4.04 Multi-Rate Signal Processing Decimation and Interpolati

4.03 Ideal, Natural & Flat-Top Sampling Techniques | Course Roadmap

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4.04 Multi-Rate Signal Processing: Decimation & Interpolation

Core Idea

Multi-rate signal processing is the branch of digital signal processing that deals with systems whose sampling rates change at different stages. Because a discrete-time sequence index $n$ must strictly be an integer, discrete scaling is non-trivial. It is achieved through Down-sampling (Decimation), which decreases the sampling rate by discarding samples, and Up-sampling (Interpolation), which increases the sampling rate by inserting zeros.

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1. The Core Philosophy of Discrete-Time Scaling

In continuous-time systems, time scaling is mathematically straightforward: we replace the independent variable $t$ with $at$. For $a>1$, the signal is compressed; for $0<a<1$, the signal is expanded.

However, in discrete-time systems, this operation is highly restricted. The independent variable is an integer sample index $n$. We cannot define a value at a fractional index like $x[1.5]$. Thus:

• A direct compression like $y[n]=x[2n]$ is possible but causes permanent data loss because we must throw away all odd-indexed samples ($x[1],x[3],\dots$).
• A direct expansion like $y[n]=x[n/2]$ is undefined for odd values of $n$ (such as $y[1]=x[0.5]$).

To resolve these physical constraints, multi-rate digital systems use decimation and interpolation to scale discrete arrays without losing mathematical consistency.

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2. Discrete-Time Down-Sampler (Decimation by $M$)

The down-sampler (also known as a sampling rate compressor) reduces the sampling rate of a discrete signal by an integer factor $M$.

                +-------+
    x[n] ------>|  | M  |------> x_d[n] = x[nM]
                +-------+

2.1 Mathematical Formulation

A down-sampler by factor $M$ keeps only every $M$-th sample of the input sequence and discards the intermediate $M-1$ samples: $x_d[n]=x[nM]\text{for }n\in \mathbb{Z}$

If the sampling period of the input signal $x[n]$ is $T$ (sampling frequency $F_s=1/T$), the sampling period of the down-sampled signal $x_d[n]$ becomes: $T^{\prime }=MT⟹F_s^{\prime }=\frac{F_s}{M}$

2.2 Spectral Effects & Digital Aliasing

In the frequency domain, shrinking a signal in the time domain expands its spectrum. Specifically, down-sampling by factor $M$ stretches the DTFT spectrum by a factor of $M$: $X_d(e^{j\omega })=\frac{1}{M}{\sum }_{i=0}^{M-1}X\left(e^{j\frac{\omega -2\pi i}{M}}\right)$

This expansion can cause the shifted copies of the spectrum to overlap, resulting in digital aliasing.

The Decimation Filter Rule

To prevent destructive aliasing during decimation, we must place a digital low-pass Anti-Aliasing Filter (or Decimation Filter) before the down-sampling block. The filter must have a normalized cutoff frequency of: ${\omega }_c=\frac{\pi }{M}$ This strips away any high-frequency components that would otherwise fold back into the baseband.

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3. Discrete-Time Up-Sampler (Interpolation by $L$)

The up-sampler (also known as a sampling rate expander) increases the sampling rate of a discrete signal by an integer factor $L$.

                +-------+
    x[n] ------>|  ^ L  |------> x_e[n]
                +-------+

3.1 Mathematical Formulation

An up-sampler by factor $L$ inserts exactly $L-1$ zero-valued samples between each consecutive pair of original samples: $x_e[n]=\{\begin{array}{ll}x\left[\frac{n}{L}\right],& n=0,\pm L,\pm 2L,\dots \\ 0,& \text{otherwise}\end{array}$

If the original sampling period is $T$, the up-sampled sequence has a sampling period of: $T^{\prime }=\frac{T}{L}⟹F_s^{\prime }=L\cdot F_s$

3.2 Spectral Effects & Imaging Distortion

In the frequency domain, expanding a signal in the time domain compresses its spectrum. Up-sampling compresses the baseband spectrum $[-\pi ,\pi ]$ by a factor of $L$: $X_e(e^{j\omega })=X(e^{j\omega L})$

Because of this compression, $L-1$ unwanted duplicates of the original spectrum (called spectral images) appear within the new interval $[-\pi ,\pi ]$.

