Hi adesign,

Here is a brief explanation for that condition.....

In a linearized SD model, consider the initial condition with no input and with e(0)=1 (loop error at n=0). Therefore, the SD output v(0) = y(0) + e(0) = 1; where y(n) is the quantizer input.

There must be at least one delay in the loop filter (this guarantees the system will have memory and will be realizable), thus y(n) cannot depend on v(n) and v(0) = e(0)=1 must hold. Considering that


V(z) = STF(z)*U(z) + NTF(z)*E(z)

And that NTF(z)= H(z) = h(0) + h(1)z^-1 + h(2)z^-2 ...

Then h(0)=1 (or H(∞)=1)

tosei

Hi adesign,

As for the e(0) initial value, I chose 1 assuming a normalized to 1V 1-bit SD modulator. However, e(0) can be any value you want. If such value is different from 0, then h(0)=1 can be demonstrated.
Concerning y(0)=0, mathematically that is the value that results from the assumption that the loop filter must guarantee at least one delay. That can be easily shown as follows:
y[z]=F[z] x U[z] + L[z] x V[z]      (1)

Where F[z] is a function affecting only the input signal and L[z] is the loop filter transfer function. If U[z] (as assumed earlier), from (1)
y[z]= L[z] x V[z]      (2)

and as stated before
v[z] = y[z] + E[z]      (3)

Now, replacing (3) in (2)
Y[z] = E[z] x L[z] / (1-L[z])      (4)

The loop filter must delay its input signal at least one clock period. That means the impulse response of L is zero, or L[\u221e]=0. This condition applied in 4 gives Y[\u221e]=0, or y(n=0)=0.

What you stated concerning the initial condition on y[n] from the practical perspective is true, but I guess (not sure about this) for guaranteeing good operation it is wise to first reset the system. This can also be understood as 0 initial condition for the loop filter (L[\u221e]=L(n=0)=0)

Hope this helps
tosei

adesign wrote on Jul 26^th, 2007, 8:40pm:


This equation does not apply to the linearized model I´m assuming (please see attached figure). This model is actually the same one that is usually assumed as you'll see below: A SD modulator  output signal is usually expressed - by means of a linear modeled - as a function of two different transfer function transfer functions: one affecting only the signal (STF) and the other one only affecting the noise (NTF)

However, expression (1) y[z]=F[z] x U[z] + L[z] x V[z] is just describing the quantizer input as a function of the input U[z] and V[z]. (recall L[z] is the loop filter transfer function and NOT the NTF. It is on the loop filter that we apply the delay≠0 for realizability).

If you replace y[z] in equation (1) with equation (3)  - which is valid by definition of linearized quantization error - you get:

V[z]=F[z]/(1-L[z]) × U[z] + 1/(1-L[z]) × E[z].

Therefore STF [z] = F[z]/(1-L[z])  and  NTF[z] = 1/(1-L[z]) = H[z] (5)

Another direct proof for the condition is just taking (5) and setting L[∞]=0 → H[∞]=1 or h(0)=1.