🚀 go-pugleaf

In article <1644@fs1.ee.ubc.ca> jthornto@ee.ubc.ca (Johan Thornton) writes:
>In article <1420@cameron.egr.duke.edu> wdp@ee.egr.duke.edu (William D. Palmer) writes:
>>How about an op-amp inverting amplifier?
>
>
>Sure... it's simple in the analog domain.  In a digital system, the solution
>is a bit more complex.  The transfer function is:
>
>      X =  -X
>
>
>-- 
>Johan Thornton
>(but my friends call me jthornto@ee.ubc.ca)

Well, sure, you all can say how easy it is, but you're stopping shy of the
hard part. Can't implement a transfer function without getting it into the
form of a difference equation! That's the challenge...


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%Well, sure, you all can say how easy it is, but you're stopping shy of the
%hard part. Can't implement a transfer function without getting it into the
%form of a difference equation! That's the challenge...

Well, I can see you guys are having fun with this, but let me make a
few comments.

First of all, 180 deg phase shift is not the same as simply inverting
the signal.  A phase shift implies a time delay of some sort.

Second of all, he *may* have meant a 180 deg phase shift for each of the
frequencies that add up to his composite signal.  The output signal will
definitely NOT look much like the input signal.  You can prove this to your
self by drawing a time axis and delaying a 100 Hz and 50 Hz signal by 180
degrees.  What you can see is similar to what happens in a typical speaker
system when an impulse sound like a drum hit is reproduced.  Most of the
time, it doesn't sound quite right because of the "phase smearing" that
results...

-- 
Todd Day  |  todd@appmag.com

todd@appmag.com (Todd Day) writes:

>%Well, sure, you all can say how easy it is, but you're stopping shy of the
>%hard part. Can't implement a transfer function without getting it into the
>%form of a difference equation! That's the challenge...

Want a difference equation ??

y(n) = -x(n)

>[...]

>First of all, 180 deg phase shift is not the same as simply inverting
>the signal.  A phase shift implies a time delay of some sort.

Oh, it does does it ??

Does a transfer function H(z) = -1 imply a time delay ??

[...]

>-- 
>Todd Day  |  todd@appmag.com
--
Wilf LeBlanc, Carleton University, Systems & Comp. Eng. Ottawa, Canada, K1S 5B6
Internet: wilf@sce.carleton.ca   UUCP: ...!uunet!mitel!cunews!sce!wilf
                Oh, cruel fate! Why do you mock me so! (H. Simpson)

In article <1991May8.022953.781@appmag.com> todd@appmag.com (Todd Day) writes:

>First of all, 180 deg phase shift is not the same as simply inverting
>the signal.  A phase shift implies a time delay of some sort.

Sorry, but a phase shift need not imply a time delay.  The amount of delay
is called the group delay and it is the slope of the graph of the delay
function.  It's really only meaningful to think of it as a delay if the
delay function is largely a straight line.

The problem is that sin waves are periodic, so you can't really say whether
on is delayed with respect to another, or advanced.  Since the signal
is composed of the sum of sin waves, if you delay (or advance) all the
sin waves by 180 degrees, then you've just inverted all their signs.  When
you sum them all together, you get -1 times the original signal.

			Bruce Karsh
			karsh@sgi.com

In article <1991May8.022953.781@appmag.com> todd@appmag.com (Todd Day) writes:
>First of all, 180 deg phase shift is not the same as simply inverting
>the signal.  A phase shift implies a time delay of some sort.

Todd, you're too smart for your own good, and so are the people talking
about Hilbert transforms, even though they are, in a sense, right.

>Second of all, he *may* have meant a 180 deg phase shift for each of the
>frequencies that add up to his composite signal.

OK, this means that |H(j omega)| should be 1 and arg(H (j omega)) should
be pi for all omega (H is the transfer function).  That means that
H(j omega) should be -1 for all omega, so simply inverting the signal
really does shift by 180 degrees at all frequencies.

Now, for the Hilbert transform people.  The Hilber transform is a filter
whose response is -j for positive frequencies and j for negative frequencies,
and it causes a 90 degree phase shift.  Now let's cascade two of them.
We square the transfer function to obtain the total transfer function.
Note that j^2 = (-j)^2 = -1, so again we get a simple inversion.

>  The output signal will
>definitely NOT look much like the input signal.  You can prove this to your
>self by drawing a time axis and delaying a 100 Hz and 50 Hz signal by 180
>degrees.

You screwed up your plot, Todd.  Otherwise you'd find that the right
answer is an inversion.  Since each individual component is negated by
a 180 degree shift, clearly the sum is.

The original poster clearly had a homework problem with a trivial answer,
and all these "experts" screwed it up.  Embarrassing...

--
Joe Buck
jbuck@janus.berkeley.edu	 {uunet,ucbvax}!janus.berkeley.edu!jbuck	

I wrote:

%Second of all, he *may* have meant a 180 deg phase shift for each of the
%frequencies that add up to his composite signal.  The output signal will
%definitely NOT look much like the input signal.

Sorry about this.  "Phase smearing" does not occur at multiples of pi/2.
I had this visualized totally wrong.  After thinking about "the sum of
infinite sinewaves", I realized that at 180, this sum does align properly
and makes the inverse.  I had gotten it confused with an argument I had
previously discussing 90 degree phase shift in speakers, which definitely
DOES smear the output.

Sorry.

-- 
Todd Day  |  todd@appmag.com