- That \(H(j\omega) = \mathcal{F}\{h(t)\}\) is a complete description of an LTI system: \(|H|\) is the gain at each frequency, \(\angle H\) the phase shift, and there is nothing else a filter can do.
- How to read a magnitude and phase response, why engineers plot them in decibels against a logarithmic frequency axis, and what the \(-3\) dB point means.
- How to get \(H(j\omega)\) straight from a differential equation or a circuit, and how to get \(h(t)\) back by partial fractions.
- The two conditions for distortionless transmission — flat magnitude and linear phase — and what group delay measures when the second one fails.
- The four ideal filters, their impulse responses, and the exact reason a brick-wall filter can never be built.
- How real filters are specified — passband ripple, transition band, stopband attenuation — and how to compute the order of a Butterworth filter that meets a specification.
- The unavoidable time-domain price of filtering: the rise-time–bandwidth product, and the ringing that a sharp cutoff always produces.
What the Frequency Response Is
Chapter 14 ended on the identity \(Y(j\omega) = H(j\omega)X(j\omega)\). Read it slowly, because everything in this chapter is an unpacking of it. The output spectrum is the input spectrum multiplied, frequency by frequency, by a fixed complex-valued function of \(\omega\). That function is the frequency response, and it is the Fourier transform of the impulse response.
One complex function of one real variable determines the system's response to every input it will ever be given. Chapter 7 said the same of \(h(t)\); the two descriptions are a transform pair and carry identical information.
The most useful consequence is the steady-state response to a sinusoid, which follows from the eigenfunction argument of Chapter 12 and needs no integration at all. Feed in \(e^{j\omega_1 t}\) and the output is \(H(j\omega_1)e^{j\omega_1 t}\): the same exponential, scaled by a complex number. Taking real parts gives the statement in the form you will use daily.
Two facts are buried in that line, and both are worth stating aloud. First, the output is a sinusoid at the same frequency: an LTI system never manufactures a frequency the input did not contain, a claim proved in Section 12-8 and now visible as a triviality — multiplying a spectrum by \(H(j\omega)\) can scale what is there but cannot put something where there was nothing. Second, the only two things a system can do to a component are change its size and shift it in time. Gain and phase. That is the complete repertoire.
Everything a filter does is therefore contained in the shape of \(|H(j\omega)|\). A curve that is large over one band and small over another is a filter, whatever circuit happens to realise it. Designing a filter means choosing that curve.
One caveat before we proceed. The analysis integral for \(H\) converges when \(h\) is absolutely integrable, which by Chapter 8 is exactly the condition for BIBO stability. An unstable system has no frequency response — its impulse response grows, the integral diverges, and there is no steady state for a sinusoid to settle into. Chapter 17 introduces the Laplace transform partly to handle that case, and the relationship there is simply \(H(j\omega) = H(s)\big|_{s=j\omega}\), valid whenever the region of convergence contains the imaginary axis.
Magnitude, Phase and Decibels
Since \(H(j\omega)\) is complex it takes two plots to display, and the pair is called the magnitude response and the phase response. For a real impulse response — every physical system — the conjugate symmetry of Section 14-3 applies, so \(|H|\) is even and \(\angle H\) is odd, and it is customary to plot only \(\omega \ge 0\).
The magnitude is almost always plotted in decibels, and the reason is not tradition but arithmetic. A cascade multiplies frequency responses; taking logarithms turns that product into a sum, so cascaded stages can be added graphically. Logarithms also compress the enormous dynamic range of a filter — a stopband attenuation of \(10^{-4}\) and a passband gain of 1 sit comfortably on the same axis once expressed as \(-80\) dB and \(0\) dB.
That last value is the one to memorise. Power is proportional to the square of amplitude, so a gain of \(1/\sqrt2\) halves the power; the frequency at which it occurs is the half-power or \(-3\) dB point, and the band between the origin and that frequency is what "bandwidth" means unless some other definition is stated. Chapter 13's essential-bandwidth calculation and this one are answering the same question with different thresholds.
