A question of PLL

C

cici

Guest
The output of my lowpass filter is a dc voltage,but the value is in
the
range of the control voltage of my VCO which is between 0 and 1v. How
to
pull down the dc voltage? I tried to use diode, not work.
Another thing is that when I connect the output of my VCO as the
feedback signal, the dc votage of output of LPF is changed(when I used
two independent
input at the PD, it is normal). what is the reason for it?
 
cici wrote:

The output of my lowpass filter is a dc voltage,but the value is in
the
range of the control voltage of my VCO which is between 0 and 1v. How
to
pull down the dc voltage? I tried to use diode, not work.
Another thing is that when I connect the output of my VCO as the
feedback signal, the dc votage of output of LPF is changed(when I used
two independent
input at the PD, it is normal). what is the reason for it?
Click to expand...
Could you provide less information, please?

--
Mike "Rocket J Squirrel" Elliott
 
Normally you will not have a DC output from a filter, unless you are making
a power supply!
You should use AC coupling- a small capacitor inline will work usually.

Cheers!

Chip Shults
My robotics, space and CGI web page - http://home.cfl.rr.com/aichip
 
Normally you will not have a DC output from a filter, unless you are making
a power supply!
You should use AC coupling- a small capacitor inline will work usually.
Click to expand...
PLL usually means Phase-Lock Loop and the low-pass filter exists to
extract the DC component of phase-detector output- and ONLY the DC
component.
 
Fred Bloggs wrote:

PLL usually means Phase-Lock Loop and the low-pass filter exists to
extract the DC component of phase-detector output- and ONLY the DC
component.
Click to expand...
This depends on your definition of DC. If, by DC, you mean a signal
that varies (with falling energy content as frequency increases) but
is unidirectional, then I agree. But if by DC you mean and an
unvarying, unidirectional signal, then I don't.

--
John Popelish
 
John Popelish wrote:
Fred Bloggs wrote:


PLL usually means Phase-Lock Loop and the low-pass filter exists to
extract the DC component of phase-detector output- and ONLY the DC
component.


This depends on your definition of DC. If, by DC, you mean a signal
that varies (with falling energy content as frequency increases) but
is unidirectional, then I agree. But if by DC you mean and an
unvarying, unidirectional signal, then I don't.
Click to expand...
It will be unvarying and unidirectional in lock:)
 
Assuming what you are locking to is unvarying.

Steve

"Fred Bloggs" <nospam@nospam.com> wrote in message
news:3FB6858D.5050809@nospam.com...
It will be unvarying and unidirectional in lock:)
Click to expand...
 
red rover wrote:
Assuming what you are locking to is unvarying.

Steve

"Fred Bloggs" <nospam@nospam.com> wrote in message
news:3FB6858D.5050809@nospam.com...

It will be unvarying and unidirectional in lock:)
Click to expand...
And that the whole system is without drift. but then you don't need a
PLL.
:)

--
John Popelish
 
John Popelish wrote:
red rover wrote:

Assuming what you are locking to is unvarying.

Steve

"Fred Bloggs" <nospam@nospam.com> wrote in message
news:3FB6858D.5050809@nospam.com...

It will be unvarying and unidirectional in lock:)
Click to expand...
Only "instantaneously"

And that the whole system is without drift. but then you don't need a
PLL.
:)
Click to expand...
A PLL can be used to demodulate FM. FM is continuously varying and so
therefore will the "DC" control voltage, i.e. the control voltage is
indeed AC.

Kevin Aylward
salesEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

http://www.anasoft.co.uk/replicators/index.html

Understanding, is itself an emotion, i.e. a feeling.
Emotions or feelings can only be "understood" by
consciousness. "Understanding" consciousness can
therefore only be understood by consciousness itself,
therefore the "hard problem" of consciousness, is
intrinsically unsolvable.

Physics is proven incomplete, that is, no
understanding of the parts of a system can
explain all aspects of the whole of such system.
 
