Take for example the RLC circuit
In the real case, we have a simple LC oscillation. A 'parasitic' resistance is present. It often uses resistor in the reception oscillating circuit, to change the quality factor. It loses usually from the signal amplitude, but gaining selectivity.
We have the following circuit:
We have the following circuit:
Now, tricky a little the problem and say that we have an extra resistor, Radd:
At any time you want to check you can use an online simulator. For example:
http://www.partsim.com/simulator#
We write the Kirchhoff's laws for summation of current and for summation electric potential.
I like such as calculations because are based on simple, intuitive and logical things:
(Voltage on coil- faraday's low combined with the definition of inductance;)
(Voltage on capacitor - C=Q/U => U=Q/C (and Q=I*T))
http://www.partsim.com/simulator#
We write the Kirchhoff's laws for summation of current and for summation electric potential.
I like such as calculations because are based on simple, intuitive and logical things:
(Voltage on coil- faraday's low combined with the definition of inductance;)
(Voltage on capacitor - C=Q/U => U=Q/C (and Q=I*T))
What we do with this system?
First get rid of integrals, to remain only with derivatives. In this way we simplified a little the problem.
The integral at capacitor must be seen in the physical sense and not mathematical. In mathematics must added and negative side of the ordinate. What is pointless because the 'negative voltage' capacitor is discharged.
I expressed in such a way that you feel these notions, with the expense of rigor.
First get rid of integrals, to remain only with derivatives. In this way we simplified a little the problem.
The integral at capacitor must be seen in the physical sense and not mathematical. In mathematics must added and negative side of the ordinate. What is pointless because the 'negative voltage' capacitor is discharged.
I expressed in such a way that you feel these notions, with the expense of rigor.
We have yet a system with three equations and three unknowns (variables)
Unknowns (variables) are current i through L, C, Radd.
Make a first replacement and we remain only two equations with two variables.
Unknowns (variables) are current i through L, C, Radd.
Make a first replacement and we remain only two equations with two variables.
Perhaps like me you expect from a 'more analytical' solution. Something more algorithmically.
But the algorithm with eigenvalue and eigenvectors is not much simple:
But the algorithm with eigenvalue and eigenvectors is not much simple:
Check now if such execution can be solved:
http://www.wolframalpha.com/input/?i=f*y%27%27+%2B+g*y%27%2Bh*y+%3D+a*b*Exp%28b*x%29
To give the following values: R=50(ohm); L=10uH; C=20pF; Radd=175(ohm);
To simplify the problem, consider V1 is a 5V sine generator 20MHz.
In this case after replacements with the numerical values, the equation becomes:
(-10u)Ic''+6000*Ic'+64Meg*Ic=[5*Sin(20Meg*2*PI*t)]' =>
(-10u)Ic''+6000*Ic'+64Meg*Ic=5*20Meg*2*PI*Sin(20Meg*2*PI*t)
Simplified by 10Meg and get:
(-1p)Ic''+0.0006*Ic'+6.4*Ic=5*2*2*PI*Sin(20Meg*2*PI*t)
-0.000 000 000 001*Ic''+0.0006*Ic'+6.4*Ic=63*Cos(125.6 *1000000*t)
Neglected first term because it has a very small contribution;
0.0006*Ic'+6.4*Ic=63*Cos(125.6 *1000000*t)
http://www.wolframalpha.com/input/?i=+-0.0006*y%27%2B6.4*y+%3D+5*Cos%28125*1000000*x%29
Ic=ct*Exp(10 600*t)+0.06m*Sin(125Meg*t)+5n*cos(125Meg*t)
How to respond to a step signal:
0.0006*Ic'+6.4*Ic=0 =>
Approximate form:
Ic=ct*E^10 600*t
In simulation this current looks like(5V gen. puls, 200ns/div graph(?)):
http://www.wolframalpha.com/input/?i=f*y%27%27+%2B+g*y%27%2Bh*y+%3D+a*b*Exp%28b*x%29
To give the following values: R=50(ohm); L=10uH; C=20pF; Radd=175(ohm);
To simplify the problem, consider V1 is a 5V sine generator 20MHz.
In this case after replacements with the numerical values, the equation becomes:
(-10u)Ic''+6000*Ic'+64Meg*Ic=[5*Sin(20Meg*2*PI*t)]' =>
(-10u)Ic''+6000*Ic'+64Meg*Ic=5*20Meg*2*PI*Sin(20Meg*2*PI*t)
Simplified by 10Meg and get:
(-1p)Ic''+0.0006*Ic'+6.4*Ic=5*2*2*PI*Sin(20Meg*2*PI*t)
-0.000 000 000 001*Ic''+0.0006*Ic'+6.4*Ic=63*Cos(125.6 *1000000*t)
Neglected first term because it has a very small contribution;
0.0006*Ic'+6.4*Ic=63*Cos(125.6 *1000000*t)
http://www.wolframalpha.com/input/?i=+-0.0006*y%27%2B6.4*y+%3D+5*Cos%28125*1000000*x%29
Ic=ct*Exp(10 600*t)+0.06m*Sin(125Meg*t)+5n*cos(125Meg*t)
How to respond to a step signal:
0.0006*Ic'+6.4*Ic=0 =>
Approximate form:
Ic=ct*E^10 600*t
In simulation this current looks like(5V gen. puls, 200ns/div graph(?)):
Checking curvature graph:
https://www.wolframalpha.com/input/?i=plot+1%2Fe^%2810660*x%29+from+x%3D1*10^-8+to+50*10^-8
!!! This text is a draft
An unfinished example and without verification.
https://www.wolframalpha.com/input/?i=plot+1%2Fe^%2810660*x%29+from+x%3D1*10^-8+to+50*10^-8
!!! This text is a draft
An unfinished example and without verification.