Question

(a) Find the Fourier transform of $$ f(\gamma, p, t)= \begin{cases}e^{-\gamma \mathrm{r}} \sin p t & t>0 \\ 0 & t<0\end{cases} $$ where \(\gamma(>0)\) and \(p\) are constant parameters. (b) The current \(I(t)\) flowing through a certain system is related to the applied voltage \(V(t)\) by the equation $$ I(t)=\int_{-\infty}^{\infty} K(t-u) V(u) d u $$ where $$ K(\tau)=a_{1} f\left(\gamma_{1}, p_{1}, \tau\right)+a_{2} f\left(\gamma_{2}, p_{2}, \tau\right) $$ The function \(f(\gamma, p, t)\) is as given in (a) and all the \(a_{i}, \gamma_{i}(>0)\) and \(p_{i}\) are fixed parameters. By considering the Fourier transform of \(I(t)\), find the relationship that must hold between \(a_{1}\) and \(a_{2}\) if the total net charge \(Q\) passed through the system (over a very long time) is to be zero for an arbitrary applied voltage.

Short answer

The relationship between the parameters is: \[ a_1 p_1 (\text{γ}_2^2 - p_2^2) = - a_2 p_2 (\text{γ}_1^2 - p_1^2) \]

Step-by-step solution

Definition of Fourier Transform

The Fourier transform of a function \( f(t) \) is given by: \[ F(u) = \frac{1}{\text{sqrt}(2\text{π})}\text{∫}_{-\text{∞}}^{\text{∞}} f(t)e^{-i2\text{π}νt} dt \]

Apply the Fourier Transform to \( f(\text{γ}, p, t) \)

We start by considering the given piecewise function: \[ f(\text{γ}, p, t)= \begin{cases} e^{-\text{γ} t} \text{sin}(pt) & t>0 \ 0 & t<0 \end{cases} \]Applying the Fourier transform to this, we only integrate from \( 0 \) to \( \text{∞} \) due to the piecewise definition.

Integrate the function

Performing the integral: \[ ∫_0^{\text{∞}} e^{-\text{γ} t} \text{sin}(pt) e^{-i2\text{π}νt} dt \]Using Euler's formula, we can write \( \text{sin}(pt) \) as \( \frac{e^{ipt} - e^{-ipt}}{2i} \).Thus, the integral becomes: \[ ∫_0^{\text{∞}} e^{-(\text{γ}+i2\text{π}ν)t} \cdot \frac{e^{ipt} - e^{-ipt}}{2i} dt \]

Simplify the integral

Separating the integral into two parts: \[ F(u) = \frac{1}{2i} \text{∫}_0^{\text{∞}} e^{-(\text{γ}+i2\text{π}ν)t+ip t} dt - \frac{1}{2i} ∫_0^{\text{∞}} e^{-(\text{γ}+i2\text{π}ν)t-ip t} dt \] Since these are exponential integrals, we can solve them directly.

Evaluate the exponential integrals

Evaluate each integral: \[ ∫_0^{\text{∞}} e^{-(\text{γ}+i2\text{π}ν+ip)t} dt = \frac{1}{\text{γ} + i (2\text{π}ν + p)} \] and \[ ∫_0^{\text{∞}} e^{-(\text{γ}+i2\text{π}ν-ip)t} dt = \frac{1}{\text{γ} + i (2\text{π}ν - p)} \]

Combine the results

Combining the results from the integrals, we get: \[ F(ν) = \frac{1}{2i} \left( \frac{1}{\text{γ} + i (2\text{π}ν + p)} - \frac{1}{\text{γ} + i (2\text{π}ν - p)} \right) \] Simplifying further, we have: \[ F(ν) = \frac{p}{ \text{γ}^2 + (2\text{π}ν)^2 - p^2 } \]

Analyze part (b) - Fourier transform of I(t)

In part (b), the current \(I(t)\) is given by: \[ I(t) = \text{∫}_{-\text{∞}}^{\text{∞}} K(t-u) V(u) du \]Fourier transform of this convolution: \( \tilde{I}(u) = \tilde{K}(u) \tilde{V}(u) \)Where \[ K(\tau) = a_1f(\text{γ}_1, p_1, \tau) + a_2f(\text{γ}_2, p_2, \tau) \]

