How to modify zeros and poles in a delta-sigma modulator loop?
One way you can achieve &quot;placeable zeros&quot; is by using a dynamic feedback. As given in the question, this is what you achieved with constant feedback on three loops:
$$G(z)=\frac{1}{z-1}$$
$$\text{NTF(z)}=\frac{1}{\left(1+\frac{\alpha}{z-1}\right)\left(1+\frac{\beta}{z-1}\right)\left(1+\frac{\gamma}{z-1}\right)}$$
$$\text{NTF(z)}=\frac{(z-1)^3}{(z-1+\alpha)(z-1+\beta)(z-1+\gamma)}$$
Now see what happens when replacing constant feedback for dynamic feedbacks with pole-zero cancelation:
$$A(z)=\alpha\frac{z-1}{z-\lambda_1}$$
$$B(z)=\beta\frac{z-1}{z-\lambda_2}$$
$$C(z)=\gamma\frac{z-1}{z-\lambda_3}$$
$$\text{NTF(z)}=\frac{1}{\left(1+A(z)G(z)\right)\left(1+B(z)G(z)\right)\left(1+C(z)G(z)\right)}$$
$$\text{NTF(z)}=\frac{1}{\left(1+\frac{\alpha}{z-\lambda_1}\right)\left(1+\frac{\beta}{z-\lambda_2}\right)\left(1+\frac{\gamma}{z-\lambda_3}\right)}$$
$$\text{NTF(z)}=\frac{(z-\lambda_1)(z-\lambda_2)(z-\lambda_3)}{(z-\lambda_1+\alpha)(z-\lambda_2+\beta)(z-\lambda_3+\gamma)}$$
With this, you have achieve three real &quot;placeable zeros&quot; at \$\lambda_1,\lambda_2,\lambda_3\$ and three real placeable poles at \$\lambda_1-\alpha,\lambda_2-\beta,\lambda_3-\gamma\$. In order to achieve complex zeros, at least one of the feedback loops would have to be of second order. I also believe you will need only two feedback loops instead of three, but perhaps other arrangements are possible:
$$A(z)=0$$
$$B(z)=\beta\frac{z-1}{z-\lambda_1}$$
$$C(z)=\gamma\frac{z-1}{\left(z-\lambda_2e^{j\theta}\right)\left(z-\lambda_2e^{-j\theta}\right)}$$
$$\text{NTF(z)}=\frac{1}{\left(1+0\right)\left(1+\frac{\beta}{z-\lambda_1}\right)\left(1+\frac{\gamma}{\left(z-\lambda_2e^{j\theta}\right)\left(z-\lambda_2e^{-j\theta}\right)}\right)}$$
$$\text{NTF(z)}=\frac{\left(z-\lambda_1\right)\left(z-\lambda_2e^{j\theta}\right)\left(z-\lambda_2e^{-j\theta}\right)}{\left(z-\lambda_1+\beta\right)\left(\left(z-\lambda_2e^{j\theta}\right)\left(z-\lambda_2e^{-j\theta}\right)+\gamma\right)}$$