Maths Formulae Transforms Translation(Time Shifting)

Translation (Time Shifting) – Delay Theorem

Understand the translation or time shifting property in Fourier transform and how it introduces phase shifts.
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Definition of Time Shifting

The Time Shifting Property, also known as the time delay or translation property, is a fundamental concept in Laplace Transforms. It states that delaying a function in the time domain by a constant 'a' corresponds to multiplying its Laplace transform in the s-domain by an exponential factor e-as. This property is essential for analyzing systems with inherent delays, such as signal propagation, transport lags in process control, and network latency.

SymbolDescription
\[ f(t) \]Original time-domain function.
\[ f(t-a)u(t-a) \]The function `f(t)` shifted (delayed) in time by 'a' units, and zero for t < a.
\[ F(s) \]The Laplace transform of the original function, `f(t)`.
\[ a \]The amount of time delay (must be positive).
\[ u(t-a) \]The Heaviside unit step function, which is 0 for `t < a` and 1 for `t ≥ a`.
\[ e^{-as} \]The exponential factor in the s-domain that represents the time delay.
\[ s \]The complex frequency variable, `s = σ + jω`.
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Key Time Shifting Formulas

\[ \mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s) \quad \text{(for } a > 0 \text{)} \]
Time Shifting Property
\[ \mathcal{L}^{-1}\{e^{-as}F(s)\} = f(t-a)u(t-a) \]
Inverse Time Shifting Property
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Diagram of Time Shifting

Time Shifting (Translation) f(t) — original starts t=0 delay t₀ ×u(t−t₀) f(t−t₀) — delayed starts t=t₀ t₀ ℒ{f(t−t₀)u(t−t₀)} = e^(−st₀) F(s)
Time Shifting: delaying f(t) by t₀ multiplies its Laplace transform by e^(−st₀)

A diagram illustrating time shifting shows two plots. The first plot displays an arbitrary function, f(t), starting at t=0. The second plot shows the same function, but its shape is translated or shifted to the right along the time axis by an amount 'a'. This new function, labeled f(t-a)u(t-a), is zero for all time t < a and identical in shape to f(t) for t ≥ a.

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Properties of Time Shifting

The time-shifting property has several important implications for system analysis:

  • Causality Preservation: The inclusion of the unit step function `u(t-a)` ensures that the output is zero before the input is applied, preserving the principle of causality in physical systems.
  • Shape Preservation: The shape of the function `f(t)` is not distorted by the time shift; it is merely translated along the time axis.
  • Magnitude Invariance: The magnitude of the frequency response, `|F(jω)|`, is unaffected by a time delay, since `|e^{-jωa}| = 1`. The delay only affects the phase.
  • Linear Phase Shift: A time delay of 'a' introduces a linear phase shift of `-ωa` to the frequency response. This means higher frequencies experience a greater phase shift for the same time delay.
  • Stability Impact: In feedback control systems, time delays (also known as transport lag) can reduce the phase margin and destabilize an otherwise stable system.
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Proof of the Time Shifting Property

The time-shifting property can be derived directly from the definition of the Laplace Transform.

\[ \mathcal{L}\{f(t-a)u(t-a)\} = \int_{0}^{\infty} f(t-a)u(t-a)e^{-st} \,dt \]
1. Start with the definition of the Laplace Transform.

The unit step function `u(t-a)` is zero for `t < a` and one for `t ≥ a`. This changes the lower limit of integration from 0 to 'a'.

\[ = \int_{a}^{\infty} f(t-a)e^{-st} \,dt \]
2. Apply the property of the unit step function.

Now, perform a change of variables. Let `τ = t - a`. This implies `t = τ + a` and `dτ = dt`. We also need to change the limits of integration: when `t = a`, `τ = 0`; when `t → ∞`, `τ → ∞`.

\[ = \int_{0}^{\infty} f(\tau)e^{-s(\tau+a)} \,d\tau \]
3. Substitute the new variable τ.

Using the properties of exponents, we can split `e^{-s(τ+a)}` into `e^{-sτ}e^{-sa}`. Since `e^{-sa}` does not depend on the variable of integration `τ`, it can be factored out of the integral.

\[ = e^{-sa} \int_{0}^{\infty} f(\tau)e^{-s\tau} \,d\tau \]
4. Factor out the constant exponential term.

