Maths Formulae Transforms Substitution(Frequency Shifting)

Substitution (Frequency Shifting) – Modulation Property

Learn the frequency shifting property of Fourier transform where multiplication by exponentials shifts the frequency spe...
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Definition of Frequency Shifting

The Frequency Shifting Property (also called s-shifting or translation property) of the Laplace Transform states that multiplying a time-domain function, f(t), by an exponential term, e-at, corresponds to a shift or translation of its Laplace Transform, F(s), in the complex frequency domain (s-domain). Specifically, the new transform becomes F(s+a). This property is fundamental for analyzing systems with exponential decay or growth, such as damped oscillations in electrical circuits and mechanical systems.

SymbolDescription
\[ f(t) \]Original Function - Base time-domain function before exponential multiplication
\[ F(s) \]Original Transform - Laplace transform of f(t)
\[ e^{-at} \]Exponential Factor - Time-domain multiplication factor
\[ F(s+a) \]Shifted Transform - s-domain function shifted by 'a'
\[ a \]Shift Parameter - Amount of s-domain translation (real constant)
\[ s \]Complex Variable - s = σ + jω in Laplace domain
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Key Formulas

\[ \mathcal{L}\{e^{-at}f(t)\} = F(s+a) \]
Frequency Shift Property (Exponential Decay)
\[ \mathcal{L}\{e^{at}f(t)\} = F(s-a) \]
Frequency Shift Property (Exponential Growth)
\[ \text{If } f(t) \leftrightarrow F(s), \text{ then } e^{-at}f(t) \leftrightarrow F(s+a) \]
Transform Pair Notation
\[ \mathcal{L}^{-1}\{F(s+a)\} = e^{-at}f(t) \]
Inverse Frequency Shift Property
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Visualizing the Shift

Frequency Shifting (Substitution s→s−a) F(s) original s = 0 ×e^(at) s → s−a F(s−a) shifted s = a shifted by a ℒ{e^(at)f(t)} = F(s−a)
Frequency Shifting: multiplying f(t) by e^(at) shifts F(s) → F(s−a) in the s-domain

This property is visualized on the complex s-plane. If the original transform F(s) has poles and zeros at certain locations, the new transform F(s+a) will have the exact same pattern of poles and zeros, but every point will be shifted to the left by 'a' units (for a positive 'a' in e-at). This shift is crucial for stability analysis, as moving poles further into the left-half of the s-plane increases system stability.

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Properties

PropertyDescription
Universal ShiftingAll poles and zeros of the original transform F(s) shift by the same amount 'a' in the s-plane.
Stability ModificationMultiplying by e<sup>-at</sup> with a > 0 shifts poles to the left, generally increasing system stability. A negative 'a' shifts poles to the right, reducing stability.
Damping CorrespondenceThe exponential factor e<sup>-at</sup> in the time domain directly introduces damping (if a > 0) or growth (if a < 0) to the system's behavior.
Inverse Transform UtilityRecognizing a shifted pattern in the s-domain, such as (s+a), allows for the direct application of the inverse property to find the time-domain function.
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Proof of the Frequency Shifting Property

The proof begins with the fundamental definition of the Laplace Transform.

\[ \mathcal{L}\{g(t)\} = \int_0^\infty g(t)e^{-st} dt \]

We want to find the transform of \(e^{-at}f(t)\). Let's substitute \(g(t) = e^{-at}f(t)\) into the definition:

\[ \mathcal{L}\{e^{-at}f(t)\} = \int_0^\infty (e^{-at}f(t))e^{-st} dt \]

Next, we combine the exponential terms using the property \(e^x e^y = e^{x+y}\):

\[ \mathcal{L}\{e^{-at}f(t)\} = \int_0^\infty f(t)e^{-(s+a)t} dt \]

Now, compare this integral to the original definition of the Laplace Transform of \(f(t)\), which is \(F(s) = \int_0^\infty f(t)e^{-st} dt\). The integral we derived is identical in form, but with \(s\) replaced by \((s+a)\). Therefore, the result is \(F(s+a)\).

\[ \int_0^\infty f(t)e^{-(s+a)t} dt = F(s+a) \]
Q.E.D.
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Worked Example

