Question

The block diagram of a two-tap high-pass FIR filter is shown below. The filter transfer function is given by \( H(z) = \frac{Y(z)}{X(z)} \).
If ratio of the maximum to minimum value of \( H(z) \) is 2 and \( |H(z)|_{\max} = 1 \), the coefficients \( \beta_0 \) and \( \beta_1 \) are _________ and _________, respectively.

• A. 0.75, –0.25
• B. 0.67, 0.33
• C. 0.60, –0.40
• D. –0.64, 0.36

Hint:

For two-tap FIR filters, maximum magnitude occurs at \( \omega = 0 \) or \( \omega = \pi \). Use simple algebra from max and min conditions to determine coefficients.

Answer 1

The Correct Option is A

Step 1: Write the filter transfer function.
For a two-tap FIR filter shown, the transfer function is:
\[ H(z) = \beta_0 + \beta_1 z^{-1} \]
In the frequency domain, substitute \( z^{-1} = e^{-j\omega} \):
\[ H(e^{j\omega}) = \beta_0 + \beta_1 e^{-j\omega} \]
Step 2: Magnitude of the frequency response.
The magnitude becomes:
\[ |H(\omega)| = |\beta_0 + \beta_1 e^{-j\omega}| \]
The maximum magnitude occurs when cosine term adds constructively:
\[ |H|_{\max} = |\beta_0 + \beta_1| = 1 \]
The minimum magnitude occurs when cosine term subtracts:
\[ |H|_{\min} = |\beta_0 - \beta_1| \]
Step 3: Use the ratio of maximum to minimum.
We are given:
\[ \frac{|H|_{\max}}{|H|_{\min}} = 2 \]
\[ \frac{1}{|\beta_0 - \beta_1|} = 2 \]
\[ |\beta_0 - \beta_1| = \frac{1}{2} \]
Step 4: Solve the equations.
We have two equations:
1. \( \beta_0 + \beta_1 = 1 \)
2. \( \beta_0 - \beta_1 = 0.5 \)
Add them:
\[ 2\beta_0 = 1.5 \quad \Rightarrow \quad \beta_0 = 0.75 \]
Substitute back:
\[ 0.75 + \beta_1 = 1 \quad \Rightarrow \quad \beta_1 = -0.25 \]
Step 5: Conclusion.
Thus, the correct coefficients are:
\[ \beta_0 = 0.75,\quad \beta_1 = -0.25 \]
Hence, the correct answer is (A).