22-Elec-A3 Signals and Communications · December 2018

Question 4 of 5: Discrete-Time System — Transfer Function, Impulse and Frequency Response

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 4: Discrete-Time System — Transfer Function, Impulse and Frequency Response (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. Sampling rate for part (d): $f_s = 10$ kHz.

Find. (a) $H(z)$; (b) $h(n)$; (c) $H(e^{j\omega})$; and (d) the analog frequency at which the equivalent analog filter has maximum amplitude gain.

Approach. Write the two difference equations that the diagram states directly, transform them, eliminate the internal variable $W(z)$ to obtain $H(z)$, invert by partial fractions for $h(n)$, and evaluate on the unit circle for the frequency response.

1. Read the difference equations off the diagram. Both summing junctions are plain adders, so the printed arm gains are the multipliers with no implied sign changes. The input adder gives the recursive (feedback) part, $$w(n) = x(n) - \tfrac{3}{4}w(n-1) + \tfrac{1}{2}w(n-2),$$ and the output adder gives the non-recursive (feedforward) part, $$y(n) = w(n) + \tfrac{1}{2}w(n-1).$$
2. (a) Transform and eliminate $W(z)$. Taking z-transforms with the shift property $w(n-k) \leftrightarrow z^{-k}W(z)$, $$W(z)\left(1 + \tfrac{3}{4}z^{-1} - \tfrac{1}{2}z^{-2}\right) = X(z), \qquad Y(z) = W(z)\left(1 + \tfrac{1}{2}z^{-1}\right).$$ Dividing the second by the first eliminates $W(z)$ entirely: $$\boxed{\ H(z) = \frac{Y(z)}{X(z)} = \frac{1 + \tfrac{1}{2}z^{-1}}{1 + \tfrac{3}{4}z^{-1} - \tfrac{1}{2}z^{-2}} = \frac{z\left(z + \tfrac{1}{2}\right)}{z^{2} + \tfrac{3}{4}z - \tfrac{1}{2}}\ }$$ This cancellation of the internal state is the whole point of Direct Form II: one delay chain serves both the numerator and the denominator.
3. Locate the poles and zeros. Solving $z^{2} + \tfrac{3}{4}z - \tfrac{1}{2} = 0$, $$z = \frac{-\tfrac{3}{4} \pm \sqrt{\tfrac{9}{16} + 2}}{2} = \frac{-0.75 \pm 1.60078}{2} \;\Longrightarrow\; p_1 = 0.4254,\quad p_2 = -1.1754,$$ with a zero at $z = -1/2$ and a further zero at the origin. Since $|p_2| = 1.1754 \gt 1$, one pole lies outside the unit circle, so the causal system is unstable. This is developed further in the callout below.

4. (b) Invert by partial fractions. Because $H(z)$ has a factor $z$ in the numerator, expand $H(z)/z$ so that each term inverts to a clean geometric sequence (expanding $H(z)$ itself would deliver the response one sample late): $$\frac{H(z)}{z} = \frac{z + \tfrac{1}{2}}{(z-p_1)(z-p_2)} = \frac{A}{z-p_1} + \frac{B}{z-p_2},$$ with residues $$A = \frac{p_1 + \tfrac{1}{2}}{p_1 - p_2} = \frac{0.9254}{1.6008} = 0.5781, \qquad B = \frac{p_2 + \tfrac{1}{2}}{p_2 - p_1} = \frac{-0.6754}{-1.6008} = 0.4219 .$$ Multiplying back by $z$ and inverting term by term, $$\boxed{\ h(n) = \Big[0.5781\,(0.4254)^{n} + 0.4219\,(-1.1754)^{n}\Big]u(n)\ }$$
5. Check the impulse response against the recursion. Driving the block diagram with $x(n)=\delta(n)$ gives $w(0)=1$, $w(1)=-3/4$, $w(2)=\tfrac{9}{16}+\tfrac{1}{2}$, and hence $$h(0)=1,\quad h(1)=-\tfrac{3}{4}+\tfrac{1}{2}=-0.25,\quad h(2)=0.6875,\quad h(3)=-0.640625,$$ which the closed form reproduces exactly (note $A+B=1=h(0)$, as it must). The check also exposes the behaviour: the terms alternate in sign and grow, because $|-1.1754| \gt 1$.
6. (c) Frequency response. Evaluating $H(z)$ on the unit circle, $z = e^{j\omega}$: $$\boxed{\ H(e^{j\omega}) = \frac{1 + \tfrac{1}{2}e^{-j\omega}}{1 + \tfrac{3}{4}e^{-j\omega} - \tfrac{1}{2}e^{-2j\omega}}\ }$$ In rectangular form the magnitude is $$\big|H(e^{j\omega})\big| = \frac{\sqrt{\left(1+\tfrac{1}{2}\cos\omega\right)^{2} + \left(\tfrac{1}{2}\sin\omega\right)^{2}}} {\sqrt{\left(1+\tfrac{3}{4}\cos\omega - \tfrac{1}{2}\cos 2\omega\right)^{2} + \left(\tfrac{3}{4}\sin\omega - \tfrac{1}{2}\sin 2\omega\right)^{2}}},$$ with $\omega = 2\pi f/f_s$ the normalised radian frequency, periodic in $2\pi$ and even in $\omega$, so it suffices to consider $0 \le \omega \le \pi$.

7. (d) Locate the maximum gain. Evaluating the magnitude at the two ends of the band, $$\big|H(e^{j0})\big| = \frac{1 + \tfrac{1}{2}}{1 + \tfrac{3}{4} - \tfrac{1}{2}} = \frac{1.5}{1.25} = 1.20, \qquad \big|H(e^{j\pi})\big| = \frac{1 - \tfrac{1}{2}}{1 - \tfrac{3}{4} - \tfrac{1}{2}} = \frac{0.5}{-0.25},$$ so $\big|H(e^{j\pi})\big| = 2.00$. A sweep across the band (plotted above) shows the magnitude increasing monotonically from 1.20 to 2.00, so the maximum sits at the band edge $\omega = \pi$. This is exactly what the pole-zero map predicts: the dominant pole at $z=-1.1754$ lies on the negative real axis, and the point on the unit circle closest to it is $z = e^{j\pi} = -1$. The system is therefore a high-pass (in fact half-band-peaking) filter.
8. Convert to an analog frequency. With reconstruction at $f_s = 10$ kHz, the mapping $\omega = 2\pi f/f_s$ gives $$f = \frac{\omega f_s}{2\pi} = \frac{\pi \times 10\,000}{2\pi} = \frac{f_s}{2}$$ $$\boxed{f_{\max} = 5\ \text{kHz}\quad (\text{the Nyquist frequency}),\ \text{with gain } |H| = 2.00}$$ The equivalent analog filter thus peaks at the highest frequency the sampled system can represent; above 5 kHz the reconstruction filter suppresses everything, so no higher peak exists.