Designer’s Guide Community

baohulu Community Member Offline Posts: 43 china	*how to implement 1/(1-az^(-1)) with SC circuit** Mar 31^st, 2008, 5:25am hi, all in SC circuit, we can always see the implement of a/(1-z^(-1)), while, can it implement 1/(1-a*z^(-1)), where a is not equal to 1? if there is some reference, please tell me thanks
Back to top	IP Logged

pancho_hideboo

Senior Fellow

Offline

Posts: 1424
Real Homeless

Re: how to implement 1/(1-a*z^(-1)) with SC circui
Reply #1 - Mar 31^st, 2008, 5:51am

baohulu wrote on Mar 31^st, 2008, 5:25am:

hi, all
�in SC circuit, we can always see the implement of a/(1-z^(-1))
, while, can it implement 1/(1-a*z^(-1)), where a is not equal to 1?
�if there is some reference, please tell me
thanks

For me, complete integral a/(1-z^(-1)) is rare.
Rather incomplete integral 1/(1-a*z^(-1)) is common.
So realize it as incomplete integral.

Also we can realize "a" as complex number.

Back to top

« Last Edit: Mar 31^st, 2008, 8:36am by pancho_hideboo »

Kita━━━━━━(ﾟ∀ﾟ)━━━━━━ !!!!!
http://www7.plala.or.jp/ungeromeppa/flash/kita.html
http://www.youtube.com/watch?v=mjIxGh55bMM&feature=related

IP Logged

baohulu Community Member Offline Posts: 43 china	*Re: how to implement 1/(1-az^(-1)) with SC circui Reply #2 -** Mar 31^st, 2008, 6:49pm then how to implement it ? use input forward feedthrough to the input? however, in the SC bandpass sigma delta adc, the resonator is usually implemented use the a/(1-z^(-1)) or az^(-1)/(1-z^(-1)), rather than 1/(1-az^(-1)) , and why?
Back to top	IP Logged

Berti

Community Fellow

Offline

Posts: 356

Re: how to implement 1/(1-a*z^(-1)) with SC circui
Reply #3 - Mar 31^st, 2008, 11:14pm

Hi,

a/(1-z^(-1)) or a*z^(-1)/(1-z^(-1)) are integrators without and with delay, respectively. Integrators are relatively straight-forward
to be implemented as a SC circuit.

1/(1-a*z^(-1)) is also called leaky-integrator (for a<1). In fact, most circuit implementations have such a transfer function. Due to
infinte gain of amplifiers etc., charge-transfer won't be ideal and you will loose a small amount of charge during each phase. This charge-loss will result in a
a transfer function equal to 1/(1-a*z^(-1)), where a can be approximated as 1-1/(Gain of amplifier).

Hope this helps, cheers

Back to top

IP Logged

pancho_hideboo

Senior Fellow

Offline

Posts: 1424
Real Homeless

Re: how to implement 1/(1-a*z^(-1)) with SC circui
Reply #4 - Mar 31^st, 2008, 11:54pm

baohulu wrote on Mar 31^st, 2008, 6:49pm:

then how to implement it ? use input forward feedthrough to the input?

The followings are not best reference, but might be helpful for you.
In [2], 1st IIR is implemented as incomplete integral(leaky integral).

[1] AUTHORS: Stephen A. Jantzi, Kenneth W. Martin, and Adel S. Sedra
TITLE: Quadrature bandpass ΔΣ modulation for digital radio
ISSUE: IEEE Journal of Solid-State Circuits, vol. 32, pp. 1935 - 1950, December 1997

[2] TITLE: A discrete-time bluetooth receiver in a 0.13μm digital CMOS process
ISSUE: IEEE International Solid-State Circuits Conference, vol. XLVII, pp. 268 - 269, February 2004

baohulu wrote on Mar 31^st, 2008, 6:49pm:

however, in the SC bandpass sigma delta adc, the resonator is usually implemented use the a/(1-z^(-1)) or a*z^(-1)/(1-z^(-1)), rather than 1/(1-a*z^(-1)) , and why?

See above [1].

Back to top

« Last Edit: Apr 1^st, 2008, 10:33am by pancho_hideboo »

Kita━━━━━━(ﾟ∀ﾟ)━━━━━━ !!!!!
http://www7.plala.or.jp/ungeromeppa/flash/kita.html
http://www.youtube.com/watch?v=mjIxGh55bMM&feature=related

IP Logged