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It is usually a combination of a Bode magnitude plot, expressing the magnitude (usually in decibels) of the frequency response, and a Bode phase plot, expressing the phase shift. As originally conceived by Hendrik Wade Bode in the 1930s, the plot is an asymptotic approximation of the frequency response, using straight line segments .
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The final step depends on the geometry of the waveguide. The easiest geometry to solve is the rectangular waveguide. In that case, the remainder of the Laplacian can be evaluated to its characteristic equation by considering solutions of the form ψ ( x , y , z , t ) = ψ 0 e i ( ω t − k z z − k x x − k y y ) . {\displaystyle \psi (x,y,z ...
The following Python code can also be used to calculate and plot the root locus of the closed-loop transfer function using the Python Control Systems Library [14] and Matplotlib [15]. import control as ct import matplotlib.pyplot as plt # Define the transfer function sys = ct .
The procedure outlined in the Bode plot article is followed. Figure 5 is the Bode gain plot for the two-pole amplifier in the range of frequencies up to the second pole position. The assumption behind Figure 5 is that the frequency f 0 dB lies between the lowest pole at f 1 = 1/(2πτ 1) and the second pole at f 2 = 1/(2πτ 2). As indicated in ...
If the transfer function of a first-order low-pass filter has a zero as well as a pole, the Bode plot flattens out again, at some maximum attenuation of high frequencies; such an effect is caused for example by a little bit of the input leaking around the one-pole filter; this one-pole–one-zero filter is still a first-order low-pass.
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We may graphically solve for as the intersection of two curves in the (,) plane: {= + = For fixed ,,, the second curve is a fixed hyperbola in the first quadrant. The first curve is a parabola with shape y = 9 16 β 2 ( z 2 ) 2 {\textstyle y={\tfrac {9}{16}}\beta ^{2}(z^{2})^{2}} , and apex at location ( 4 3 β ( ω 2 − α ) , δ 2 ω 2 ...