Topic: Digital Signal Processing
No of stages(S) in direct form FIR filter is given as $S=$ integer $(K*F_s/\Delta f)$ Where Fs = Sampling Frequency, $\Delta f=$ Filter transition band, $K=3$ (assume). If $x(n)$ is signal with frequency range 0-2.4 MHz and sampled at $F_s=400$ MHz and it is filtered by ... $x(n) \rightarrow LPF(1) \rightarrow$ Decimation by 50 $\rightarrow LPF(2) \rightarrow$ Decimation by 2 $\rightarrow Y_{Filter}(n)$. Assumptions: Both filters are having flat passbands and stopbands. Passband Frequency LPF(1): 1.8 MHz, Stopband Frequency LPF(1): 4 MHz. Passband Frequency LPF(2): 1.8 MHz, Stopband Frequency LPF(2): 2 MHz. Passband attenuation of both filters = 0 dB and stop band attenuation of both filters is infinity. Calculate total no. of stages $S_{LPF1}+S_{LPF2}$
- A. 120
- B. 545
- C. 555
- D. 665
Answer: D. 665
Explanation: For LPF(1), the sampling frequency $F_{s1} = 400$ MHz. The transition band $\Delta f_1 = \text{Stopband} - \text{Passband} = 4 - 1.8 = 2.2$ MHz. Number of stages $S_{LPF1} = \frac{3 \times 400}{2.2} = 545.45 \approx 545$. After decimation by 50, the new sampling frequency $F_{s2} = 400 / 50 = 8$ MHz. For LPF(2), the transition band $\Delta f_2 = 2 - 1.8 = 0.2$ MHz. Number of stages $S_{LPF2} = \frac{3 \times 8}{0.2} = 120$. Total stages = $S_{LPF1} + S_{LPF2} = 545 + 120 = 665$.