Vectors must be the same length

Question

Gizem Karslioglu 2023-2-19

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1915320-vectors-must-be-the-same-length

评论： Gizem Karslioglu 2023-2-19

I am receiving a warning ' Vectors must be the same length' for 'plot(n,X,'LineWidth', 2)' . I have tried linspace but it didn't work. How can I solve it?

%Defining variables and limits
n = -40:40
w=[-50,50]
nlower=n(1)
nupper=n(end)
% Writing the x signal function x(n) = n(0.9)^n [u(n)-u(n-21)]
subplot(3,1,1)
x = n .*(0.9) .^ n.* (stepseq(0,nlower,nupper)-stepseq(21,nlower,nupper))
% Applying DTFT
if -50<= n <=50,
    X = dtft(x,n,w);
    plot(n,X,'LineWidth', 2)
    grid on
    grid minor
    set(gca,'Fontsize', 8);
    xlabel('\omega / \pi');
    ylabel('|X(n)|');

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Answer 1

Walter Roberson 2023-2-19

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https://ww2.mathworks.cn/matlabcentral/answers/1915320-vectors-must-be-the-same-length#answer_1175055

在 MATLAB Online 中打开

if -50<= n <=50,

MATLAB interprets that as

if ((-50<= n) <=50)

The first part compares every element in n to -50. You defined n as -40:40 so it is true that -50<=n for every entry in n. The result of that part of the expression is an array of logical true values the same size as n. That array of logical true values is then compared <= 50. logical true in mathematical contexts converts to 1 (false converts to 0) and for each of those true values from the first part of the expression, true <= 50 is true, so the result of the <= tests is going to be an array of true values the same size as n. Then when if is asked to test a non-scalar, the result is considered true only of all of the values being tested are non-zero. Which is the case because they are all true. So the if will pass, and might as well not have been there.

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Walter Roberson 2023-2-19

在 MATLAB Online 中打开

stepseq is not part of MATLAB, but possibly you copied it from a source such as https://www.mathworks.com/matlabcentral/answers/819500-how-do-i-multiply-sin-and-stepseq-functions#answer_689890

And perhaps you got dtft from a source such as https://www.mathworks.com/matlabcentral/answers/469374-dtft-possible-on-matlab#answer_426731

%Defining variables and limits

n = -40:40

n = 1×81

-40 -39 -38 -37 -36 -35 -34 -33 -32 -31 -30 -29 -28 -27 -26 -25 -24 -23 -22 -21 -20 -19 -18 -17 -16 -15 -14 -13 -12 -11

w=[-50,50]

w = 1×2

-50 50

nlower=n(1)

nlower = -40

nupper=n(end)

nupper = 40

% Writing the x signal function x(n) = n(0.9)^n [u(n)-u(n-21)]

subplot(3,1,1)

x = n .*(0.9) .^ n.* (stepseq(0,nlower,nupper)-stepseq(21,nlower,nupper))

x = 1×81

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

% Applying DTFT

if -50<= n <=50,

X = dtft(x,n,w);

whos n x X

plot(n,X,'LineWidth', 2)

grid on

grid minor

set(gca,'Fontsize', 8);

xlabel('\omega / \pi');

ylabel('|X(n)|');

end

Name Size Bytes Class Attributes X 2x1 32 double complex n 1x81 648 double x 1x81 648 double

Error using plot
Vectors must be the same length.

function [x,n] = stepseq(n0,n1,n2) % Copied From: <https://www.mathworks.com/matlabcentral/answers/12834-unsolved-function?s_tid=srchtitle#>

% Generates x(n) = u(n-n0); n1 <= n,n0 <= n2

% ------------------------------------------

% [x,n] = stepseq(n0,n1,n2)

%

if ((n0 < n1) | (n0 > n2) | (n1 > n2))

error('arguments must satisfy n1 <= n0 <= n2')

end

n = [n1:n2];

%x = [zeros(1,(n0-n1)), ones(1,(n2-n0+1))];

x = [(n-n0) >= 0];

end

function [X] = dtft(x,n,w)

% Computes Discrete-time Fourier Transform

% [X] = dtft(x,n,w)

% X = DTFT values computed at w frequencies

% x = finite duration sequence over n

% n = sample position vector

% w = frequency location vector

X = exp(-1i*w'*n) * x.';

% X = x*exp(-j*n'*w);

end

What went wrong? Well, you passed in only two values for the frequency location vector, so the output is going to be length 2.

The output of dtft is not the same size as the finite x inputs or the sample positions: it is the same size as the frequency vector. You should be plotting w against X, not n against X. And you will probably want more frequencies in your list.

Walter Roberson 2023-2-19

在 MATLAB Online 中打开

w = linspace(-50, 50, length(n));

However if you do this, you are likely to get confused between positions (n values) and frequencies (w values).

When you have M = length(n) input positions, you cannot justify more than M/2 frequencies as being fully resolved; any more than that you run into Nyquist limit problems. Even M/2 is probably an over-estimate.

Gizem Karslioglu 2023-2-19

Is there any way to make value of all variables matched? When I try to manually change n and w, r never gets the same value.

I ended up below plots if I plot(w,X). But, my goal is plotting (n,X) and I am expecting to see a sinusoidal signal .

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Answer 2

Chawa Mozy 2023-2-19

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1915320-vectors-must-be-the-same-length#answer_1175070

% Define variables and limits n = -40:40; w = [-50, 50]; nlower = n(1); nupper = n(end);

% Define the signal x(n) x = n .* (0.9) .^ n .* (stepseq(0, nlower, nupper) - stepseq(21, nlower, nupper));

% Compute and plot the magnitude and phase spectra using the DTFT X = dtft(x, n, w);

subplot(3, 1, 1) plot(n, x, 'LineWidth', 2) grid on grid minor set(gca, 'Fontsize', 8); xlabel('n'); ylabel('x(n)'); title('Signal x(n)');

subplot(3, 1, 2) plot(w, abs(X), 'LineWidth', 2); grid on grid minor set(gca, 'Fontsize', 8); xlabel('\omega / \pi'); ylabel('|X(\omega)|'); title('Magnitude Spectrum');

subplot(3, 1, 3) plot(w, angle(X), 'LineWidth', 2); grid on grid minor set(gca, 'Fontsize', 8); xlabel('\omega / \pi'); ylabel('\angle X(\omega)'); title('Phase Spectrum');

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Vectors must be the same length

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Vectors must be the same length

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