The frequency axis is logarithmic for the same reason: it makes power-law behaviour into straight lines. A response falling as \(1/\omega\) loses 20 dB every time the frequency multiplies by ten — one decade — and appears as a line of constant slope. A response falling as \(1/\omega^{n}\) has slope \(-20n\) dB per decade, or equivalently \(-6n\) dB per octave, since doubling is \(20\log_{10}2 = 6.02\) dB. Section 15-6 draws such a plot in full.
Frequency Response from a Differential Equation
Chapter 9 described LTI systems by constant-coefficient differential equations, and Chapter 14 turned differentiation into multiplication by \(j\omega\). Putting the two together gives the frequency response of any such system without solving anything. Transform every term of
Read the coefficients off the differential equation and the frequency response is written. No initial conditions appear, because \(H\) describes the forced response only — the transform assumes the system has been running forever.
Three things can be read off such a ratio immediately, and they are usually all a first sketch requires. The DC gain is \(H(0) = b_0/a_0\), obtained by setting \(\omega = 0\). The high-frequency behaviour is governed by the leading terms: for \(\omega\) large, \(H \approx (b_M/a_N)(j\omega)^{M-N}\), so the response rolls off at \(20(N-M)\) dB per decade. And the poles — the roots of the denominator — control everything in between, since \(|H|\) rises wherever \(j\omega\) approaches a pole and falls wherever it approaches a zero.
The simplest case is worth doing in full because it recurs constantly. A series \(RC\) circuit with the output taken across the capacitor obeys \(RC\,\dot{y} + y = x\), by Kirchhoff's voltage law and \(i = C\dot{y}\). Transforming gives
The same expression can be written down without the differential equation at all, by treating the circuit as a voltage divider between an impedance \(R\) and an impedance \(1/j\omega C\) — which is legitimate precisely because the exponential \(e^{j\omega t}\) is an eigenfunction, so the algebra of phasors is the algebra of frequency responses. Section 15-6 examines this response in detail.
Going the other way is equally routine. Given \(H(j\omega)\) as a ratio of polynomials, expand it in partial fractions and invert term by term with the Chapter 13 pair \(1/(a+j\omega) \leftrightarrow e^{-at}u(t)\). Worked Example 2 does this, and the method — factor, expand, invert — is the one Chapter 19 will formalise with the Laplace transform, where it also copes with initial conditions and unstable systems.
Distortionless Transmission and Group Delay
Before asking how to change a signal, it is worth asking what it means to leave one alone. A channel is distortionless if its output is the input, possibly amplified and possibly delayed, but otherwise identical in shape: \(y(t) = K\,x(t - t_d)\). Section 12-7 introduced the idea; the properties of Chapter 14 now let us state exactly what such a channel's frequency response must look like. Transform both sides, using linearity and the time-shift property:
Constant magnitude and phase proportional to frequency. Both are required, and both only over the band the signal actually occupies — outside it, \(H\) may do whatever it likes.
Failing the first condition is amplitude distortion: some components are amplified more than others and the waveform's shape changes even if every component arrives on time. Failing the second is phase distortion, and it is the subtler of the two, because the magnitude spectrum — the thing a spectrum analyser shows you — is completely unaffected. Every frequency is present in the right amount; they simply do not line up.
Why must the phase be linear rather than merely small? Because a delay of \(t_d\) shifts a component at \(\omega\) by \(\omega t_d\) radians, and the shifts must be in proportion to frequency if all components are to be delayed by the same time. A constant phase shift applied to every frequency is not a delay at all; it is a reshaping.
When the phase is not linear, two derived quantities measure what is happening. The phase delay is the delay experienced by a single sinusoid; the group delay is the delay experienced by the envelope of a narrow band of frequencies, and it is the one that matters for pulses and modulated signals.
For a pure delay, \(\theta = -\omega t_d\), and both quantities equal \(t_d\) at every frequency — the definition and the special case agree. When they do not agree, or when either varies with frequency, the channel is dispersive: different parts of the signal travel at different speeds and a pulse launched cleanly arrives smeared. Optical fibre disperses; so does a long cable; so does any filter with a curved phase response, which is nearly all of them.