The Interpolation Filter Rule

The zero-inserted sequence $x_e[n]$ has a rough, staircase-like shape. To eliminate the high-frequency spectral images and interpolate the actual intermediate signal values, we must pass $x_e[n]$ through an Interpolation Filter (a digital low-pass filter) with:

1. Cutoff Frequency: ${\omega }_c=\frac{\pi }{L}$
2. Passband Gain: $L$ (to compensate for the amplitude drop caused by zero-insertion)

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4. Comprehensive Multi-Rate Comparison Table

The physical and mathematical characteristics of the two multi-rate operators are summarized below:

Feature	Down-sampling ($\downarrow M$)	Up-sampling ($\uparrow L$)
Mathematical Equation	$x_d[n]=x[nM]$	$x_e[n]=x[n/L]$ (for $n=kL$)
Time Domain Effect	Discards $M-1$ samples	Inserts $L-1$ zeros
Information Volume	Data loss occurs	Zero-padding (no new info yet)
New Sampling Period	$T^{\prime }=MT$ (slower)	$T^{\prime }=T/L$ (faster)
Spectral Effect	Spectrum expands by $M$	Spectrum compresses by $L$
Associated Distortion	Digital Aliasing	Spectral Imaging
Associated Filter	Anti-Aliasing Filter (before $\downarrow M$)	Anti-Imaging Filter (after $\uparrow L$)
Filter Cutoff & Gain	${\omega }_c=\frac{\pi }{M}$, $\text{Gain}=1$	${\omega }_c=\frac{\pi }{L}$, $\text{Gain}=L$

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5. Comprehensive Worked Examples (The Exam Killers)

5.1 Example 1: The 10-Mark 2024 KUET Exam Classic [PYQ 2024]

Question: Find out the up-sampling sequence $x_u[n]$ and down-sampling sequence $x_d[n]$ from the input signal sequence $x[n]=[1,6,11,14]$ for a factor of 3.

Step-by-Step Solution:

Scenario A: Default Origin Assumption (Leftmost Element is $n=0$)

Under the standard convention, if the origin is not explicitly marked with an arrow ($\uparrow$) or underline ($\underline{}$), the first element is located at the origin $n=0$: $x[0]=1,x[1]=6,x[2]=11,x[3]=14$ All other indices are zero: $x[n]=0$ for $n<0$ or $n\ge 4$.

1. Down-sampling by Factor $M=3$: The formula is: $x_d[n]=x[3n]$ Let’s map individual indices $n$:

• $n=0⟹x_d[0]=x[0]=1$
• $n=1⟹x_d[1]=x[3]=14$
• $n=2⟹x_d[2]=x[6]=0$
• $n=-1⟹x_d[-1]=x[-3]=0$

Thus, the down-sampled sequence is: ${\mathbf{x}}_{\mathbf{d}}[\mathbf{n}]=[\underline{1},14]$

2. Up-sampling by Factor $L=3$: The up-sampler inserts $L-1=2$ zeros between each consecutive sample of the original sequence:

• $x_u[0]=x[0]=1$
• $x_u[1]=0$
• $x_u[2]=0$
• $x_u[3]=x[1]=6$
• $x_u[4]=0$
• $x_u[5]=0$
• $x_u[6]=x[2]=11$
• $x_u[7]=0$
• $x_u[8]=0$
• $x_u[9]=x[3]=14$

Thus, the up-sampled sequence is: ${\mathbf{x}}_{\mathbf{u}}[\mathbf{n}]=[\underline{1},0,0,6,0,0,11,0,0,14]$

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Scenario B: Alternative Origin Assumption (Origin at $n=0$ is marked, e.g., at value $11$)

If the examiner defines a different origin during the exam, say at the third sample ($11$): $x[-2]=1,x[-1]=6,x[\underline{0}]=11,x[1]=14$

