FIR filter structure in Signal Processing

Introduction

An FIR filter helps clean or change signals by mixing past input values with fixed weights.

To remove noise from a recorded sound or sensor data.

To smooth out sudden jumps in temperature readings.

To shape the frequency of a music signal for better sound.

To delay a signal without changing its shape.

To create simple digital effects in audio processing.

Syntax

Signal Processing

y[n] = b0*x[n] + b1*x[n-1] + b2*x[n-2] + ... + bM*x[n-M]

y[n] is the output at time n.

x[n] is the input at time n, and b0, b1, ..., bM are fixed filter weights called coefficients.

Examples

This is a simple 5-point moving average filter that smooths the input signal.

Signal Processing

y[n] = 0.2*x[n] + 0.2*x[n-1] + 0.2*x[n-2] + 0.2*x[n-3] + 0.2*x[n-4]

A 3-point FIR filter with different weights to emphasize recent inputs more.

Signal Processing

y[n] = 0.5*x[n] + 0.3*x[n-1] + 0.2*x[n-2]

Sample Program

This code creates a noisy sine wave and smooths it using a 5-point FIR filter (moving average). The plot shows how the filter cleans the noise.

Signal Processing

import numpy as np
import matplotlib.pyplot as plt

# Input signal: noisy sine wave
fs = 100  # samples per second
T = 1    # seconds
x = np.linspace(0, T, fs*T, endpoint=False)
signal = np.sin(2 * np.pi * 5 * x) + 0.5 * np.random.randn(len(x))

# FIR filter coefficients (simple moving average of length 5)
coeffs = np.ones(5) / 5

# Apply FIR filter using convolution
filtered_signal = np.convolve(signal, coeffs, mode='same')

# Plot input and filtered signals
plt.figure(figsize=(8,4))
plt.plot(x, signal, label='Noisy input')
plt.plot(x, filtered_signal, label='Filtered output', linewidth=2)
plt.xlabel('Time [s]')
plt.ylabel('Amplitude')
plt.title('FIR Filter Structure Example')
plt.legend()
plt.tight_layout()
plt.show()

Important Notes

The FIR filter output depends only on current and past input values, never on past outputs.

FIR filters are always stable and have a linear phase, meaning they do not distort signal timing.

The length of the filter (number of coefficients) controls how much smoothing or shaping happens.

Summary

An FIR filter mixes past input values with fixed weights to change a signal.

It is useful for noise removal, smoothing, and shaping signals.

The filter output is a weighted sum of current and past inputs only.

Practice

1. What does an FIR filter primarily use to process a signal?

easy

A. Random noise to modify the signal

B. Past output values and feedback loops

C. A fixed set of past input values and weights

D. Only the current input value without memory

Solution

1. Step 1: Understand FIR filter basics

   An FIR filter uses a fixed number of past input values combined with fixed weights (coefficients) to produce the output.

2. Step 2: Compare options with FIR definition

   A fixed set of past input values and weights correctly describes this process. Options B and C describe other filter types or incorrect concepts. Only the current input value without memory ignores the memory aspect.

3. Final Answer:

   A fixed set of past input values and weights -> Option C

4. Quick Check:

   FIR = past inputs + fixed weights [OK]

Hint: FIR filters use past inputs, not past outputs [OK]

Common Mistakes:

• Confusing FIR with IIR filters that use feedback
• Thinking FIR uses only current input
• Assuming FIR filters add noise intentionally

2. Which of the following is the correct way to represent the output y[n] of an FIR filter with coefficients h[k] and input x[n]?

easy

A. y[n] = sum(x[k] * y[n-k]) for k=0 to N-1

B. y[n] = sum(h[k] * y[n-k]) for k=0 to N-1

C. y[n] = h[n] * x[n]

D. y[n] = sum(h[k] * x[n-k]) for k=0 to N-1

Solution

1. Step 1: Recall FIR filter formula

   The output y[n] is the sum of products of filter coefficients h[k] and past inputs x[n-k] from k=0 to N-1.

