Fourier Transform and Its Applications - Lab Report

Abstract

In general, integral transforms are very useful and productive tools for solving problems involving various types of ordinary differential equations and partial differential equations (PDEs). The Fourier transform, in particular, finds extensive applications in mathematics, engineering, physical sciences, and, most notably, in signal and image processing. The Fourier transform can be viewed as a special case of the Laplace transform and an extension of Fourier series. This report aims to explore how the Fourier transform is applied to circuit analysis to obtain circuit responses, elucidate the differential equations involved in circuit problems, and demonstrate how the Fourier transform can effectively solve these ordinary differential equations.

Introduction

The Fourier transform is a mathematical technique that transforms a function from the time domain to the frequency domain. It accomplishes this by decomposing any function into a sum of sinusoidal functions, revealing hidden information in the time domain that can be expressed in the frequency domain. The Fourier transform is an extension of Fourier series for non-periodic functions and is a special case of the Laplace transform.

In the Fourier transform, the absolute value represents the frequency content of the original function, while the complex argument represents the phase offset of the basic sinusoidal component at that frequency. Fourier theory has widespread applications in various fields, including solving differential equations and PDEs, signal processing, and the analysis of light and radiation from surface currents.

The Fourier transform of a function is represented as:

F(ω) = ∫_-∞^∞ f(t) * e^-iωt dt

Eq. (1)

And its inverse transform is represented as:

f(t) = (1 / (2π)) * ∫_-∞^∞ F(ω) * e^iωt dω

Eq. (2)

The Fourier transform possesses several special properties that facilitate its applications.

Ordinary differential equations (ODEs) are integral to describing various physical phenomena in engineering and science, including physics, biology, chemistry, and circuits. This report focuses on how the Fourier transform can be applied to analyze circuits, specifically addressing the natural and step responses of various circuit configurations.

Problem Statement

In AC circuits, first-order circuits, such as RC (resistor-capacitor) and RL (resistor-inductor) circuits, have their current and voltage described by first-order differential equations. These equations are instrumental in analyzing the natural and step responses of first-order circuits. The following equations describe the natural response of RL and RC circuits:

For RL circuits:

L * di(t)/dt + Ri(t) = 0

Eq. (3)

After solving Eq. (3) to find i(t):

i(t) = (V₀/R) * e^{(-t / τ)}

Eq. (4)

For RC circuits:

RC * dv(t)/dt + v(t) = 0

Eq. (5)

The step response for RL and RC circuits pertains to the circuit's behavior after applying a sudden DC voltage or current source. This section discusses the problem of finding the generated current and voltage in first-order circuits. Using mathematical techniques involving differentiation and integration, the following equations describe the step response of RL and RC circuits:

Step response of RL circuits:

di(t)/dt + (R/L) * i(t) = 0

Eq. (6)

Step response of RC circuits:

dv(t)/dt + (1/RC) * v(t) = 0

Eq. (7)

For RLC (resistor-inductor-capacitor) circuits, we will consider two simple structures: parallel and series RLC circuits. To determine the natural and step responses, we will solve the associated ordinary differential equations. For parallel RLC circuits, the natural response involves finding the voltage across parallel branches, while the step response deals with the sudden voltage applied to the circuits and the resulting current in parallel branches. In series RLC circuits, the natural response requires finding the current flowing through the connected series elements, while the step response focuses on the current generated due to a sudden DC voltage source.

For the natural response of parallel RLC circuits, the first step is to derive the differential equation for voltage, leading to:

L * d²v(t)/dt² + R * dv(t)/dt + (1/C) * v(t) = 0

Eq. (8)

After further mathematical derivations and integrations, we arrive at the following equation:

L * d²v(t)/dt² + R * dv(t)/dt + (1/C) * v(t) = 0

Eq. (9)

This equation is of the second order, as it involves both capacitors and inductors. It cannot be solved by separating variables. To proceed, we assume a solution of the form v(t) = A * e^st. Substituting this into Eq. (9) yields:

A * [(s² + (R/L) * s + 1/(LC))] * e^st = 0

Eq. (10)

This results in the characteristic equation with roots:

s_1,2 = (-R/2L) ± √[(R/L)² - (4/(LC))]

Eq. (11)

Eq. (12)

To determine the value of A, we consider:

A = V₀ / √[(1 - (s₁/s₂))² + (π/2)²]

Eq. (13)

With this information, the equation for the natural response of parallel RLC circuits can be expressed as:

v(t) = V₀ * e^-Rt/(2L) * [cos(ω_nt + δ)]

Eq. (14)

In the step response of parallel RLC circuits, the focus shifts to finding the current in the inductive branch, which does not approach zero over time. Solving this requires a second-order differential equation:

L * d²i(t)/dt² + R * di(t)/dt + i(t) = 0

Eq. (15)

Since s = σ + jω, and for step responses, σ = 0, we have:

jωL * d²i(t)/dt² + R * jω * di(t)/dt + i(t) = 0

Eq. (16)

Furthermore, when the voltage is a sinusoidal waveform, the response is similar to finding the step response in an RL circuit. However, in this case, the voltage is a time-varying sinusoidal voltage, as opposed to a constant DC voltage. Applying Kirchhoff's voltage law to Eq. (16) yields the following ordinary differential equation:

L * d²i(t)/dt² + R * di(t)/dt + i(t) = V_m * sin(ω_mt)

Eq. (18)

Where:

V_m = Amplitude of the sinusoid

ω_m = Angular frequency in radians per second

ω_mt = Argument of the sinusoid

Solving this differential equation for i(t) results in:

i(t) = I_T * (1 - e^-σt) * sin(ω_dt + δ)

Eq. (20)

Where:

I_T = Maximum current amplitude

σ = Damping factor

ω_d = Damped angular frequency

The first term in Eq. (20) represents the transient component of the current, while the second term is the steady-state component of the solution.

