Continuous and Discrete-Time Signals: Fourier, Sampling, z-Transform and Filters

The last third of Section 2 is where the Electronics & Communication paper stops looking at circuits and starts looking at signals, and it is the part that feeds Communications and Digital Circuits directly, so it repays study twice over. The organising idea is a single one repeated in four dialects: any signal can be written as a sum of complex exponentials, and an LTI system treats each exponential independently. For a periodic continuous signal that sum is the Fourier series, a discrete set of harmonics whose amplitudes decay in a pattern that the waveform's symmetry predicts before you compute anything. For an aperiodic one it becomes the Fourier transform, a continuum, and with it the reciprocity that governs the whole subject — narrow in time means wide in frequency, always, with no exceptions and no way round it. Then the chapter crosses from the continuous world to the discrete one, and the crossing is the Nyquist sampling theorem: sample faster than twice the highest frequency present and nothing is lost, sample slower and high frequencies fold down onto low ones permanently, because an alias is not a corruption you can filter out afterwards — it is a genuine low-frequency component that no longer differs from the real thing in any measurable way. On the discrete side the same four dialects reappear. The DTFT is the Fourier transform of a sequence and is periodic in 2π, which is the frequency-domain shadow of sampling. The DFT is what a computer can actually compute: N samples in, N numbers out, with a resolution of f_s/N and a leakage penalty whenever the signal does not fit a whole number of cycles into the window. The z-transform is to sequences what Laplace is to functions, and it carries one extra piece of information that trips candidates up every year — the region of convergence, without which the same algebraic expression describes two entirely different signals, one causal and unstable, the other anti-causal and stable. The chapter closes on filter design, where the two families divide cleanly: FIR filters can be made exactly linear phase and are always stable, IIR filters meet a given magnitude specification with far fewer coefficients and cannot be linear phase, and knowing which trade-off a question is testing is most of the answer.

Fourier Series: Symmetry Tells You the Answer First

A periodic signal of period T is a sum of harmonics of f_0 = 1/T: in trigonometric form x(t) = a_0 + Σ [a_n cos(nω_0 t) + b_n sin(nω_0 t)], and in the exponential form that everything later uses, x(t) = Σ c_n e^(jnω_0 t) with c_n = (1/T)∫_T x(t)e^(-jnω_0 t) dt. The integrals are examinable but rarely the point; what is examinable and quick is reading the structure off the waveform before integrating anything, because three symmetries each kill half the work:

  • Even symmetry, x(-t) = x(t): all b_n = 0, so the series is cosines only and every c_n is real.
  • Odd symmetry, x(-t) = -x(t): all a_n = 0 including the mean, so the series is sines only and every c_n is purely imaginary.
  • Half-wave symmetry, x(t + T/2) = -x(t): all even harmonics vanish, including the DC term, whatever else is true of the waveform.

The square wave alternating between +1 and -1 has all three of the properties that matter — zero mean, odd, and half-wave symmetric — so before computing a single integral you know the answer contains only odd sine harmonics. Doing the integral gives amplitudes 4/(nπ) for odd n: 1.273 for the fundamental, 0.424 for the third, 0.255 for the fifth, 0.182 for the seventh. The decay is only 1/n, which is slow, and that slowness has a cause worth naming: a discontinuity in the waveform forces a 1/n envelope. Smooth the corners and the harmonics fall off faster; a triangular wave, which is continuous but has kinked slope, decays as 1/n². So you can estimate the harmonic content of an unfamiliar waveform from how many times you have to differentiate it before an impulse appears.

🎯 A square wave needs infinitely many harmonics and 95 per cent of it is in four
Parseval's theorem for a Fourier series says the average power of the signal equals the sum of the average powers of its harmonics — the harmonics are orthogonal, so their powers add even though their amplitudes do not. Test it on the unit square wave, whose average power is trivially 1 because x²= 1 everywhere. Each harmonic of amplitude A_n = 4/(nπ) carries power A_n²/2 = 8/(n²π²), so the total is (8/π²) Σ_(n odd) 1/n², and since Σ_(n odd) 1/n² = π²/8 the sum is exactly 1 ✓. Now read the partial sums, which is where the insight is. The fundamental alone carries 8/π² = 0.8106, so 81.1 per cent of a square wave's power sits in a single sine wave. Adding the third takes it to 90.1 per cent, the fifth to 93.3, the seventh to 94.96. Four terms, and you have 95 per cent of the power of a waveform that formally requires infinitely many. But now look at the shape rather than the power: a four-term approximation still has visible ripple near each edge, and adding terms narrows that ripple without reducing its height — the Gibbs phenomenon, an overshoot of about 9 per cent that never goes away at any finite number of terms. Two lessons, and they pull in opposite directions, which is exactly why both are examined. If you care about power — a heating element, an RMS meter, a power-factor calculation — a handful of harmonics is plenty, and this is why harmonic distortion is quoted as a percentage of total power. If you care about shape — a clock edge, a pulse, a data symbol — the tail matters enormously, and truncating the spectrum of a square wave is precisely what a bandlimited channel does to a digital signal, which is where intersymbol interference comes from.

