Use units of measurement with Symbolic Math Toolbox™. This page shows how to define units, use units in equations (including differential equations), and verify the dimensions of expressions.
Load units by using symunit
.
u = symunit;
Specify a unit by using u.
unit
. For example,
specify a distance of 5
meters, a weight of 50
kilograms, and a speed of 10
kilometers per hour. In displayed output,
units are placed in square brackets []
.
d = 5*u.m w = 50*u.kg s = 10*u.km/u.hr
d = 5*[m] w = 50*[kg] s = 10*([km]/[h])
Tip
Use tab expansion to find names of units. Type u.
,
press Tab, and continue typing.
Units are treated like other symbolic expressions and can be used in any standard operation or function. Units are not automatically simplified, which provides flexibility. Common alternate names for units are supported. Plurals are not supported.
Add 500
meters and 2
kilometers. The resulting
distance is not automatically simplified.
d = 500*u.m + 2*u.km
d = 2*[km] + 500*[m]
Simplify d
by using simplify
. The
simplify
function automatically chooses the unit to simplify to.
d = simplify(d)
d = (5/2)*[km]
Instead of automatically choosing a unit, convert d
to a specific unit
by using unitConvert
. Convert d
to meters.
d = unitConvert(d,u.m)
d = 2500*[m]
There are more unit conversion and unit system options. See Unit Conversions and Unit Systems.
Find the speed if the distance d
is crossed in 50
seconds. The result has the correct units.
t = 50*u.s; s = d/t
s = 50*([m]/[s])
By default, temperatures are assumed to represent differences and not absolute
measurements. For example, 5*u.Celsius
is assumed to represent a
temperature difference of 5 degrees Celsius. This assumption allows arithmetical operations on
temperature values.
To represent absolute temperatures, use kelvin, so that you do not have to distinguish an absolute temperature from a temperature difference.
Convert 23
degrees Celsius to kelvin, treating it first as a
temperature difference and then as an absolute temperature.
u = symunit; T = 23*u.Celsius; diffK = unitConvert(T,u.K)
diffK = 23*[K]
absK = unitConvert(T,u.K,'Temperature','absolute')
absK = (5923/20)*[K]
In longer expressions, visually checking for units is difficult. You can check the dimensions of expressions automatically by verifying the dimensions of an equation.
First, define the kinematic equation , where v
represents velocity, a
represents acceleration, and s
represents distance. Assume
s
is in kilometers and all other units are in SI base units. To
demonstrate dimension checking, the units of a
are intentionally
incorrect.
syms v v0 a s u = symunit; eqn = (v*u.m/u.s)^2 == (v0*u.m/u.s)^2 + 2*a*u.m/u.s*s*u.km
eqn = v^2*([m]^2/[s]^2) == v0^2*([m]^2/[s]^2) + (2*a*s)*(([km]*[m])/[s])
Observe the units that appear in eqn
by using
findUnits
. The returned units show that both kilometers and meters are
used to represent distance.
findUnits(eqn)
ans = [ [km], [m], [s]]
Check if the units have the same dimensions (such as length or time) by using
checkUnits
with the 'Compatible'
input. MATLAB® assumes symbolic variables are dimensionless. checkUnits
returns logical 0
(false
), meaning the units are
incompatible and not of the same physical dimensions.
checkUnits(eqn,'Compatible')
ans = logical 0
Looking at eqn
, the acceleration a
has incorrect
units. Correct the units and recheck for compatibility again. eqn
now has
compatible units.
eqn = (v*u.m/u.s)^2 == (v0*u.m/u.s)^2 + 2*a*u.m/u.s^2*s*u.km; checkUnits(eqn,'Compatible')
ans = logical 1
Now, to check that each dimension is consistently represented by the same unit, use
checkUnits
with the 'Consistent'
input.
checkUnits
returns logical 0
(false
) because meters and kilometers are both used to represent distance
in eqn
.
checkUnits(eqn,'Consistent')
ans = logical 0
Convert eqn
to SI base units to make the units consistent. Run
checkUnits
again. eqn
has both compatible and
consistent units.
eqn = unitConvert(eqn,'SI')
eqn = v^2*([m]^2/[s]^2) == v0^2*([m]^2/[s]^2) + (2000*a*s)*([m]^2/[s]^2)
checkUnits(eqn)
ans = struct with fields: Consistent: 1 Compatible: 1
After you finish working with units and only need the dimensionless equation or
expression, separate the units and the equation by using
separateUnits
.
