# How to generate a vector of a shifted impulse function?

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FW on 27 Jul 2020
Commented: Star Strider on 27 Jul 2020
Hello, Suppose we have a time vector x=0:0.1: 50. I would like to have a delta function at a non-zero position, say at 25 with unit height (or any other scaled version of it).
MATLAB has a function d = dirac(x)
It generates dirac at x=0. If we write, d=dirac(x-25), it does not shift the impulse function like the H=heaviside(t-25) translates the heaviside function at 25.
I tried differentiating the translated heaviside function but I get 0.5 0.5 at the desired location instead of 1 at 25, no matter what the sampling frequency is.
Is there are a better way to do
(a) Generate a vector unit delta at a non-zero position
(b) Differentiate translated Heaviside and get a shifted delta at the desired position.
Thanks.

Star Strider on 27 Jul 2020
If we write, d=dirac(x-25), it does not shift the impulse function like the H=heaviside(t-25) translates the heaviside function at 25.
It does, actually.
Consider:
x = 0:0.1:50;
d = dirac(x - 25);
nzdidx = find(d>0) % Index
dnzd = d(nzdidx) % Value
producing:
nzdidx =
251
dnzd =
Inf
So it will not appear on the plot, since it has infinite amplitude and 0 width, integrating to an area of 1.
.

Star Strider on 27 Jul 2020
As always, my pleasure!
I am not certain what you are doing.
Try this:
syms x
h = heaviside(x - 1)
dh = diff(h)
to get:
dh =
dirac(x - 1)
(I am using R2020a, although I doube that there would be any version differences.)
.
FW on 27 Jul 2020
Strider, I was trying to demonstrate the effect of convolution of a Lorentzian with a shifted delta to a person. I was stumped as to how one can generate a displaced delta vector in MATLAB. Thanks for your assistance. I have MATLAB 2019b. I can download 2020 but I don't think it is worth changing.
Star Strider on 27 Jul 2020
As always, my pleasure!
As for upgrading to R2020a, see the Release Notes to see if it would be of any benefit to you. (Note that Update 4 is current.)

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