Fourier Transform: The Reciprocity of Time and Frequency

Let the period of a periodic signal grow without bound and the harmonics crowd together until the discrete spectrum becomes a continuous one. That limit is the Fourier transform, X(f) = ∫ x(t)e^(-j2πft) dt, with the inverse x(t) = ∫ X(f)e^(j2πft) df. Its properties are more useful in the exam than the integral, because most questions are solved by recognising a transform pair and applying one property to it. Time shift multiplies the transform by e^(-j2πf t_0), changing only the phase and leaving |X(f)| untouched — which is why a delayed signal has the same magnitude spectrum and why magnitude spectra cannot tell you about timing. Time scaling by a scales the frequency axis by 1/a and the amplitude by 1/|a|. Convolution in time is multiplication in frequency, and vice versa, which is the single most used property in the whole section: it turns a filter from an integral into a product. And duality says every pair works both ways, so the transform of a rectangle is a sinc and the transform of a sinc is a rectangle.

💡 A one-millisecond pulse cannot be squeezed into less than a kilohertz
Take a rectangular pulse of unit height and duration τ = 1 ms. Its transform is X(f) = τ sinc(fτ) = τ sin(πfτ)/(πfτ), which has value τ = 1 ms at DC — the area of the pulse, as it must be, since X(0) = ∫x(t)dt — and its first null at f = 1/τ = 1 kHz, with a main lobe running from -1 kHz to +1 kHz. Now shorten the pulse to 0.5 ms and the first null moves out to 2 kHz: halving the duration doubles the bandwidth, exactly. There is no waveform, no clever coding and no filter that escapes this, because it is a property of the transform and not of the pulse: the product of time extent and bandwidth is bounded below. Think of it the way a musician does. A single struck note held for a long time has a definite pitch, and you can name it. The same note clipped to a few milliseconds becomes a click, and the shorter you make it the less it sounds like any pitch at all — the energy has spread across the spectrum because there is no longer enough time to establish a frequency. That is the same statement as the sinc null moving outward, and it is the reason radar designers trade range resolution against transmitter bandwidth, the reason a 1 Gbit/s link needs at least hundreds of megahertz of channel, and the reason a fast logic edge radiates: a 1 ns rise time has significant content past 300 MHz whether the designer wanted it there or not. Two related facts sit on the same principle. The transform of an impulse is flat across all frequencies, which is the extreme of the trade — zero duration, infinite bandwidth — and is exactly why the impulse response characterises an LTI system completely: an impulse tests every frequency at once. And a pure sinusoid, which has infinite duration, transforms to a pair of impulses at ±f_0 with zero width, the opposite extreme.

Nyquist Sampling: Replication, and Why an Alias Is Permanent

Sampling a continuous signal at intervals of T_s = 1/f_s is, mathematically, multiplication by an impulse train. Multiplication in time is convolution in frequency, and the transform of an impulse train is another impulse train, so the spectrum of the sampled signal is the original spectrum replicated at every multiple of f_s. That one sentence contains the whole theorem. If the original signal is bandlimited to f_m and f_s > 2f_m, the replicas do not overlap, an ideal lowpass filter of cutoff f_s/2 recovers the original exactly, and nothing has been lost. If f_s < 2f_m the replicas overlap, the overlapping parts add, and the sum cannot be unmixed. The Nyquist rate is 2f_m; the Nyquist frequency is f_s/2, and mixing up those two names is a reliable way to lose a mark on an otherwise correct calculation.

