How to fix this error in interp1?

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)));  

[Jan]

zoom y lim = [-5,-1]

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

[Huijia Ma]

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

Answer 1

[Chunru]

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

[Huijia Ma]

Thank you very much! So, i want to get a figure like this, how to do this?

[Chunru]

See the updated.

[Huijia Ma]

really appreciate you

[Huijia Ma]

could you please tell me: if i want to plot a figure to show the gain(H = tf([R*C,0], [R*C,1])) of filter like this, how to do that

part of the code like this:

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;

[Chunru]

First find two measurement points:

[~, i1] = min(abs(f_vec - F1));
[~, i2] = min(abs(f_vec - F2));

Then plot the measurement response:

hold on
plot(fvec([i1 i2]), 20*log10(abs(resp([i1 i2]))), 'ro');   % check if u need log10

[Huijia Ma]

sorry, i don't what u mean? could you please tell me how to plot the green line. thank you

[Huijia Ma]

could you please explain what the min(abs(f_vec - F1)) mean? thanks

[Chunru]

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]

I see. Thank u very much. Do you know how to plot the green line