Question E7F14:
Which of the following would allow a digital signal processing filter to create a sharper filter response?  

Choices:  
A: Higher data rate  
B: More taps  
C: Lower Q  
D: Double-precision math routines    

Correct Answer is  B: More taps  

Correct answer: B — More taps

Why B is correct:
- In a DSP FIR filter, the number of taps $N$ is the filter length. More taps give a longer impulse response, which allows a narrower transition band (sharper skirts) and better control of the frequency response.
- FIR output: $y[n] = \sum_{k=0}^{N-1} h[k]\,x[n-k]$. Increasing $N$ increases the degrees of freedom to shape $H(e^{j\omega})$ more steeply.

Useful design relation (Kaiser/windowed-sinc estimate):
$$N \approx \frac{A - 8}{2.285\,\Delta\omega}$$
where:
- $A$ = desired stopband attenuation in dB
- $\Delta\omega$ = normalized transition width in rad/sample

This shows $N \propto 1/\Delta\omega$: more taps ⇒ smaller transition width ⇒ sharper filter.

Why the other choices are incorrect:
- A: Higher data rate  
  Increasing the sampling frequency by itself does not sharpen the filter’s normalized response. For a fixed set of coefficients, the transition band in Hz actually widens proportionally with sample rate. You do not get a sharper filter just by sampling faster.

- C: Lower Q  
  Q is an analog/IIR resonator concept. Lower Q means broader, less selective response (the opposite of sharper). FIR “sharpness” is not improved by lowering Q.

- D: Double-precision math routines  
  Higher numerical precision reduces coefficient quantization and rounding noise but does not make the ideal filter skirts steeper. Sharpness is determined primarily by the filter design (e.g., number of taps and window/optimization), not arithmetic precision.

Rules of thumb:
- Doubling the number of taps roughly halves the transition width.
- For a given window family, increasing $N$ primarily narrows the transition band; achieving higher stopband attenuation also depends on the window/optimization parameters.