⚠️ The alias lands exactly on a real component, and a 2 V tone becomes 3 V
Let x(t) = 3cos(2π·1000t) + 2cos(2π·3000t) + cos(2π·5000t). The highest frequency is 5 kHz so the Nyquist rate is 10 kHz, and we shall deliberately sample at 8 kHz. Fold each component with f_alias = |f - k f_s|, choosing the k that lands the result in [0, f_s/2]. The 1 kHz stays at 1 kHz. The 3 kHz stays at 3 kHz. The 5 kHz becomes |5000 - 8000| = 3000 Hz — it lands exactly on top of the genuine 3 kHz component. And it does not merely land nearby, it becomes indistinguishable, which you can verify sample by sample: the samples of the 5 kHz tone are cos(2π·5n/8), and since 5/8 = 1 - 3/8, that equals cos(2πn - 2π·3n/8) = cos(2π·3n/8), the samples of the 3 kHz tone, term for term. So after reconstruction through an ideal 4 kHz lowpass you get 3cos(2π·1000t) + 3cos(2π·3000t) — the 3 kHz component has grown from amplitude 2 to amplitude 3, and the 5 kHz component has vanished. Now the part that matters. No filter applied after sampling can undo this, because there is nothing left to separate: the two components produced identical sample values, so the information distinguishing them was destroyed at the instant of sampling and not encoded anywhere. This is why the anti-aliasing filter goes before the sampler and is not optional, and why a specification always reads bandlimit, then sample in that order. Sample the same signal at 12 kHz instead and all three components return unchanged at 1, 3 and 5 kHz, because 12 kHz clears the 10 kHz Nyquist rate. One more trap on the same page, and it is the reason the theorem says strictly greater than. Sample sin(2π·1000t) at exactly 2 kHz, the Nyquist rate itself: the samples are sin(πn) = 0 for every n, so a perfectly good sinusoid samples to a string of zeros and reconstructs as silence. Sampling at the Nyquist rate is not enough; you need to exceed it, which in practice means leaving a guard band — audio CDs bandlimit to 20 kHz and sample at 44.1 kHz, giving the anti-aliasing filter 2.05 kHz of transition width in which to fall from passband to stopband.

Reconstruction is the other half of the theorem, and the ideal recipe is x(t) = Σ x[n] sinc((t - nT_s)/T_s): put a sinc function of unit height at every sample, scaled by that sample, and add them. It works because every sinc is zero at all the other sample instants, so each sample is reproduced exactly while the sincs interpolate between them. It is also unrealisable, because a sinc extends infinitely in both directions and therefore needs the entire future of the signal. What real converters use instead is the zero-order hold — hold each sample constant until the next one arrives — which is a rectangle of width T_s and therefore has a sinc(fT_s) frequency response. That response is not flat: at f = f_s/4 it has fallen to 0.900 (-0.91 dB) and at f = f_s/2 to 2/π = 0.6366, which is -3.92 dB. So a plain ZOH digital-to-analogue converter attenuates the top of its own band by nearly 4 dB, and any audio or instrumentation converter that matters therefore follows it with a sinc-compensation filter that rises across the band to cancel exactly that droop. This is the sort of number GATE asks for directly, and it is one line of arithmetic once you recognise that a hold is a convolution with a rectangle.

DTFT and DFT: Periodic in 2π, and What a Computer Can Actually Do

The DTFT of a sequence is X(e^(jω)) = Σ x[n]e^(-jωn), a function of a continuous frequency variable ω measured in radians per sample — and it is periodic in 2π, always, for every sequence. That periodicity is not a mathematical curiosity, it is aliasing seen from the frequency side: ω = 2π corresponds to f = f_s, and the replication of the spectrum at multiples of f_s is exactly the statement that the DTFT repeats. So the entire information content of a discrete-time spectrum lives in -π ≤ ω < π, with ω = 0 being DC and ω = π being the highest representable frequency, f_s/2. Take x[n] = (0.5)^n u[n], whose DTFT is 1/(1 - 0.5e^(-jω)). At ω = 0 it is 2, at ω = π/2 its magnitude is 0.894, and at ω = π it is 2/3 = 0.667 — a gentle lowpass, with a DC-to-Nyquist gain ratio of exactly 3.

The DTFT is continuous in ω and therefore not computable. The DFT fixes that by sampling it: given N samples, X[k] = Σ_(n=0)^(N-1) x[n] e^(-j2πkn/N) for k = 0 to N-1. N numbers in, N numbers out, and the k-th output corresponds to the frequency k f_s/N, so the bin spacing — the frequency resolution — is f_s/N and depends only on how long you observed, never on how fast you sampled. Work a four-point example fully, because every property is visible in it. For x[n] = {1, 2, 3, 4} with N = 4, the twiddle factors are 1, -j, -1, j and the transform comes out X = {10, -2 + j2, -2, -2 - j2}. Read three things off it. X[0] = 10 is the sum of the samples, always. X[1] and X[3] are complex conjugates, as they must be for a real input — conjugate symmetry means half the output is redundant. And Parseval holds in the discrete form Σ|x[n]|² = (1/N)Σ|X[k]|²: here 1 + 4 + 9 + 16 = 30 and (100 + 8 + 4 + 8)/4 = 30 ✓.