Methodology

After identifying the problem of solving circuit equations in the frequency domain, several techniques can be employed for circuit analysis. These techniques include Kirchhoff's laws, series and parallel simplifications, voltage and current division, Thevenin and Norton theorems, node and mesh analysis, and the phasor method.

To find the response of a circuit, the Fourier transform is utilized, as it generalizes the phasor technique to handle non-periodic functions. Applying the Fourier transform to circuits with non-sinusoidal excitations is analogous to using phasor techniques for sinusoidal excitations.

Where:

F(ω) = Fourier transform of voltage, current, or impedance

After applying the Fourier transform, circuit techniques can be employed to obtain the unknown responses (voltage and current). Finally, if the response in the time domain is required, the inverse Fourier transform is applied.

The transfer function is defined as the ratio of the output response to the input excitation:

H(ω) = V_out(ω) / V_in(ω)

It is important to note that this is analogous to H(s) in Laplace transform, with the difference being the replacement of s with jω.

Comparing Fourier transform and Laplace transform:

Pros:

1. Fourier transform can be used for functions defined for all time, while the Laplace transform is one-sided, with integral limits for positive time only.
2. Fourier transform is applicable to signals that have no physical realizable counterparts and cannot be solved using the Laplace transform.
3. Fourier transform is particularly useful for steady-state problems.
4. Fourier transform provides more extensive frequency characteristics for signals.

Cons:

1. Laplace transform is more widely used and applied to a broader range of functions than the Fourier transform.
2. Fourier transform is not commonly employed for problems involving initial conditions.

Application

Now, let's apply the Fourier transform to a circuit with a non-sinusoidal excitation. Consider the following example:

Find v(t) in the circuit for A.

Example:

Solution:

First, utilize the current division technique:

I₁ = (G₂ / (G₁ + G₂)) * I_S

As G₂ = 1/R₂ and G₁ = 1/R₁, we have:

I₁ = (R₁ / (R₁ + R₂)) * I_S

As a result, the voltage drop across R₁ is:

V_R1 = R₁ * I₁

Using the inverse Fourier transform, we can express the voltage V_R1 in the time domain as follows:

V_R1(t) = (R₁ / (R₁ + R₂)) * I_S * ∫_-∞^∞ A(ω) * e^jωt dω

Applying the shift property of the impulse function:

V_R1(t) = (R₁ / (R₁ + R₂)) * I_S * ∫_-∞^∞ A(ω) * e^{j(ω - ω₀)t} dω

Where:

ω₀ = angular frequency corresponding to the impulse function

Now, we can simplify the expression further:

V_R1(t) = (R₁ / (R₁ + R₂)) * I_S * A(ω₀) * ∫_-∞^∞ e^{j(ω - ω₀)t} dω

The integral in the above equation represents the Fourier transform of the impulse function at the angular frequency ω₀. Therefore, we can write it as:

V_R1(t) = (R₁ / (R₁ + R₂)) * I_S * A(ω₀) * F^-1[δ(ω - ω₀)]

Where F^-1 represents the inverse Fourier transform, and δ(ω - ω₀) is the impulse function at frequency ω₀. Therefore, we obtain:

V_R1(t) = (R₁ / (R₁ + R₂)) * I_S * A(ω₀) * δ(t)

This expression provides the voltage across resistor R₁ in the time domain in response to the non-sinusoidal excitation. The impulse function δ(t) indicates that the voltage is present only at t = 0, which aligns with the nature of the impulse function.

Conclusion

In conclusion, the Fourier transform is a powerful mathematical tool with diverse applications in mathematics, engineering, physical sciences, and signal processing. It allows us to analyze functions in the frequency domain and extract valuable information that may not be apparent in the time domain. This report explored the application of the Fourier transform in circuit analysis, specifically addressing the natural and step responses of various circuit configurations.

We discussed how the Fourier transform can be used to solve differential equations associated with RL and RC circuits, as well as more complex RLC circuits. The methodology involved employing the Fourier transform to transform circuit equations into the frequency domain, where they can be analyzed more effectively. We also compared the Fourier transform to the Laplace transform, highlighting their respective advantages and disadvantages.

Finally, we applied the Fourier transform to a circuit with non-sinusoidal excitation, demonstrating its utility in solving real-world circuit problems. The inverse Fourier transform allowed us to obtain the time-domain response of the circuit, emphasizing the practicality of this mathematical technique in engineering applications.

The Fourier transform continues to be a valuable tool for engineers and scientists, enabling them to gain deeper insights into complex systems and solve challenging problems across various fields.