special sequence.

1. Formula
   Observation

   the bottom text box presents the property of the unit sample sequence as a caution for scaling transformations

the scaling transformation property of the unit sample sequence is a special and important case within the general scaling transformation rules for discrete signals, requiring particular attention to its distinction from continuous signals.

1. Audio
   Observation

   The instructor demonstrates how to find the new indices by setting $n = 2n$'.

1. Derivation for sequence compression $x(2n)$
2. Derivation for the expanded sequence $x(n/2)$.

1. Formula
   Observation

   Text in red box on screen

1. Scaling property of the unit sample sequence

1. Audio
   Observation

   We discard all non-integer points.

1. Scaling of discrete sequences
2. Mistakenly assuming interpolation or retention is possible at non-integer points

1. Audio
   Observation

   It is equivalent to an even-number extraction process.

1. Even-index sampling property of the compressed sequence $x(2n)$

1. Audio
   Observation

   Here we divide, so over here expansion must involve multiplication.

1. Index mapping for the expanded sequence $x(n/2)$

1. Formula
   Observation

   The discrete-time impulse function of an equals the discrete-time impulse function of n.

1. Amplitude remains unchanged under scaling of the unit sample sequence
2. Mistakenly believing that scaling changes the impulse amplitude.

1. Audio
   Observation

   The instructor demonstrates in detail the evaluation process for $x(n/2)$.

1. Scaling transformation of discrete-time signals
2. derivation of values for discrete signal expansion transformation
3. Example of discrete signal scaling expansion

1. Formula
   Observation

   The text box at the bottom explicitly states $\delta$(an) = $\delta(n)$.

1. Scaling transformation property of the unit sample sequence
2. mistakenly believing that scaling transformation alters the signal amplitude