⚠️ Two and a half cycles in the window, and suddenly every bin is non-zero
Sample cos(2π · 2000t) at f_s = 8 kHz and take N = 8 points. The bin spacing is 8000/8 = 1000 Hz, so 2 kHz is exactly bin 2 — and the DFT is clean: X[2] = X[6] = 4 and every other bin is exactly zero. That is what a textbook plot looks like, and it happens because the 8-sample window contains exactly 2 complete cycles. Now change the tone to 2500 Hz, which puts 2.5 cycles in the same window. The DFT magnitudes become {1.00, 1.19, 2.80, 2.40, 1.00, 2.40, 2.80, 1.19}: no bin is zero, the peak is split between bins 2 and 3, and neither peak reaches the correct amplitude. Nothing has gone wrong with the arithmetic. This is spectral leakage, and its cause is a single fact worth internalising: the DFT implicitly assumes the N samples repeat periodically. Two and a half cycles repeated end to end has a discontinuity at the join, and a discontinuity has broadband content — the same 1/n tail that the square wave taught us. So leakage is not a defect of the DFT, it is the DFT correctly reporting the spectrum of the periodic signal you implicitly gave it. Two cures, and they do different jobs. Windowing — multiplying the samples by a taper such as Hamming or Hann that goes smoothly to zero at both ends — removes the discontinuity and so cuts the leakage far from the peak, at the cost of widening the peak itself. Lengthening the record improves the resolution: N = 16 at the same 8 kHz gives 500 Hz bins, so 2500 Hz lands exactly on bin 5 and the leakage disappears entirely. Note which knob does which, because it is a favourite examinable distinction: resolution is f_s/N and is set by the observation time alone, so zero-padding a short record to a longer transform length interpolates the display and reveals no new detail whatever. And one more thing to keep straight: the FFT is not a different transform, it is an algorithm that computes the same DFT in O(N log N) instead of O(N²) operations, so nothing above changes when you use one.

One consequence of that implicit periodicity is worth its own paragraph, because it is the commonest DFT bug in real code and a standard GATE question. Multiplying two DFTs corresponds not to ordinary convolution but to circular convolution, in which the tail of the result wraps round onto the beginning. Take x = {1, 2} and h = {1, 1}. Their linear convolution is {1, 3, 2}, of length 2 + 2 - 1 = 3. Compute it with 2-point DFTs and you get {3, 3} — the trailing 2 has wrapped onto index 0, giving 1 + 2 = 3. Use N = 3 and you get {1, 3, 2}, correct; use N = 4 and you get {1, 3, 2, 0}, correct with a padding zero. So the rule is simply N ≥ L_x + L_h - 1, achieved by zero-padding both sequences before transforming, and it is the reason fast convolution routines pad to the next power of two.

The z-Transform: Why the Region of Convergence Is Not Optional

The z-transform X(z) = Σ x[n]z^(-n) is to sequences what the Laplace transform is to functions: it turns difference equations into algebra, delays into multiplications by z^(-1), and convolution into multiplication. Setting z = e^(jω) — evaluating on the unit circle — recovers the DTFT, exactly as setting s = jω recovered the frequency response from H(s). The mapping between the two planes is the one to hold in mind: the left half of the s plane maps to the inside of the unit circle, the imaginary axis maps to the circle itself, and the right half maps outside. So every stability statement transfers with one substitution, and a causal discrete system is stable exactly when all its poles lie strictly inside the unit circle.

