How to fix this error in interp1?

2 次查看（过去 30 天）
Huijia Ma
Huijia Ma 2021-11-16
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)));
  2 个评论
Jan
Jan 2021-11-18
zoom y lim = [-5,-1]
This is no valid Matlab code: variables cannot contain spaces.
Huijia Ma
Huijia Ma 2021-11-18
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

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Chunru
Chunru 2021-11-16
编辑：Chunru 2021-11-17
Ran in:
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
  9 个评论
显示 7更早的评论
Chunru
Chunru 2021-11-20
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
Huijia Ma 2021-11-20
I see. Thank u very much. Do you know how to plot the green line

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Plot Customization 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by