suppose a first order FIR filter H(z)=1+az^-1 has white noise of unit variance (zero mean) ) passed through it. Then the output power (variance) will be (1+a^2) X1 which is always greater than the input power?
This is daft FIR power
Started by ●May 23, 2011
Reply by ●May 24, 20112011-05-24
On May 23, 10:52�pm, HardySpicer <gyansor...@gmail.com> wrote:> suppose a first order FIR filter > > H(z)=1+az^-1 > > has white noise of unit variance (zero mean) �) passed through it. > > Then the output power (variance) will be > > (1+a^2) X1 > > which is always greater than the input power?Noise always makes its own power. Consider a series-wound motor. There will be kT Johnson noise circulating in the windings, and since torque id proportional to current squared, it will always be in the same direction. A well designed motor with good bearings should turn on its own. The hotter it gets, the faster it goes. Jerry -- Forgive me. It's late and I'm giddy.
Reply by ●May 24, 20112011-05-24
On May 24, 4:52�am, HardySpicer <gyansor...@gmail.com> wrote:> suppose a first order FIR filter > > H(z)=1+az^-1 > > has white noise of unit variance (zero mean) �) passed through it. > > Then the output power (variance) will be > > (1+a^2) X1 > > which is always greater than the input power?Yes? Rune
Reply by ●May 24, 20112011-05-24
>On May 24, 4:52=A0am, HardySpicer <gyansor...@gmail.com> wrote: > suppose a first order FIR filter > > H(z)=3D1+az^-1 > > has white noise of unit variance (zero mean) =A0) passed through it. > > Then the output power (variance) will be > > (1+a^2) X1 > > which is always greater than the input power?Try considering H(z) = 2, i.e., a simple scale on the input. Mark
Reply by ●May 24, 20112011-05-24
On 05/23/2011 07:52 PM, HardySpicer wrote:> suppose a first order FIR filter > > H(z)=1+az^-1 > > has white noise of unit variance (zero mean) ) passed through it. > > Then the output power (variance) will be > > (1+a^2) X1 > > which is always greater than the input power?Given that H(z) not only has power gain, but has DC gain greater than 1, why are you surprised? -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●May 24, 20112011-05-24
On May 25, 3:43�am, Tim Wescott <t...@seemywebsite.com> wrote:> On 05/23/2011 07:52 PM, HardySpicer wrote: > > > suppose a first order FIR filter > > > H(z)=1+az^-1 > > > has white noise of unit variance (zero mean) �) passed through it. > > > Then the output power (variance) will be > > > (1+a^2) X1 > > > which is always greater than the input power? > > Given that H(z) not only has power gain, but has DC gain greater than 1, > why are you surprised? > > -- > > Tim Wescott > Wescott Design Serviceshttp://www.wescottdesign.com > > Do you need to implement control loops in software? > "Applied Control Theory for Embedded Systems" was written for you. > See details athttp://www.wescottdesign.com/actfes/actfes.htmlOk fair enough. Daft question. So where does the excess power go in a digital filter when you attenuate?
Reply by ●May 24, 20112011-05-24
On May 24, 3:09�pm, HardySpicer <gyansor...@gmail.com> wrote:> On May 25, 3:43�am, Tim Wescott <t...@seemywebsite.com> wrote: > > > > > > > > > > > On 05/23/2011 07:52 PM, HardySpicer wrote: > > > > suppose a first order FIR filter > > > > H(z)=1+az^-1 > > > > has white noise of unit variance (zero mean) �) passed through it. > > > > Then the output power (variance) will be > > > > (1+a^2) X1 > > > > which is always greater than the input power? > > > Given that H(z) not only has power gain, but has DC gain greater than 1, > > why are you surprised? > > > -- > > > Tim Wescott > > Wescott Design Serviceshttp://www.wescottdesign.com > > > Do you need to implement control loops in software? > > "Applied Control Theory for Embedded Systems" was written for you. > > See details athttp://www.wescottdesign.com/actfes/actfes.html > > Ok fair enough. Daft question. So where does the excess power go in a > digital filter > when you attenuate?The same place it came from. How long will your flashlight shine when powered by a pageful of sample magnitudes? Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●May 24, 20112011-05-24
On May 24, 2:43�pm, Jerry Avins <j...@ieee.org> asked:> How long will your flashlight shine when > powered by a pageful of sample magnitudes?Oh, shoot, and here I was all set to solve the energy crisis very simply (and thus win a Nobel prize) by just doing arithmetic left shifts on each sample value. You spoilsport, Jerry, you!
Reply by ●May 24, 20112011-05-24
On 05/24/2011 12:09 PM, HardySpicer wrote:> On May 25, 3:43 am, Tim Wescott<t...@seemywebsite.com> wrote: >> On 05/23/2011 07:52 PM, HardySpicer wrote: >> >>> suppose a first order FIR filter >> >>> H(z)=1+az^-1 >> >>> has white noise of unit variance (zero mean) ) passed through it. >> >>> Then the output power (variance) will be >> >>> (1+a^2) X1 >> >>> which is always greater than the input power? >> >> Given that H(z) not only has power gain, but has DC gain greater than 1, >> why are you surprised? >> >> -- >> >> Tim Wescott >> Wescott Design Serviceshttp://www.wescottdesign.com >> >> Do you need to implement control loops in software? >> "Applied Control Theory for Embedded Systems" was written for you. >> See details athttp://www.wescottdesign.com/actfes/actfes.html > > Ok fair enough. Daft question. So where does the excess power go in a > digital filter > when you attenuate?The bit bucket heats up. Actually Richard Feynman wrote a book on the thermodynamics of computing, finding the lower bound of the power consumption of a digital processor from strictly thermodynamic concerns. It turns out that computers today waste energy in orders of magnitude higher than the thermodynamic limits. So, if you _did_ manage to build a digital filter that operated at close to the thermodynamic minimum, I assume that the extra information that you're discarding would go to heating up something. Actually, if I'm not mistaken, the extra information that you're discarding would _require_ energy to be put _into_ your system, as the thermodynamic 'cost' of computation isn't the computation itself, -- that's a constant entropy operation as long as it's infinitely reversible -- the energy is required when the machine needs to _forget_ its state to make room for new calculations. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●May 25, 20112011-05-25
>Ok fair enough. Daft question. So where does the excess power go in a >digital filter >when you attenuate?You need to keep in mind that once you have digital data, the relative power levels are fixed (noise and signal.) No matter how you scale the data (gain or attenuation,) the same information is still there (other than the effects of quantization - if you scale by very large or very small numbers you may lose information.) There is no "power" in this signal any more, it is just numbers. These numbers are somewhat arbitrary, though ordered (in amplitude and likely time) based on whatever was originally digitized. Mark
