Impedance Diagram of Low Pass Filter

[BIGEYE]

I am to draw an impedance diagram of a low pass filter and indicate the
phase angle.
Is the phase angle found by:
1) -arctan(?RC)
or
2) -arctan(Xc/R)

Appreciate if someone could advise which one of the above is correct.

TIA

 

[John Nice]

I am to draw an impedance diagram of a low pass filter and indicate the
phase angle.
Is the phase angle found by:
1) -arctan(?RC)
or
2) -arctan(Xc/R)

Appreciate if someone could advise which one of the above is correct.

Have you thought of trying a textbook?

 

[Paul Hovnanian P.E.]

Have you thought of trying a textbook?

What are we charging for homework answers these days?

 

[BIGEYE]

A simple answer 1 or 2 would be fine, the notes we have been handed says to
use -arctan(wRC), however, I was unsure of this thinking back to my old ac
theory days. I came over an article on the net which says to
use -arctan(Xc/R).
There are a lot of other questions associated with this question, I will
work everything out. All I need to know is which formula is correct. Or
don't you know?
Maybe I should ask elsewhere.
TIA

 

[Don Kelly]

I am to draw an impedance diagram of a low pass filter and indicate the
phase angle.
Is the phase angle found by:
1) -arctan(?RC)
or
2) -arctan(Xc/R)

Appreciate if someone could advise which one of the above is correct.

TIA

-----------------

The phase angle depends on the frequency. IF, as you have not indicated, you
are looking for a half power point, there are specific expressions which can
easily be worked out. Your question is somewhat vague. I would suggest that
you do your homework as asked. This might require some effort on your part
but it is worthwhile doing.

 

[no_one]

That's what happened when everyone started using calculators instead of
slide rules!

No, I learned it just fine with a calculator in 1977-1982. Pick another
reason.

 

[daestrom]

I found it a little more worrying that they don't seem to be teaching
reduction of the units expressed within equations. eg If someone needs
something to be a ratio and it reduces to a value in volts, they have got
the equation wrong...

Ah yes. "Unit analysis" was actually taught as a course for half semester
in "my day". By tagging the units on to every thing, and knowing what items
were ratios or 'pure numbers', we could cancel out the various units and
figure out the necessary conversions.

daestrom

 

[daestrom]

No, I learned it just fine with a calculator in 1977-1982. Pick another
reason.

I know I'll "stir up a hornet's nest" with this, but I think it's been the
conversion to the metric system. Here's why. In the old 'English' system,
one thing you always had to do is express the answer in the requested units.
So part of learning how to do all sorts of calculations is how to carry the
units with each item, and conversions from one form of units to another.
With the odd/awkward units of the 'English' system ( ft-lbf, hp, BTU,
etc...), we just naturally learned to always carry the units along with each
item. So we had to learn all those darn conversions :-(. But we also
learned to keep track of units and dimensions (length, time, energy, etc...)


With the metric system, quite often you can 'get the right answer' without
keeping track of units, simply because most of the conversions are a factor
of '1'. Take 12 Newtons, through a distance of 5 meters, how many Joules??
Nowadays, someone is just as likely to say "12*5=60" instead of "12N * 5m *
(1 J / N-m) = 60 J".

But in "English", take 12 lbf through a distance of 15 ft, how many BTU???
(12lbf * 15ft) * (1BTU / 778 ft-lbf) = 0.231 BTU. You just *have* to keep
track of units in order to know what conversions need to be applied.

Now, I like the metric system as much as the next person. It *is* easier to
work many things with it. But I still carry all the 'units' through, just
because "that's the way I learnt it".

As for the OP, trying to take the arctan of something, the 'something' has
to be a dimensionless number. R*C would give you units of ohms-Farads,
there is no way to take the arctan(ohm-Farad). Think about it some more,
and you'll see the answer.

daestrom

 

[Don Kelly]

I know I'll "stir up a hornet's nest" with this, but I think it's been the
conversion to the metric system. Here's why. In the old 'English'
system, one thing you always had to do is express the answer in the
requested units. So part of learning how to do all sorts of calculations
is how to carry the units with each item, and conversions from one form of
units to another. With the odd/awkward units of the 'English' system (
ft-lbf, hp, BTU, etc...), we just naturally learned to always carry the
units along with each item. So we had to learn all those darn conversions
:-(. But we also learned to keep track of units and dimensions (length,
time, energy, etc...)

With the metric system, quite often you can 'get the right answer' without
keeping track of units, simply because most of the conversions are a
factor of '1'. Take 12 Newtons, through a distance of 5 meters, how many
Joules?? Nowadays, someone is just as likely to say "12*5=60" instead of
"12N * 5m * (1 J / N-m) = 60 J".

But in "English", take 12 lbf through a distance of 15 ft, how many BTU???
(12lbf * 15ft) * (1BTU / 778 ft-lbf) = 0.231 BTU. You just *have* to keep
track of units in order to know what conversions need to be applied.

Now, I like the metric system as much as the next person. It *is* easier
to work many things with it. But I still carry all the 'units' through,
just because "that's the way I learnt it".

As for the OP, trying to take the arctan of something, the 'something' has
to be a dimensionless number. R*C would give you units of ohms-Farads,
there is no way to take the arctan(ohm-Farad). Think about it some more,
and you'll see the answer.

daestrom