The engineering response depends on the application. Human hearing is remarkably insensitive to moderate phase distortion, so audio filters are usually designed for magnitude alone. A data receiver is not: dispersion smears one symbol into the next and causes intersymbol interference, so communication systems either use filters with deliberately linear phase — the reason linear-phase FIR filters of Chapter 25 are so heavily used — or add an equaliser to flatten the group delay after the fact.
Ideal Filters and Why They Cannot Be Built
The most obvious thing to ask of a filter is that it pass one band perfectly and reject everything else absolutely. That specification defines the four ideal filters, whose magnitude responses are rectangles: lowpass, highpass, bandpass and bandstop. They are the reference against which real filters are measured, and they are also, exactly and provably, unbuildable.
Take the ideal lowpass, \(H(j\omega) = 1\) for \(|\omega| \lt \omega_c\) and zero elsewhere, and find its impulse response by the synthesis equation. The integral is elementary because the integrand is an exponential over a finite range:
This is Chapter 13's sinc pair, arriving from the system side. And it is fatal, for a reason that is visible in one glance at the plot: \(h(t)\) is non-zero for \(t \lt 0\). The filter responds before it is excited. Feed an impulse in at \(t = 0\) and the output has already been ringing for all previous time — the system is not causal, and by Chapter 8's criterion no causal system can have this frequency response.
The obvious repair fails. Delaying the impulse response by \(t_d\) — which costs only the linear phase \(e^{-j\omega t_d}\), and by Section 15-4 introduces no distortion at all — shifts the sinc to the right but does not truncate it. However large \(t_d\) is, the tails of \(\sin(\omega_c(t-t_d))/(\pi(t-t_d))\) still extend to \(t \to -\infty\), because a sinc never actually reaches zero. Delay makes an ideal filter approximately realisable, in the sense that the pre-response can be made negligibly small and then truncated; it never makes it exactly realisable.
The obstruction is fundamental rather than technical, and it has a name. The Paley–Wiener condition states that a causal system's magnitude response cannot vanish identically over any band of non-zero width. A filter may make \(|H|\) as small as you please, but it may not make it exactly zero over an interval. Perfect rejection and causality are incompatible.
The other three ideal filters follow from the lowpass by the properties of Chapter 14, and it is more economical to derive them than to memorise them. A highpass is everything the lowpass rejects, so \(H_{\text{HP}} = 1 - H_{\text{LP}}\) and therefore \(h_{\text{HP}}(t) = \delta(t) - \sin(\omega_c t)/\pi t\). An ideal bandpass centred at \(\omega_0\) is a lowpass shifted to \(\pm\omega_0\), so by the modulation property its impulse response is \(2\cos(\omega_0 t)\sin(\omega_c t)/\pi t\) — a sinc envelope filled in with the carrier. A bandstop is one minus a bandpass. All four are non-causal for the same reason.
The First-Order Lowpass, Properly
Set the ideal aside and examine a filter that can actually be built out of two components. Section 15-3 derived its frequency response; here we extract everything it has to say.
Three regions tell the whole story. Well below the corner, \(\omega \ll \omega_c\), the magnitude is essentially 1 and the phase essentially zero: the filter passes the signal untouched. Well above it, \(\omega \gg \omega_c\), the magnitude falls as \(\omega_c/\omega\) and the phase approaches \(-90^{\circ}\). At the corner itself, \(\omega = \omega_c\), the magnitude is exactly \(1/\sqrt2\) and the phase exactly \(-45^{\circ}\) — the half-power point, which is why \(\omega_c = 1/RC\) is called the filter's bandwidth.
Plotted in decibels against \(\log\omega\), the two asymptotes are straight lines: a horizontal line at 0 dB and a line of slope \(-20\) dB/decade, meeting at \(\omega_c\). The true curve hugs them both and is at its furthest from their intersection exactly at the corner, where it lies \(3.01\) dB below.
The impulse response confirms that this filter is buildable: inverting \(H\) with the standard pair gives \(h(t) = (1/RC)\,e^{-t/RC}u(t)\), which is zero for \(t \lt 0\) as causality demands, and absolutely integrable as stability demands. Its step response is \(s(t) = (1 - e^{-t/RC})u(t)\), the familiar exponential charging curve of a capacitor, and Section 15-8 uses it to make the bandwidth–speed trade-off quantitative.
