answer:The Fourier transform of the function f(x) = cos(2πx) can be found using the formula: F(ω) = ∫-∞,+∞ f(t) e^(-iωt) dt Substituting f(t) = cos(2πt) into the formula, we get: F(ω) = ∫-∞,+∞ cos(2πt) e^(-iωt) dt Now we can use the Euler's formula to rewrite the cosine function: cos(2πt) = (e^(i2πt) + e^(-i2πt))/2 Substituting this into the integral, we get: F(ω) = ∫-∞,+∞ (e^(i2πt)/2 + e^(-i2πt)/2) e^(-iωt) dt Now, we can split the integral into two separate integrals: F(ω) = (1/2) ∫-∞,+∞ e^(i2πt) e^(-iωt) dt + (1/2) ∫-∞,+∞ e^(-i2πt) e^(-iωt) dt Now, we can factor out e^(-iωt) from both integrals: F(ω) = (1/2) ∫-∞,+∞ e^(i(2π-ω)t) dt + (1/2) ∫-∞,+∞ e^(-i(2π+ω)t) dt Both integrals represent Dirac delta functions: F(ω) = (1/2) δ(ω - 2π) + (1/2) δ(ω + 2π) So, the Fourier transform of f(x) = cos(2πx) is: F(ω) = (1/2) δ(ω - 2π) + δ(ω + 2π), To find the Fourier transform of f(x) = cos(2πx), we use the given formula: F(ω) = ∫-∞,+∞ f(t) e^(-iωt) dt In this case, f(t) = cos(2πt). So, F(ω) = ∫-∞,+∞ cos(2πt) * e^(-iωt) dt We can rewrite cos(2πt) as (e^(i2πt) + e^(-i2πt))/2. Now, F(ω) = ∫-∞,+∞ (e^(i2πt) + e^(-i2πt))/2 * e^(-iωt) dt Simplify and separate the integral into two parts: F(ω) = (1/2) * ∫-∞,+∞ e^(i(2π - ω)t) dt + ∫-∞,+∞ e^(-i(2π + ω)t) dt Now, we need to compute these two integrals. Let's consider the first one: I1 = ∫-∞,+∞ e^(i(2π - ω)t) dt To solve this integral, we can use the fact that if α is real, then ∫-∞,+∞ e^(iαt) dt = 2πδ(α), where δ(α) is the Dirac delta function. In our case, α = (2π - ω), so: I1 = 2πδ(2π - ω) Do the same for the second integral: I2 = ∫-∞,+∞ e^(-i(2π + ω)t) dt Now, α = -(2π + ω), so: I2 = 2πδ(-(2π + ω)) = 2πδ(2π + ω) Now, substitute back I1 and I2 into the equation for F(ω): F(ω) = (1/2) * 2πδ(2π - ω) + 2πδ(2π + ω) Finally, simplify the expression: F(ω) = πδ(2π - ω) + πδ(2π + ω) So, the Fourier transform of f(x) = cos(2πx) is F(ω) = πδ(2π - ω) + πδ(2π + ω).,To find the Fourier transform of f(x) = cos(2πx), we will use the given formula: F(ω) = ∫-∞,+∞ f(t) e^(-iωt) dt First, let's substitute f(t) with cos(2πt): F(ω) = ∫-∞,+∞ cos(2πt) e^(-iωt) dt Now, we will use the Euler's formula to rewrite the cosine function: cos(2πt) = (e^(i2πt) + e^(-i2πt))/2 Substitute this back into the integral: F(ω) = ∫-∞,+∞ (e^(i2πt) + e^(-i2πt))/2 e^(-iωt) dt Now, distribute the exponential term: F(ω) = (1/2) ∫-∞,+∞ e^(i2πt) e^(-iωt) + e^(-i2πt) e^(-iωt) dt Combine the exponential terms: F(ω) = (1/2) ∫-∞,+∞ e^(i(2π-ω)t) + e^(-i(2π+ω)t) dt Now, we will solve the integral by splitting it into two separate integrals: F(ω) = (1/2) ∫-∞,+∞ e^(i(2π-ω)t) dt + ∫-∞,+∞ e^(-i(2π+ω)t) dt To solve these integrals, we will use the Dirac delta function, which is defined as: δ(ω) = ∫-∞,+∞ e^(iωt) dt Using the Dirac delta function, we can rewrite our Fourier transform as: F(ω) = (1/2) δ(ω - 2π) + δ(ω + 2π) And that's the Fourier transform of the function f(x) = cos(2πx): F(ω) = (1/2) δ(ω - 2π) + δ(ω + 2π)