The remaining integral is, by definition, the Laplace Transform of `f(t)`, which is `F(s)`.

\[ \mathcal{L}\{f(t-a)u(t-a)\} = e^{-as}F(s) \]
5. Final result.
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Worked Example

Find the Laplace transform of the function `g(t) = (t-2)^3 u(t-2)`.
  1. Identify the base function `f(t)` and the time delay `a`. In this case, the function is a shifted version of `f(t) = t^3`, and the delay is `a = 2`.
  2. Find the Laplace transform of the base function, `F(s) = L{f(t)}`. Using the standard transform for `t^n`, we get `F(s) = L{t^3} = 3! / s^{3+1} = 6 / s^4`.
  3. Apply the time-shifting property: `L{f(t-a)u(t-a)} = e^{-as}F(s)`.
  4. Substitute the values for `a` and `F(s)` into the property: `G(s) = e^{-2s} * (6 / s^4)`.
\[ G(s) = \frac{6e^{-2s}}{s^4} \]
Find the inverse Laplace transform of `G(s) = (e^{-5s}) / (s^2 + 9)`.
  1. Identify the exponential delay term `e^{-as}` and the base transform `F(s)`. Here, `e^{-as} = e^{-5s}`, so the delay `a = 5`. The base transform is `F(s) = 1 / (s^2 + 9)`.
  2. Find the inverse transform of the base function, `f(t) = L^{-1}{F(s)}`. We recognize `F(s)` as the transform of a sine function. We rewrite it as `F(s) = (1/3) * (3 / (s^2 + 3^2))`. Therefore, `f(t) = (1/3)sin(3t)`.
  3. Apply the inverse time-shifting property: `L^{-1}{e^{-as}F(s)} = f(t-a)u(t-a)`.
  4. Substitute `f(t)` and `a` to find the final answer: `g(t) = (1/3)sin(3(t-5))u(t-5)`.
\[ g(t) = \frac{1}{3}\sin(3(t-5))u(t-5) \]
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Applications

Control Systems with Transport Delays: In process control, such as in chemical plants or manufacturing lines, there is often a delay between when a controller takes action (e.g., opening a valve) and when the result is measured (e.g., a change in temperature or flow rate downstream). The time-shifting property is crucial for modeling this 'transport lag' to design stable and effective feedback controllers.

Communication Systems: The property is used to analyze signal propagation delays in networks, satellite links, and fiber optic cables. It helps in understanding and mitigating issues like latency, echo, and synchronization in communication protocols.

Circuit Analysis: In high-frequency electronics, the finite speed of light causes noticeable propagation delays as signals travel along transmission lines or PCB traces. The time-shifting property allows engineers to model these delays and analyze their impact on signal integrity and timing.

Signal Processing: Echoes and reverberations in audio signals can be modeled as time-delayed and attenuated versions of the original signal. This property is fundamental to designing echo cancellation algorithms and audio effects.

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Real-World Examples

A radar system sends out a pulse `p(t) = u(t) - u(t - 10^{-6})`, representing a pulse of 1 microsecond duration. The pulse reflects off an aircraft and returns to the receiver after a delay of 200 microseconds (2 x 10<sup>-4</sup> s). Find the Laplace transform of the received signal, `r(t)`.
  1. The received signal `r(t)` is the original pulse `p(t)` delayed by `a = 2 * 10^{-4}` s. So, `r(t) = p(t - 2*10^{-4})`.
  2. First, find the transform of the original pulse, `P(s) = L{u(t) - u(t-10^{-6})}`. Using the standard transform for a step function and the time-shifting property for the second term, `P(s) = 1/s - (e^{-10^{-6}s})/s = (1 - e^{-10^{-6}s})/s`.
  3. Now, apply the time-shifting property to `P(s)` for the round-trip delay `a = 2 * 10^{-4}`. `R(s) = e^{-as}P(s)`.
  4. Substitute the expressions for `a` and `P(s)`: `R(s) = e^{-2*10^{-4}s} * (1 - e^{-10^{-6}s})/s`.
\[ R(s) = \frac{e^{-2 \times 10^{-4}s}(1 - e^{-10^{-6}s})}{s} \]
Water flows through a 15-meter pipe at a velocity of 3 m/s. A heater at the pipe's entrance instantly increases the water temperature by 20°C, which can be modeled as a step input `ΔT_in(t) = 20u(t)`. What is the Laplace transform of the temperature change measured by a sensor at the end of the pipe?
  1. Calculate the time delay `a` for the water to travel the length of the pipe. `a = distance / velocity = 15 m / 3 m/s = 5 s`.
  2. The temperature change at the outlet, `ΔT_out(t)`, is the input signal delayed by `a=5` seconds: `ΔT_out(t) = 20u(t-5)`.
  3. Find the Laplace transform of the base function, `f(t) = 20u(t)`. This is `F(s) = 20/s`.
  4. Apply the time-shifting property, `L{f(t-a)u(t-a)} = e^{-as}F(s)`. Substitute `a=5` and `F(s)=20/s`.
\[ \Delta T_{out}(s) = \frac{20e^{-5s}}{s} \]
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Real-World Scenarios

green at t=t₀ (delay)
Traffic Signal Timing
A green traffic light at time t₀ is modelled as u(t−t₀) — a delayed step function. The Laplace translation theorem ℒ{u(t−t₀)} = e^(−st₀)/s lets engineers compute signal timing and phase offsets for coordinated traffic flow.
SRC DST propagation delay τ x(t−τ) ↔ e^(−sτ)X(s) QoS modelling
Network Latency Modelling
Network engineers model packet delays as time shifts: signal x(t) arriving after delay τ becomes x(t−τ). The Laplace translation theorem e^(−sτ)X(s) is used in queuing theory and Quality of Service analysis.
release at t₀ e^(−st₀)·C(s)
Timed Drug Release
Extended-release capsules deliver medication after a delay t₀ — modelled as a delayed step input. Pharmacologists use ℒ{f(t−t₀)u(t−t₀)} = e^(−st₀)F(s) to predict concentration profiles and optimise dosing schedules.