Compare the two filters honestly. The ideal lowpass has a perfect passband, a perfect stopband and no transition band, and cannot exist. The \(RC\) lowpass has a passband that droops, a stopband that deepens by only 20 dB per decade, a transition band a decade wide — and costs two components. Everything between these extremes is the subject of filter design.
Specifications and the Butterworth Family
Since a brick wall is unobtainable, a real specification must state how much imperfection is tolerable. The standard form does so with four numbers: the passband edge \(\omega_p\) and the maximum attenuation \(A_p\) permitted anywhere inside the passband; the stopband edge \(\omega_s\) and the minimum attenuation \(A_s\) required everywhere beyond it. Between \(\omega_p\) and \(\omega_s\) lies the transition band, where nothing is demanded.
The design problem is then: find the lowest-order — that is, the cheapest — rational \(H(j\omega)\) whose magnitude threads the mask. Several standard families solve it, each making a different compromise, and the one to know first is the Butterworth, defined by a magnitude-squared response with no free parameters beyond its order and its cutoff.
Monotonic — no ripple anywhere — and maximally flat at the origin: the first \(2n-1\) derivatives of \(|H|^{2}\) vanish at \(\omega = 0\). At \(\omega = \omega_c\) the response is \(1/\sqrt2\) for every \(n\), so the cutoff is always the \(-3\) dB point, and far beyond it the response falls at \(20n\) dB per decade.
The order is what you buy. Setting \(n=1\) recovers the \(RC\) filter of Section 15-6; \(n=2\) gives a two-pole section falling at 40 dB/decade; \(n=8\) falls at 160 dB/decade and approaches a brick wall as closely as anyone normally needs. Each increment costs another pole, meaning another energy-storage element in an analogue realisation or another delay in a digital one.
How much order does a given mask require? Write the attenuation in decibels at the two edges and solve. At \(\omega_s\) we need \(10\log_{10}\!\big[1+(\omega_s/\omega_c)^{2n}\big] \ge A_s\), and at \(\omega_p\) we need the corresponding expression to be at most \(A_p\). Eliminating \(\omega_c\) between the two gives the standard order formula:
Read the formula rather than memorising it and the economics of filter design become clear. The numerator grows with the attenuation you demand; the denominator grows with the ratio of the two band edges. A narrow transition band — \(\omega_s\) barely above \(\omega_p\) — makes the denominator small and the required order large. Sharpness is expensive, and it is expensive in proportion to the logarithm of the band-edge ratio, which is why halving a transition band costs far more than halving a stopband requirement. Worked Example 5 puts numbers on this.
The other families trade ripple for sharpness. A Chebyshev filter permits equal ripple in the passband and, in exchange, reaches a given attenuation at a lower order than Butterworth. An elliptic filter ripples in both bands and is the sharpest of all for a given order. A Bessel filter gives up magnitude sharpness entirely in return for maximally flat group delay, which makes it the choice whenever pulse shape matters more than frequency selectivity.
| Family | Passband | Stopband | Bought with | Typical use |
|---|---|---|---|---|
| Butterworth | Maximally flat | Monotonic | Higher order | General purpose |
| Chebyshev I | Equal ripple | Monotonic | Passband ripple | Sharp cutoff, cheap |
| Chebyshev II | Flat | Equal ripple | Stopband ripple | Flat passband needed |
| Elliptic | Equal ripple | Equal ripple | Ripple in both | Lowest order possible |
| Bessel | Flat, gentle | Slow rolloff | Poor selectivity | Pulses, linear phase |
Bandwidth, Rise Time and Ringing
Filtering is always described in the frequency domain, but its cost is paid in the time domain, and two consequences follow so reliably that they are used as rules of thumb by every practising engineer.
