22-Elec-A3 Signals and Communications · December 2019

Question 4 of 5: FIR System from its Impulse Response — Structure, Transfer Function, Response and Stability

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

Question 4: FIR System from its Impulse Response — Structure, Transfer Function, Response and Stability (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. $h(n) = u(n) - u(n-4) + \tfrac{1}{2}\delta(n-4)$, with $u(\cdot)$ the unit step and $\delta(\cdot)$ the unit sample. Sampling frequency $f_s = 50$ kHz for part (d). Input for part (e): a symmetric triangular pulse $x(n) = 1 - |n-6|/6$ on $0 \le n \le 12$, zero elsewhere.

Find. A realisable block diagram; the input–output difference equation; $H(z)$; the frequency response $H(e^{j\omega})$ expressed in hertz at $f_s = 50$ kHz; the output for the triangular input; and a stability verdict with justification.

Approach. Evaluate $h(n)$ sample by sample to expose it as a five-tap FIR filter, then read the structure, difference equation and transfer function directly off the tap values; convolve for part (e) and apply the absolute-summability test for part (f).

1. Evaluate the impulse response sample by sample. The difference $u(n)-u(n-4)$ is a rectangular window that equals 1 for $n = 0,1,2,3$ and 0 elsewhere; the extra term adds $\tfrac12$ at $n = 4$ only. Hence $$h(n) = \{\,1,\ 1,\ 1,\ 1,\ \tfrac{1}{2}\,\}\quad \text{for } n = 0,1,2,3,4;\qquad h(n)=0 \text{ otherwise}$$ The response has finite length (five samples), so the system is an FIR filter of order 4 — there is no feedback anywhere in it.
2. (a) Draw the block diagram. An FIR filter of order 4 needs four unit delays and five tap multipliers, summed into a common output rail. The tap gains are exactly the impulse-response values $1, 1, 1, 1, \tfrac12$.

3. (b) Write the difference equation. Convolution with a finite $h(n)$ is a weighted sum of delayed inputs: $$y(n) = \sum_{k=0}^{4} h(k)\,x(n-k)$$ $$\boxed{\;y(n) = x(n) + x(n-1) + x(n-2) + x(n-3) + \tfrac{1}{2}x(n-4)\;}$$ The filter is thus a running sum over four samples with a half-weighted fifth tap — a crude low-pass (smoothing) operation.
4. (c) Write the transfer function. Taking the z-transform term by term, $x(n-k) \leftrightarrow z^{-k}X(z)$, so $$\boxed{\;H(z) = 1 + z^{-1} + z^{-2} + z^{-3} + \tfrac{1}{2}z^{-4}\;}$$ Equivalently $H(z) = \dfrac{z^4 + z^3 + z^2 + z + \tfrac12}{z^4}$: all four poles sit at the origin, which is the defining fingerprint of an FIR system, and the region of convergence is the entire z-plane except $z = 0$.
5. (d) Obtain the frequency response at $f_s = 50$ kHz. Evaluating $H(z)$ on the unit circle, $z = e^{j\omega}$ with the normalised radian frequency $\omega = 2\pi f/f_s$, $$H\!\left(e^{j\omega}\right) = 1 + e^{-j\omega} + e^{-j2\omega} + e^{-j3\omega} + \tfrac{1}{2}e^{-j4\omega}$$ Summing the first four terms as a geometric series and factoring out the linear-phase centre gives the compact plottable form $$\boxed{\;H\!\left(e^{j\omega}\right) = e^{-j\,3\omega/2}\,\frac{\sin(2\omega)}{\sin(\omega/2)} + \tfrac{1}{2}e^{-j4\omega},\qquad \omega = \frac{2\pi f}{50\,000}\;}$$ valid for $0 \le f \le 25$ kHz (the folding frequency). At DC the running sum gives the maximum gain $H(1) = 4.5$; the response then falls away with nulls near the zeros of $\sin(2\omega)$, confirming low-pass behaviour.
6. (e) Convolve the triangular input with $h(n)$. The input is the 13-sample triangle $x(n) = \{0,\ \tfrac16,\ \tfrac26,\ \tfrac36,\ \tfrac46,\ \tfrac56,\ 1,\ \tfrac56,\ \tfrac46,\ \tfrac36,\ \tfrac26,\ \tfrac16,\ 0\}$ for $n = 0 \ldots 12$. Since $x$ has length 13 and $h$ has length 5, the output has length $13 + 5 - 1 = 17$, spanning $0 \le n \le 16$. Applying $y(n) = x(n)+x(n-1)+x(n-2)+x(n-3)+\tfrac12 x(n-4)$ sample by sample:

   $n$	0	1	2	3	4	5	6	7	8
   $y(n)$	0	1/6	1/2	1	5/3	29/12	19/6	43/12	11/3

   $n$	9	10	11	12	13	14	15	16
   $y(n)$	41/12	17/6	25/12	4/3	3/4	1/3	1/12	0

   $$\boxed{\;y(n)_{\max} = \tfrac{11}{3} = 3.667 \text{ at } n = 8;\qquad y(n)=0 \text{ outside } 0 \le n \le 16\;}$$ Two checks confirm the arithmetic. The area identity requires $\sum y = \left(\sum x\right)\left(\sum h\right) = 6 \times 4.5 = 27$, and the tabulated values do sum to 27. The output is also slightly asymmetric — its peak sits at $n = 8$ rather than at the input peak $n = 6$ — because the filter's half-weighted final tap shifts the centre of mass of $h(n)$ to $n \approx 1.78$.

7. (f) Test bounded-input bounded-output stability. A discrete LTI system is BIBO stable if and only if its impulse response is absolutely summable. Here $$\sum_{n=-\infty}^{\infty}|h(n)| = 1+1+1+1+\tfrac{1}{2} = 4.5 \lt \infty$$ $$\boxed{\text{The system is BIBO stable, with peak gain } \textstyle\sum|h(n)| = 4.5}$$ The justification is stronger than a mere inequality: every FIR filter with finite tap values is BIBO stable, because a finite sum of finite terms is finite. Consistently, all four poles of $H(z)$ lie at $z = 0$, well inside the unit circle. The bound is tight — an input bounded by $|x(n)| \le 1$ can produce an output as large as 4.5 in magnitude.