How to fix this error in interp1?

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Huijia Ma 2021-11-16

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1588039-how-to-fix-this-error-in-interp1

评论： Huijia Ma 2021-11-20

this is my code, i want to zoom in the db from -5 to -1, there is a problem that error using interp1 X mush be a vector, could you please help me?

R = 1e3; % resistor value [Ohms]
C = 1e-6; % Capacitor value [Farads]
H = tf([R*C,0], [R*C,1]);
% plot a bode plot (3dB frequency should be at 1/RC = 1000 rads/sec.
[mag,phase,wout] = bode(H);  
zoom y lim = [-5,-1]; % y-axis range for zoom in section
min max w zoom = interp1(20*log10(mag),wout,zoom y lim);
indexOfInterest = (wout > min(min max w zoom)) & (wout < max(min max w zoom));
plot(wout(indexOfInterest),20*log10(mag(indexOfInterest)));  

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Jan 2021-11-18

在 MATLAB Online 中打开

zoom y lim = [-5,-1]

This is no valid Matlab code: variables cannot contain spaces.

Huijia Ma 2021-11-18

在 MATLAB Online 中打开

if i want to plot a figure to show the gain(H = tf([R*C,0], [R*C,1])) of filter like this (green line), how to do that

part of the code is:

y_vec = lsim(H,x_vec,t_vec);
% Compute the input and outputs in the frequency domain
in_f = fft(x_vec)/length(x_vec);
out_f = fft(y_vec)/length(x_vec);
resp = out_f./in_f;
resp(abs(out_f)<1e-4) = nan;

some variables are mean as follows: x_vec in this figure ocmbine two signals, but i just want to plot a figure that the gain changes with frequency

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回答（1 个）

Chunru 2021-11-16

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1588039-how-to-fix-this-error-in-interp1#answer_832724

编辑：Chunru 2021-11-17

在 MATLAB Online 中打开

R = 1e3; % resistor value [Ohms]

C = 1e-6; % Capacitor value [Farads]

H = tf([R*C,0], [R*C,1]);

% plot a bode plot (3dB frequency should be at 1/RC = 1000 rads/sec.

[mag,phase,wout] = bode(H);

wout = squeeze(wout);

mag = squeeze(mag);

plot(wout, 20*log10(mag), 'b:');

zoom_y_lim = (-5:.1:-1); % y-axis range for zoom in section

w_zoom = interp1(20*log10(mag), wout, zoom_y_lim);

whos

Name Size Bytes Class Attributes C 1x1 8 double H 1x1 1281 tf R 1x1 8 double mag 55x1 440 double phase 1x1x55 440 double w_zoom 1x41 328 double wout 55x1 440 double zoom_y_lim 1x41 328 double

hold on

plot(w_zoom, zoom_y_lim, 'r-', 'Linewidth', 2);

ylim([-8 0])

xlim( [0 1]*1e4)

figure

semilogx(wout, 20*log10(mag), 'b'); grid on

ax = axes('Pos', [.4 .2 .4 .4], 'Box', 'on');

semilogx(w_zoom, zoom_y_lim); grid on

Let's assume that f_vec=[0 1 2 3 4 5 6 7]; F1=4.99; min(abs(f_vec-F1)) is to find the closest point in f_vec to F1. In this case, it is f_vec(6)=5 which is closest to 4.99. The whole statement of [~, i1]=min(abs(f_vec-F1)) will assign the index 6 to i1.

Huijia Ma 2021-11-20