The first concerns speed. The step response of the first-order lowpass is \(s(t) = 1 - e^{-t/RC}\), and its rise time — conventionally the interval between the 10% and 90% levels — is obtained by solving \(1 - e^{-t/RC} = 0.1\) and \(0.9\) and subtracting:
Its half-power bandwidth in hertz is \(B = \omega_c/2\pi = 1/(2\pi RC)\), so the two are inversely proportional and their product is a pure number:
A system cannot respond faster than its bandwidth permits. An oscilloscope with a 100 MHz bandwidth cannot display an edge faster than about 3.5 ns, no matter how fast the edge really was — the instrument's own rise time dominates.
The constant 0.35 belongs to the first-order response specifically; other filter shapes give values between roughly 0.3 and 0.45, and the rule is used with that precision. What does not change is the inverse proportionality, which is Chapter 14's scaling property in disguise: a narrow spectrum forces a slow time response, always.
The second consequence concerns shape. Sharpening a filter's cutoff makes its impulse response longer and oscillatory — the sinc tails of Section 15-5 — and convolving a step with an oscillatory impulse response makes the output overshoot and ring. This is precisely the Gibbs phenomenon of Chapter 11, now produced by a filter rather than by truncating a series, and the mechanism is identical: an abruptly truncated spectrum reassembles into a waveform with an overshoot that does not diminish as the cutoff rises. An ideal lowpass overshoots a step by about 9% of the jump however high its cutoff frequency, which is why filters intended for pulse work are deliberately given gentle, Bessel-like cutoffs.
Filtering in Practice
Four applications show what the theory buys, and each one is an instance of \(Y = HX\) with a particular \(H\) in mind.
Separating signal from noise. If the wanted signal occupies \(|\omega| \lt W\) and the interference lies outside that band, a lowpass filter removes the interference and leaves the signal essentially untouched. The condition is the whole story: filtering separates signals only where their spectra do not overlap. Broadband noise sitting on top of a baseband signal can be reduced only in the part of the band the signal does not occupy, and noise inside the signal band cannot be removed by any linear filter whatsoever. The most common instance of the successful case is the notch filter — a bandstop with a very narrow rejection band placed on 50 Hz mains hum, or its harmonics — which works because the interference is confined to a band the signal barely uses.
Selecting one channel. Section 14-2 showed that modulation places a message at \(\pm\omega_c\). A receiver applies a bandpass filter centred on the wanted carrier and wide enough to admit both sidebands, rejecting every other station. The filter's quality factor \(Q = \omega_0/B\), where \(B\) is the \(-3\) dB bandwidth, measures its selectivity; a tuned RLC section with \(Q = 100\) at 1 MHz has a 10 kHz bandwidth, comfortably enough for an AM channel and narrow enough to reject the neighbour.
Anti-alias filtering. Before a signal may be sampled it must be band-limited, and the lowpass that enforces this is the most consequential filter in any digital system. Chapter 20 proves why it is needed and Chapter 21 shows what happens when it is omitted; the point to carry forward from here is that the anti-alias filter must do its work before the sampler, because no processing afterwards can undo aliasing.
Equalisation. If a channel distorts with a known response \(H_{\text{ch}}\), a filter with \(H_{\text{eq}} = 1/H_{\text{ch}}\) placed in cascade restores the signal, since the product is unity. Section 14-7 already noted the limitation: where \(|H_{\text{ch}}|\) is small the equaliser must have large gain, which amplifies noise, and where \(H_{\text{ch}}\) is exactly zero no equaliser exists at all. Practical equalisers therefore flatten the response only over the band where the channel has something left to recover.
Every one of these is realised today far more often in software than in hardware. The concepts transfer directly: Chapter 16 develops the discrete-time frequency response, Chapter 25 designs discrete-time filters, and the vocabulary of passband, stopband, transition band and order is unchanged. What changes is that \(H\) becomes periodic in frequency — the direct consequence of the discrete-time frequency wrap-around first met in Chapter 3.
Worked Examples
Problem. An \(RC\) lowpass has \(R = 10\ \text{k}\Omega\) and \(C = 0.1\ \mu\text{F}\). Find the steady-state output for \(x(t) = 5\cos(100t) + 5\cos(1000t) + 5\cos(10{,}000t)\) volts, and comment on the distortion.
