Why do I get an error on line 17?

2 次查看(过去 30 天)
Porgs
Porgs 2018-5-7
评论: KSSV 2018-5-7
clc
deviation = (1) .^ 0.5;
n = deviation .* randn(100000, 1); % noise
t = [-5: 0.01: 4.99]; % time
p = 5 .* sin(20 .* pi .* t + (pi / 4)); % pulse p(t)
T = (2 * 1500) / 1000;
t2 = t - T;
p2 = 5 * sin(20 .* pi .* t2 + (pi / 4));
r = p2 + n;
x = randn(length(t), 1);
i = (t - T);
y = linfilt(x, i, length(t));
figure(1);
subplot(3, 1, 1);
plot(t, p);
title('Plot 4a: Pulse p(t)'); % pulse p(t)
xlabel('Time');
ylabel('Amplitude');
grid;
subplot(3, 1, 2);
plot(t2, r);
title('Plot 4b: Received Signal r(t)'); % received signal r(t)
xlabel('Time');
ylabel('Amplitude');
axis([-5 4.99 min(y) max(y)]);
grid;
[C, tau] = xcorr(r,p); % cross-correlation R_rp(tau)
subplot(3, 1, 3);
plot(tau, C);
title('Plot 3c: Cross-Correlation R_rp(tau)');
grid;
  2 个评论
Walter Roberson
Walter Roberson 2018-5-7
We do not know which is line 17. Also, linfit() is not a MATLAB routine. There are several linfit() in the File Exchange, but we would not know which one of those you are using.
KSSV
KSSV 2018-5-7
?What is this function linfilt?

请先登录,再进行评论。

回答(0 个)

类别

Help CenterFile Exchange 中查找有关 Graphics Object Identification 的更多信息

Community Treasure Hunt

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

Start Hunting!

Translated by