POWER SYSTEM ANALYSIS+DESIGN-EBK >I<
POWER SYSTEM ANALYSIS+DESIGN-EBK >I<
6th Edition
ISBN: 9781337259170
Author: Glover
Publisher: INTER CENG
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a b 1 ΚΩΣ 56002 82092 470Ω Rab, Rbc, Rde d e O 470Ω Σ 5 Ω 25$ 5602 3 4 Ω
MY code is experiencing a problem as I want to show both the magnitude ratio on low pass, high pass, and bandbass based on passive filters: Code: % Define frequency range for the plot f = logspace(1, 5, 500); % Frequency range from 10 Hz to 100 kHz w = 2*pi*f; % Angular frequency   % Parameters for the filters (you can modify these) R = 1e3; % Resistance in ohms (1 kOhm) C = 1e-6; % Capacitance in farads (1 uF) L = 10e-3; % Inductance in henries (10 mH)   % Transfer function for Low-pass filter: H_low = 1 / (1 + jωRC) H_low = 1 ./ (1 + 1i*w*R*C);   % Transfer function for High-pass filter: H_high = jωRC / (1 + jωRC) H_high = 1i*w*R*C ./ (1 + 1i*w*R*C);   % Transfer function for Band-pass filter: H_band = jωRC / (1 + jωL/R + jωRC) H_band = 1i*w*R*C ./ (1 + 1i*w*L/R + 1i*w*R*C);   % Plot magnitude responses figure; subplot(3,1,1); semilogx(f, 20*log10(abs(H_low))); % Low-pass filter title('Magnitude Response of Low-pass Filter'); xlabel('Frequency (Hz)'); ylabel('Magnitude (dB)'); grid…
*10. For the network of Fig. 7.83, determine: a. Ip. b. VDS. c. VD. d. Vs. 20 V 2.2 ΚΩ ID -4 V IDSS = 4.5 mA VDS Vp = -5V 0.68 ΚΩ
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