Solution. First the corner: \(RC = (10^{4})(10^{-7}) = 10^{-3}\ \text{s}\), so \(\omega_c = 1/RC = 1000\) rad/s, or \(f_c = 159.2\) Hz. Now evaluate \(|H|\) and \(\angle H\) at each input frequency using \(|H| = [1+(\omega/\omega_c)^{2}]^{-1/2}\) and \(\angle H = -\arctan(\omega/\omega_c)\).
| \(\omega\) (rad/s) | \(\omega/\omega_c\) | \(|H|\) | \(\angle H\) | Output amplitude |
|---|---|---|---|---|
| 100 | 0.1 | 0.995 | \(-5.71^{\circ}\) | 4.98 V |
| 1000 | 1 | 0.707 | \(-45.0^{\circ}\) | 3.54 V |
| 10 000 | 10 | 0.0995 | \(-84.3^{\circ}\) | 0.50 V |
so that
The filter has done its job: the highest tone is down by a factor of ten, 20 dB, exactly as the \(-20\) dB/decade asymptote predicts one decade past the corner. But notice the phases. Distortionless transmission requires \(\angle H\) proportional to \(\omega\); here the phase goes \(-5.71^{\circ}\), \(-45^{\circ}\), \(-84.3^{\circ}\) while the frequency goes up by factors of ten. The phase delays \(-\theta/\omega\) are \(0.997\) ms, \(0.785\) ms and \(0.147\) ms respectively — every component is delayed by a different amount, so even the two surviving tones no longer stand in their original relationship. An \(RC\) filter distorts in phase as well as amplitude, and no choice of \(R\) and \(C\) prevents it.
Problem. A system obeys \(\dfrac{d^{2}y}{dt^{2}} + 5\dfrac{dy}{dt} + 6y = 3x(t)\). Find \(H(j\omega)\), the DC gain, the high-frequency rolloff, and \(h(t)\).
Solution. Replace each derivative of \(y\) by \((j\omega)^{k}Y\) and each derivative of \(x\) likewise:
The DC gain is \(H(0) = 3/6 = 0.5\). The denominator degree exceeds the numerator degree by two, so \(|H|\) falls as \(1/\omega^{2}\) at high frequency — a rolloff of 40 dB/decade, twice as steep as the \(RC\) filter. Factoring the denominator gives \((j\omega+2)(j\omega+3)\), so the poles lie at \(-2\) and \(-3\), both with negative real part, and the system is stable — which is what entitled us to write down a frequency response at all.
For \(h(t)\), expand in partial fractions. With \(v = j\omega\), \(3/[(v+2)(v+3)]\) has residue \(3/(3-2) = 3\) at \(v = -2\) and \(3/(2-3) = -3\) at \(v = -3\):
Check with the area rule of Section 14-9: \(H(0)\) must equal \(\int h\,dt = 3(\tfrac12 - \tfrac13) = \tfrac12\), which matches the DC gain computed directly. Note also \(h(0) = 0\): a system whose denominator exceeds its numerator by two degrees cannot respond instantaneously.
Problem. An ideal lowpass has a cutoff of 1 kHz, so \(\omega_c = 2000\pi\) rad/s. Evaluate its impulse response at \(t = 0\) and at \(t = -0.25\) ms, and comment.
Solution. The impulse response is \(h(t) = \sin(\omega_c t)/\pi t\). At the origin the expression is indeterminate; taking the limit with \(\sin x \approx x\) gives \(h(0) = \omega_c/\pi = 2000\). At \(t = -2.5\times10^{-4}\) s the argument is \(\omega_c t = 2000\pi(-2.5\times10^{-4}) = -\pi/2\), so
The filter produces an output of 1273 units a quarter of a millisecond before the impulse arrives — some 64% of its peak value. This is not a small violation of causality that might be engineered away; it is a response comparable in size to the main one, occurring before the cause. The filter cannot be built.
Delaying the response by \(t_d\) hides the problem without solving it. Choosing \(t_d = 5\) ms shifts the whole sinc right by 5 ms, and the pre-response then lives at \(t \lt 0\) only in the tails, where \(|h| \lt 1/(\pi|t|)\) has fallen below about 60 — roughly 3% of the peak. Truncating what remains gives a causal filter whose response is close to ideal, at the cost of 5 ms of delay and a slightly rippled passband. That is the standard windowed-sinc design of Chapter 25, met here as a compromise rather than a technique.