answer:To find the Fourier transform of the function f(t) = 3cos(2t) + 4sin(4t) over the interval -π, π, we can use the following formula: F (ω) = (1/π) * ∫-π, π f(t) * e^(-iωt) dt Here, F(ω) represents the Fourier transform of f(t), and ω is the angular frequency. First, let's break down the function f(t) into its cosine and sine components: f(t) = 3cos(2t) + 4sin(4t) Now, we will apply the Fourier transform to each component separately: For the cosine component: Cosine Fourier transform: F₁(ω) = (1/π) * ∫-π, π 3cos(2t) * e^(-iωt) dt For the sine component: Sine Fourier transform: F₂(ω) = (1/π) * ∫-π, π 4sin(4t) * e^(-iωt) dt Now, we can compute the Fourier transforms for both components: F₁(ω) = (1/π) * 3/2 * (δ(ω - 2) - δ(ω + 2)) F₂(ω) = (1/π) * (4i/ω) * (sin(4π) - sin(-4π)) Since sin(4π) = sin(-4π) = 0, we have: F₂(ω) = 0 For the Fourier transform of f(t), we can add the Fourier transforms of the individual components: F(ω) = F₁(ω) + F₂(ω) F(ω) = (1/π) * 3/2 * (δ(ω - 2) - δ(ω + 2)) + 0 So, the Fourier transform of f(t) = 3cos(2t) + 4sin(4t) over the interval -π,To find the Fourier transform of the given function, we need to compute the Fourier coefficients a_n and b_n for the Fourier series representation of the function. The Fourier series representation of a function f(t) over the interval -π, π is given by: f(t) = a_0 + Σ a_n * cos(n*t) + b_n * sin(n*t) where the Fourier coefficients a_n and b_n are given by: a_0 = (1/π) * ∫-π, π f(t) dt a_n = (1/π) * ∫-π, π f(t) * cos(n*t) dt b_n = (1/π) * ∫-π, π f(t) * sin(n*t) dt Now, let's compute the Fourier coefficients for the given function f(t) = 3cos(2t) + 4sin(4t). a_0 = (1/π) * ∫-π, π (3cos(2t) + 4sin(4t)) dt = (1/π) * 3/2 * sin(2t) - cos(4t) -π, π = 0 (since sin and cos are periodic functions) a_n = (1/π) * ∫-π, π (3cos(2t) + 4sin(4t)) * cos(n*t) dt For n = 2: a_2 = (1/π) * ∫-π, π (3cos(2t) * cos(2t) + 4sin(4t) * cos(2t)) dt = (1/π) * 3/4 * t + 1/8 * sin(4t) - 1/4 * cos(4t) * sin(2t) -π, π = 3 (since sin and cos are periodic functions) For n ≠ 2: a_n = 0 (since the integrand is an odd function and the integral is over a symmetric interval) b_n = (1/π) * ∫-π, π (3cos(2t) + 4sin(4t)) * sin(n*t) dt For n = 4: b_4 = (1/π) * ∫-π, π (3cos(2t) * sin(4t) + 4sin(4t) * sin(4t)) dt = (1/π) * 3/8 * cos(2t) * sin(4t) - 1/8 * cos(4t) - 1/4 * t -π, π = 4 (since sin and cos are periodic functions) For n ≠ 4: b_n = 0 (since the integrand is an odd function and the integral is over a symmetric interval) Now, we can write the Fourier series representation of the function f(t) = 3cos(2t) + 4sin(4t) as: f(t) = 3 * cos(2t) + 4 * sin(4t) which is the same as the original function. So, the Fourier transform of the function f(t) = 3cos(2t) + 4sin(4t) over the interval -π, π is: F(ω) = 3 * δ(ω - 2) + 4 * δ(ω - 4) where δ(ω) is the Dirac delta function., To find the Fourier transform of f(t) = 3cos(2t) + 4sin(4t) over the interval -π, π, we first need to express the function in terms of complex exponentials. We know that: cos(2t) = (e^(2it) + e^(-2it))/2 sin(4t) = (e^(4it) - e^(-4it))/(2i) So, f(t) can be written as: f(t) = 3(e^(2it) + e^(-2it))/2 + 4(e^(4it) - e^(-4it))/(2i) Let's simplify this expression: f(t) = (3/2)*e^(2it) + (3/2)*e^(-2it) + i*(4/2)*e^(4it) - i*(4/2)*e^(-4it) f(t) = (3/2)*e^(2it) + (3/2)*e^(-2it) + 2i*e^(4it) - 2i*e^(-4it) Now, we can find the Fourier transform F(ω) using the definition: F(ω) = ∫_{-π}^{π} f(t)*e^(-iωt) dt Substitute f(t) into the equation: F(ω) = ∫_{-π}^{π} (3/2)*e^(2it) + (3/2)*e^(-2it) + 2i*e^(4it) - 2i*e^(-4it) * e^(-iωt) dt Next, integrate term by term. We'll use integration by substitution for each term. For the first term: Let u = 2it, then du/dt = 2i, and dt = du/(2i). The integral becomes: ∫ (3/2)*e^u * e^(-iωt) * (du/(2i)) from -2π to 2π Solve for u in terms of t: u = 2it -> t = -iu/(2i) -> t = u/(2i) The limits of integration become -π*i to π*i. So, the first integral is: (3/2) * (1/(2i)) * ∫ from -π*i to π*i e^(u*(1-iω/2)) du Now, integrate with respect to u: (3/(4i)) * e^(u*(1-iω/2)) / (1-iω/2) from -π*i to π*i Evaluate at the limits: (3/(4i)) * e^((π*i)*(1-iω/2)) / (1-iω/2) - e^((-π*i)*(1-iω/2)) / (1-iω/2) Simplify the expression: (3/(4i)) * 2i * (sin(π - ωπ/2) + i*cos(π - ωπ/2)) / (1 - iω/2) Factor out 2i: 3 * (sin(π - ωπ/2) + i*cos(π - ωπ/2)) / (2 - iω) Do the same process for the remaining terms, then add them together. The final result should be: F(ω) = 3*((sin(π - ωπ/2) + i*cos(π - ωπ/2)) / (2 - iω)) + ((sin(π + ωπ/2) + i*cos(π + ωπ/2)) / (-2 - iω)) - ((sin(4π - ωπ/2) + i*cos(4π - ωπ/2)) / (2 - iω)) + ((sin(4π + ωπ/2) + i*cos(4π + ωπ/2)) / (-2 - iω))

answer:To find the Fourier Transform of f(t) = 3cos(4πt) + 2sin(6πt), we can use Euler's formula to express the cosine and sine functions as complex exponentials. Euler's formula: e^(jθ) = cos(θ) + jsin(θ) So, cos(4πt) = Re{e^(j4πt)} sin(6πt) = Im{e^(j6πt)} Now, let's rewrite f(t): f(t) = 3Re{e^(j4πt)} + 2jIm{e^(j6πt)} To simplify this expression, we will multiply it by another complex exponential: e^(-jπt) * f(t) = 3Re{e^(j4πt) * e^(-jπt)} + 2jIm{e^(j6πt) * e^(-jπt)} = 3Re{e^(j3πt)} - 2jIm{e^(j5πt)} Now, we can find the Fourier Transform of f(t) using the following property of Fourier Transforms: If F(ω) is the Fourier Transform of f(t), then G(ω) = F(ω - ω0) is the Fourier Transform of g(t) = e^(jω0t) * f(t). Using this property, we can say that the Fourier Transform of e^(j3πt) is G1(ω) = F1(ω - 3π) and the Fourier Transform of e^(j5πt) is G2(ω) = F2(ω - 5π). Since our function g(t) has both a real and an imaginary part, we need to find their Fourier Transforms separately. Real part: G1(ω) = 3F1(ω - 3π) To find F1(ω), we need to recall that the Fourier Transform of cos(2πat) is: Fourier Transform of cos(2πat) = δ(ω - a) + δ(ω + a), where δ is the Dirac delta function. So, the Fourier Transform of cos(4πt) is: F1(ω) = δ(ω - 4) + δ(ω + 4) Now, we can find G1(ω): G1(ω) = 3F1(ω - 3π) = 3δ(ω - (3π + 4)) + δ(ω - (3π - 4)) Imaginary part: -2jG2(ω) = -2jF2(ω - 5π) To find F2(ω), we need to recall that the Fourier Transform of sin(2πbt) is: Fourier Transform of sin(2πbt) = jδ(ω - b) - δ(ω + b) So, the Fourier Transform of sin(6πt) is: F2(ω) = jδ(ω - 6) - δ(ω + 6) Now, we can find G2(ω): G2(ω) = -0.5jδ(ω - (5π + 6)) - δ(ω - (5π - 6)) Finally, combining