Topic: Networks, Signals and Systems
Two systems $h1[n]=A(b_{1})^{n}u[n]$ and $h2[n]=A(b_{2})^{n}u[n]$ are cascaded. If the effective $H(z)=\frac{4}{4-z^{-2}}$ find a possible value of $(A,b_{1},b_{2})$
- A. $A=1/2$, $b_{1}=1/2$, $b_{2}=-1/2$
- B. $A=2$, $b_{1}=2$, $b_{2}=-2$
- C. $A=1$, $b_{1}=1/2$, $b_{2}=-1/2$
- D. $A=1$, $b_{1}=1$, $b_{2}=-1$
Answer: C. $A=1$, $b_{1}=1/2$, $b_{2}=-1/2$
Explanation: The z-transforms of the individual systems are $H_1(z) = \frac{A}{1-b_1z^{-1}}$ and $H_2(z) = \frac{A}{1-b_2z^{-1}}$.
For cascaded systems, the effective transfer function is their product:
$H(z) = H_1(z) H_2(z) = \frac{A^2}{(1-b_1z^{-1})(1-b_2z^{-1})}$
We are given $H(z) = \frac{4}{4-z^{-2}} = \frac{1}{1-0.25z^{-2}}$.
Factoring the denominator using the difference of squares: $1-0.25z^{-2} = (1-0.5z^{-1})(1+0.5z^{-1})$.
Equating the two expressions for $H(z)$:
$A^2 = 1 \Rightarrow A = 1$ (or $-1$)
$(1-b_1z^{-1})(1-b_2z^{-1}) = (1-0.5z^{-1})(1+0.5z^{-1})$.
This implies $b_1 = 0.5$ and $b_2 = -0.5$. Therefore, a possible set of values is $(1, 1/2, -1/2)$.