when and how to use dot multiplication

Question

Matt Thomas 2022-5-13

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评论： John D'Errico 2023-8-12

ok, for my first practice homeowrk, since i have no idea what i am doing in matlab, i was asked to plot the response of a RLC equation that is in the time domain. I got the correct answer, but im not sure what the syntax is doing and if it is needed.

q0=10;R=60;L=9;C=0.0005;

t=linspace(0,0.8)

q=q0*exp(-R*t/(2*L)).*cos(sqrt(1/(L*C)-(R/(2*L))^2)*t);

plot(t,q)

in the equation q, there is the .* operation. Is that even needed if a matrix was not even used? im confused as when to use .* vs *

i know if i had

a=[1 2 3]

b=[5 6 7]

a.*b

would return

5 12 21

but is .* needed in my above q equation?

thanks

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Answer 1

Voss 2022-5-13

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在 MATLAB Online 中打开

"but is .* needed in my above q equation?"

Yes.

t is a vector, and q0, R, L, and C are all scalars.

q0=10;R=60;L=9;C=0.0005; % scalars
t=linspace(0,0.8); % 1-by-100 vector

Consider the expressions to the left and to the right of the .*, which both include t and some scalars:

q0*exp(-R*t/(2*L)) % vector the size of t
ans = 1×100
   10.0000    9.7342    9.4755    9.2237    8.9786    8.7400    8.5077    8.2816    8.0615    7.8472    7.6387    7.4357    7.2381    7.0457    6.8584    6.6762    6.4987    6.3260    6.1579    5.9942    5.8349    5.6799    5.5289    5.3820    5.2389    5.0997    4.9642    4.8323    4.7038    4.5788
cos(sqrt(1/(L*C)-(R/(2*L))^2)*t) % vector the size of t
ans = 1×100
    1.0000    0.9931    0.9726    0.9386    0.8917    0.8326    0.7620    0.6808    0.5904    0.4917    0.3864    0.2757    0.1612    0.0444   -0.0729   -0.1892   -0.3029   -0.4125   -0.5164   -0.6131   -0.7015   -0.7801   -0.8480   -0.9043   -0.9481   -0.9788   -0.9961   -0.9996   -0.9894   -0.9655

They are both vectors the size of t, so when you multiply those two vectors, you want to do it element-wise, which is what .* does

q=q0*exp(-R*t/(2*L)).*cos(sqrt(1/(L*C)-(R/(2*L))^2)*t)
q = 1×100
   10.0000    9.6672    9.2155    8.6574    8.0065    7.2767    6.4825    5.6385    4.7592    3.8588    2.9513    2.0497    1.1664    0.3131   -0.5000   -1.2633   -1.9688   -2.6095   -3.1798   -3.6753   -4.0929   -4.4309   -4.6887   -4.8668   -4.9668   -4.9916   -4.9446   -4.8303   -4.6538   -4.4210

giving you another vector the size of t.

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Voss 2022-5-13

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"10 adnd 1 share the same place holder so multiply them out and 9.7 with .9 9.4 with .97 etc etc"

Exactly.

You want to calculate q for each t here:

q=q0*exp(-R*t/(2*L)).*cos(sqrt(1/(L*C)-(R/(2*L))^2)*t);

and you don't want any element of the result q to come from any "mixing" of different elements of t - it wouldn't make sense. But that's what matrix multiplication (*) would do if you use it, by the definition of matrix multiplication. Instead, you want all the elements of t to be used independently of each other, so that's element-wise (.*), as you describe.

[For scalars being multiplied by anything (i.e., another scalar, a vector, a matrix, or even a higher-dimensional array), it doesn't matter if you use * or .* because there's only one thing it can mean: multiply every element of the thing (scalar, vector, matrix, nd-array) by the scalar. That's why the other occurrences of * in the expression for q are ok - they all involve multiplying two scalars or multiplying 1 scalar and 1 vector. (Any or all of them could be .* instead and that would be ok too.) Only the .* between the exp() and the cos() expressions involves multiplying two vectors.]

Voss 2022-5-13

在 MATLAB Online 中打开

"so a scalar is just a single varible defined as a single number such as a=3?"

That's right, since we've only been talking about numbers. But technically you can have scalar (and vector, and matrix, and higher-dimensional arrays) other things besides numbers, e.g., a scalar character:

c = 'a'
c = 'a'

or a scalar string:

s = "Alice"
s = "Alice"

or a scalar cell array:

C = {'Bob'}
C = 1×1 cell array
    {'Bob'}

or a scalar struct:

S = struct('name','Bob','age',24)
S = struct with fields:
    name: 'Bob'
     age: 24

etc. Or you can have vectors of those things:

c = 'abc'
c = 'abc'
s = ["Alice" "Bob" "Charlie"]
s = 1×3 string array
    "Alice"    "Bob"    "Charlie"
C = {'Alice' 'Bob' 'Charlie'}
C = 1×3 cell array
    {'Alice'}    {'Bob'}    {'Charlie'}
S = struct('name',{'Alice' 'Bob' 'Charlie'},'age',{22 24 34})
S = 1×3 struct array with fields:
    name
    age

Or 2-dimensional (or higher) arrays of those things (or any other type of thing) too.

m = magic(5) % matrix of numbers
m = 5×5
    17    24     1     8    15
    23     5     7    14    16
     4     6    13    20    22
    10    12    19    21     3
    11    18    25     2     9
c = num2str(m) % 2d character array (not usually called a "matrix of characters" but it really is)
c = 5×18 char array
    '17  24   1   8  15'
    '23   5   7  14  16'
    ' 4   6  13  20  22'
    '10  12  19  21   3'
    '11  18  25   2   9'
C = arrayfun(@num2str,m,'UniformOutput',false) % 2d cell array (again, not usually called a "cell matrix" but it is)
C = 5×5 cell array
    {'17'}    {'24'}    {'1' }    {'8' }    {'15'}
    {'23'}    {'5' }    {'7' }    {'14'}    {'16'}
    {'4' }    {'6' }    {'13'}    {'20'}    {'22'}
    {'10'}    {'12'}    {'19'}    {'21'}    {'3' }
    {'11'}    {'18'}    {'25'}    {'2' }    {'9' }
S = struct('value',C) % 2d struct array
S = 5×5 struct array with fields:
    value

So the terms "scalar", "vector", "matrix", "array" don't imply anything about the class of the variable, only the dimensionality.

a scalar is 1-by-1
a vector is n-by-1 (column vector) or 1-by-n (row vector), for some n
a matrix is m-by-n, for some m and n (a lot of times people use the term "matrix" to imply a matrix of numbers, but it doesn't have to be)
an array is a generic term that can refer to an object of any dimensionality

More info about this terminology: https://www.mathworks.com/matlabcentral/answers/505251-scalar-vs-matrix-vs-vector-vs-array

"thanks again for all the help. I will most likely be an active member on here."

You're welcome! (for the help), and welcome to MATLAB Answers. By the way, if someone here helps you out or solves your problem or answers your question, it's courteous to Vote for their answer and/or click "Accept This Answer". (And in the case of accepting an answer it helps future visitors who have the same problem/question know that the given solution works.)