Interpolation

Interpolation in digital signal processing refers to the process of increasing the sampling rate of a discrete-time signal by inserting new samples between existing ones. It is a fundamental operation in multirate DSP systems, used wherever signals must be converted to a higher sampling rate for further processing, digital-to-analog conversion, or transmission. Understanding interpolation requires grasping both the frequency-domain implications of upsampling and the role of the anti-imaging filter.

Core Concept: What Interpolation Does

When a discrete-time signal x[n] sampled at rate Fs is upsampled by an integer factor L, the first step is to create a new sequence v[m] where every L-th sample equals x[n] and the remaining L-1 samples between them are set to zero. This operation is called upsampling or zero-insertion. The output rate becomes L times Fs, but no new information has been created yet.

The problem with simple zero-insertion is visible in the frequency domain. The spectrum of the upsampled signal V(e^jw) contains L-1 replicas of the original spectrum, known as images, located at multiples of 2pi/L. These images are artifacts and must be eliminated. The anti-imaging lowpass filter removes these spectral copies while preserving the original baseband spectrum, completing the interpolation process.

The lowpass filter used in interpolation has a cutoff frequency of pi/L (in normalized frequency) and a passband gain equal to L. The gain of L compensates for the L-fold reduction in average signal amplitude caused by zero-insertion, ensuring the output amplitude matches the input signal level correctly.

Mathematical Expression

The upsampling operation is defined as:

v[m] = x[m/L] if m is a multiple of L, and v[m] = 0 otherwise

In the frequency domain, the upsampled signal has spectrum:

V(e^jw) = X(e^jwL)

This compresses the original spectrum along the frequency axis by L, causing it to repeat L times between 0 and 2pi. The anti-imaging lowpass filter with frequency response H(e^jw) = L for |w| less than pi/L and 0 elsewhere removes all images. The final interpolated output is:

Y(e^jw) = H(e^jw) . V(e^jw) = X(e^jwL) . H(e^jw)

This means the output spectrum equals the original spectrum scaled by L within the baseband and zero outside, which is exactly what a bandlimited interpolation should produce.

Practical Understanding

Interpolation is encountered whenever a signal must interface with a system running at a higher clock rate. A common example is in digital audio, where a signal recorded at 44.1 kHz must be converted to 192 kHz for a high-resolution audio interface. Rather than performing analog conversion at both rates, digital interpolation inserts new computed samples to bridge the rate difference entirely in the digital domain.

The quality of interpolation depends heavily on the quality of the anti-imaging lowpass filter. An ideal brick-wall filter is theoretically perfect but practically unrealizable. Real implementations use FIR filters with a very high order to approximate the ideal, often combined with polyphase filter structures for computational efficiency. In GATE and university exams, the ideal filter is assumed unless stated otherwise.

A critical insight for exams: the output of the upsampler alone is not interpolated. Only after the lowpass filter is applied does true interpolation occur. Many students confuse upsampling with interpolation, but upsampling is just one step inside the full interpolation chain.

Given:
Input signal x[n] sampled at Fs = 8 kHz
Interpolation factor L = 4
Anti-imaging LPF cutoff = pi/L (normalized)

Why this formula applies:
Interpolation increases sampling rate by integer L using upsampling + LPF.
LPF cutoff = pi/L removes spectral images at multiples of 2*pi/L.

Formula:
Output rate = L x Fs
LPF cutoff (normalized) = pi / L
LPF passband gain = L

Substitution:
Output rate = 4 x 8000
LPF cutoff = pi / 4 = 0.25*pi rad/sample
Passband gain = 4

Calculation:
Output sampling rate = 32,000 Hz = 32 kHz
Maximum signal frequency preserved = Fs/2 = 4 kHz (original bandwidth retained)
Images at 8, 16, 24 kHz are suppressed by LPF

Final Answer:
Output rate = 32 kHz
Anti-imaging LPF cutoff = 0.25*pi rad/sample
Filter gain in passband = 4 (to restore amplitude after zero-insertion)

Exam Tip: In GATE, the anti-imaging LPF cutoff for interpolation by L is always pi/L in normalized frequency, with passband gain equal to L. Do not confuse with decimation where the LPF cutoff is pi/M and gain is 1.

Mechanism Summary

• Upsampling inserts L-1 zero-valued samples between each original sample, increasing the sequence length and output rate by L.
• Zero-insertion causes L periodic spectral replicas (images) of the original spectrum to appear in the frequency domain, which are unwanted artifacts.
• The anti-imaging lowpass filter with normalized cutoff pi/L and passband gain L removes these images by zeroing all frequency content above pi/L.
• The gain of L in the filter compensates for the amplitude reduction caused by inserting L-1 zeros per original sample.
• The ideal interpolation result is equivalent to evaluating the sinc-interpolation formula to find the exact bandlimited value at each new sample position.

Quick Revision

• Interpolation increases sampling rate by integer L using upsampling (zero insertion) followed by an anti-imaging lowpass filter.
• LPF cutoff for interpolation = pi/L (normalized), passband gain = L.
• Upsampling alone is NOT interpolation. The LPF is mandatory to eliminate spectral images.
• Spectrum of upsampled signal: V(e^jw) = X(e^jwL), which compresses and repeats the spectrum L times.
• Output rate = L x input rate. Output bandwidth of signal remains the same as input.
• Exam trap: confusing interpolation (increase rate) with decimation (decrease rate). Decimation uses LPF with gain 1, not L.
• Real-world example: CD audio (44.1 kHz) interpolated to 176.4 kHz uses L = 4 with LPF at cutoff 22.05 kHz.