Problem. A channel has \(|H(j\omega)| = 1\) for all frequencies of interest and phase \(\theta(\omega) = -10^{-5}\omega^{2}\) radians. Find the output for \(x(t) = \cos(100t) + \cos(300t)\), and the group delay at each frequency.
Solution. The magnitude is flat, so no component changes size. Each is shifted by the phase at its own frequency: \(\theta(100) = -10^{-5}(10^{4}) = -0.1\) rad and \(\theta(300) = -10^{-5}(9\times10^{4}) = -0.9\) rad. Hence
Write each term as a delayed cosine to see what has happened: \(\cos(100t-0.1) = \cos\big[100(t - 0.001)\big]\) and \(\cos(300t-0.9) = \cos\big[300(t-0.003)\big]\). The first component is delayed by 1 ms, the second by 3 ms. They arrive at different times, so the sum is not the input delayed — it is a different waveform. The peak amplitude of \(x\) is 2, occurring when both cosines peak together; in \(y\) that coincidence no longer happens, and the peak is smaller.
The group delay quantifies the effect: \(\tau_g = -d\theta/d\omega = 2\times10^{-5}\omega\), giving 2 ms at \(\omega = 100\) and 6 ms at \(\omega = 300\). It varies across the band, which is the signature of a dispersive channel. Had \(\theta\) been \(-0.002\omega\) instead, both delays would have been 2 ms at every frequency and the output would have been \(x(t-0.002)\) exactly — same magnitude response, utterly different result.
Problem. Design a Butterworth lowpass with at most 1 dB attenuation at 1 kHz and at least 30 dB attenuation at 3 kHz. Find the required order, the cutoff frequency, and the attenuation actually achieved at 3 kHz.
Solution. Apply the order formula with \(A_p = 1\) dB, \(A_s = 30\) dB, \(\omega_s/\omega_p = 3\). The two bracket terms are \(10^{3} - 1 = 999\) and \(10^{0.1} - 1 = 0.2589\):
Order must be an integer, so round up: a fourth-order filter, four poles, falling at 80 dB/decade. Now place the cutoff. The passband condition \(10\log_{10}[1+(\omega_p/\omega_c)^{8}] = 1\) requires \((\omega_p/\omega_c)^{8} = 0.2589\), so \(\omega_p/\omega_c = 0.2589^{1/8} = 0.845\) and \(f_c = 1000/0.845 = 1184\) Hz.
Finally, check the stopband. At 3 kHz, \(\omega_s/\omega_c = 3000/1184 = 2.534\), and \(2.534^{8} = 1698\), giving an attenuation of \(10\log_{10}(1+1698) = 32.3\) dB. The specification asked for 30 dB and we have 32.3 dB, the surplus being the price of rounding \(n\) up from 3.76. Had the transition band been tighter — say 30 dB required at 1.5 kHz instead of 3 kHz — the denominator would have become \(2\log_{10}1.5 = 0.352\) and the order would have jumped to \(n = 11\). Sharpness is expensive.
Problem. A square wave of amplitude \(\pm 1\) V and fundamental 1 kHz, with the odd symmetry of Chapter 10, is passed through an ideal lowpass with a 4 kHz cutoff. Find the output and its peak value.
Solution. From Chapter 10 the square wave is \(x(t) = \frac{4}{\pi}\sum_{k \text{ odd}}\frac{1}{k}\sin(k\omega_0 t)\) with \(\omega_0 = 2\pi(1000)\). Its spectrum is a set of lines at 1, 3, 5, 7, ... kHz, and the filter passes those below 4 kHz with gain 1 and kills the rest. Only the 1 kHz and 3 kHz lines survive:
To find the peak, differentiate: \(y' \propto \cos\omega_0 t + \cos3\omega_0 t = 2\cos2\omega_0 t\,\cos\omega_0 t\), which vanishes when \(\omega_0 t = \pi/4\). There \(\sin\omega_0 t = \sin 3\omega_0 t = 1/\sqrt2\), so
The output overshoots the 1 V level by 20%, and it does so at \(\omega_0 t = \pi/4\), that is one eighth of a period after the edge. This is the Gibbs phenomenon of Chapter 11 produced by a filter: truncating a spectrum abruptly always overshoots. Admitting more harmonics sharpens the edges but does not remove the overshoot, which settles towards about 9% of the 2 V jump — roughly 1.18 V — no matter how high the cutoff is raised. If flat tops matter more than fast edges, the cure is a gentler filter, not a wider one.