⚠️ The same expression, two different signals, and only the ROC tells them apart
Compute the z-transform of x[n] = 2^n u[n], a growing causal exponential. The geometric series Σ (2/z)^n converges only for |z| > 2, and sums to 1/(1 - 2z^(-1)) = z/(z - 2). Now compute the z-transform of x[n] = -2^n u[-n-1], an anti-causal sequence that grows as you go backwards in time. The sum runs over negative n, converges only for |z| < 2, and comes to exactly the same expression, z/(z - 2). Two completely different signals, one algebraic formula. The only thing that distinguishes them is the region of convergence, and that is why a z-transform is not fully specified without one. The consequences are the examinable part. The first signal is causal but unstable: its ROC |z| > 2 excludes the unit circle, so its DTFT does not exist and a bounded input can produce an unbounded output. The second is stable but non-causal: its ROC |z| < 2 contains the unit circle, so it has a perfectly good DTFT, and it is unrealisable in real time for a different reason — it needs future input. Three rules turn this into a decision procedure. The ROC of a causal sequence is the region outside the outermost pole; the ROC of an anti-causal sequence is inside the innermost pole; and a sequence is stable if and only if its ROC includes the unit circle |z| = 1. Put the last two together and you get the familiar textbook criterion as a special case: for a causal system, ROC outside the outermost pole must include |z| = 1, which forces every pole to satisfy |p| < 1. But notice what the general statement buys you that the special case does not. A pole at |z| = 2 is fatal for a causal system and perfectly acceptable for an anti-causal one, and a two-sided sequence can have poles both inside and outside while still being stable, provided the annular ROC between them straddles the unit circle. GATE asks this as *given H(z) and the ROC, is the system causal, stable, both or neither* — four boxes, and the ROC is the only thing that fills them in.

Two workhorse pairs cover most z-transform questions. a^n u[n] ↔ z/(z - a) with ROC |z| > |a|, and n a^n u[n] ↔ az/(z - a)², which is where repeated poles come from. Inversion is by partial fractions plus the table, exactly as in Laplace, with one extra habit: write H(z) in powers of z^(-1) when you want the difference equation and in powers of z when you want the poles, because the two forms hide each other's information. H(z) = z/(z - 0.5) is 1/(1 - 0.5z^(-1)), which reads directly as y[n] = 0.5y[n-1] + x[n], a one-multiply one-add recursion whose impulse response is 0.5^n u[n], summing to 2 and therefore stable — and its single pole at 0.5 is comfortably inside the unit circle, saying the same thing a second way.

FIR and IIR Filter Design: Two Different Bargains

A digital filter is a difference equation, and the syllabus splits the design problem the way the implementations split. An FIR filter has a finite impulse response and no feedback: y[n] = Σ_(k=0)^(M) b_k x[n-k]. It has no poles other than at the origin, so it is unconditionally stable — you cannot make an FIR filter unstable by choosing bad coefficients — and if its coefficients are symmetric it has exactly linear phase. An IIR filter feeds output back: y[n] = Σ b_k x[n-k] - Σ a_k y[n-k]. It has genuine poles, so it must be checked for stability, it cannot have linear phase (a linear-phase causal filter with poles would need a mirror-image pole outside the unit circle), and in exchange it meets a given magnitude specification with far fewer coefficients — often an order of magnitude fewer.

The standard FIR route is windowing. Start from the ideal filter, take its inverse DTFT, and truncate. For an ideal lowpass with cutoff ω_c the ideal impulse response is h_d[n] = sin(ω_c n)/(πn) with h_d[0] = ω_c/π, a sinc that extends infinitely in both directions and is therefore neither causal nor finite. At ω_c = π/4 its central values are h_d[0] = 0.2500, h_d[±1] = 0.2251, h_d[±2] = 0.1592, h_d[±3] = 0.0750, h_d[±4] = 0 and h_d[±5] = -0.0450. Keep a symmetric window of them, shift to make it causal, and you have a linear-phase FIR filter. But truncating abruptly is multiplying by a rectangular window, and a rectangle has that 1/n spectral tail again, so the result has a peak stopband ripple of about -21 dB that does not improve as you lengthen the filter — lengthening narrows the transition band and leaves the ripple height alone, which is the Gibbs phenomenon reappearing in a new costume. The cure is to taper the window.

Standard windows for FIR design: the trade is always ripple against transition width, with M the filter length
WindowMain lobe widthPeak side lobeMinimum stopband attenuation
Rectangular4π/M−13 dB−21 dB
Hanning8π/M−31 dB−44 dB
Hamming8π/M−41 dB−53 dB
Blackman12π/M−57 dB−74 dB

Read that table as one sentence: every window buys stopband attenuation by widening the transition band, and the only way to have both is a longer filter. Rectangular gives the narrowest transition and the worst ripple; Blackman gives 74 dB of attenuation and pays with a main lobe three times as wide. The other point worth noting is which linear-phase FIR types can do which job. A symmetric filter of odd length (Type I) can realise a lowpass, highpass, bandpass or bandstop; a symmetric filter of even length (Type II) is forced to have a zero at z = -1, that is at ω = π, so it cannot be a highpass. And the linear phase itself costs a delay of (M - 1)/2 samples: a 21-tap filter at f_s = 8 kHz delays everything by 10 samples, which is 1.25 ms — a fixed, harmless lag in an audio path and a disqualifying one in a feedback control loop.

