plot discrete time domain signals
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Given n=[-1:0.01:10]. Plot the discrete time domain signal: y[n]=e^(-n)*u(n).
Below is my program but i'm not sure if it's correct since this is my fisrt time learning Matlab. Please help. Thanks!
n=-1:0.01:10;
y=exp(-n).*heaviside(n);
stem(t,y)
1 个评论
Be careful using heaviside for u[n]. In control systems and signal processing the function u[n] is the unit step function that is (typically) defined as
u[n] = 1 for n >= 0
u[n] = 0 for n < 0.
However, the default defintion of the heaviside function in Matlab has heaviside(0) = 0.5., which is clearly seen in the plot. Is that the desired answer?
Also, it looks a bit peculiar to have non-integer values of n. Please make sure that's the correct problem statement.
采纳的回答
Paul Hoffrichter
2021-2-14
I recommend using following substitutions:
plot(t,y)
axis( [-1 10 0 1])
6 个评论
Jinquan Li
2021-2-14
Paul Hoffrichter
2021-2-14
编辑：Paul Hoffrichter
2021-2-14
Your program looks OK. It is your choice (or an instructor's choice) as to whether you should use plot or stem. Here is an explanation of the differences.
With your x-limits, you cannot make out the stem distinct lines. If you zoom in, you will be able to see the stem vertical lines better.
If you want to use plot and also see the distinct discrete y-values, you can do this so that each (t,y) value has a dot that are connected with lines to the next dot.
plot(t,y, '-b.)
Jinquan Li
2021-2-14
Paul Hoffrichter
2021-2-14
Your very welcome!
Paul Hoffrichter
2021-2-15
Just be careful as to how u(n) is defined. As @Paul mentioned, "heaviside(0) = 0.5., which is clearly seen in the plot. Is that the desired answer?". As you know, e^(-0) is 1, not 0.5, so it is up to you to decide whether your u(n) is defined the way you want it to be defined. If it is a unit step function, where u(0) is 1, then using the heaviside without adjustments leads to an incorrect value for y(n) when n is 0.
Paul Hoffrichter
2021-2-15
Take a look at
The following defines the myStep function so that myStep(0) is 0 instead of 0.5.
myStep = @(n) (n>0);
If you wanted a value of 1 at n = 0, then you could use this instead:
myStep1 = @(n) (n>=0);
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