Chapter Summary
\(H(j\omega) = \mathcal{F}\{h(t)\}\), and \(Y = HX\). A sinusoid in gives \(|H|\) times a sinusoid out, shifted by \(\angle H\).
An LCCDE gives \(H\) as a ratio of polynomials in \(j\omega\): DC gain \(b_0/a_0\), rolloff \(20(N-M)\) dB/decade.
Flat \(|H|\) and linear \(\angle H\) over the signal's band. Curved phase means a group delay that varies — dispersion.
A brick wall has \(h(t) = \sin\omega_c t/\pi t\), non-zero before \(t=0\). Causality forbids exact rejection over a band.
Butterworth: \(|H|^{2} = 1/[1+(\omega/\omega_c)^{2n}]\), \(-3\) dB at \(\omega_c\), \(20n\) dB/decade, order set by the band-edge ratio.
\(t_r \approx 0.35/B\), and a sharp cutoff always rings. Selectivity, speed and clean edges cannot all be had at once.
Problems
Problems 1 to 3 exercise the frequency response itself; 4 and 5 concern distortion and delay; 6 to 8 are design questions. Quote a magnitude and a phase for every frequency you evaluate, and state the units of every bandwidth you compute.
- A system has \(h(t) = 4e^{-4t}u(t)\). Find \(H(j\omega)\), the DC gain, and the \(-3\) dB bandwidth in both rad/s and hertz. What is the steady-state output for \(x(t) = 2 + 3\cos(4t)\)?
- For the series \(RL\) circuit with output taken across the resistor, show that \(H(j\omega) = 1/(1+j\omega L/R)\) and identify the cutoff. Then find the response of the same circuit with the output taken across the inductor, and say which of the four ideal filter types it approximates.
- A system obeys \(\dot{y} + 3y = \dot{x} - x\). Find \(H(j\omega)\), sketch \(|H|\), and show that \(|H(j\omega)| \to 1\) at high frequency while \(H(0) = -1/3\). What kind of filter is this?
- A channel has \(|H| = 2\) and \(\angle H = -0.004\omega\) over \(0 \lt \omega \lt 5000\) rad/s. Find the output for \(x(t) = \cos(1000t) + 0.5\cos(3000t)\), and state whether the transmission is distortionless. Repeat with \(\angle H = -0.5\) rad, a constant, and explain the difference.
- Find the phase delay and the group delay of the first-order lowpass \(1/(1+j\omega/\omega_c)\) as functions of \(\omega\). Evaluate both at \(\omega = 0\), \(\omega_c\) and \(10\omega_c\), and say which components of a pulse are delayed most.
- Show that the ideal bandpass with passband \(\omega_1 \lt |\omega| \lt \omega_2\) has impulse response \(h(t) = \big[\sin\omega_2 t - \sin\omega_1 t\big]/\pi t\), by writing it as the difference of two ideal lowpass filters.
- A Butterworth lowpass must attenuate by no more than 3 dB at 2 kHz and by at least 40 dB at 8 kHz. Find the required order and the attenuation actually achieved at 8 kHz. Then state the order needed if the stopband edge is moved to 4 kHz.
- An oscilloscope has a 3 dB bandwidth of 200 MHz and is used to measure a pulse whose true 10–90% rise time is 1.2 ns. Estimate the rise time displayed, given that the measured and true rise times combine approximately as \(t_{\text{disp}}^{2} = t_{\text{true}}^{2} + t_{\text{scope}}^{2}\). Is the instrument adequate?