🧠 Prewarp before you design, or your 1 kHz cutoff arrives at 953 Hz
IIR filters are almost never designed from scratch; they are converted from a well-understood analogue prototype — Butterworth, Chebyshev, elliptic — and there are two conversions. Impulse invariance samples the analogue impulse response, h[n] = T h_a(nT), which preserves the shape of the response but aliases: the analogue prototype's tail folds back at f_s/2, so the method is unusable for a highpass or bandstop, whose responses never die away. The bilinear transform substitutes s = (2/T)(z-1)/(z+1), which maps the entire jω axis onto the unit circle exactly once. Nothing aliases, ever — but the mapping is nonlinear, ω_analogue = (2/T)tan(ω_digital/2), so the frequency axis is warped, compressed increasingly as you approach f_s/2. Put numbers on the consequence. Suppose f_s = 8 kHz and you want a digital cutoff at 1 kHz, that is ω_d = 2π(1000)/8000 = π/4 = 0.7854 rad/sample. Design an analogue prototype naively at 1 kHz and transform it, and the cutoff you actually get is ω_d = 2 arctan(ω_a T/2) = 2 arctan(0.3927) = 0.7484 rad, which is 953 Hz — low by 4.7 per cent, and the error grows as the cutoff approaches Nyquist. The fix is one line, done before designing anything: prewarp the specification by ω_a = (2/T)tan(ω_d/2) = 16000 tan(π/8) = 6627.4 rad/s, which is 1054.8 Hz. Design the analogue prototype for 1054.8 Hz, transform it, and the digital cutoff lands on 1000 Hz exactly. Notice that the constant 2/T cancels out of the round trip entirely, which is why some texts set T = 1 or T = 2 and nothing changes — the warping depends only on the ratio of the frequency to the sampling rate. Two closing comparisons worth having ready. For a given magnitude specification an IIR filter typically needs five to ten times fewer coefficients than an equivalent FIR, which is decisive on a small microcontroller. But the FIR is unconditionally stable, has exactly linear phase, is insensitive to coefficient quantisation, and cannot oscillate on its own — and those four properties are why an audio crossover, an image filter or any application where waveform shape matters usually ends up FIR despite the cost.

Key takeaways

  • Symmetry answers a Fourier series question before you integrate. Even means cosines only, odd means sines only, half-wave symmetry kills every even harmonic. A discontinuity forces a 1/n envelope, so 81 per cent of a square wave's power is in the fundamental and 95 per cent is in four terms — while the shape still needs the whole tail.
  • Narrow in time is wide in frequency, with no exceptions: a 1 ms pulse has its first spectral null at 1 kHz and a 0.5 ms pulse at 2 kHz. That single reciprocity explains the flat spectrum of an impulse, the zero width of a sinusoid, and why fast edges radiate.
  • Sampling replicates the spectrum at every multiple of f_s, so an alias is not noise you can filter out — sampling 1, 3 and 5 kHz at 8 kHz makes the 5 kHz tone identical to the 3 kHz one, sample for sample, and the 2 V component reconstructs as 3 V. Bandlimit first, then sample, and exceed 2f_m strictly: a sine sampled at exactly twice its frequency can read all zeros.
  • A zero-order-hold DAC is a rectangle, so its response is sinc(fT_s): −3.92 dB of droop at f_s/2, which is why real converters follow the hold with a sinc-compensation filter rather than an ideal interpolator no one can build.
  • The DFT assumes its N samples repeat. Two whole cycles in the window give a single clean bin; 2.5 cycles leak into every bin. Resolution is f_s/N and is fixed by observation time alone, so windowing suppresses leakage and lengthening the record improves resolution — zero-padding does neither. Multiplying DFTs gives circular convolution, so pad to N ≥ L_x + L_h - 1.
  • z/(z - 2) is both a causal unstable signal and an anti-causal stable one; only the ROC decides which. Causal means ROC outside the outermost pole, anti-causal means inside the innermost, and stable means the ROC contains |z| = 1. The familiar all poles inside the unit circle is just those three rules applied to a causal system.
  • FIR is unconditionally stable and can be exactly linear phase, at the price of (M-1)/2 samples of delay and many coefficients; IIR meets the same magnitude spec with five to ten times fewer and can never be linear phase. Windowing trades stopband depth for transition width — rectangular −21 dB, Hamming −53 dB, Blackman −74 dB. And prewarp before designing an IIR filter by the bilinear transform: wanting 1 kHz at f_s = 8 kHz from a prototype designed at 1 kHz lands you at 953 Hz, while designing at 16000 tan(π/8) = 6627 rad/s = 1054.8 Hz lands exactly on 1 kHz. Impulse invariance warps nothing but aliases, so it cannot make a highpass.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A periodic voltage has a total RMS value of 10 V. Its DC component is 6 V and its fundamental has an RMS value of 6 V. What is the combined RMS value of all the harmonics above the fundamental?