1. Down-sampling by Factor $M=3$: $x_d[n]=x[3n]$

• $n=0⟹x_d[0]=x[0]=11$
• $n=1⟹x_d[1]=x[3]=0$
• $n=-1⟹x_d[-1]=x[-3]=0$

Thus, the down-sampled sequence collapses to: ${\mathbf{x}}_{\mathbf{d}}[\mathbf{n}]=[\underline{11}]$

2. Up-sampling by Factor $L=3$: Insert 2 zeros between each consecutive element, maintaining the origin at value $11$: ${\mathbf{x}}_{\mathbf{u}}[\mathbf{n}]=[1,0,0,6,0,0,\underline{11},0,0,14]$

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5.2 Example 2: The 2022 Exam Sequence [PYQ 2022]

Question: If $x[n]=[2,6,\underline{0},3,1,2]$, find the compressed (down-sampled) and expanded (up-sampled) sequences for a factor of 3.

Step-by-Step Solution:

1. Extract the Index Map relative to the underlined origin: $x[-2]=2,x[-1]=6,x[\underline{0}]=0,x[1]=3,x[2]=1,x[3]=2$

2. Calculate Down-sampling ($M=3$): $x_d[n]=x[3n]$

   • $n=0⟹x_d[0]=x[0]=0$
   • $n=1⟹x_d[1]=x[3]=2$
   • $n=-1⟹x_d[-1]=x[-3]=0$ (out of bounds)

   Therefore, the down-sampled sequence is: ${\mathbf{x}}_{\mathbf{d}}[\mathbf{n}]=[\underline{0},2]$

3. Calculate Up-sampling ($L=3$): Insert $L-1=2$ zeros between each consecutive pair of elements, keeping the origin aligned at the original sample $x[0]=0$: ${\mathbf{x}}_{\mathbf{e}}[\mathbf{n}]=[2,0,0,6,0,0,\underline{0},0,0,3,0,0,1,0,0,2]$

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6. Common Mistakes That Cost Marks

The Origin Alignment Trap

Do not always assume that the first element in a bracket is the origin! Always check the array for an arrow ($\uparrow$) or underline ($\underline{}$). If you down-sample without mapping the indices relative to $n=0$ first, your entire output array will shift, resulting in a zero-mark evaluation on the numerical.

The Anti-Imaging Filter Gain Oversight

When asked to draw the block diagram of an interpolator, students often forget that up-sampling scales down the signal amplitude. The matching digital low-pass reconstruction filter must contain a scaling gain factor of $L$ to restore the signal’s original physical voltage range.

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7. PYQ Bank — Verbatim Questions & Answer Plans

7.1 PYQ 2024 Question 7c [10 Marks]

Question: Find out the up-sampling sequence $x_u(n)$ and down-sampling sequence $x_d(n)$ from the input signal sequence $x(n)=[1,6,11,14]$ for the discrete-time system under a factor of 3.

• Answer Plan:
  1. State the mathematical down-sampling and up-sampling equations.
  2. Write out the step-by-step coordinate substitutions showing how index values map under both Scenario A (default leftmost origin) and Scenario B (alternative marked origin).
  3. Provide the final bolded output arrays clearly labeling the $n=0$ coordinate.

7.2 PYQ 2021 [5 Marks]

Question: Define (i) Discrete time up-sampler; (ii) Discrete time down-sampler.

• Answer Plan:
  1. Provide the formal time-domain equations for both operators.
  2. Draw the standard block diagrams ($\uparrow L$ and $\downarrow M$).
  3. Explain the physical effect of both blocks in the time domain (zero-insertion vs. sample compression).

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8. Self-Check Before Moving On

• Can you define down-sampling and up-sampling mathematically and graphically?
• Do you know why an anti-aliasing filter is required before a down-sampler, and what its cutoff frequency must be?
• Can you explain why up-sampling creates spectral images, and how the interpolation filter removes them?
• Can you reliably execute index mappings on discrete arrays with non-zero time origins?

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Source: (k.Deergha Rao) signals and systems.pdf (Section 6.6), 1.05 The Sampling Theorem Bridge & Multi-Rate Processing.md, pyq catagorised (Table 7).