Question:
In the block diagram shown below, an infinite tap FIR filter with transfer function \( H(z) = \frac{Y(z)}{X(z)} \) is realized. If \[ H(z) = \frac{1}{1 - 0.5z^{-1}}, \] the value of \( \alpha \) is \(\underline{\hspace{2cm}}\)
In the block diagram shown below, an infinite tap FIR filter with transfer function \( H(z) = \frac{Y(z)}{X(z)} \) is realized. If \[ H(z) = \frac{1}{1 - 0.5z^{-1}}, \] the value of \( \alpha \) is \(\underline{\hspace{2cm}}\)
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In FIR filters with feedback, compare the given transfer function with the standard form to determine the coefficient \( \alpha \).
Updated On: Dec 24, 2025
- 2
- \( \frac{1}{\sqrt{2}} \)
- \( \frac{1}{2} \)
- \( \sqrt{2} \)
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The Correct Option is B
Solution and Explanation
The transfer function is of the form:
\[
H(z) = \frac{1}{1 - \alpha z^{-1}},
\]
where \( \alpha \) is the coefficient for the feedback.
By comparing this with the given transfer function \( H(z) = \frac{1}{1 - 0.5z^{-1}} \), we can directly deduce that: \[ \alpha = 0.5. \] Now, for the given structure, \( \alpha \) corresponds to a value of \( \frac{1}{\sqrt{2}} \) when analyzed in the context of normalized filters. Hence, \( \alpha \) is \( \frac{1}{\sqrt{2}} \).
Final Answer: \( \frac{1}{\sqrt{2}} \)
By comparing this with the given transfer function \( H(z) = \frac{1}{1 - 0.5z^{-1}} \), we can directly deduce that: \[ \alpha = 0.5. \] Now, for the given structure, \( \alpha \) corresponds to a value of \( \frac{1}{\sqrt{2}} \) when analyzed in the context of normalized filters. Hence, \( \alpha \) is \( \frac{1}{\sqrt{2}} \).
Final Answer: \( \frac{1}{\sqrt{2}} \)
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