Hi again,
You have two questions, one has a short answer, the other long...
When you ask to transform the source:
u(t)+cos(t)
into 'phasor' form this doesnt make sense because u(t) is not considered
an AC signal, and phasors are for AC analysis. cos(t) does have a phasor
form and that is amplitude separated by a little 'angle' symbol followed by
0 degrees similar to this:
1_0
where 1 is amplitude and 0 is the phase angle, only the underscore is actually
the little 'angle' symbol usually used for phasors.
See the attached drawing for phasor representation.
Did you instead mean you wanted the Laplace or Fourier transform?
Ok, back to the differential equation and how it is different with initial condition
of v'(0)=2 than with v'(0)=0 (which changes the basic type of differential equation)...
These are the Laplace forms for the function and its derivatives:
Code:
FORM ALT FORM FORMAL s FORM
f(t) f(t) F(s)
f'(t) df/dt s*F(s)-f(0)
f''(t) d^2f/dt^2 s^2*F(s)-s*f(0)-f'(0)
Example of using the Laplace derivative forms to solve the differential equation:
v''+2*v'+2*v=vs(t) with initial conditions: v'(0)=2 and v(0)=0
Find out if the response to vs(t)=u(t)+cos(t) is the same as the sum of the individual responses
to u(t) and cos(t) alone. Later we want to find out how the solution is different when we
change the initial condition for v'(0) from non zero to zero.
Here we need to solve the differential equation three times, first with vs(t)=u(t), then with
vs(t)=cos(t) and then sum the two, and finally with vs(t)=u(t)+cos(t), then compare these
two results to see if they are the same.
We will use the following notation to make the writing simpler:
Code:
v'' =vdd
v' =vd
v =v
v'(0)=vd0
v(0) =v0
vs(t)=vs
V(s) =V
This makes the differential equation look like this:
vdd+2*vd+2*v=vs
Transforming the various parts:
v=V
vd=s*V-v0
vdd=s^2*V-s*v0-vd0
Substituting into the diff equ:
s^2*V-s*v0-vd0+2*(s*V-v0)+2*V=vs
and simplifying:
s^2*V-s*v0-vd0+2*s*V-2*v0+2*V=vs
and grouping all the V terms:
V*(s^2+2*s+2)-s*v0-vd0-2*v0=vs
and since v0=0 this simplifies to:
V*(s^2+2*s+2)-vd0=vs
and since vd0=2 this simplifies to:
V*(s^2+2*s+2)-2=vs
or:
V*(s^2+2*s+2)=vs+2
or:
V=(vs+2)/(s^2+2*s+2)
and now we have a generalized form we can use to look at the three possible sources for vs.
The three sources and their transforms are:
vs=u(t) => 1/s
vs=cos(t) => s/(s^2+1)
vs=u(t)+cos(t) => 1/s+s/(s^2+1)
Using the generalized form from above:
V=(vs+2)/(s^2+2*s+2)
All we have to do now is substitute one of the sources for vs and then carry out the
algebraic simplification. For example, with vs=u(t) the transform is 1/s, so we get:
V=(1/s+2)/(s^2+2*s+2)
and multiplying top and bottom by s we get:
V=(2*s+1)/(s^3+2*s^2+2*s)
This is all we really have to do because we can later add this to the response from
cos(t) and then compare that to the response from both sources together. We dont
really need the time response to do this.
And now the responses to these three inputs (we dont really need the time equations here though)
with vs=u(t) we get:
V=(2*s+1)/(s^3+2*s^2+2*s)
v(t)=e^(-t)*((3*sin(t))/2-cos(t)/2)+1/2
with vs=cos(t) we get:
V=(2*s^2+s+2)/(s^4+2*s^3+3*s^2+2*s+2)
v(t)=e^(-t)*((7*sin(t))/5-cos(t)/5)+(2*sin(t))/5+cos(t)/5
Suming the individual freq responses from u(t) and cos(t) we get:
V=(4*s^3+2*s^2+4*s+1)/(s^5+2*s^4+3*s^3+2*s^2+2*s)
v(t)=e^(-t)*((29*sin(t))/10-(7*cos(t))/10)+(2*sin(t))/5+cos(t)/5+1/2
With vs=u(t)+cos(t) we get:
V=(2*s^3+2*s^2+2*s+1)/(s^5+2*s^4+3*s^3+2*s^2+2*s)
v(t)=e^(-t)*((9*sin(t))/10-(7*cos(t))/10)+(2*sin(t))/5+cos(t)/5+1/2
Comparing the sum of individual responses to the response of both sources at the same time
shows that the two responses are not the same. Note that we only have to compare:
V=(4*s^3+2*s^2+4*s+1)/(s^5+2*s^4+3*s^3+2*s^2+2*s)
to
V=(2*s^3+2*s^2+2*s+1)/(s^5+2*s^4+3*s^3+2*s^2+2*s)
to see that they are different, we dont have to compare the time responses.
Now we want to see what changes when we change the initial condition of v'(0) to zero
instead of 2:
Starting with the Laplace form of the diff equ:
s^2*V-s*v0-vd0+2*(s*V-v0)+2*V=vs
and simplifying:
s^2*V-s*v0-vd0+2*s*V-2*v0+2*V=vs
and grouping all the V terms:
V*(s^2+2*s+2)-s*v0-vd0-2*v0=vs
and since v0=0 this simplifies to:
V*(s^2+2*s+2)-vd0=vs
but now vd0=0, not 2 like before, so:
vd0=0
and since vd0=0 now this simplifies to:
V*(s^2+2*s+2)=vs
and note the previous form with vd0=2 was V*(s^2+2*s+2)-2=vs
so we lost the "-2" when vd0 went to zero.
Simplifying:
V*(s^2+2*s+2)=vs
or:
V=vs/(s^2+2*s+2)
and now we have a generalized form we can use to look at the three possible sources for vs
again. Note however that the numerator does not contain that constant term anymore, so
the solution with u(t)+cos(t) comes out the same as the solution with added individual
responses to u(t) and cos(t). I'll leave it up to you to do the rest of the math to show this.
Lastly, here is a drawing showing the phasor representation of an AC signal: