Part 3 · Chapter 12

Frequency Spectra and System Response

Chapter 10 turned a periodic signal into a list of complex numbers and Chapter 11 gave that list its algebra; this chapter finally asks what a system does to the list, and discovers that an LTI system can only rescale the harmonics it is given — it can never invent a new one.

Signals and Systems Prof. Mithun Mondal Reading time ≈ 50 min
i What you'll learn
  • How to read a line spectrum — the magnitude and phase plots of \(a_k\) — and why the two together lose nothing at all from \(x(t)\).
  • The power spectrum \(|a_k|^2\), and how it turns Parseval's relation into a working definition of bandwidth.
  • The eigenfunction property: why \(e^{j\omega t}\) survives an LTI system unchanged in form, and how that single fact defines the frequency response \(H(j\omega)\).
  • The rule that governs the rest of the book: \(b_k = H(jk\omega_0)\,a_k\) — the system multiplies each harmonic and creates none.
  • Filtering as deliberate reshaping of a spectrum, the four ideal filter types, and the honest behaviour of a real \(RC\) network.
  • Distortionless transmission, and why it demands flat magnitude and phase that is linear in \(\omega\), not phase that is constant.
  • Phase delay and group delay, and the harmonic-distortion test that exposes a system which is not linear.
Section 12-1

Two Pictures of One Signal

By the end of Chapter 11 we could compute the Fourier coefficients of almost any periodic signal without ever performing the analysis integral. What we had not yet asked is the obvious question: having gone to the trouble of producing the numbers \(a_k\), what are they good for?

The first answer is that they are a picture. The synthesis equation \(x(t) = \sum_k a_k e^{jk\omega_0 t}\) says the signal is completely determined by the sequence \(\{a_k\}\); the analysis equation says the sequence is completely determined by the signal. Neither description is more fundamental than the other. They are two coordinate systems for the same object, and moving between them destroys nothing — the same way that giving a point by its Cartesian coordinates and by its polar coordinates are equally complete.

What changes is what is easy to see. In the time-domain picture you read off when things happen. In the frequency-domain picture you read off which oscillations are present and how strongly, and that is precisely the information a filter, an amplifier, a transmission line or a loudspeaker acts on. A plot of \(|a_k|\) against frequency is called the line spectrum of the signal, because a periodic signal contains energy only at the isolated frequencies \(0, \pm\omega_0, \pm 2\omega_0, \dots\) — its spectrum is a picket fence of lines, never a continuum.

Why "lines". Periodicity is a very strong constraint. A signal that repeats exactly every \(T\) seconds can only be built from oscillations that also repeat every \(T\) seconds, and those are exactly the harmonics of \(\omega_0 = 2\pi/T\). Nothing between them is allowed. When we abandon periodicity in Chapter 13, the constraint disappears and the lines merge into a continuous curve — that curve is the Fourier transform.
Section 12-2

Magnitude and Phase Spectra

Each \(a_k\) is a complex number, so it takes two real plots to display the whole sequence. Writing \(a_k = |a_k|e^{j\angle a_k}\), we plot \(|a_k|\) against \(k\omega_0\) as the magnitude spectrum and \(\angle a_k\) against \(k\omega_0\) as the phase spectrum. Both are needed. The magnitude spectrum alone tells you how much of each oscillation is present but not how the oscillations are lined up in time, and it is the alignment that decides the shape of the waveform.

For a real signal the two plots are not independent, because Chapter 11 established the conjugate symmetry \(a_{-k} = a_k^{*}\). Taking modulus and argument of that statement gives the symmetry that every spectrum in this book obeys.

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Spectra of a real signal
\[ |a_{-k}| = |a_k| \qquad\qquad \angle a_{-k} = -\angle a_k \]

The magnitude spectrum of a real signal is an even function of frequency and the phase spectrum is odd. Half the plot is therefore redundant, which is why the negative-frequency lines are often omitted entirely.

