22-Elec-A3 Signals and Communications · May 2018

Question 4 of 5: Discrete-Time LTI System — Delayed-Step Response and Frequency Response

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

Notes on this paper

Paper format. Engineers Canada national examination 16-Elec-A3 — Signals and Communications, May 2018. Closed book (one approved Casio or Sharp calculator permitted), 3 hours, 4 pages. Five questions, all compulsory, all of equal value (20 marks each). All five are solved in full below, every sub-part answered.

Reference texts. B. P. Lathi & Z. Ding, Modern Digital and Analog Communication Systems, 5th ed. (AM modulation and detection, PCM quantisation, frequency conversion); B. P. Lathi, Linear Systems and Signals, 2nd ed. (Fourier series of pulse trains, signal power); A. V. Oppenheim & A. S. Willsky, Signals and Systems, 2nd ed. (discrete-time LTI systems, convolution, DTFT); J. G. Proakis & D. G. Manolakis, Digital Signal Processing, 4th ed. (difference equations, frequency response); S. Haykin, Communication Systems, 5th ed. (envelope and coherent detection, mixers and image products).

Check: a note on the Question 3 figure. The plotted message and the printed equation agree: the plotted waveform has period 1 in normalised time with peaks of $+1.5$ at $t = 0, \pm 1, \pm 2$ and troughs of $-1.5$ midway between them, which is exactly $m(t) = \cos(2\pi f_m t) + \tfrac{1}{2}\cos(6\pi f_m t)$ evaluated with $f_m t$ as the abscissa. The solution below therefore uses the printed equation, and the peak value $|m|_{\max} = 1.5$ read from it. The paper's "$f_m = 4$ Kz" is read as 4 kHz.

Question 4: Discrete-Time LTI System — Delayed-Step Response 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.

Quantity	Symbol	Value
Impulse response	$h(n)$	$e^{-\alpha n}u(n)$
Input	$x(n)$	$u(n-n_0)$, delayed unit step
Delay	$n_0$	fixed positive integer (4 in part b)
Decay parameter	$\alpha$	symbolic ($\alpha = 1$ in part b); $\alpha > 0$ assumed

Find. A closed-form expression for $y(n)$ for general $\alpha$ and $n_0$; the plotted response for $\alpha = 1$, $n_0 = 4$; and the frequency response $H(e^{j\omega})$.

Approach. Convolve directly — the delayed step turns the convolution sum into a finite geometric series — sum it in closed form, evaluate the first several samples for the plot, and obtain the frequency response as the DTFT of $h(n)$, which is another geometric series.

Set up the convolution. For an LTI system, $$y(n) = \sum_{k=-\infty}^{\infty} h(k)\,x(n-k) = \sum_{k=-\infty}^{\infty} e^{-\alpha k}u(k)\,u(n-k-n_0).$$ The first step function restricts $k \ge 0$; the second requires $n - k - n_0 \ge 0$, i.e. $k \le n - n_0$. The two together give a finite sum from $k = 0$ to $k = n - n_0$, which is empty (and $y = 0$) whenever $n < n_0$. The system is causal and the input starts at $n_0$, so nothing can happen before $n_0$ — a sanity check worth stating.
Sum the geometric series. For $n \ge n_0$, writing $r = e^{-\alpha}$, $$y(n) = \sum_{k=0}^{n-n_0} r^{k} = \frac{1 - r^{\,n-n_0+1}}{1-r}.$$ Restoring $r = e^{-\alpha}$ gives the answer to part (a): $$\boxed{\,y(n) = \frac{1 - e^{-\alpha (n-n_0+1)}}{1 - e^{-\alpha}}\; u(n-n_0)\,}$$ Two limits confirm it. At $n = n_0$ the numerator is $1 - e^{-\alpha}$ and $y(n_0) = 1$ — the very first sample of the step sees only $h(0) = 1$. As $n\to\infty$, $y \to 1/(1-e^{-\alpha})$, which is $\sum_k h(k) = H(e^{j0})$, the DC gain, exactly as a step response must settle to.
Evaluate for $\alpha = 1$, $n_0 = 4$ (part b). Here $e^{-1} = 0.36788$ and $1 - e^{-1} = 0.63212$, so $y(n) = \left[1 - e^{-(n-3)}\right]/0.63212$ for $n \ge 4$: $$y(4)=1.000,\quad y(5)=1.368,\quad y(6)=1.503,\quad y(7)=1.553,\quad y(8)=1.571,\quad y(9)=1.578,\quad y(10)=1.581,$$ approaching the limit $1/(1-e^{-1}) = 1.582$. The samples for $n \le 3$ are identically zero. The response is a delayed, saturating build-up: each new sample adds a term $e^{-\alpha}$ times smaller than the last, so the curve is within 1% of its final value by $n = 8$ (four samples after the step arrives, i.e. about four time constants since $\alpha = 1$ means one time constant per sample).