Find the Laplace Transform of \( f(t) = e^{-2t}\cos(5t) \).
  1. Identify the base function, which is \( g(t) = \cos(5t) \).
  2. Find the known Laplace Transform of the base function: \( \mathcal{L}\{\cos(5t)\} = G(s) = \frac{s}{s^2 + 5^2} = \frac{s}{s^2 + 25} \).
  3. Identify the exponential factor, \( e^{-2t} \). This corresponds to a frequency shift with \( a = 2 \).
  4. Apply the frequency shifting property: \( \mathcal{L}\{e^{-at}g(t)\} = G(s+a) \).
  5. Substitute every \( s \) in \( G(s) \) with \( (s+2) \).
\[ F(s) = G(s+2) = \frac{(s+2)}{(s+2)^2 + 25} \]
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Applications

Control System Stability: Used in pole placement design. By introducing exponential factors, engineers can shift the poles of a system's transfer function to desired locations in the s-plane to ensure stability and achieve a specific transient response (e.g., faster settling time, less overshoot).

Circuit Analysis with Damping: Essential for analyzing RLC circuits. The resistance (R) in the circuit introduces an exponential decay term (damping). The frequency shifting property allows for straightforward analysis of the circuit's transient and steady-state response to inputs.

Mechanical System Damping: Models damped spring-mass systems, such as shock absorbers in a car. The damping component introduces an exponential decay to the oscillations, which is perfectly described by the frequency shifting property when transforming the equations of motion.

Signal Processing: In analyzing modulated signals, such as in AM radio, the carrier wave is a sinusoid multiplied by the information signal. While this is more directly related to the Fourier Transform, the underlying principle is analogous. It is also used in applying exponential windowing functions to signals to analyze specific portions of their frequency content.

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Real-World Examples

An RLC series circuit has a current response given by \( i(t) = 5e^{-10t}\sin(100t) \) Amperes after a switch is closed. Find the Laplace transform \( I(s) \) of this current.
  1. The base function is \( g(t) = \sin(100t) \), and the constant multiplier is 5.
  2. The transform of the base function is \( \mathcal{L}\{\sin(100t)\} = \frac{100}{s^2 + 100^2} \).
  3. The exponential term is \( e^{-10t} \), so the shift is \( a=10 \).
  4. Apply the property \( \mathcal{L}\{5e^{-10t}\sin(100t)\} = 5 \times \frac{100}{(s+10)^2 + 100^2} \).
\[ I(s) = \frac{500}{(s+10)^2 + 10000} \]
The displacement of a car's shock absorber after hitting a bump is modeled by \( x(t) = 0.1e^{-4t}\cos(12t) \) meters. Determine the transform \( X(s) \) of the displacement.
  1. The base function is \( g(t) = \cos(12t) \) with a 0.1 multiplier.
  2. The transform of the base function is \( \mathcal{L}\{\cos(12t)\} = \frac{s}{s^2 + 12^2} \).
  3. The damping factor is \( e^{-4t} \), so the shift is \( a=4 \).
  4. Apply the property, replacing \( s \) with \( (s+4) \): \( X(s) = 0.1 \times \frac{(s+4)}{(s+4)^2 + 144} \).
\[ X(s) = \frac{0.1(s+4)}{(s+4)^2 + 144} \]
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Real-World Scenarios

message × carrier e^(jω₀t)×m(t)→M(ω−ω₀) shift to carrier ω₀
AM Radio Modulation
AM radio multiplies the audio signal m(t) by a carrier e^(jω₀t) to shift the audio spectrum to the carrier frequency ω₀ — the Laplace substitution s→s−jω₀. Different stations use different carrier frequencies.
× Local Osc. IF RF × e^(jω_LO t) → IF s-substitution s→s−jω_LO
Superheterodyne Receiver
All AM/FM/TV receivers use frequency shifting: the incoming RF signal is multiplied by a local oscillator to shift it to a fixed Intermediate Frequency (IF) for filtering — a direct application of the Laplace substitution theorem.
f(t)=cos(ωt) × e^(−at) damped cos e^(−at)cos(ωt)→F(s+a) pole at s = −a+jω
Damped Oscillator
Damped oscillations like e^(−at)cos(ωt) appear in shock absorbers, RLC circuits, and building sway. The substitution s→s+a shifts the pure cosine transform, placing the pole at s = −a + jω in the s-plane.

