answer:The Fourier transform is not necessary to determine the total impedance of the network in this case. Instead, we can simply add the impedance values of the components together if they are connected in series, or use the formula for parallel impedance if they are connected in parallel. Assuming the components are connected in series, the total impedance (Z_total) can be calculated by adding the impedance values of each component: Z_total = (50 + j10) + (120 - j40) + (80 + j30) Z_total = 50 + j10 + 120 - j40 + 80 + j30 Z_total = (50 + 120 + 80) + j(10 - 40 + 30) Z_total = 250 + j0 ohms The total impedance of the network at a frequency of 400 Hz is 250 + j0 ohms if the components are connected in series. If the components are connected in parallel, the total impedance can be calculated using the formula for parallel impedance: 1/Z_total = 1/(50 + j10) + 1/(120 - j40) + 1/(80 + j30) To simplify the calculations, we can find the complex conjugate of each impedance and multiply both the numerator and denominator by it: 1/Z_total = [(50 - j10) + (120 + j40) + (80 - j30)] / [(50 + j10)(120 - j40)(80 + j30)] Now, we can multiply the numerators and denominators: Numerator = 50 - j10 + 120 + j40 + 80 - j30 = 250 Denominator = (50 + j10)(120 - j40)(80 + j30) To find the product of the denominators, we can multiply the complex numbers step by step: (50 + j10)(120 - j40) = 6000 - 2000j + 120j - 400j^2 = 6000 - 1880j + 400 (6400 - 1880j)(80 + j30) = 512000 + 192000j - 56400j - 21120j^2 = 512000 + 135600j + 21120 Now, we can divide the numerator by the denominator: Z_total = 250 / (532120 + 135600j) To find the magnitude and phase of the total impedance, we can use the polar form: Magnitude = |Z_total| = 250 / sqrt(532120^2 + 135600^2) ≈ 0.000467 ohms Phase = arctan(135600 / 532120) ≈ -14.5 degrees The total impedance of the network at a frequency of 400 Hz is approximately 0.000467 ohms at an angle of -14.5 degrees if the components are connected in parallel.

answer:To find the output of the network, we first need to find the Laplace transform of the input signal f(t) = 5cos(3t) + 2sin(4t). The Laplace transform of cos(at) is given by s/(s^2 + a^2) and the Laplace transform of sin(bt) is given by b/(s^2 + b^2). So, the Laplace transform of f(t) is: F(s) = 5 * (s/(s^2 + 3^2)) + 2 * (4/(s^2 + 4^2)) F(s) = 5s/(s^2 + 9) + 8/(s^2 + 16) Now, we need to find the output in the Laplace domain by multiplying the transfer function H(s) with the input F(s): Y(s) = H(s) * F(s) Y(s) = ((2s+3)/(s^2+2s+5)) * (5s/(s^2 + 9) + 8/(s^2 + 16)) Now, we need to find the inverse Laplace transform of Y(s) to get the output y(t) in the time domain. We can do this by performing partial fraction decomposition on Y(s) and then finding the inverse Laplace transform of each term. After performing partial fraction decomposition, we get: Y(s) = A/(s^2 + 9) + B/(s^2 + 16) + Cs/(s^2 + 2s + 5) + Ds/(s^2 + 9) + Es/(s^2 + 16) Now, we need to find the coefficients A, B, C, D, and E by equating the numerators of the original Y(s) and the decomposed Y(s). After solving for these coefficients, we get: A = 1.25, B = 0.5, C = 1, D = 1, E = -0.5 So, the decomposed Y(s) is: Y(s) = 1.25/(s^2 + 9) + 0.5/(s^2 + 16) + s/(s^2 + 2s + 5) + s/(s^2 + 9) - 0.5s/(s^2 + 16) Now, we can find the inverse Laplace transform of each term: y(t) = 1.25 * cos(3t) + 0.5 * cos(4t) + e^(-t) * cos(2t) + sin(3t) - 0.5 * sin(4t) So, the output of the network for the given input signal is: y(t) = 1.25cos(3t) + 0.5cos(4t) + e^(-t)cos(2t) + sin(3t) - 0.5sin(4t)

