22-Elec-A3 Signals and Communications · December 2017

Question 2 of 5: Discrete-Time System — Transfer Function, Impulse Response and Pulse Response

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

Question 2: Discrete-Time System — Transfer Function, Impulse Response and Pulse 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. The direct-form-II structure of Figure 2.1, with feedback gains $-3/4$ on $w(n-1)$ and $+1/2$ on $w(n-2)$, and a single feedforward gain $+1/2$ on $w(n-1)$. Both summing junctions are drawn as plain adders (no inversion marks), so the printed arm gains are the multipliers themselves. The input in part (c) is the two-sample rectangular pulse $x(n) = u(n) - u(n-2)$.

Find. (a) $H(z)$; (b) $h(n)$ in closed form; (c) $y(n)$ for the two-sample pulse.

[Figure not reproduced: Figure 2.1 — The block diagram redrawn. The left summing junction forms the intermediate sequence w(n); the right one forms y(n). See the official exam paper.]

Approach. Read the two difference equations off the diagram, transform them to obtain $H(z)$ as a ratio of polynomials in $z^{-1}$, invert by partial fractions on $H(z)/z$ to get $h(n)$, then exploit the fact that the pulse is just two shifted impulses so that $y(n) = h(n) + h(n-1)$.

1. (a) Write the node equations. Let $w(n)$ be the signal leaving the left summing junction. That junction adds the input and the two feedback arms; the right junction adds $w(n)$ and the feedforward arm: $$w(n) = x(n) - \tfrac{3}{4}\,w(n-1) + \tfrac{1}{2}\,w(n-2), \qquad y(n) = w(n) + \tfrac{1}{2}\,w(n-1)$$ This is the canonical direct-form-II arrangement: the delay chain is shared, the feedback arms build the denominator and the feedforward arms build the numerator.
2. Transform and eliminate $W(z)$. Taking z-transforms of both equations, $$W(z)\Bigl(1 + \tfrac{3}{4}z^{-1} - \tfrac{1}{2}z^{-2}\Bigr) = X(z), \qquad Y(z) = W(z)\Bigl(1 + \tfrac{1}{2}z^{-1}\Bigr)$$ Dividing the second by the first eliminates the internal variable and gives the transfer function $$\boxed{\,H(z) = \frac{Y(z)}{X(z)} = \frac{1 + \tfrac{1}{2}z^{-1}}{1 + \tfrac{3}{4}z^{-1} - \tfrac{1}{2}z^{-2}} = \frac{z^{2} + 0.5z}{z^{2} + 0.75z - 0.5}\,}$$ The corresponding overall difference equation is $y(n) + \tfrac34 y(n-1) - \tfrac12 y(n-2) = x(n) + \tfrac12 x(n-1)$.
3. Locate the poles and zeros. The zeros are the roots of $z(z + 0.5)$, namely $z = 0$ and $z = -0.5$. The poles satisfy $z^{2} + 0.75z - 0.5 = 0$, so $$p_{1,2} = \frac{-3 \pm \sqrt{41}}{8} \;\Longrightarrow\; p_1 = 0.42539, \qquad p_2 = -1.17539$$ Since $|p_2| = 1.175 \gt 1$, one pole lies outside the unit circle. The system as drawn is therefore unstable — not BIBO stable, and its impulse response will diverge. This is an honest reading of the diagram, and the impulse response computed below confirms it by matching the direct recursion sample for sample.
Figure 2.4 — Pole-zero map. Crosses are poles, circles zeros; p₂ sits well outside the unit circle, so the system is unstable.
4. (b) Partial-fraction expansion. Because the numerator and denominator have equal degree, expand $H(z)/z$ rather than $H(z)$ — this keeps the inverse transform aligned at $n = 0$ instead of one sample late: $$\frac{H(z)}{z} = \frac{z + 0.5}{(z - p_1)(z - p_2)} = \frac{A}{z - p_1} + \frac{B}{z - p_2}$$ with residues $$A = \frac{p_1 + 0.5}{p_1 - p_2} = \frac{0.92539}{1.60078} = 0.5781, \qquad B = \frac{p_2 + 0.5}{p_2 - p_1} = \frac{-0.67539}{-1.60078} = 0.4219$$ A quick check: $A + B = 1.0000$, which must equal $h(0)$ for a transfer function whose numerator and denominator are both monic.
5. Invert to obtain the impulse response. Multiplying back by $z$ and using the pair $z/(z-p) \leftrightarrow p^{n}u(n)$, $$\boxed{\,h(n) = \bigl[\,0.5781\,(0.42539)^{n} + 0.4219\,(-1.17539)^{n}\,\bigr]u(n)\,}$$ The first samples are $h(0) = 1$, $h(1) = -0.25$, $h(2) = 0.6875$, $h(3) = -0.6406$, $h(4) = 0.8242$. Substituting these into the original recursion reproduces them exactly, which confirms both the reading of the diagram and the residues. The alternating, growing envelope is the signature of the negative pole of magnitude 1.175.
Figure 2.2 — Impulse response h(n). The alternating samples grow without bound: the pole at z = -1.175 lies outside the unit circle.
6. (c) Decompose the input. The pulse $x(n) = u(n) - u(n-2)$ is unity at $n = 0$ and $n = 1$ and zero elsewhere, so $$x(n) = \delta(n) + \delta(n-1) \quad\Longleftrightarrow\quad X(z) = 1 + z^{-1}$$ By linearity and time invariance the output is simply the impulse response added to a one-sample-delayed copy of itself: $$y(n) = h(n) + h(n-1)$$
7. Evaluate the pulse response. Combining terms for $n \ge 1$, $$\boxed{\,y(n) = A\,p_1^{\,n-1}(p_1 + 1) + B\,p_2^{\,n-1}(p_2 + 1), \quad n \ge 1; \qquad y(0) = 1\,}$$ Numerically $y(0) = 1$, $y(1) = 0.75$, $y(2) = 0.4375$, $y(3) = 0.0469$, $y(4) = 0.1836$, after which the $(-1.175)^{n}$ term takes over and the response grows without bound. Note the factor $(p_2 + 1) = -0.17539$, which is small: the two-sample pulse partially cancels the unstable mode at first, which is why the early samples look deceptively well behaved before the divergence asserts itself.
Figure 2.3 — Response to the two-sample pulse x(n) = u(n) − u(n−2), obtained as h(n) + h(n−1).