so, here is a very simple model. here

The CCDE is given, but I was trying to derive it on my own and now I am stuck.

first of all, the reverse y[n] when goes through the delayed system, it turns to y[n-M]. then when it passes through the leaky integrator system, $y'[n-M] = (1 - \lambda) * y[n-M] + \lambda * y'[n-M-1]$

combining two, $y[n] = x[n] + \alpha * y'[n-M]$ $ = x[n] + \alpha * (1 - \lambda) * y[n - M] + \alpha * \lambda * y'[n-M-1]$

Now, I am stuck. Can anyone help me with it? Did I understand something wrong?


1 Answer 1


It's probably more straightforward to first compute the total transfer function in terms of $H(z)$, then plug in the actual expression for $H(z)$, and from that write down the difference equation:

$$Y(z)=X(z)+\alpha z^{-M}H(z)Y(z)\tag{1}$$

From $(1)$ we can derive the total transfer function:

$$\frac{Y(z)}{X(z)}=\frac{1}{1-\alpha z^{-M}H(z)}\tag{2}$$

Plugging $H(z)=(1-\lambda)/(1-\lambda z^{-1})$ into $(2)$ gives

$$\begin{align}\frac{Y(z)}{X(z)}&=\frac{1}{1-\alpha z^{-M}\frac{1-\lambda}{1-\lambda z^{-1}}}\\&=\frac{1-\lambda z^{-1}}{1-\lambda z^{-1}-\alpha(1-\lambda)z^{-M}}\tag{3}\end{align}$$

Eq. $(3)$ is equivalent to

$$Y(z)=\big(1-\lambda z^{-1}\big)X(z)+\big(\lambda z^{-1}+\alpha(1-\lambda)z^{-M}\big)Y(z)\tag{4}$$

from which you can directly see the corresponding difference equation.

  • $\begingroup$ Thanks a lot. Now I understand how to solve similar kind of problems. At least I think so. $\endgroup$
    – Rima
    Commented Feb 24, 2021 at 9:37

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