cascaded electrical systems and transfer function

[PG1995]

Hi

Could you please help me with this query? Thank you.

Regards
PG

 

[Winterstone]

Let us set:

R1C1=T1
R2C2=T2

It is easy to show that

(T1+T2)^2/4*T1*T2 always is equal or larger than unity (equal to unity only for T1=T2).

Now,
because the sum in brackets is even larger due to R1C2, the inequality is always fulfilled.

 

[PG1995]

Thank you, WS.

I hope I'm following you correctly.

Isn't it possible to prove it the way I was trying to do here at the bottom? Kindly let me know. Thanks.

Regards
PG

 

[Winterstone]

In your calculation, it is NOT correct that

(K1+K2)^2 > 4*K1*K2. For K1=K2 both sides are equal.

 

[MrAl]

Thank you, WS.

I hope I'm following you correctly.

Isn't it possible to prove it the way I was trying to do here at the bottom? Kindly let me know. Thanks.

Regards
PG

Hi,

There is a much simpler proof here.

Given a denominator:
Denom=a*s^2+b*s+1

if the roots are real then the square root of the discriminant is real, but if the roots are not real then the square root of the discriminant is imaginary. And the square root of the discriminant is real when the discriminant is greater than zero (equal to zero is not always considered a valid solution although we could include that too).
The discriminant again is:
d=b^2-4*a*1=b^2-4*a

which translates for this problem to:
d=(C2*R2+C2*R1+C1*R1)^2-4*C1*C2*R1*R2

which as you can see is a proof in itself. IF d is ever negative then the roots are not real.

 

[PG1995]

Thank you, WS, MrAl.

In your calculation, it is NOT correct that

(K1+K2)^2 > 4*K1*K2. For K1=K2 both sides are equal.

I hope this time I'm interpreting what you said correctly but I'm still stuck. Could you please help me? Thanks.

 

[MrAl]

Hello again,


I think i see where you are trying to go with this.

The idea is to try to force this to be an inequality that is compared to zero, as: equation>0 or equation<0 and to make equation always positive or always negative.

Setting the following variables:
T11=R1*C1
T22=R2*C2
T12=R1*C2

noting that the first number of the 'T' part is the resistor number and the second number of the 'T' part is the capacitor number.
And also note that all three of these are always positive because all the R's and C's must always be positive in a real circuit with regular physical components.

The original equation is:
T22^2+2*T12*T22+2*T11*T22+T12^2+2*T11*T12+T11^2>4*T11*T22

and subtracting the right side we get:
T22^2+2*T12*T22+2*T11*T22+T12^2+2*T11*T12+T11^2-4*T11*T22>0

so we've accomplished one goal, we got an equation that is all on one side compared to zero. But the difficulty that comes up is that we have a subtraction in the left hand side, and that means it's hard to tell when this equation holds. It turns out though that all we have to do is factor the left hand side and we end up with an equation that must always be positive that is compared to zero.

The original equation after the subtraction then factors into:
(T22-T11)^2+T12*(2*T22+T12+2*T11)>0

and from this we can see that we moved the negative part of the equation under the square so that combined part must always be positive. So we now have it in a form:
equation>0

where equation is always positive, so we've proved it except for the case where all three new variables are equal to zero, but that's a topological impossibility so we're ok.


Visualization:

Imagine taking two chairs placing them together back to back, then move them apart a few feet so they are still back to back but separated by a distance. Then take a large bed sheet and tack one corner of the sheet to one chair back (on the top) and another diagonal corner on the other chair back in the same place as the other chair tack was placed. The sheet if it was thick enough (like a rubber sheet) would form a valley with two peaks at the diagonal chair corners. It would look like a hammock with a pointed head and toe area (these points are the points of the other two corners of the sheet that are not tacked). If the valley touched the floor that would be for T12=0. As T12 is increased above zero, the far loose corner of the sheet gets lifted higher and higher while the forefront corner goes up too but by less. So as we increase T12 the far corner goes up a lot and the forefront corner goes up less, so the entire hammock is tilted. To get this right we'd also have to lift the two chairs but the main point though is that the entire sheet no matter what the value of T12 is (as long as it is positive of course) always stays above zero for T12>0.
In the above visualization, the hammock is formed by holding T12 constant and letting T11 and T22 vary from zero to some high values. With any T12 greater than zero and T11 and T22 also greater than zero we always get a positive value so all points on the sheet are always above zero.

 

[PG1995]

Thanks a lot, MrAl.

So, it's proved that the inequality always holds and hence the roots are real.

Regards
PG