That redundancy is what licenses the one-sided or amplitude–phase form. Pair the term at \(+k\omega_0\) with its partner at \(-k\omega_0\) and use \(z + z^{*} = 2\,\mathrm{Re}\{z\}\):

Collapsing a conjugate pair into one real cosine
\[ a_k e^{jk\omega_0 t} + a_{-k}e^{-jk\omega_0 t} = a_k e^{jk\omega_0 t} + \big(a_k e^{jk\omega_0 t}\big)^{*} = 2\,\mathrm{Re}\Big\{|a_k|e^{j(k\omega_0 t + \angle a_k)}\Big\} = 2|a_k|\cos\!\big(k\omega_0 t + \angle a_k\big) \]

Summing over \(k \ge 1\) and keeping the DC term separately gives the form an instrument would report:

One-sided amplitude–phase form
\[ x(t) = a_0 + \sum_{k=1}^{\infty} 2|a_k|\cos\!\big(k\omega_0 t + \angle a_k\big) \]

So the physical amplitude of the \(k\)-th harmonic is \(2|a_k|\), not \(|a_k|\): the two-sided spectrum splits every real oscillation into a conjugate pair and gives half of it to each. Forgetting the factor of two is the commonest slip in spectrum questions, and it always shows up as an answer that is exactly half the right size.

The phase spectrum has one behaviour worth memorising, because it recurs in every chapter from here to the end of the book. Chapter 11's time-shifting property says the coefficients of \(x(t-t_0)\) are \(a_k e^{-jk\omega_0 t_0}\). The magnitudes are untouched; the phases each pick up \(-k\omega_0 t_0\), a term proportional to the harmonic number. A pure delay is therefore invisible in the magnitude spectrum and appears in the phase spectrum as a straight line through the origin whose slope is the delay. Hold on to that; Section 12-7 is built on it.

ω |a_k| −2ω₀ −ω₀ ω₀ 2ω₀ 2 1
Magnitude spectrum — even in \(\omega\)
ω ∠a_k +π/2 +π/3 −π/3 −π/2 −2ω₀ −ω₀ ω₀ 2ω₀
Phase spectrum — odd in \(\omega\)
Section 12-3

The Power Spectrum and Bandwidth

Chapter 11 proved Parseval's relation for periodic signals: the average power of \(x(t)\) equals \(\sum_k |a_k|^2\). Read as a picture rather than an equation, it says the total power is a sum of contributions, one per spectral line, and that each line's contribution is \(|a_k|^2\). Plotting \(|a_k|^2\) against \(k\omega_0\) therefore shows how the signal's power is distributed across frequency, and that plot is the power spectrum.

Because the magnitude spectrum is even, the physical harmonic at \(k\omega_0\) carries \(|a_k|^2 + |a_{-k}|^2 = 2|a_k|^2\) — the same factor-of-two bookkeeping as before, and consistent with the amplitude-\(2|a_k|\) cosine having mean-square value \((2|a_k|)^2/2 = 2|a_k|^2\).

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Power spectrum
\[ P = \frac{1}{T}\int_{T}|x(t)|^2\,dt = \sum_{k=-\infty}^{\infty}|a_k|^2, \qquad P_k^{\text{harm}} = 2|a_k|^2 \ \ (k \ge 1) \]

Power adds across harmonics with no cross terms, because distinct harmonics are orthogonal over a period. This is the single fact that makes "how much of the signal lives below 5 kHz" a meaningful question.

It also makes bandwidth definable. A periodic signal built from infinitely many harmonics has, strictly, infinite bandwidth; in practice the high harmonics carry so little power that discarding them changes nothing audible, visible or measurable. The essential bandwidth is the frequency below which some agreed fraction — commonly 95% or 99% — of the total power lies.

Take the symmetric square wave of Chapter 10, of amplitude \(\pm 1\), whose harmonic amplitudes are \(4/(\pi k)\) for odd \(k\). Its total power is 1 W into a \(1\,\Omega\) reference, and the power in the \(k\)-th harmonic is \(\tfrac12\big(4/\pi k\big)^2 = 8/(\pi^2k^2)\). Accumulating these gives a table that is worth remembering, because it explains why square waves survive being filtered so brutally.

Harmonics retainedPower in this harmonicCumulative powerFraction of total
Fundamental only\(8/\pi^2 = 0.8106\)0.810681.06%
Through the 3rd\(8/9\pi^2 = 0.0901\)0.900690.06%
Through the 5th\(8/25\pi^2 = 0.0324\)0.933193.31%
Through the 7th\(8/49\pi^2 = 0.0165\)0.949694.96%
Through the 9th\(8/81\pi^2 = 0.0100\)0.959695.96%
Power converges long before shape does. Four harmonics capture 95% of a square wave's power, yet Chapter 11 showed that the partial sum still overshoots each edge by 9% no matter how many terms are taken. Power is a mean-square measure and is blind to what happens at isolated instants; the Gibbs ears live exactly at those instants. Both statements are true, and they answer different questions.
Section 12-4

Complex Exponentials Are Eigenfunctions

Everything so far has been about signals. Now bring in a system, and the reason for all this machinery finally appears.

Chapter 7 established that an LTI system is completely described by its impulse response \(h(t)\), acting through the convolution integral \(y(t) = \int h(\tau)x(t-\tau)\,d\tau\). Feed it a single everlasting complex exponential \(x(t) = e^{j\omega t}\) and watch what the integral does.

Pushing \(e^{j\omega t}\) through the convolution integral
\[ y(t) = \int_{-\infty}^{\infty} h(\tau)\,e^{j\omega(t-\tau)}\,d\tau = e^{j\omega t}\underbrace{\int_{-\infty}^{\infty} h(\tau)\,e^{-j\omega\tau}\,d\tau}_{\text{a number, not a function of } t} \]

The step that matters is trivial algebra: \(e^{j\omega(t-\tau)} = e^{j\omega t}e^{-j\omega\tau}\), and the factor \(e^{j\omega t}\) does not involve the integration variable, so it comes straight out. What is left inside is a definite integral over \(\tau\) whose value depends on \(\omega\) but not on \(t\). Call it \(H(j\omega)\).

The output is the input again, multiplied by a complex constant. In the language of linear algebra, \(e^{j\omega t}\) is an eigenfunction of every LTI system and \(H(j\omega)\) is the corresponding eigenvalue. No other family of signals does this: a step comes out of an \(RC\) network as an exponential, a triangle comes out rounded, but \(e^{j\omega t}\) comes out as \(e^{j\omega t}\), rescaled and rotated, and nothing else.

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The frequency response
\[ e^{j\omega t} \;\longrightarrow\; H(j\omega)\,e^{j\omega t}, \qquad H(j\omega) = \int_{-\infty}^{\infty}h(\tau)e^{-j\omega\tau}\,d\tau \]

In discrete time the identical argument on \(y[n]=\sum_k h[k]x[n-k]\) with \(x[n]=e^{j\Omega n}\) gives \(H(e^{j\Omega}) = \sum_{k}h[k]e^{-j\Omega k}\), and \(H(e^{j\Omega})\) is periodic in \(\Omega\) with period \(2\pi\) for the reason Chapter 3 gave.

Look at the integral defining \(H(j\omega)\). It takes the impulse response and produces a function of frequency; it is, in fact, exactly the Fourier transform of \(h(t)\), which is why Chapter 13 can be read as the general theory of the operation we have just performed on one particular signal. For now, treat \(H(j\omega)\) as what it plainly is: a complete description of the system, one complex number per frequency, whose modulus is the gain the system applies at that frequency and whose argument is the phase shift.

Two practical remarks. First, if \(h(t)\) is real then \(H(-j\omega) = H^{*}(j\omega)\), so the frequency response of a real system obeys the same even-magnitude, odd-phase symmetry as the spectrum of a real signal. Second, the integral must converge for \(H(j\omega)\) to exist at all, and \(\int|h(\tau)|d\tau \lt \infty\) is exactly the BIBO stability condition of Chapter 8. Unstable systems have no frequency response, which is the mathematics agreeing with the obvious physics: you cannot measure the steady-state gain of a system that never reaches steady state.

Section 12-5

The Response to a Periodic Input

We now have the two halves. A periodic input is a sum of complex exponentials with known weights; each exponential passes through the system multiplied by a known number. Linearity does the rest.

Superposing the eigenfunction responses
\[ x(t) = \sum_{k=-\infty}^{\infty} a_k e^{jk\omega_0 t} \;\longrightarrow\; y(t) = \sum_{k=-\infty}^{\infty} a_k\,H(jk\omega_0)\,e^{jk\omega_0 t} \]

The output is a sum of the very same exponentials, so it is periodic with the same period \(T\), and its Fourier coefficients can be read off by inspection.

