How on earth to graph discrete time signals?

Question

Alex Miller 2021-9-8

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https://ww2.mathworks.cn/matlabcentral/answers/1449524-how-on-earth-to-graph-discrete-time-signals

编辑： Paul 2021-9-8

I'm trying to plot this discrete time function in MATLAB. I missed the last several of my classes due to sickness and as such have quite literally no idea what I'm doing. This is as far as I've gotten-

n = -5:1:5;

x = ((1 / 7) ^ -n) * heaviside(-n) + ((2 / 3) ^ n) * heaviside(n) + ((3/5) ^ n) * heaviside(n);

figure(1);

plot(t,y, '-b')

axis( [-1 10 0 1]);

This has been slowly cobbled together from other forum posts, and as such I don't understand it and it doesn't work. I'm pretty quickly running out of things to try. Any help?

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Answer 1

Star Strider 2021-9-8

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The only problems were in not using element-wise operations (use the ‘dot operator’ with the exponentiation an d multiplication operators), and presenting the correct arguments to plot. With those corrections (the only ones I made to the posted code), the original code works. See Array vs. Matrix Operations to understand about element-wise operations.

n = -5:1:5;

x = ((1 / 7) .^ -n) .* heaviside(-n) + ((2 / 3) .^ n) .* heaviside(n) + ((3/5) .^ n) .* heaviside(n);

figure(1);

plot(n, x, '-b')

Another way to plot this is to use the stem fucntion —

figure(2);

stem(n, x, '-b', 'filled')

That sometimes makes more sense when plotting discrete values.

..

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Star Strider 2021-9-8

Thank you!

(For the record, I’m a physician. That’s what we do!)

.

Paul 2021-9-8

编辑：Paul 2021-9-8

在 MATLAB Online 中打开

One needs to be careful using the heaviside() function for discrete time problems. Unless the sympref is changed from the default, we see that

heaviside(0)
ans = 0.5000

Consequently, the code returns 1.5 at n=0.

n = -5:1:5;
x = ((1 / 7) .^ -n) .* heaviside(-n) + ((2 / 3) .^ n) .* heaviside(n) + ((3/5) .^ n) .* heaviside(n);
[n;x]
ans = 2×11
   -5.0000   -4.0000   -3.0000   -2.0000   -1.0000         0    1.0000    2.0000    3.0000    4.0000    5.0000
    0.0001    0.0004    0.0029    0.0204    0.1429    1.5000    1.2667    0.8044    0.5123    0.3271    0.2094

However, it's possible, perhaps even likely, that @Alex Miller's course uses the typical convention (at least in signals and controls) that u[0] = 1, in which case the result should be x[0] = 3. In either case, beware of using the heaviside() function for discrete time problems; make sure that heavisde(0) is consistent with the definition of the unit step function in the problem statement.

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