Forums

Average Power

Started by HardySpicer October 9, 2007
Ok suppose we have an R-C network - low pass filter with tf

G(s)=1/(1+sT) where T=CR.  We can easily calculate the output power
spectrum if this is driven by zero-mean white noise as

Phiyy(w)=mag(G(jw))^2 x power of input noise

and the area under the PSD is the total average power.

Now let's do the same thing digitally...and produce an equivalent
digital filter G(z)


The output periodogram is

Phiyy(theta)=mag(G(exp(jtheta))^2 x input noise power and the area is
the total output power.

Now, assuming the two output power values at the output are the same,
where does the missing power go? ie the output power not equal input
power.


Hardy

HardySpicer wrote:
> Ok suppose we have an R-C network - low pass filter with tf > > G(s)=1/(1+sT) where T=CR. We can easily calculate the output power > spectrum if this is driven by zero-mean white noise as > > Phiyy(w)=mag(G(jw))^2 x power of input noise > > and the area under the PSD is the total average power. > > Now let's do the same thing digitally...and produce an equivalent > digital filter G(z) > > > The output periodogram is > > Phiyy(theta)=mag(G(exp(jtheta))^2 x input noise power and the area is > the total output power. > > Now, assuming the two output power values at the output are the same, > where does the missing power go? ie the output power not equal input > power.
Power is a convenient fiction in digital systems. Analog filters raise a similar question not so easily answered. Suppose a broadband signal with the spectrum of noise (flat over some band of interest) delivering power to a resistive load, Now suppose a lossless (R-L) filter interposed between the source and load that blocks half the spectrum. The filter doesn't get hot, so where does the blocked power go? Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
On Oct 10, 5:15 pm, Jerry Avins <j...@ieee.org> wrote:
> HardySpicer wrote: > > Ok suppose we have an R-C network - low pass filter with tf > > > G(s)=3D1/(1+sT) where T=3DCR. We can easily calculate the output power > > spectrum if this is driven by zero-mean white noise as > > > Phiyy(w)=3Dmag(G(jw))^2 x power of input noise > > > and the area under the PSD is the total average power. > > > Now let's do the same thing digitally...and produce an equivalent > > digital filter G(z) > > > The output periodogram is > > > Phiyy(theta)=3Dmag(G(exp(jtheta))^2 x input noise power and the area is > > the total output power. > > > Now, assuming the two output power values at the output are the same, > > where does the missing power go? ie the output power not equal input > > power. > > Power is a convenient fiction in digital systems. Analog filters raise a > similar question not so easily answered. Suppose a broadband signal with > the spectrum of noise (flat over some band of interest) delivering power > to a resistive load, Now suppose a lossless (R-L) filter interposed > between the source and load that blocks half the spectrum. The filter > doesn't get hot, so where does the blocked power go? > > Jerry > -- > Engineering is the art of making what you want from things you can get. > =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF= =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF Yes but without R the energy just gets transfered back and forth. With an R there is always i^2R losses. Hardy
>> Analog filters raise a similar question not so easily answered. >> The filter doesn't get hot, so where does the blocked power go?
That's easy. It gets reflected to the source if I think in terms of S-parameters (waves), or it doesn't load the source if I think of it as an impedance. This doesn't answer the OP's question, though... -mn
On 10 Okt, 03:10, HardySpicer <gyansor...@gmail.com> wrote:
> Ok suppose we have an R-C network - low pass filter with tf > > G(s)=1/(1+sT) where T=CR. We can easily calculate the output power > spectrum if this is driven by zero-mean white noise as > > Phiyy(w)=mag(G(jw))^2 x power of input noise > > and the area under the PSD is the total average power. > > Now let's do the same thing digitally...and produce an equivalent > digital filter G(z) > > The output periodogram is > > Phiyy(theta)=mag(G(exp(jtheta))^2 x input noise power and the area is > the total output power. > > Now, assuming the two output power values at the output are the same, > where does the missing power go? ie the output power not equal input > power.