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The central rule of this chapter
\[ b_k = H(jk\omega_0)\,a_k \qquad\qquad\text{(discrete time: } b_k = H\big(e^{jk2\pi/N}\big)a_k\text{)} \]

An LTI system acts on a periodic signal by multiplying its spectrum, line by line, by samples of the frequency response taken at the harmonic frequencies. The convolution of Chapter 7 has become a multiplication.

Three consequences follow immediately, and they are the reason engineers think in the frequency domain at all.

The first is that an LTI system creates no new frequencies. Whatever appears at the output was already present at the input; the system can amplify a harmonic, attenuate it, shift its phase, or kill it entirely by having \(H(jk\omega_0) = 0\), but it cannot put power at a frequency where \(a_k\) was zero. The set of frequencies is fixed by the input; only the weights are the system's business.

The second is that magnitudes multiply and phases add: \(|b_k| = |H(jk\omega_0)||a_k|\) and \(\angle b_k = \angle H(jk\omega_0) + \angle a_k\). The two spectral plots of the output are obtained from those of the input by a pointwise product and a pointwise sum — no integrals, no convolution, no differential equation.

The third is that the whole calculation is local in frequency. What happens to the seventh harmonic does not depend in any way on what happens to the third. That independence is what allows a system to be specified, designed and tested one frequency at a time, and it is precisely what a network analyser exploits when it sweeps a sinusoid across a band and records amplitude ratio and phase shift.

A word about steady state. The eigenfunction argument assumed the exponential had been applied since \(t=-\infty\), so \(b_k = H(jk\omega_0)a_k\) describes the steady-state response. If the input is switched on at \(t=0\) the output also contains a transient from the system's natural modes, which for a stable system decays and leaves exactly this periodic answer behind. Chapter 9 computed such transients directly; Chapter 19 will let the Laplace transform handle both parts at once.
QuantityInputSystemOutput
Coefficient\(a_k\)\(H(jk\omega_0)\)\(b_k = H(jk\omega_0)a_k\)
Harmonic amplitude\(2|a_k|\)gain \(|H(jk\omega_0)|\)\(2|H(jk\omega_0)||a_k|\)
Harmonic phase\(\angle a_k\)shift \(\angle H(jk\omega_0)\)\(\angle a_k + \angle H(jk\omega_0)\)
Harmonic power\(2|a_k|^2\)\(|H(jk\omega_0)|^2\)\(2|H(jk\omega_0)|^2|a_k|^2\)
Frequencies present\(k\omega_0\)\(k\omega_0\), never anything else
Section 12-6

Filtering: Reshaping a Spectrum on Purpose

If a system multiplies each harmonic by \(H(jk\omega_0)\), then designing \(|H(j\omega)|\) to be large over one band of frequencies and small over another is a way of selecting which parts of a signal survive. A system built for that purpose is a filter, and the frequency band it passes is its passband, the band it rejects its stopband.

The four idealised shapes are the vocabulary of the subject. An ideal lowpass filter has \(|H| = 1\) for \(|\omega| \lt \omega_c\) and \(0\) beyond; an ideal highpass is its complement; an ideal bandpass passes \(\omega_1 \lt |\omega| \lt \omega_2\); an ideal band-stop removes exactly that band. Each is defined on both positive and negative frequencies, symmetrically, because a real filter acting on a real signal must produce a real output.

−ω_c ω_c LOWPASS 1 −ω_c ω_c HIGHPASS ω₁ ω₂ BANDPASS ω₁ ω₂ BAND-STOP |H(jω)| — each response is even in ω, as it must be for a real filter
The four ideal magnitude responses

These shapes are idealisations, and Chapter 15 will show that none of them can be built: a magnitude response that is exactly zero over a band forces an impulse response that is non-zero for \(t \lt 0\), so an ideal filter is non-causal. That does not make them useless. They are the target a real design approximates, and they make the arithmetic of "which harmonics get through" completely transparent.