the real and imaginary parts, we get the Fourier Transform of f(t): F(ω) = G1(ω) + G2(ω) = 3δ(ω - (3π + 4)) + δ(ω - (3π - 4)) - 0.5jδ(ω - (5π + 6)) - δ(ω - (5π - 6)),To find the Fourier Transform of the given function, we'll use the definition of the Fourier Transform: F(ω) = ∫f(t) * e^(-jωt) dt, where j is the imaginary unit and ω is the angular frequency. Given function: f(t) = 3cos(4πt) + 2sin(6πt) Let's find the Fourier Transform of each term separately and then add them. For the first term: 3cos(4πt) F1(ω) = ∫3cos(4πt) * e^(-jωt) dt = 3∫cos(4πt) * e^(-jωt) dt Now, we use the Euler's formula: cos(x) = (e^(jx) + e^(-jx))/2 F1(ω) = 3/2∫(e^(j4πt) + e^(-j4πt)) * e^(-jωt) dt = 3/2∫e^(j(4π-ω)t) + e^(-j(4π+ω)t) dt Now, we integrate each term separately: F1(ω) = 3/2 (1/j(4π-ω)) * e^(j(4π-ω)t) - (1/j(4π+ω)) * e^(-j(4π+ω)t) evaluated from -∞ to ∞ For the second term: 2sin(6πt) F2(ω) = ∫2sin(6πt) * e^(-jωt) dt = 2∫sin(6πt) * e^(-jωt) dt Now, we use the Euler's formula: sin(x) = (e^(jx) - e^(-jx))/(2j) F2(ω) = ∫(e^(j6πt) - e^(-j6πt)) * e^(-jωt) / (2j) dt = 1/j∫e^(j(6π-ω)t) - e^(-j(6π+ω)t) dt Now, we integrate each term separately: F2(ω) = 1/j (1/j(6π-ω)) * e^(j(6π-ω)t) + (1/j(6π+ω)) * e^(-j(6π+ω)t) evaluated from -∞ to ∞ Now, we add F1(ω) and F2(ω) to get the Fourier Transform of the given function: F(ω) = F1(ω) + F2(ω) F(ω) = 3/2 (1/j(4π-ω)) * e^(j(4π-ω)t) - (1/j(4π+ω)) * e^(-j(4π+ω)t) + 1/j (1/j(6π-ω)) * e^(j(6π-ω)t) + (1/j(6π+ω)) * e^(-j(6π+ω)t) This is the Fourier Transform of the given function f(t) = 3cos(4πt) + 2sin(6πt).,The Fourier Transform of a continuous-time signal can be calculated using the following formula: F(ω) = ∫f(t) * e^(-jωt) dt, where ω represents the angular frequency and j is the imaginary unit. For the given function f(t) = 3cos(4πt) + 2sin(6πt), let's calculate the Fourier Transform one by one. F1(ω) = ∫3cos(4πt) * e^(-jωt) dt Using Euler's formula, cos(4πt) can be rewritten as (1/2)(e^(j4πt) + e^(-j4πt)). So, the integral becomes: F1(ω) = (3/2)∫ e^(j4πt)e^(-jωt) + e^(-j4πt)e^(-jωt) dt Now, using distribution property of real and imaginary parts, we get: F1(ω) = (3/2)∫ e^(-j(ω-4π)t) dt + ∫ e^(-j(ω+4π)t) dt By solving the integrals, F1(ω) becomes: F1(ω) = (3/2)(2πδ(ω - 4π) + 2πδ(ω + 4π)) Since F(ω) = 2πδ(ω - 4π) - (1/4)δ(ω + 4π)δ(ω - 4π) F2(ω) = ∫2sin(6πt) * e^(-jωt) dt By using Euler's formula to convert sin(6πt) to appropriate complex exponentials, we can write sin(6πt) = (1/2j)(e^(j6πt) - e^(-j6πt)). Then, the integral becomes: F2(ω) = (1/j)∫ e^(j6πt)e^(-jωt) - e^(-j6πt)e^(-jωt) dt

answer:To find the Fourier transform of a function, we first need to define the Fourier transform. In general, the Fourier transform of a function f(t) is defined as: F(ω) = ∫ff(t)e^{-jωt}dt where F(ω) is the Fourier transform of the function f(t), ω is the angular frequency, and j is the imaginary unit. Now, let's find the Fourier transform of the given function f(t) = e^{-3|t|}. Since the function is even (i.e., f(t) = f(-t)), we can simplify the Fourier transform as follows: F(ω) = ∫∞(-∞)f(t)e^{-jωt}dt Since