Highway Traffic Flow: On a long highway, a change in traffic density (like a slowdown) caused by an incident at one point propagates down the road like a wave. The traffic conditions observed by a driver several miles downstream are a time-delayed version of the conditions at the source of the incident.

Satellite Internet Latency: When using satellite internet, there is a noticeable delay in activities like video conferencing. This is because the signal must travel from your computer to a satellite in orbit and back down to the destination server. This round-trip time is a physical delay that is modeled using the time-shifting property to analyze network performance.

Assembly Line Production: In a factory, a product moves from one station to the next on a conveyor belt. An action performed at Station A (e.g., painting a part) will only be visible at Station B after the part has traveled for a certain amount of time. This transport delay is fundamental to scheduling and synchronizing operations on the assembly line.

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Related Laplace Properties

The Time Shifting property is often confused with the Frequency Shifting property. They describe dual operations between the time and s-domains.

PropertyTime Domain Operations-Domain Operation
Time ShiftingDelay the signal: `f(t-a)u(t-a)`Multiply transform by `e^{-as}`: `e^{-as}F(s)`
Frequency ShiftingMultiply signal by `e^{-at}`: `e^{-at}f(t)`Shift the transform: `F(s+a)`
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Common Mistakes

⚠️ Forgetting the Unit Step Function `u(t-a)`: A common error is to transform `f(t-a)` directly. The step function `u(t-a)` is crucial because it enforces causality by ensuring the function is zero for `t < a`. The property is strictly `L{f(t-a)u(t-a)}`.
⚠️ Incorrectly Handling the Shifted Argument: To find the transform of a function like `t*u(t-2)`, you cannot directly apply the theorem. You must first rewrite the time-domain function in terms of `(t-2)`, i.e., `(t-2+2)u(t-2) = (t-2)u(t-2) + 2u(t-2)`, and then transform each term separately.
⚠️ Confusing `f(t-a)u(t-a)` with `f(t)u(t-a)`: The property applies when the argument of the function `f` is shifted. The transform of `f(t)u(t-a)` (a truncated function) is not simply `e^{-as}F(s)` and requires a different method to solve.
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Study Strategy

1 🧠 Grasp the Core Concepts
  • Focus on the 'Definition of Time Shifting' to understand what shifting a function in the time domain, f(t-a), physically represents.
  • Study the 'Diagram of Time Shifting' to visually connect the original function f(t) with its delayed (right shift) or advanced (left shift) versions.
  • Review the 'Proof of the Time Shifting Property' to build a deeper, foundational understanding of why the exponential term appears in the frequency domain.
  • Compare the time shifting property with 'Related Transform Properties' like frequency shifting to understand their distinct effects.
2 ✍️ Commit Formulas to Memory
  • Write out the primary 'Key Time Shifting Formula', L{f(t-a)u(t-a)} = e^(-as)F(s), at least ten times.
  • Create flashcards that distinguish between the time delay formula and the related formula for L{e^(at)f(t)}.
  • Pay special attention to the role of the Heaviside step function, u(t-a), which is crucial for a causal shift.
  • Memorize the sign convention: a minus in the time domain (t-a, a delay) corresponds to a minus in the exponent of the s-domain (e^-as).
3 🏋️ Reinforce with Practice Problems
  • Follow the provided 'Worked Example' step-by-step, then cover the solution and solve it yourself to check your understanding.
  • Find practice problems that require you to first express a piecewise function using Heaviside functions before applying the transform.
  • Actively watch for the pitfalls mentioned in the 'Common Mistakes' section, such as transforming f(t) instead of f(t-a).
  • Practice inverse transforms where an e^(-as) term is present, requiring you to apply the time-shifting property in reverse.
4 🌍 Connect to Real-World Applications
  • Read the 'Applications' and 'Real-World Examples' sections to connect the abstract formula to signal processing, control systems, and communication.
  • Analyze a 'Real-World Scenario', such as a delayed input signal to an electronic circuit, and identify where the time shift occurs.
  • Try to model a simple physical delay, like an echo in an audio signal, using a function and its time-shifted version.
  • Explain to a peer how time shifting is essential for analyzing systems where an action or signal does not start at t=0.
By systematically understanding, memorizing, practicing, and applying the time-shifting formula, you can master the analysis of dynamic systems with delays.

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