A filter you can build is the single \(RC\) lowpass network of Chapter 9, whose frequency response follows from the impedance divider or, equivalently, from applying the eigenfunction property to its differential equation:

The \(RC\) lowpass filter
\[ RC\frac{dy}{dt} + y = x \;\Longrightarrow\; H(j\omega) = \frac{1}{1 + j\omega RC} = \frac{1}{1 + j\omega/\omega_c}, \qquad \omega_c = \frac{1}{RC} \]

Substituting \(x = e^{j\omega t}\), \(y = H e^{j\omega t}\) into the differential equation turns \(d/dt\) into multiplication by \(j\omega\) and reduces calculus to algebra — a foretaste of the whole of Part 5. Its magnitude and phase are

Magnitude and phase of the \(RC\) lowpass
\[ |H(j\omega)| = \frac{1}{\sqrt{1+(\omega/\omega_c)^2}}, \qquad \angle H(j\omega) = -\arctan\!\big(\omega/\omega_c\big) \]

At \(\omega = \omega_c\) the magnitude is \(1/\sqrt{2} = 0.707\) and the phase is \(-45^\circ\). Since power goes as the square of magnitude, half the power is lost at that frequency — hence the names half-power frequency, \(-3\,\)dB frequency and cutoff, all for the same \(\omega_c\). Beyond cutoff the magnitude falls as \(\omega_c/\omega\), a decade of frequency costing a factor of ten: a gentle roll-off compared to the vertical wall of the ideal filter, but real.

The figure below shows the whole calculation in one picture for a square wave whose fundamental sits exactly at the cutoff. The input lines are multiplied, one at a time, by the height of the response curve above them.

ω₀3ω₀ 5ω₀7ω₀ INPUT SPECTRUM 1.273 × 1 .707 ω_c=ω₀5ω₀ |H(jω)| = ω₀3ω₀ 5ω₀7ω₀ OUTPUT SPECTRUM 0.900 every line is scaled by the height of the curve above it — nothing moves sideways
Filtering is a line-by-line multiplication of the spectrum
Why a lowpass filter "smooths". Chapter 11 showed that a signal's high harmonics are the ones that build its sharp edges — differentiating a signal multiplies \(a_k\) by \(jk\omega_0\), so discontinuities are exactly what makes the coefficients decay slowly. Attenuating the high harmonics therefore rounds the corners. Smoothing in time and lowpass filtering in frequency are not two effects; they are one effect seen in two coordinate systems.
Section 12-7

Distortionless Transmission

Sometimes the object is not to reshape a signal but to move it — down a cable, through an amplifier, across a radio link — with its shape intact. Perfect transmission cannot mean "output equals input", because any physical channel imposes some delay and usually some gain. The honest definition allows both and forbids everything else.

Definition of distortionless transmission
\[ y(t) = K\,x(t - t_d), \qquad K \gt 0,\ t_d \ge 0 \]

What frequency response does that demand? Apply the definition harmonic by harmonic. If \(x\) has coefficients \(a_k\), then Chapter 11's shifting property gives \(y\) the coefficients \(K a_k e^{-jk\omega_0 t_d}\). Comparing with \(b_k = H(jk\omega_0)a_k\) and reading off at every frequency:

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Condition for distortionless transmission
\[ H(j\omega) = K e^{-j\omega t_d} \quad\Longleftrightarrow\quad |H(j\omega)| = K \ \text{ and }\ \angle H(j\omega) = -\omega t_d \]

Flat magnitude across the whole band occupied by the signal, and phase that is a straight line through the origin with slope \(-t_d\). Violating the first gives amplitude distortion; violating the second gives phase distortion.

The phase requirement is the one that surprises people, so it is worth saying plainly why constant phase will not do. A phase shift of \(\theta\) at frequency \(\omega\) is a time shift of \(\theta/\omega\) seconds — the same angle at a higher frequency corresponds to a shorter delay. A system that shifts every harmonic by \(-90^\circ\) delays the fundamental by a quarter of a period, the third harmonic by a twelfth of a period, and the fifth by a twentieth. The harmonics arrive out of step, and a waveform assembled from misaligned harmonics is not the waveform you sent, even though every amplitude is untouched. Only phase proportional to frequency delays all harmonics equally, because then \(\theta/\omega = t_d\) is the same for all of them.

ω K |H(jω)| flat over the signal band
Magnitude — constant, so no harmonic is favoured
ω constant slope = −t_d ∠H(jω)
Phase — linear, not merely small

Two measures make the idea quantitative. The phase delay \(t_p(\omega) = -\angle H(j\omega)/\omega\) is the delay experienced by a single sinusoid of frequency \(\omega\). The group delay \(t_g(\omega) = -\,d\,\angle H(j\omega)/d\omega\) is the delay experienced by the envelope of a narrow band of frequencies centred on \(\omega\). For a distortionless channel both equal \(t_d\) at every frequency, since the phase is \(-\omega t_d\) and its slope is \(-t_d\).