2. Step 2: Check each option

   y[n] = sum(h[k] * x[n-k]) for k=0 to N-1 matches the FIR formula. y[n] = sum(h[k] * y[n-k]) for k=0 to N-1 uses past outputs, which is IIR. y[n] = h[n] * x[n] is just a single multiplication, not a sum. y[n] = sum(x[k] * y[n-k]) for k=0 to N-1 mixes inputs and outputs incorrectly.

3. Final Answer:

   y[n] = sum(h[k] * x[n-k]) for k=0 to N-1 -> Option D

4. Quick Check:

   FIR output = weighted sum of past inputs [OK]

Hint: FIR sums weighted past inputs, not outputs [OK]

Common Mistakes:

• Using past outputs instead of inputs
• Forgetting to sum over all coefficients
• Confusing multiplication with convolution

3. Given the FIR filter coefficients h = [0.2, 0.5, 0.3] and input signal x = [1, 2, 3, 4], what is the output y[2] (third output value)?

medium

A. 2.3

B. 1.9

C. 3.0

D. 1.5

Solution

1. Step 1: Write the formula for y[2]

   y[2] = h[0]*x[2] + h[1]*x[1] + h[2]*x[0] = 0.2*3 + 0.5*2 + 0.3*1

2. Step 2: Perform the calculation

   0.2*3 = 0.6, 0.5*2 = 1.0, 0.3*1 = 0.3; sum = 1.9

3. Final Answer:

   1.9 -> Option B

4. Quick Check:

   0.2*3 + 0.5*2 + 0.3*1 = 1.9 [OK]

Hint: Multiply and sum coefficients with past inputs [OK]

Common Mistakes:

• Mixing coefficient order
• Wrong indexing of input signal
• Ignoring zero padding for early outputs

4. Identify the error in this FIR filter implementation snippet:

h = [0.1, 0.2, 0.3]
x = [1, 2, 3, 4]
y = []
for n in range(len(x)):
    s = 0
    for k in range(len(h)):
        s += h[k] * x[n+k]
    y.append(s)

medium

A. Indexing x with n+k causes out-of-range error

B. Filter coefficients h should be reversed

C. Output y should be initialized with zeros

D. The outer loop should run over len(h) instead of len(x)

Solution

1. Step 1: Analyze indexing in inner loop

   The code uses x[n+k], but when n is near the end of x, n+k can exceed x's length causing an error.

2. Step 2: Confirm correct indexing for FIR

   FIR uses x[n-k], not x[n+k]. Using n+k moves forward, causing out-of-range and wrong filtering.

3. Final Answer:

   Indexing x with n+k causes out-of-range error -> Option A

4. Quick Check:

   Index x with n-k, not n+k [OK]

Hint: Use x[n-k] indexing to avoid errors [OK]

Common Mistakes:

• Using forward indexing n+k instead of backward n-k
• Not handling boundary conditions
• Confusing coefficient order

5. You want to design an FIR filter to smooth a noisy signal by averaging the last 4 input samples equally. Which coefficient vector h should you use?

hard

A. [0.25, 0.25, 0.25, 0.25]

B. [1, 0, 0, 0]

C. [0.1, 0.2, 0.3, 0.4]

D. [0.4, 0.3, 0.2, 0.1]

Solution

1. Step 1: Understand smoothing by averaging

   Smoothing by averaging means each of the last 4 inputs contributes equally to the output.

2. Step 2: Choose equal coefficients summing to 1

   Equal weights for 4 samples means each coefficient is 1/4 = 0.25, so h = [0.25, 0.25, 0.25, 0.25].

3. Final Answer:

   [0.25, 0.25, 0.25, 0.25] -> Option A

4. Quick Check:

   Equal weights sum to 1 for smoothing [OK]

Hint: Equal weights summing to 1 smooth signals [OK]

Common Mistakes:

• Using unequal weights for averaging
• Not normalizing coefficients to sum 1
• Choosing coefficients that emphasize recent inputs only