Figure Q4.1 — the step response y(n): identically zero for n < 4, then a saturating geometric build-up to 1/(1 - e^-α).

Frequency response (part c). The frequency response is the DTFT of the impulse response, $$H(e^{j\omega}) = \sum_{n=-\infty}^{\infty} h(n)e^{-j\omega n} = \sum_{n=0}^{\infty} \left(e^{-\alpha}e^{-j\omega}\right)^{n},$$ which is a geometric series convergent because $|e^{-\alpha}e^{-j\omega}| = e^{-\alpha} < 1$ for $\alpha > 0$ (so the system is BIBO stable). Summing, $$\boxed{\,H(e^{j\omega}) = \frac{1}{1 - e^{-\alpha}e^{-j\omega}}\,}$$ Equivalently $H(z) = 1/(1 - e^{-\alpha}z^{-1})$, a single pole at $z = e^{-\alpha}$ inside the unit circle. Separating magnitude and phase, $$\left|H(e^{j\omega})\right| = \frac{1}{\sqrt{1 - 2e^{-\alpha}\cos\omega + e^{-2\alpha}}},\qquad \angle H(e^{j\omega}) = -\arctan\!\frac{e^{-\alpha}\sin\omega}{1 - e^{-\alpha}\cos\omega}.$$ For $\alpha = 1$ this runs from $|H| = 1/(1-e^{-1}) = 1.582$ at DC down to $1/(1+e^{-1}) = 0.731$ at $\omega = \pi$ — a gentle first-order low-pass with about 6.7 dB of tilt across the band, consistent with the smoothing seen in the step response.

Figure Q4.2 — magnitude of the frequency response for α = 1: a first-order low-pass shape falling from 1/(1 - e⁻¹) = 1.582 at DC to 1/(1 + e⁻¹) = 0.731 at ω = π.

Part	Quantity	Result
(a)	Output	$y(n) = \dfrac{1-e^{-\alpha(n-n_0+1)}}{1-e^{-\alpha}}u(n-n_0)$
(a)	Steady-state value	$1/(1-e^{-\alpha})$
(b)	Samples, $\alpha=1$, $n_0=4$	0 for $n\le3$; 1.000, 1.368, 1.503, 1.553, 1.571, … → 1.582
(c)	Frequency response	$H(e^{j\omega}) = 1/(1-e^{-\alpha}e^{-j\omega})$
(c)	Magnitude at DC / at $\pi$ ($\alpha=1$)	1.582 / 0.731

Topic. Discrete-time LTI systems: convolution with a delayed step, geometric-series closed forms, and the DTFT of a one-pole impulse response.

Key relations. $y = h * x$; $\sum_{k=0}^{N} r^k = (1-r^{N+1})/(1-r)$; $H(e^{j\omega}) = \sum h(n)e^{-j\omega n}$; DC gain $= \sum h(n) = H(e^{j0})$; stability requires $|e^{-\alpha}| < 1$.

Why this works. A step input turns convolution into a running sum of the impulse response, and when that impulse response is a geometric sequence the running sum has a closed form. The delay $n_0$ does nothing except shift the whole result, because LTI systems are time-invariant — solve the undelayed problem and translate. The same geometric sum, evaluated on the unit circle instead of over time, produces the frequency response, which is why the step response's final value and the DC gain are the same number arrived at two ways. That coincidence is the cheapest available check on the algebra.

Common pitfall. Getting the upper limit of the sum wrong by one: the number of terms is $n - n_0 + 1$, not $n - n_0$, and dropping the $+1$ makes $y(n_0) = 0$ instead of 1, which contradicts the system's causal response to the first input sample. Also common is writing $H$ as $1/(1-e^{-\alpha n})$, carrying the time index $n$ into a frequency-domain expression where it cannot appear.

Source. A. V. Oppenheim & A. S. Willsky, Signals and Systems, 2nd ed., Ch. 2 and §5.1 (discrete convolution, DTFT); J. G. Proakis & D. G. Manolakis, Digital Signal Processing, 4th ed., Ch. 2 and 4.

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