Your base assumption is wrong: Digital systems don't behave like analog systems. The physics of the analog world dictates two important conditions which do not apply in the digital world: - There is always friction present in the physical world, which dissipate energy in terms of heat. - One is limited to only observe real-valued signals but need imaginary components to make the energy computations go up. In the digital world there are no dissipation terms, only numerical accuracy issues, and one does not need the complex/valued maths to make the energy computations work. In essence, the two troublesome issues from the physical world, what energy is concerned, just don't exist in the digital world. Which means you will have to look elsewhere for sources for discrepancy, the most common sources being certain normalization factors in the most popular implementations of the FFT. Rune
On Oct 10, 9:38 pm, Rune Allnor <all...@tele.ntnu.no> wrote:
> On 10 Okt, 03:10, HardySpicer <gyansor...@gmail.com> wrote: > > > > > Ok suppose we have an R-C network - low pass filter with tf > > > G(s)=1/(1+sT) where T=CR. We can easily calculate the output power > > spectrum if this is driven by zero-mean white noise as > > > Phiyy(w)=mag(G(jw))^2 x power of input noise > > > and the area under the PSD is the total average power. > > > Now let's do the same thing digitally...and produce an equivalent > > digital filter G(z) > > > The output periodogram is > > > Phiyy(theta)=mag(G(exp(jtheta))^2 x input noise power and the area is > > the total output power. > > > Now, assuming the two output power values at the output are the same, > > where does the missing power go? ie the output power not equal input > > power. > > Your base assumption is wrong: Digital systems don't > behave like analog systems. > > The physics of the analog world dictates two important > conditions which do not apply in the digital world: > > - There is always friction present in the physical world, > which dissipate energy in terms of heat. > - One is limited to only observe real-valued signals > but need imaginary components to make the energy > computations go up. > > In the digital world there are no dissipation terms, > only numerical accuracy issues, and one does not need > the complex/valued maths to make the energy computations > work. > > In essence, the two troublesome issues from the physical > world, what energy is concerned, just don't exist in the > digital world. > > Which means you will have to look elsewhere for sources > for discrepancy, the most common sources being certain > normalization factors in the most popular implementations > of the FFT. > > Rune
Ok let's take a simpler example - an A/D, factor of 0.5 and a D/A. Suppose a sine-wave of unit amplitude goes in and if there are no other losses we get a sine wave of amplitude 0.5 out. Input power is 0.5 and output power is 0.25 as measured over a 1 Ohm resistor. If we had an analogue system the power would be disipated as heat via a resistor divider. In the digital world the power still has to be disipated somewhere. If we have a gain instead of attenuation - say x2 then this extra power flows in from the power supply in an analogue system and must also do in a digital system. Hardy
HardySpicer wrote:
> On Oct 10, 5:15 pm, Jerry Avins <j...@ieee.org> wrote:
>> ... Analog filters raise a >> similar question not so easily answered. Suppose a broadband signal with >> the spectrum of noise (flat over some band of interest) delivering power >> to a resistive load, Now suppose a lossless (R-L) filter interposed >> between the source and load that blocks half the spectrum. The filter >> doesn't get hot, so where does the blocked power go?
...
> Yes but without R the energy just gets transfered back and forth. With > an R there is always i^2R losses.
Yes what? Transfered from where to where? If the power is delivered by the source and doesn't reach the load, where does it go? (That's an obfuscating question.) Jerry -- Engineering is the art of making what you want from things you can get. &macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
HardySpicer wrote:

   ...

> Ok let's take a simpler example - an A/D, factor of 0.5 and a D/A. > Suppose a sine-wave of unit amplitude goes in and if there are no > other losses we get a sine wave of amplitude 0.5 out. Input power is > 0.5 and output power is 0.25 as measured over a 1 Ohm resistor. If we > had an analogue system the power would be disipated as heat via a > resistor divider. In the digital world the power still has to be > disipated somewhere. If we have a gain instead of attenuation - say x2 > then this extra power flows in from the power supply in an analogue > system and must also do in a digital system.
You're still barking up the wrong tree. "Power" in digital systems is a convenient fiction. suppose instead of multiplying by 0.5, we multiplied by 2.0. Would that be the foundation of perpetual motion? How many joules does a bit represent? Jerry -- Engineering is the art of making what you want from things you can get. &macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
On Oct 10, 8:48 am, HardySpicer <gyansor...@gmail.com> wrote:
> On Oct 10, 9:38 pm, Rune Allnor <all...@tele.ntnu.no> wrote: > > > > > On 10 Okt, 03:10, HardySpicer <gyansor...@gmail.com> wrote: > > > > Ok suppose we have an R-C network - low pass filter with tf > > > > G(s)=1/(1+sT) where T=CR. We can easily calculate the output power > > > spectrum if this is driven by zero-mean white noise as > > > > Phiyy(w)=mag(G(jw))^2 x power of input noise > > > > and the area under the PSD is the total average power. > > > > Now let's do the same thing digitally...and produce an equivalent > > > digital filter G(z) > > > > The output periodogram is > > > > Phiyy(theta)=mag(G(exp(jtheta))^2 x input noise power and the area is > > > the total output power. > > > > Now, assuming the two output power values at the output are the same, > > > where does the missing power go? ie the output power not equal input > > > power. > > > Your base assumption is wrong: Digital systems don't > > behave like analog systems. > > > The physics of the analog world dictates two important > > conditions which do not apply in the digital world: > > > - There is always friction present in the physical world, > > which dissipate energy in terms of heat. > > - One is limited to only observe real-valued signals > > but need imaginary components to make the energy > > computations go up. > > > In the digital world there are no dissipation terms, > > only numerical accuracy issues, and one does not need > > the complex/valued maths to make the energy computations > > work. > > > In essence, the two troublesome issues from the physical > > world, what energy is concerned, just don't exist in the > > digital world. > > > Which means you will have to look elsewhere for sources > > for discrepancy, the most common sources being certain > > normalization factors in the most popular implementations > > of the FFT. > > > Rune > > Ok let's take a simpler example - an A/D, factor of 0.5 and a D/A. > Suppose a sine-wave of unit amplitude goes in and if there are no > other losses we get a sine wave of amplitude 0.5 out. Input power is > 0.5 and output power is 0.25 as measured over a 1 Ohm resistor. If we > had an analogue system the power would be disipated as heat via a > resistor divider. In the digital world the power still has to be > disipated somewhere. If we have a gain instead of attenuation - say x2 > then this extra power flows in from the power supply in an analogue > system and must also do in a digital system. > > Hardy
The input power is not 0.5 watts. The power doesn't actually get *delivered* to the A/D. An ideal A/D would have infinite input impedance and would receive no power, but in practice it will be some very small value. There's a difference between actual power delivered to the load and the "power into 1 ohm" value. Likewise, on the output, the output power need not be 0.25 watts. If you're using the D/A to drive a 1 ohm resistor (and it can source that much current), then yes, you would have that much output power. The current would be sourced by the power supply of the output amplifier. In the digital signal processing realm, all you have are numbers (bits). Multiplying one of those numbers by a value doesn't do anything physical at all. Jason
cincydsp@gmail.com wrote:
(snip on power in analog and digital signals)

> The input power is not 0.5 watts. The power doesn't actually get > *delivered* to the A/D. An ideal A/D would have infinite input > impedance and would receive no power, but in practice it will be some > very small value.
To properly terminate the input (say it is at the end of a long transmission line) it will need the appropriate load resistance. That could be an external resistor or the resistor tree in the A/D converter.
> There's a difference between actual power delivered > to the load and the "power into 1 ohm" value. Likewise, on the output, > the output power need not be 0.25 watts. If you're using the D/A to > drive a 1 ohm resistor (and it can source that much current), then > yes, you would have that much output power. The current would be > sourced by the power supply of the output amplifier. In the digital > signal processing realm, all you have are numbers (bits). Multiplying > one of those numbers by a value doesn't do anything physical at all.
Now consider an analog filter. Signal coming in, Op-amp with appropriate resistors, capacitors, and power supply, signal going out. The incoming power is dissipated in the impedance matching circuit on the input, the output power comes from the Op-amp and its power source. It doesn't sound so different to me. -- glen