I read in sci.electronics.design that red rover
<NOnatpressSPAM@sympatico.ca> wrote (in <62Atb.50240$xI2.1521927@news20.
bellglobal.com>) about 'A question of PLL', on Sat, 15 Nov 2003:

Steve

"Fred Bloggs" <nospam@nospam.com> wrote in message
news:3FB6858D.5050809@nospam.com...

It will be unvarying and unidirectional in lock:)



Assuming what you are locking to is unvarying.
[Top-posted response re-positioned.]
Click to expand...

More significantly, assuming that what you are *locking* is unvarying,
which is normally not the case.
--
Regards, John Woodgate, OOO - Own Opinions Only. http://www.jmwa.demon.co.uk
Interested in professional sound reinforcement and distribution? Then go to
http://www.isce.org.uk
PLEASE do NOT copy news posts to me by E-MAIL!
 
Kevin Aylward wrote:
John Popelish wrote:

red rover wrote:

Assuming what you are locking to is unvarying.

Steve

"Fred Bloggs" <nospam@nospam.com> wrote in message
news:3FB6858D.5050809@nospam.com...

It will be unvarying and unidirectional in lock:)


Only "instantaneously"


And that the whole system is without drift. but then you don't need a
PLL.
:)


A PLL can be used to demodulate FM. FM is continuously varying and so
therefore will the "DC" control voltage, i.e. the control voltage is
indeed AC.
Click to expand...
Okay- put it this way, the LPF is attempting to output a constant DC:)
 
John Woodgate wrote:
I read in sci.electronics.design that red rover
NOnatpressSPAM@sympatico.ca> wrote (in <62Atb.50240$xI2.1521927@news20.
bellglobal.com>) about 'A question of PLL', on Sat, 15 Nov 2003:


Steve

"Fred Bloggs" <nospam@nospam.com> wrote in message
news:3FB6858D.5050809@nospam.com...

It will be unvarying and unidirectional in lock:)



Assuming what you are locking to is unvarying.

[Top-posted response re-positioned.]

More significantly, assuming that what you are *locking* is unvarying,
which is normally not the case.
Click to expand...
Actually, these days, the majority of applications for PLL have to be
clock recovery, for digital and/or serial comm apps, and communications
frequency synthesis, by sheer bulk- and it is locking to a constant
reference- the LPF is usually the PI(proportional-integral) active
filter with infinite gain at DC.
 
In article <8x$KKcAd9zt$Ew5p@jmwa.demon.co.uk>,
jmw@jmwa.demon.contraspam.yuk says...
I read in sci.electronics.design that red rover
NOnatpressSPAM@sympatico.ca> wrote (in <62Atb.50240$xI2.1521927@news20.
bellglobal.com>) about 'A question of PLL', on Sat, 15 Nov 2003:


Steve

"Fred Bloggs" <nospam@nospam.com> wrote in message
news:3FB6858D.5050809@nospam.com...

It will be unvarying and unidirectional in lock:)



Assuming what you are locking to is unvarying.
[Top-posted response re-positioned.]

More significantly, assuming that what you are *locking* is unvarying,
which is normally not the case.
Click to expand...
What about clock multipliers (frequency synthesizers)? There are
a few billion of these made each year. Ok, many of these are
modulated for spread-spectrum, but...

--
Keith
 
cici wrote:
The output of my lowpass filter is a dc voltage,but the value is in
the
range of the control voltage of my VCO which is between 0 and 1v. How
to
pull down the dc voltage? I tried to use diode, not work.
Another thing is that when I connect the output of my VCO as the
feedback signal, the dc votage of output of LPF is changed(when I used
two independent
input at the PD, it is normal). what is the reason for it?
Click to expand...
Well, the the control voltage serves to shift the oscillation
frequency. If the reference has a fractional frequency offset from the
nominal VCO frequency, the filter corrects this with a (somewhat) DC
value. When you connect two independant references to the phase
detector, there is no feedback in the system. If there is a frequency
offset between the two reference the phase between them will be
constantly drifting and either saturate the PD or lead to some sort of
sawtooth in the phase detector output.


--
Jason Rosinski
Zarlink Semiconductor
Jason.Rosinski@zarlink.com
 
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