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Derivation of closed-loop transfer function of the type-II PLL

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I am reading on the section on derivation of closed-loop transfer function of the type-II PLL

I found that expression (9.14) is not the differentiated version. It should have the apostrophe ( ' or ') character to imply that it is the result of differentiating Eq. (9.13) with respect to time

In this case, I really suspect if expression (9.15) is correct or not since H(S) = (9.14) * (Kvco / s) / (1 + (9.14) * (Kvco / s) )

Could anyone suggest if I overlook something ?

Note: The book screenshots below are from Razavi "RF microelectronics" 2nd edition

Ob4hYxr.png


BfuTQNN.png


dZ7QMov.png
 

u(t) is a unit step function. tu(t) is a ramp function. The derivative of a ramp function is a unit step function. the laplace transform of a unit step function is 1/s.

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edit:
in your derivation, you can ignore the impulse function because it only exists at the origin an basically what's left is L{u(t)} -> 1/s
 

rmanalo said:
u(t) is a unit step function. tu(t) is a ramp function. The derivative of a ramp function is a unit step function. the laplace transform of a unit step function is 1/s.

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edit:
in your derivation, you can ignore the impulse function because it only exists at the origin an basically what's left is L{u(t)} -> 1/s
Click to expand...

I have already derived all you have stated in your post.

Please check carefully my question.

I found that expression (9.14) is not the differentiated version. It should have the apostrophe ( ' or ') character to imply that it is the result of differentiating Eq. (9.13) with respect to time

In this case, I really suspect if expression (9.15) is correct or not since H(S) = (9.14) * (Kvco / s) / (1 + (9.14) * (Kvco / s) )
Click to expand...
 

It should have the apostrophe ( ' or ') character to imply that it is the result of differentiating Eq. (9.13) with respect to time
Click to expand...

Differentiating Eq. (9.13) with respect to time, normalizing to ΔФ, and taking the Laplace Transform, we have
Click to expand...

Eq. (9.14) is in fact the result of differentiating Eq. (9.13) and then transformed into its laplace equivalent. i.e. Vcont'(t) -> Vcont(S).
 

Vcont'(t) -> Vcont(S)
Click to expand...

Did you use any laplace properties ? What happen to the apostrophe ( ' or ') character AFTER laplace transform ?
 

Did you use any laplace properties ?
Click to expand...
laplace table.jpg
unit step functions is 1/s

What happen to the apostrophe ( ' or ') character AFTER laplace transform ?
Click to expand...
I think you're mixing up time domain and s domain. Besides, the apostrophe you're referring to is just a symbol describing dVcont(t)/dt.
 

laplace transform of x′(t) gives s*X(s)
Click to expand...

the exact statement is x'(t)=s*X(S)-x(0)

Vcont'(t) -> Vcont(S)
Click to expand...

I apologize. I did not mean its exact equivalent but its transformation from time domain to s domain.

Please check very carefully your own post statement
Click to expand...
proof.jpg
 
rmanalo said:
the exact statement is x'(t)=s*X(S)-x(0)



I apologize. I did not mean its exact equivalent but its transformation from time domain to s domain.


View attachment 148537
Click to expand...

However, how are you going to arrive at expression (9.15) using your handwritten computation result ?
 

how are you going to arrive at expression (9.15)
Click to expand...

RF circuits is not my field. I only assumed your problem is with (9.14). But Fig. 9.27 is in feedback which explains eq 9.15.
 

Your handwritten computation result just proves otherwise, that expression (9.14) is not correct. Please double check very carefully again.

Vcont(s) has the denominator of quadratic power while expression (9.14) only have denominator of linear power

But Fig. 9.27 is in feedback which explains eq 9.15.
Click to expand...

I really suspect if expression (9.15) is correct or not since H(S) = (9.14) * (Kvco / s) / (1 + (9.14) * (Kvco / s) )
 

Like I said before, RF circuits is not my field. I'm only familiar with the math.

Vcont(s) has the denominator of quadratic power while expression (9.14) only have denominator of linear power
Click to expand...
PLLs are a topic where RF touches control engineering.
Click to expand...
I agree with him. And in control engineering they view the system in blocks where each block contributes poles or zeroes depending on how they are modeled
e.g. X(S)->[block contributing pole or zero]->Y(S).
Here I assume the VCO block contributes a pole, i.e. KVCO/s.

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I took a quick scan on razavi's book. Try looking at Eq.(8.177) if this checks out.
 

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