Correct answer is (c) |v(n-1)| ≤ \(\frac{(1/2).2^{-b}}{1-|a|}\)
The best explanation: Since the quantization product av(n-1) is obtained by rounding, it follows that the quantization error is bounded as
|Qr[av(n-1)]-av(n-1)| ≤ \((\frac{1}{2}).2^{-b}\)
Where ‘b’ is the number of bits (exclusive of sign) used in the representation of the pole ‘a’ and v(n). Consequently, we get
|v(n-1)|-|av(n-1)| ≤ \((\frac{1}{2}).2^{-b}\)
and hence
|v(n-1)| ≤ \(\frac{(\frac{1}{2}).2^{-b}}{1-|a|}\) .