For the \(RC\) lowpass they are not equal, and neither is constant. With \(\angle H = -\arctan(\omega/\omega_c)\),

Phase and group delay of the \(RC\) lowpass
\[ t_p(\omega) = \frac{\arctan(\omega/\omega_c)}{\omega}, \qquad t_g(\omega) = \frac{1/\omega_c}{1+(\omega/\omega_c)^2} \]

Taking \(RC = 1\) ms, so \(\omega_c = 1000\) rad/s, gives the numbers below. At low frequency both delays tend to \(RC = 1\) ms; by the cutoff they have separated, and by \(5\omega_c\) the group delay has all but vanished while the phase delay is still a quarter of a millisecond. Different harmonics are delayed by different amounts — the channel disperses the signal — which is phase distortion made arithmetic.

\(\omega\) (rad/s)\(\angle H\)Phase delay \(t_p\)Group delay \(t_g\)
100\(-5.71^\circ\)0.997 ms0.990 ms
500\(-26.57^\circ\)0.927 ms0.800 ms
1000 \((=\omega_c)\)\(-45.00^\circ\)0.785 ms0.500 ms
2000\(-63.43^\circ\)0.554 ms0.200 ms
5000\(-78.69^\circ\)0.275 ms0.038 ms
How much phase distortion is tolerable? It depends entirely on what the signal is for. The ear is largely insensitive to modest phase distortion, so audio filters are specified almost entirely on magnitude. A digital receiver deciding whether a pulse was a one or a zero is extremely sensitive to it, because dispersion smears each pulse into its neighbours. The same filter can therefore be excellent equipment for one job and unusable for the other.
Section 12-8

When New Frequencies Appear

The rule \(b_k = H(jk\omega_0)a_k\) has a contrapositive that is more useful in the laboratory than the rule itself. If a system produces output power at a frequency the input did not contain, that system is not LTI. There is no exception and no special case: the derivation used only linearity and time invariance, so a frequency that was not in the input can only come from the failure of one of those two properties.

The commonest culprit is a mild nonlinearity. Suppose an amplifier is very nearly linear but has a small cubic term, \(y(t) = x(t) + \varepsilon x^3(t)\), and drive it with a pure cosine \(x(t) = A\cos\omega_0 t\). Expanding the cube with \(\cos^3\theta = \tfrac34\cos\theta + \tfrac14\cos 3\theta\):

A cubic nonlinearity manufactures a third harmonic
\[ y(t) = \Big(A + \tfrac34\varepsilon A^3\Big)\cos\omega_0 t \;+\; \tfrac14\varepsilon A^3\cos 3\omega_0 t \]

A frequency that was nowhere in the input has appeared at the output, its size growing as the cube of the drive level — which is why distortion in an amplifier worsens sharply as it is pushed hard. The standard figure of merit collects all the unwanted harmonics into one number.

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Total harmonic distortion
\[ \text{THD} = \frac{\sqrt{A_2^2 + A_3^2 + A_4^2 + \cdots}}{A_1} \]

\(A_k\) is the measured amplitude of the \(k\)-th harmonic when the input is a single sinusoid at \(\omega_0\). For an LTI system every \(A_k\) with \(k \ge 2\) is exactly zero, so THD is a direct measurement of how far from linear a device is.

Time variation is the other route to new frequencies, and it is not a defect but a tool. Multiplying a signal by a carrier, \(y(t) = x(t)\cos\omega_c t\), is a linear operation but a time-varying one, and Chapter 11's multiplication property showed it shifts every spectral line by \(\pm\omega_c\). Every radio transmitter ever built depends on that shift, and Chapter 28 will treat it properly. The lesson is that "no new frequencies" requires both halves of LTI, and that giving up either one buys a capability worth having.

Section 12-9

Worked Examples

1 Reading a spectrum off a signal

Problem. For \(x(t) = 2 + 4\cos(100\pi t + \pi/3) + 2\sin(200\pi t)\), find the exponential Fourier coefficients, sketch the magnitude and phase spectra, and compute the average power.