    1. 5.29 V
    2. 4.00 V
    3. 8.00 V
    4. 2.83 V
    Show answer

    Answer: A — 5.29 V

    Harmonics of a periodic waveform are mutually orthogonal, so their mean-square values add even though their amplitudes do not. Taking the total mean square as 100, the DC term contributes 36 and the fundamental 36, leaving 100 - 36 - 36 = 28 for everything above, whose RMS value is the square root of 28 = 5.29 V. The answer 4.00 V comes from subtracting the RMS values directly, 10 - 6, which would only be legitimate if the components added in phase; they do not, and that is the whole content of the theorem. The 8.00 V value is the square root of 100 - 36 and forgets that the fundamental also has to be removed.

  2. A finite-energy signal is delayed by a fixed time before being transmitted. What happens to its Fourier transform?

    1. Only its magnitude changes, being scaled by the delay
    2. Both its magnitude and its phase change
    3. Only its phase changes, acquiring a term linear in frequency, while the magnitude is untouched
    4. Neither changes, because a pure delay carries no information
    Show answer

    Answer: C — Only its phase changes, acquiring a term linear in frequency, while the magnitude is untouched

    Delaying a signal by t0 multiplies its Fourier transform by exp(-j·2·pi·f·t0), a factor of unit magnitude, so the magnitude spectrum is unchanged and the phase gains a term proportional to frequency. That is why a magnitude spectrum alone can never reveal timing, and why two signals that look identical on a spectrum analyser may arrive at quite different instants. The idea that nothing at all changes is the tempting answer for anyone who reasons only from the unchanged magnitude plot, but the linear phase term is precisely where the delay is recorded, and it is what a group-delay measurement reads back.

  3. A 2.5 kHz sinusoid is sampled at 20 kHz and a 32-point discrete Fourier transform is taken of the samples. On which bin does the tone land, and is the transform free of leakage?

    1. Bin 8, and the transform is free of leakage
    2. Bin 4, and the transform is free of leakage because the window holds a whole number of cycles
    3. The tone falls exactly on a bin, yet leakage still spreads energy into every other bin
    4. Between bins 3 and 4, so the peak is split between them
    Show answer

    Answer: B — Bin 4, and the transform is free of leakage because the window holds a whole number of cycles

    The bin spacing is the sampling rate divided by the transform length, 20000/32 = 625 Hz, so 2500 Hz falls on 2500/625 = bin 4 exactly, with its conjugate at bin 28. Whether leakage appears depends on a second check: the record holds 32 times 2500/20000 = 4 complete cycles, so repeating the window end to end produces no discontinuity and every other bin is exactly zero. Bin 8 comes from dividing by a spacing of 312.5 Hz, that is from using twice the transform length. The belief that a transform always leaks misses the point that leakage is the correct spectrum of the periodic extension the transform implicitly assumes, and that extension is smooth here.

  4. The sequences x = {1, 2, 3} and h = {1, 1} are transformed with three-point discrete Fourier transforms, the transforms are multiplied point by point, and the product is inverse transformed. What sequence comes out?

    1. {1, 3, 5}
    2. {1, 3, 5, 3}
    3. {4, 5, 3}
    4. {4, 3, 5}
    Show answer

    Answer: D — {4, 3, 5}

    Multiplying transforms of length N gives circular convolution, not the ordinary kind. The linear convolution of {1, 2, 3} with {1, 1} is {1, 3, 5, 3}, of length 3 + 2 - 1 = 4, which will not fit in three points, so the trailing 3 wraps round onto index 0 and adds to the 1 already there, giving {4, 3, 5}. Answering {1, 3, 5, 3} is the tempting mistake of reporting the linear result, which is what point-by-point multiplication would give only if the transform length were at least 4; padding both sequences to N = 4 before transforming is exactly the fix, and it is why fast convolution routines pad to the next convenient length.