answer:To determine the overall frequency response of the network, we need to multiply the transfer functions of each filter since they are connected in series: H(w) = H1(w) * H2(w) * H3(w) * H4(w) * H5(w) * H6(w) Now, let's calculate the magnitude and phase response at the frequency of 3 Hz (i.e., w = 6π). H(6π) = H1(6π) * H2(6π) * H3(6π) * H4(6π) * H5(6π) * H6(6π) H1(6π) = 2cos(4 * 6π) = 2cos(24π) = 2 H2(6π) = sin(2 * 6π) = sin(12π) = 0 H3(6π) = 3cos(6 * 6π) = 3cos(36π) = 3 H4(6π) = sin(4 * 6π) = sin(24π) = 0 H5(6π) = 4cos(8 * 6π) = 4cos(48π) = 4 H6(6π) = sin(6 * 6π) = sin(36π) = 0 H(6π) = 2 * 0 * 3 * 0 * 4 * 0 = 0 The overall frequency response of the network at 3 Hz (w = 6π) is 0. Since the magnitude response is 0, there is no need to calculate the phase response as it will not affect the output signal. The network completely attenuates the signal at 3 Hz.

answer:To find the amplitude and phase angle of the input signal that will result in a maximum output amplitude of 10, we first need to find the magnitude of the transfer function H(ω). The magnitude of H(ω) can be found using the Pythagorean theorem: |H(ω)| = √((5cos(2ω))^2 + (3sin(4ω))^2) |H(ω)| = √(25cos^2(2ω) + 9sin^2(4ω)) Now, we want to find the maximum value of |H(ω)|. To do this, we can differentiate |H(ω)| with respect to ω and set the result to zero: d|H(ω)|/dω = 0 However, finding the derivative of the square root function can be quite complicated. Instead, we can find the maximum value of the function inside the square root, which will also give us the maximum value of |H(ω)|: d(25cos^2(2ω) + 9sin^2(4ω))/dω = 0 Now, we can find the derivative: d(25cos^2(2ω))/dω = -100cos(2ω)sin(2ω) d(9sin^2(4ω))/dω = 72sin(4ω)cos(4ω) So, the derivative of the function inside the square root is: -100cos(2ω)sin(2ω) + 72sin(4ω)cos(4ω) = 0 Now, we need to find the value of ω that satisfies this equation. This can be done using numerical methods or by observation. In this case, we can observe that when ω = π/4, the equation is satisfied: -100cos(π/2)sin(π/2) + 72sin(π)cos(π) = 0 Now that we have the value of ω that maximizes |H(ω)|, we can find the maximum value of |H(ω)|: |H(π/4)| = √(25cos^2(π/2) + 9sin^2(2π)) |H(π/4)| = √(25(0)^2 + 9(1)^2) |H(π/4)| = √9 |H(π/4)| = 3 Now, we want the output amplitude to be 10. Since the output amplitude is the product of the input amplitude and the transfer function magnitude, we can find the input amplitude A: A * |H(π/4)| = 10 A * 3 = 10 A = 10/3 So, the input amplitude is 10/3. Now, we need to find the phase angle of the input signal. Since the transfer function is a linear combination of cosines and sines, the phase angle will be the same as the phase angle of the transfer function at ω = π/4: H(π/4) = 5cos(π/2) + 3sin(π) H(π/4) = 5(0) + 3(0) H(π/4) = 0 Since H(π/4) = 0, the phase angle of the input signal is the same as the phase angle of the transfer function, which is 0. In conclusion, the amplitude and phase angle of the input signal that will result in a maximum output amplitude of 10 are 10/3 and 0, respectively.