Question 5 of 5: Transfer Function and Realisation of a Two-Tap FIR Filter

Question 5: Transfer Function and Realisation of a Two-Tap FIR Filter (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Difference equation $y(n) = x(n) - \tfrac12 x(n-1)$; sampling frequency $f_s = 10$ kHz (sampling interval $T_s = 100$ µs).

Find. The transfer function $H(z)$, the frequency response in a form accurate enough to plot, the impulse response, and a minimum-delay realisation.

Approach. Take the $z$-transform of the difference equation to get $H(z)$, evaluate it on the unit circle to get $H(e^{j\omega})$, reduce the magnitude to a closed form free of complex arithmetic, and read the impulse response straight off the coefficients.

1. Transform the difference equation (part a). Applying the $z$-transform and the delay property $x(n-1) \leftrightarrow z^{-1}X(z)$: $$Y(z) = X(z) - \tfrac12 z^{-1}X(z) = \left(1 - \tfrac12 z^{-1}\right)X(z),$$ $$\boxed{\,H(z) = \frac{Y(z)}{X(z)} = 1 - \tfrac12 z^{-1} = \frac{z - 0.5}{z}\,}$$ This is a finite impulse response (FIR) filter with a single zero at $z = 0.5$ and a trivial pole at the origin, so it is stable for any input and, since the zero lies inside the unit circle, minimum-phase.
2. Evaluate on the unit circle. The frequency response is $H(z)$ at $z = e^{j\omega}$ with $\omega = 2\pi f/f_s$: $$H(e^{j\omega}) = 1 - \tfrac12 e^{-j\omega} = \left(1 - \tfrac12\cos\omega\right) + j\,\tfrac12\sin\omega.$$
3. Reduce the magnitude to a plottable closed form. Squaring and adding the two parts, and using $\cos^2 + \sin^2 = 1$: $$|H|^2 = \left(1 - \tfrac12\cos\omega\right)^2 + \tfrac14\sin^2\omega = 1 - \cos\omega + \tfrac14,$$ $$\boxed{\,\left|H(e^{j\omega})\right| = \sqrt{1.25 - \cos\omega}, \qquad \angle H = \arctan\!\frac{0.5\sin\omega}{1 - 0.5\cos\omega}\,}$$ with $\omega = 2\pi f/(10^4)$ radians per sample. This form contains no complex arithmetic, which is exactly what the question means by “accurate so that we can use it to create a plot”.
4. Tabulate the response across the baseband. Substituting $\omega = 2\pi f/f_s$:

   $f$ (kHz)	0	1.25	2.5	3.75	5.0
   $\omega$ (rad/sample)	0	$\pi/4$	$\pi/2$	$3\pi/4$	$\pi$
   $|H|$	0.500	0.769	1.118	1.383	1.500

   The filter is a gentle high-pass: it attenuates dc by a factor of two and boosts the folding frequency $f_s/2 = 5$ kHz by a factor of 1.5. The dc value is confirmed independently by $H(1) = 1 - 0.5 = 0.5$, and the Nyquist value by $H(-1) = 1 + 0.5 = 1.5$.

5. Impulse response (part b). The impulse response is the inverse transform of $H(z)$; since $H(z)$ is a polynomial in $z^{-1}$, the coefficients are the samples: $$\boxed{\,h(n) = \delta(n) - \tfrac12\,\delta(n-1) \ \Longrightarrow\ h(0) = 1,\quad h(1) = -0.5,\quad h(n) = 0 \text{ otherwise}\,}$$ The response has length 2 and dies out after one sample, which is the defining property of an FIR filter. As a check, $\sum_n h(n) = 0.5$, which must equal (and does equal) the dc gain $|H|$ at $\omega = 0$.
6. Minimum-delay realisation (part c). The difference equation needs exactly one past value of the input, $x(n-1)$, and no past value of the output. One $z^{-1}$ element is therefore both sufficient and necessary — a two-tap FIR filter cannot be built with none. The direct-form structure is: tap the input, delay it once, scale by $-\tfrac12$, and add to the undelayed input.

Note that the transposed direct form and the direct form are identical here, and both use one delay; there is no canonical-form saving to be had because the filter has no feedback path whose storage could be shared.