Bungee Jumping: The motion of a bungee jumper after the initial fall is a classic example of a damped oscillation. The jumper oscillates up and down, but the elasticity of the cord and air resistance cause the amplitude of each bounce to decrease exponentially over time. This motion is precisely modeled by a sine or cosine function multiplied by a decaying exponential.

Musical Instrument Decay: When a guitar string is plucked or a piano key is struck, the note produced does not last forever. The sound's amplitude decays over time due to energy dissipation. This decay is often exponential, meaning the sound wave can be modeled as a sinusoidal function (the note's pitch) multiplied by a term like e-at, where 'a' determines how quickly the sound fades out.

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Common Function Transformations

The frequency shifting property can be applied to any function with a known Laplace transform. The table below shows its effect on several common transform pairs.

Original Function, f(t)Original Transform, F(s)Shifted Function, e<sup>-at</sup>f(t)Shifted Transform, F(s+a)
\[ u(t) \] (unit step)\[ \frac{1}{s} \]\[ e^{-at}u(t) \]\[ \frac{1}{s+a} \]
\[ t^n \]\[ \frac{n!}{s^{n+1}} \]\[ e^{-at}t^n \]\[ \frac{n!}{(s+a)^{n+1}} \]
\[ \sin(\omega t) \]\[ \frac{\omega}{s^2 + \omega^2} \]\[ e^{-at}\sin(\omega t) \]\[ \frac{\omega}{(s+a)^2 + \omega^2} \]
\[ \cos(\omega t) \]\[ \frac{s}{s^2 + \omega^2} \]\[ e^{-at}\cos(\omega t) \]\[ \frac{s+a}{(s+a)^2 + \omega^2} \]
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Common Mistakes

⚠️ Sign Errors: A common mistake is mixing up the signs. Remember that multiplication by e-at in the time domain leads to a shift of (s+a) in the frequency domain. Conversely, e+at leads to a shift of (s-a).
⚠️ Incomplete Substitution: When applying the shift, you must replace *every* instance of 's' in the original transform F(s) with '(s+a)'. Forgetting to substitute one of the 's' terms is a frequent error, especially in more complex fractions like the transform of cosine.
💡 Forgetting the Inverse: The property is just as useful for inverse transforms. When you see a term like (s+a) consistently appearing where 's' normally would be, recognize it as a frequency shift. First, perform the inverse transform as if the variable were just 's', then multiply the resulting time-domain function by e-at.
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Study Strategy

1 🔍 Grasp the Fundamentals
  • Review the 'Definition of Frequency Shifting' to understand what it means to shift a signal's frequency components.
  • Study the 'Visualizing the Shift' section to connect the mathematical formula to a graphical representation in the frequency domain.
  • Read through the 'Properties' to learn how frequency shifting interacts with other transform properties like linearity and time scaling.
  • Absorb the 'Proof of the Frequency Shifting Property' to build a deeper, foundational understanding of why the formula works.
2 🧠 Commit to Memory
  • Write down the primary 'Key Formulas' for both the forward and inverse transforms multiple times without looking.
  • Create flashcards for the 'Common Function Transformations' involving frequency shifting (e.g., modulated sine/cosine).
  • Verbally explain the formula to a peer, focusing on the role of the complex exponential term e^(jω₀t).
  • Link the frequency shift 'ω₀' in the time domain multiplication to the corresponding shift in the frequency domain F(ω - ω₀).
3 ✍️ Sharpen Your Skills
  • Follow the 'Worked Example' step-by-step, ensuring you understand the logic behind each calculation.
  • Solve additional problems that apply the frequency shift property to basic signals like rectangular pulses or impulse functions.
  • Actively look for the 'Common Mistakes' described on the formula page in your own work and learn how to avoid them.
  • Try to derive the transforms of modulated signals (like cos(ω₀t)f(t)) from scratch using the frequency shifting property and Euler's formula.
4 📡 Connect to Applications
  • Read the 'Applications' section and identify how frequency shifting is crucial in areas like AM radio and communications.
  • Analyze the 'Real-World Scenarios' to understand how modulating a carrier wave is a direct use of this property.
  • Find a simple 'Real-World Example,' such as tuning a radio, and explain the underlying frequency shifting process.
  • Explore how frequency shifting is related to other 'Related Formulas' like filtering and multiplexing in signal processing systems.
Mastering frequency shifting unlocks the core principles of modern communication systems and signal processing.

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