22-Elec-A3 Signals and Communications · May 2016

Question 5 of 6: Discrete-Time System — Realisation, Transfer Function and Step Response

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

Question 5: Discrete-Time System — Realisation, Transfer Function and Step 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. A causal first-order recursive (IIR) system, $y(n)=\tfrac13 y(n-1)+2x(n)$, initially at rest ($y(n)=x(n)=0$ for $n\lt 0$).

Find. (a) a block-diagram realisation, (b) $H(z)$, (c) $h(n)$ and (d) the unit-step response $y(n)$.

Approach. Read the realisation straight off the difference equation, take the z-transform of both sides with zero initial conditions, invert the single-pole transfer function by inspection, and obtain the step response either as a geometric sum or from $Y(z)=H(z)\,z/(z-1)$.

1. Draw the realisation (part a). The equation says: scale the present input by 2, add one-third of the previous output, and call the result the present output. That maps onto one multiplier on the input path, one adder, one unit delay $z^{-1}$ in the feedback path, and one multiplier of $1/3$ in that path. Only one delay element is needed, which is the minimum for a first-order system (a canonical Direct Form II realisation).

2. Transform the difference equation (part b). With zero initial conditions the delay property is $y(n-1)\leftrightarrow z^{-1}Y(z)$, so $$Y(z)=\tfrac13 z^{-1}Y(z)+2X(z)\;\Longrightarrow\;Y(z)\left[1-\tfrac13z^{-1}\right]=2X(z),$$ giving $$\boxed{\;H(z)=\frac{Y(z)}{X(z)}=\frac{2}{1-\tfrac13 z^{-1}}=\frac{2z}{z-\tfrac13}, \qquad |z|\gt \tfrac13\;}$$ The single pole at $z=1/3$ lies well inside the unit circle, so the causal system is BIBO stable; there is a zero at the origin. The DC gain is $H(z)\big|_{z=1}=2/(1-\tfrac13)=3$, a number that will reappear in part (d).
3. Invert to get the impulse response (part c). $H(z)$ is already in the form of the standard causal pair $a^{n}u(n)\leftrightarrow 1/(1-az^{-1})$ with $a=1/3$, so $$\boxed{\;h(n)=2\left(\tfrac13\right)^{n}u(n)\;}$$ i.e. $h(0)=2$, $h(1)=2/3$, $h(2)=2/9,\dots$ — a decaying exponential. Running the difference equation directly with $x(n)=\delta(n)$ reproduces exactly these values, which is the cheapest possible check. The response is absolutely summable ($\sum|h(n)|=2/(1-\tfrac13)=3\lt\infty$), confirming stability.
4. Step response as a running sum (part d). For $x(n)=u(n)$ the convolution sum becomes a finite geometric series, $$y(n)=\sum_{k=0}^{n}h(k)=2\sum_{k=0}^{n}\left(\tfrac13\right)^{k} =2\,\frac{1-\left(\tfrac13\right)^{n+1}}{1-\tfrac13} =3\left[1-\left(\tfrac13\right)^{n+1}\right].$$ Simplifying the bracket, $3\left(\tfrac13\right)^{n+1}=\left(\tfrac13\right)^{n}$, so $$\boxed{\;y(n)=\left[3-\left(\tfrac13\right)^{n}\right]u(n)\;}$$
5. Confirm the answer at both ends. At $n=0$ the formula gives $y(0)=3-1=2$, which matches the difference equation directly ($y(0)=2x(0)=2$); at $n=1$ it gives $y(1)=3-\tfrac13=8/3=2.667$, matching $\tfrac13(2)+2$. As $n\to\infty$ the transient $(1/3)^{n}$ vanishes and $y(\infty)=3$, exactly the DC gain found in step 2. The response therefore climbs monotonically from 2 to 3 with a per-sample decay factor of $1/3$ — it is within 1 % of final value by $n=5$.