  5. A z-transform has poles at z = 0.5 and z = 2, and its region of convergence is the annulus 0.5 < |z| < 2. What can be said about the sequence it represents?

    1. It is two-sided and BIBO stable, but not causal
    2. It is causal and BIBO stable
    3. It is causal but not BIBO stable
    4. It is anti-causal and not BIBO stable
    Show answer

    Answer: A — It is two-sided and BIBO stable, but not causal

    Three rules settle it. A causal sequence has its region of convergence outside the outermost pole, which here would be |z| > 2; an anti-causal one has it inside the innermost pole, |z| < 0.5. An annulus lying between two poles belongs to neither and therefore to a two-sided sequence. Stability is decided separately: the region contains the unit circle |z| = 1, so the discrete-time Fourier transform exists and the sequence is absolutely summable. The answer that it is causal and stable is what you get by reading only the pole at 0.5 and applying the familiar all-poles-inside rule, which is the special case of these three rules for a causal sequence and does not apply here.

  6. A causal discrete-time filter obeys the difference equation y[n] = 0.6·y[n-1] + x[n] - x[n-1]. What is the magnitude of its frequency response at the highest representable frequency, omega = pi radians per sample?

    1. 0.00
    2. 0.80
    3. 2.50
    4. 1.25
    Show answer

    Answer: D — 1.25

    Rewrite the recursion as a transfer function first: collecting terms gives H(z) = (1 - z inverse)/(1 - 0.6 z inverse). Evaluating on the unit circle at omega = pi means z = -1, so z inverse = -1 and H = (1 + 1)/(1 + 0.6) = 2/1.6 = 1.25. The value 0.00 is the response at zero frequency, where the numerator zero at z = 1 kills the output entirely, and it is the tempting answer for anyone who substitutes z = 1 out of habit; that zero is what makes this a highpass rather than a lowpass. The value 0.80 is the reciprocal of the right answer and follows from inverting the ratio of numerator to denominator.

  7. Two cosinusoids, one at 1.8 kHz and one at 8.2 kHz, are each sampled at 10 kHz by the same ideal sampler. What relationship do the two sample sequences bear to one another?

    1. They differ in sign at every odd-numbered sample
    2. They are identical, sample for sample
    3. They have equal magnitudes but are ninety degrees apart in phase
    4. They are unrelated, because 8.2 kHz is below the sampling rate
    Show answer

    Answer: B — They are identical, sample for sample

    Write the samples of the higher tone as cos(2·pi·8200·n/10000). Since 8200/10000 = 1 - 1800/10000, that is cos(2·pi·n - 2·pi·1800·n/10000), and cosine is even and 2·pi periodic, so it equals cos(2·pi·1800·n/10000) term for term. The two sequences are therefore the same numbers, and no filter applied after the sampler can separate them, because the information that distinguished them was destroyed at the instant of sampling. Assuming they are unrelated because 8.2 kHz is under the 10 kHz sampling rate is the classic error: what matters is whether the tone is under half the rate, and 8.2 kHz is not.

  8. A designer must choose between a non-recursive structure and a recursive one for a lowpass filter that will run on a small processor. Which of the following statements bearing on that choice are correct?

    1. A causal filter whose finite impulse response is symmetric about its centre has exactly linear phase
    2. A filter with a finite impulse response cannot be made unstable by any choice of its coefficients
    3. For a given magnitude specification a recursive filter generally needs more coefficients than a non-recursive one
    4. Converting an analogue prototype by the bilinear transform folds the prototype response back at half the sampling rate
    Show answer

    Answer: A — A causal filter whose finite impulse response is symmetric about its centre has exactly linear phase; B — A filter with a finite impulse response cannot be made unstable by any choice of its coefficients

    Coefficient symmetry forces the frequency response to be a real function multiplied by a pure delay term, which is exactly linear phase, and the delay costs half the filter length in samples. A non-recursive filter has no poles away from the origin, so no coefficient choice can move a pole outside the unit circle and it is unconditionally stable. The claim about coefficient counts is the wrong way round: a recursive design meets the same magnitude mask with far fewer coefficients, often an order of magnitude fewer, which is why it wins on a small processor. Folding at half the sampling rate is the defect of impulse invariance; the bilinear transform maps the whole imaginary axis onto the unit circle once and aliases nothing, at the price of a warped frequency axis.