On Sat, 02 May 2009 11:36:55 -0700, wooopik wrote:
> they use z^-1 not z
>
> its about eq (7)
>
> for single reservoir in time domain whe have y(t)=1/K*exp(- t/K) and
> x(t)=t
> i have no ide how they arrived until eq (7) but it seem to be working
> because i
> have coherent results
>
> i should try a inverse z transform but it a little to hard , i dont have
> any program for symbolic operations
I don't see where they define the single reservoir response -- I think
that the time-domain model that you are using is simpler than what
they're using -- they seem to imply that the reservoir response has a
significant delay component which (a) kinda matches a reservoir that's
physically large compared to it's outlet size, and (b) matches my memory
of how the water rises in most storms (i.e. it isn't at it's maximum
immediately after a whopping big shower; instead it peaks a bit later).
--
http://www.wescottdesign.com
Reply by ●May 2, 20092009-05-02
they use z^-1 not z
its about eq (7)
for single reservoir in time domain whe have
y(t)=1/K*exp(- t/K) and x(t)=t
i have no ide how they arrived until eq (7)
but it seem to be working because i
have coherent results
i should try a inverse z transform but it a little
to hard , i dont have any program for
symbolic operations
Reply by ●May 2, 20092009-05-02
they use z^-1 not z
its eq (7)
as i say before single reservoire Nash - cascade is
y(t)=1/K*exp(- t/K) and x(t)=t
for discrete time values t=T*n T=1 n=1 .. N
and there is one little mistake in eq (7)
in { } should be 1 + exp... in numerator
because i have obtain the same results as in Table 1 with +
so it seem to be workig
i should try the inverse z transform but its a little
to hard , i dont have any program for
symbolic operations :]
Reply by Tim Wescott●May 2, 20092009-05-02
On Sat, 02 May 2009 01:18:08 -0700, wooopik wrote:
> Hi
>
> i have a free online article : root selection methods in flood analysis
> http://www.hydrol-earth-syst-sci.net/7/151/2003/hess-7-151-2003.html and
> there is a z transform for the unit hydrograph for a single reservoir
> (Nash-cascade) i can't figure out how its made !
>
> in the time domaine single reservoir is y=1/K*exp(- t/K)
> general form for n- reservoirs
> y=[ ( 1/( K*(n-1)! ) )* (x/K).^(n-1).* exp(-t/K) ]
>
> where N=100
> t=1,2,3 ... N
> K=storage const
> n=1 single reservoire
>
> Y(z^-1)=H(z^-1)X(z^-1)
>
> and in the article there two mysterious statments meaby with mistakes ?
>
> H(z^-1)=[ ( exp(t/K) -1 )/t ] * exp(-t/K) and below (its multiplicated
> or its a general form??) { ( 1-[exp(-t/K)z^-1]^N )/ ( 1-exp(-t/K)z^-1 )
> }
>
> how???
> because if you see the analytical form of y it should be a simple
> transform for the exp set c= 1/K
> then
> y=c *exp (- ct)
> Z{y}=Z{c *exp (- ct)}=c* Z{exp (- ct)} = c/(1-exp (- ct)z^-1) = Y (z^-1)
> Z{x}=Z{t}= (z^-1)/( (1-z^-1)^2 ) = X(z^-1)
>
> and H(z^-1) is simple division of Y(z^-1)/X(z^-1)
>
> someone can werify this for me ? ;)
>
> with regards
> Wojtek
I tried, but there's too many distractions in the article. Post equation
numbers and maybe I'll try again.
Keep in mind that they may be using non-traditional forms.
--
http://www.wescottdesign.com
Reply by ●May 2, 20092009-05-02
Hi
i have a free online article : root selection methods in flood
analysis
http://www.hydrol-earth-syst-sci.net/7/151/2003/hess-7-151-2003.html
and there is a z transform for the unit hydrograph for a single
reservoir
(Nash-cascade) i can't figure out how its made !
in the time domaine single reservoir is
y=1/K*exp(- t/K)
general form for n- reservoirs
y=[ ( 1/( K*(n-1)! ) )* (x/K).^(n-1).* exp(-t/K) ]
where N=100
t=1,2,3 ... N
K=storage const
n=1 single reservoire
Y(z^-1)=H(z^-1)X(z^-1)
and in the article there two mysterious statments
meaby with mistakes ?
H(z^-1)=[ ( exp(t/K) -1 )/t ] * exp(-t/K)
and below (its multiplicated or its a general form??)
{ ( 1-[exp(-t/K)z^-1]^N )/ ( 1-exp(-t/K)z^-1 ) }
how???
because if you see the analytical form of y
it should be a simple transform for the exp
set c= 1/K
then
y=c *exp (- ct)
Z{y}=Z{c *exp (- ct)}=c* Z{exp (- ct)} = c/(1-exp (- ct)z^-1) = Y
(z^-1)
Z{x}=Z{t}= (z^-1)/( (1-z^-1)^2 ) = X(z^-1)
and H(z^-1) is simple division of Y(z^-1)/X(z^-1)
someone can werify this for me ? ;)
with regards
Wojtek