[eqn,units] = separateUnits(eqn)
eqn = v^2 == v0^2 + 2000*a*s units = 1*([m]^2/[s]^2)
You can return the original equation with units by multiplying eqn
with
units
and expanding the result.
expand(eqn*units)
ans = v^2*([m]^2/[s]^2) == v0^2*([m]^2/[s]^2) + (2000*a*s)*([m]^2/[s]^2)
To calculate numeric values from your expression, substitute for symbolic variables using
subs
, and convert to numeric values using double
or vpa
.
Solve eqn
for v
. Then find the value of
v
where v0 = 5
, a = 2.5
, and
s = 10
. Convert the result to double.
v = solve(eqn,v); v = v(2); % choose the positive solution vSol = subs(v,[v0 a s],[5 2.5 10]); vSol = double(vSol)
vSol = 223.6627
Use units in differential equations just as in standard equations. This section shows how to use units in differential equations by deriving the velocity relations v = v0 + at and starting from the definition of acceleration .
Represent the definition of acceleration symbolically using SI units. Given that the
velocity V
has units, V
must be differentiated with
respect to the correct units as T = t*u.s
and not just
t
.
syms V(t) a u = symunit; T = t*u.s; % time in seconds A = a*u.m/u.s^2; % acceleration in meters per second eqn1 = A == diff(V,T)
eqn1(t) = a*([m]/[s]^2) == diff(V(t), t)*(1/[s])
Because the velocity V
is unknown and does not have units,
eqn1
has incompatible and inconsistent units.
checkUnits(eqn1)
ans = struct with fields: Consistent: 0 Compatible: 0
Solve eqn1
for V
with the condition that the initial
velocity is v0. The result is the equation v(t) = v0 + at.
syms v0 cond = V(0) == v0*u.m/u.s; eqn2 = V == dsolve(eqn1,cond)
eqn2(t) = V(t) == v0*([m]/[s]) + a*t*([m]/[s])
Check that the result has the correct dimensions by substituting
rhs(eqn2)
into eqn1
and using
checkUnits
.
checkUnits(subs(eqn1,V,rhs(eqn2)))
ans = struct with fields: Consistent: 1 Compatible: 1
Now, derive . Because velocity is the rate of change of distance, substitute
V
with the derivative of distance S
. Again, given that
S
has units, S
must be differentiated with respect to
the correct units as T = t*u.s
and not just t
.
syms S(t) eqn2 = subs(eqn2,V,diff(S,T))
eqn2(t) = diff(S(t), t)*(1/[s]) == v0*([m]/[s]) + a*t*([m]/[s])
Solve eqn2
with the condition that the initial distance covered is
0
. Get the expected form of S
by using
expand
.
cond2 = S(0) == 0; eqn3 = S == dsolve(eqn2,cond2); eqn3 = expand(eqn3)
eqn3(t) = S(t) == t*v0*[m] + ((a*t^2)/2)*[m]
You can use this equation with the units in symbolic workflows. Alternatively, you can
remove the units by returning the right side using rhs
, separating units
by using separateUnits
, and using the resulting unitless
expression.
[S units] = separateUnits(rhs(eqn3))
S(t) = (a*t^2)/2 + v0*t units(t) = [m]
When you need to calculate numeric values from your expression, substitute for symbolic
variables using subs
, and convert to numeric values using
double
or vpa
.
Find the distance traveled in 8
seconds where v0 =
20
and a = 1.3
. Convert the result to double.
S = subs(S,[v0 a],[20 1.3]); dist = S(8); dist = double(dist)
dist = 201.6000
checkUnits
| findUnits
| isUnit
| newUnit
| separateUnits
| symunit2str
| unitConversionFactor
| unitConvert