f(t) = e^{-3|t|}, we can split the integral into two parts for positive and negative values of t: F(ω) = ∫∞(0)e^{-3t}e^{-jωt}dt + ∫∞(0)e^{3t}e^{-jωt}dt Now, we know that: ∫∞(0)e^{-3t}e^{-jωt}dt = (1/(3+jω))e^{-3t} evaluated from 0 to ∞ = (1/(3+jω)) And ∫∞(0)e^{3t}e^{-jωt}dt = (1/(3-jω))e^{3t} evaluated from 0 to ∞ = (1/(3-jω)) Since the given function f(t) is an even function, we can reformulate the Fourier transform as: F(ω) = 2∫∞(0)f(t)cos(ωt)dt = (2/π)π(1/(3+jω) + 1/(3-jω)) F(ω) = (4/(3π))(1/(3+jω) + 1/(3-jω)) So, the Fourier transform of the function f(t) = e^{-3|t|} is: , To find the Fourier transform of f(t) = e^{-3|t|}, we'll first consider two cases: when t geq 0 and when t < 0. For t geq 0, we have f(t) = e^{-3t}. The Fourier Transform for this case is given by: F(omega) = int_{-infty}^{infty} f(t) e^{-iomega t} dt = int_{0}^{infty} e^{-(3+iomega)t} dt Now, let's solve the integral: F(omega) = left-frac{1}{3+iomega} e^{-(3+iomega)t}right_0^{infty} As t to infty, the exponential term goes to zero, so we get: F(omega) = -frac{1}{3+iomega} cdot 0 - left(-frac{1}{3+iomega} cdot 1right) = frac{1}{3+iomega} For t < 0, we have f(t) = e^{3t}. The Fourier Transform for this case is given by: F(omega) = int_{-infty}^{infty} f(t) e^{-iomega t} dt = int_{-infty}^{0} e^{3t} e^{-iomega t} dt Combining the exponents, we get: F(omega) = int_{-infty}^{0} e^{(3-iomega)t} dt Now, let's solve the integral: F(omega) = leftfrac{1}{3-iomega} e^{(3-iomega)t}right_{-infty}^{0} As t to -infty, the exponential term goes to zero, so we get: F(omega) = frac{1}{3-iomega} cdot 1 - left(frac{1}{3-iomega} cdot 0right) = frac{1}{3-iomega} Since f(t) is an even function (f(-t) = f(t)), its Fourier Transform is real and even. Therefore, we can write the Fourier Transform as: F(omega) = frac{1}{2}left(frac{1}{3-iomega} + frac{1}{3+iomega}right) Now, we'll combine the terms in the denominator using conjugates: F(omega) = frac{1}{2}left(frac{1}{3-iomega} cdot frac{3+iomega}{3+iomega} + frac{1}{3+iomega} cdot frac{3-iomega}{3-iomega}right) Simplifying further, we get: F(omega) = frac{1}{2}left(frac{3+iomega + 3-iomega}{(3^2 + omega^2)} right) = frac{3}{3^2 + omega^2} Thus, the Fourier Transform of f(t) = e^{-3|t|} is F(omega) = frac{3}{3^2 + omega^2}.,To find the Fourier transform of the function f(t) = e^{-3|t|}, we'll use the definition of the Fourier transform: F(ω) = ∫_{-∞}^{∞} f(t) e^{-jωt} dt Since f(t) = e^{-3|t|} has different expressions for t < 0 and t > 0, we'll split the integral into two parts: F(ω) = ∫_{-∞}^{0} e^{3t} e^{-jωt} dt + ∫_{0}^{∞} e^{-3t} e^{-jωt} dt Now, we'll integrate each part separately: For the first integral, let's substitute u = 3t - jωt: ∫_{-∞}^{0} e^{3t} e^{-jωt} dt = (1/3) ∫_{-∞}^{0} e^u du = (1/3) e^u_{-∞}^{0} = (1/3) (1 - 0) = 1/3 For the second integral, let's substitute v = -3t - jωt: ∫_{0}^{∞} e^{-3t} e^{-jωt} dt = (-1/3) ∫_{0}^{∞} e^v dv = (-1/3) -e^v_{0}^{∞} = (-1/3) (0 - 1) = 1/3 Now, we'll add the two parts together: F(ω) = 1/3 + 1/3 = 2/3 Therefore, the Fourier transform of the function f(t) = e^{-3|t|} is F(ω) = 2/3.