Frequency Domain Analysis

Introduction

Frequency domain analysis is one of the most powerful perspectives in signals and systems engineering. Instead of asking "what happens to a signal over time?", we ask "what frequencies are present and how does the system treat each frequency?" This shift in viewpoint transforms complex convolution integrals into simple multiplications and reveals information about signal bandwidth, filter behavior, and system stability that would be nearly impossible to see in the time domain alone.

For B.Tech students, frequency domain thinking is the key skill that separates textbook knowledge from engineering intuition. When you look at a system and immediately think about its frequency response — which frequencies it passes, which it blocks, and how much phase shift it introduces — you are thinking like a practicing engineer.

The Frequency Response of LTI Systems

For an LTI system with impulse response $h(t)$, the frequency response is:

$$H(j\omega) = \int_{-\infty}^{\infty} h(t) e^{-j\omega t} dt$$

This is simply the Fourier transform of the impulse response. When a sinusoidal input $x(t) = Ae^{j\omega_0 t}$ enters the system, the output is:

$$y(t) = A \cdot H(j\omega_0) \cdot e^{j\omega_0 t}$$

The system scales the amplitude by $|H(j\omega_0)|$ and shifts the phase by $\angle H(j\omega_0)$, but does NOT change the frequency. This is the eigensignal property that makes frequency analysis so powerful.

Magnitude and Phase Response

The frequency response is a complex function that decomposes into:

$$H(j\omega) = |H(j\omega)| \cdot e^{j\phi(\omega)}$$

Magnitude response $|H(j\omega)|$: gain applied to each frequency. Values > 1 amplify, values < 1 attenuate.

Phase response $\phi(\omega) = \angle H(j\omega)$: phase shift applied to each frequency.

Group delay: $\tau_g(\omega) = -\frac{d\phi}{d\omega}$ — the time delay experienced by a narrow band of frequencies centered at $\omega$. For distortion-free transmission, group delay must be constant (linear phase).

Input-Output in Frequency Domain

The convolution theorem states:

$$y(t) = x(t) * h(t) \quad \Leftrightarrow \quad Y(j\omega) = X(j\omega) \cdot H(j\omega)$$

This transforms the complex convolution integral into simple multiplication! To find the output spectrum:

1. Compute (or look up) the input spectrum $X(j\omega)$
2. Compute (or look up) the system frequency response $H(j\omega)$
3. Multiply: $Y(j\omega) = X(j\omega) \cdot H(j\omega)$
4. If needed, inverse transform to find $y(t)$

Bandwidth and Filtering

Signal Bandwidth

The bandwidth of a signal is the range of frequencies it occupies. For a baseband signal:

$$B = f_{max} - 0 = f_{max} \text{ Hz}$$

Common bandwidth definitions:

• 3-dB bandwidth: frequencies where power drops to half (-3 dB from peak)
• Null-to-null bandwidth: between first spectral zeros
• 99% energy bandwidth: contains 99% of signal energy

Filter Bandwidth

A filter's bandwidth is its passband width:

• Low-pass filter: $B_{LP} = f_c$ (cutoff frequency)
• Band-pass filter: $B_{BP} = f_2 - f_1$ (upper minus lower cutoff)
• Quality factor: $Q = f_0/B_{BP}$ (center frequency divided by bandwidth) — higher Q means narrower, more selective filter

Frequency Response from Transfer Function

For systems described by rational transfer functions:

$$H(s) = \frac{N(s)}{D(s)} = \frac{b_m s^m + \cdots + b_0}{a_n s^n + \cdots + a_0}$$

The frequency response is obtained by substituting $s = j\omega$:

$$H(j\omega) = \frac{N(j\omega)}{D(j\omega)}$$

Example: RC low-pass filter with $H(s) = \frac{1}{1+sRC}$:

$$H(j\omega) = \frac{1}{1+j\omega RC}$$

$$|H(j\omega)| = \frac{1}{\sqrt{1+(\omega RC)^2}}, \quad \phi(\omega) = -\arctan(\omega RC)$$

At $\omega = 1/RC$ (cutoff): $|H| = 1/\sqrt{2}$ (-3 dB) and $\phi = -45°$.

Bode Plot Representation

Bode plots display frequency response on logarithmic axes:

• Magnitude: $20\log_{10}|H(j\omega)|$ in dB vs. $\log(\omega)$
• Phase: $\angle H(j\omega)$ in degrees vs. $\log(\omega)$

Advantages of Bode plots:

• Products of transfer functions become sums in dB (cascade systems)
• Standard slopes: first-order pole/zero gives ±20 dB/decade, second-order gives ±40 dB/decade
• Asymptotic approximations allow quick sketching

Distortion-Free Transmission

A system transmits a signal without distortion if:

$$H(j\omega) = Ke^{-j\omega t_0}$$

This means:

• Constant magnitude: $|H(j\omega)| = K$ (all frequencies amplified equally)
• Linear phase: $\phi(\omega) = -\omega t_0$ (constant time delay for all frequencies)

In practice, achieving this exactly is impossible over infinite bandwidth, but it can be approximated over the signal's bandwidth.

Worked Example

Problem: An input signal $x(t) = 3\cos(100t) + 2\cos(500t)$ passes through a system with $H(j\omega) = \frac{10}{10+j\omega}$. Find the output.

Solution:

At $\omega = 100$: $H(j100) = \frac{10}{10+j100} = \frac{10}{\sqrt{10100}}e^{-j\arctan(10)} \approx 0.0995e^{-j84.3°}$

At $\omega = 500$: $H(j500) = \frac{10}{10+j500} = \frac{10}{\sqrt{250100}}e^{-j\arctan(50)} \approx 0.02e^{-j88.9°}$

Output: $y(t) \approx 0.299\cos(100t - 84.3°) + 0.04\cos(500t - 88.9°)$

The system is low-pass — it significantly attenuates the 500 rad/s component more than the 100 rad/s component.

Key Takeaways

• Frequency response $H(j\omega)$ completely characterizes how an LTI system modifies each frequency
• Convolution in time becomes multiplication in frequency: $Y = X \cdot H$
• Magnitude response shows gain/attenuation; phase response shows time delay per frequency
• Bandwidth quantifies the useful frequency range of signals and systems
• Bode plots provide intuitive visualization using logarithmic scales
• Distortion-free transmission requires flat magnitude and linear phase over the signal bandwidth
• Group delay $\tau_g = -d\phi/d\omega$ reveals frequency-dependent timing distortion