Compute Fourier transform of K(τ)

The Fourier transform of \( K(τ) \) is: \[ 𝑲(ν) = a_1\frac{p_1}{ \text{γ}_1^2 + (2\text{π}ν)^2 - p_1^2 } + a_2\frac{p_2|}{γ_2^2 + (2\text{π}ν)^2 - p_2^2 } \]

Condition for zero net charge Q

The total net charge \(Q\) is given by: \[ Q = I(ν=0) = K(ν=0) V(ν=0) \]. Therefore, set \(Q = 0 \): \[ 𝑲(0) = 0 \] From the previous results, this simplifies to:\[ a_1\frac{p_1}{γ_1^2 - p_1^2} + a_2\frac{p_2}{γ_2^2 - p_2^2} = 0\]

Solve for the relationship between a_1 and a_2

Solving the above equation: \[ a_1 p_1 (\text{γ}_2^2 - p_2^2) + a_2 p_2 (\text{γ}_1^2 - p_1^2) = 0 \]Hence, \[ a_1 p_1 (\text{γ}_2^2 - p_2^2) = - a_2 p_2 (\text{γ}_1^2 - p_1^2) \]

Key concepts

Understanding Piecewise Functions

Piecewise functions are defined by different expressions depending on the interval of the input. In our exercise, the function is given as:\[ f(\text{γ}, p, t)= \begin{cases} e^{-\text{γ} t} \text{sin}(pt) & t>0 \ 0 & t<0 \end{cases} \]This means for positive values of t, our function is the product of an exponential decay function and a sine function. When t is negative, the function value is zero.Piecewise functions are crucial in modeling scenarios where different rules apply to different conditions, such as in our case, for t greater than and less than zero. This property is helpful in physics and engineering to describe systems with varying behaviors over time.

Applying the Convolution Integral in Physics

Convolution integrals are fundamental in systems analysis. They combine two functions to get a third function showing how the shape of one is modulated by the other. Mathematically, this is defined as:\[ I(t) = \int_{-\infty}^{\infty} K(t-u) V(u) \, du \]In our case, I(t) describes the current in a system based on the applied voltage V(t) and a kernel function K(\tau). The kernel K(\tau) represents the impulse response of the system given by:\[ K(\tau) = a_1 f(\gamma_1, p_1, \tau) + a_2 f(\gamma_2, p_2, \tau) \]The convolution integral helps analyze how different frequencies in the applied voltage V(u) affect the resulting current I(t). This is particularly useful in signal processing and electrical engineering.

Net Charge Analysis in Electrical Systems

Net charge analysis involves calculating the total charge that flows through a system over time. In this exercise, the net charge Q is given by integrating the current over all time, which is equivalent to setting the frequency domain equivalent to zero:\[ Q = I(u=0) = K(u=0) V(u=0) \]To achieve zero net charge, we need the total Fourier transform of the current at zero frequency to be zero:\[ K(0) = 0 \]Plugging in the values from the Fourier transform of K(\tau), we derive the condition:\[ a_1\frac{p_1}{\gamma_1^2 - p_1^2} + a_2\frac{p_2}{\gamma_2^2 - p_2^2} = 0 \]This ensures that when the applied voltage retains arbitrary characteristics, the total net charge across the system remains balanced.

Solving Exponential Integrals

Exponential integrals appear frequently in physics, especially when dealing with decaying processes or oscillatory functions. In our problem, we encounter these integrals while applying the Fourier transform to the piecewise function defined:\[ \int_0^{\infty} e^{-(\text{γ}+i2\text{π}u)t} \text{sin}(pt) \, dt \]Using Euler's formula, we simplify the sine function and separate the integral into two parts with integrals of the form:\[ \int_0^{\infty} e^{(-a)t} \, dt = \frac{1}{a} \]where a is a complex number derived from the combination of parameters (\text{γ}, p, and u). Solving these integrals with complex exponents gives us the components needed to derive the Fourier transform, which is fundamental for frequency domain analysis in signal processing and systems dynamics.