Solution. The two oscillations are at \(100\pi\) and \(200\pi\) rad/s, so \(\omega_0 = 100\pi\) rad/s (\(f_0 = 50\) Hz, \(T = 20\) ms) and the second term is the second harmonic. Expand each term with Euler's formula:

Working
\[ 4\cos(\omega_0 t + \tfrac{\pi}{3}) = 2e^{j\pi/3}e^{j\omega_0 t} + 2e^{-j\pi/3}e^{-j\omega_0 t}, \qquad 2\sin(2\omega_0 t) = -j\,e^{j2\omega_0 t} + j\,e^{-j2\omega_0 t} \]

Reading off the coefficients: \(a_0 = 2\), \(a_1 = 2e^{j\pi/3}\), \(a_{-1} = 2e^{-j\pi/3}\), \(a_2 = -j = e^{-j\pi/2}\), \(a_{-2} = j = e^{j\pi/2}\), and \(a_k = 0\) otherwise.

The magnitude spectrum has lines of height 2 at \(\omega = 0, \pm\omega_0\) and height 1 at \(\pm 2\omega_0\) — even, as required. The phase spectrum has \(+\pi/3\) at \(+\omega_0\), \(-\pi/3\) at \(-\omega_0\), \(-\pi/2\) at \(+2\omega_0\) and \(+\pi/2\) at \(-2\omega_0\) — odd, as required. These are the two plots drawn in Section 12-2.

Parseval gives the power directly: \(P = \sum_k |a_k|^2 = 2^2 + 2^2 + 2^2 + 1^2 + 1^2 = 14\) W. Checking in the time domain, the DC term contributes \(2^2 = 4\), the 4 V cosine contributes \(4^2/2 = 8\), and the 2 V sine contributes \(2^2/2 = 2\); the total is 14 W, as it must be.

2 A square wave through an \(RC\) lowpass

Problem. A symmetric square wave of amplitude \(\pm1\) V and fundamental frequency 1 kHz drives an \(RC\) lowpass whose cutoff is also 1 kHz. Find the amplitude and phase of the first four output harmonics.

Solution. From Chapter 10 the square wave is \(x(t) = \frac{4}{\pi}\big[\sin\omega_0 t + \tfrac13\sin3\omega_0 t + \tfrac15\sin5\omega_0 t + \cdots\big]\), so the input harmonic amplitudes are \(4/(\pi k)\) for odd \(k\). With \(\omega_c = \omega_0\), the response at the \(k\)-th harmonic is

Working
\[ H(jk\omega_0) = \frac{1}{1+jk}, \qquad |H| = \frac{1}{\sqrt{1+k^2}}, \qquad \angle H = -\arctan k \]

Multiplying each input amplitude by \(|H|\) and adding \(\angle H\) to each phase:

\(k\)Input amplitude\(|H(jk\omega_0)|\)Output amplitudePhase shift
11.27320.70710.9003\(-45.00^\circ\)
30.42440.31620.1342\(-71.57^\circ\)
50.25460.19610.0499\(-78.69^\circ\)
70.18190.14140.0257\(-81.87^\circ\)

so \(y(t) \approx 0.900\sin(\omega_0 t - 45^\circ) + 0.134\sin(3\omega_0 t - 71.6^\circ) + 0.050\sin(5\omega_0 t - 78.7^\circ) + 0.026\sin(7\omega_0 t - 81.9^\circ)\). The third harmonic was 33.3% of the fundamental at the input and is only \(0.1342/0.9003 = 14.9\%\) of it at the output. The square edges have been rounded, and the arithmetic says exactly by how much.

3 How much power the filter passes

Problem. For the system of Example 2, find the average power at the input and output, and the fraction of the output power carried by the fundamental.

Solution. A \(\pm1\) V square wave has \(x^2(t) = 1\) at every instant, so its average power is exactly 1 W. In the frequency domain the same total is assembled from \(8/(\pi^2k^2)\) per odd harmonic, and \(\sum_{k \text{ odd}} 8/(\pi^2k^2) = (8/\pi^2)(\pi^2/8) = 1\) — a satisfying consistency check on Parseval.

At the output each harmonic power is scaled by \(|H(jk\omega_0)|^2 = 1/(1+k^2)\):

Working
\[ P_y = \sum_{k \text{ odd}} \frac{8}{\pi^2 k^2}\cdot\frac{1}{1+k^2} = 0.4053 + 0.0090 + 0.0012 + 0.0003 + \cdots = 0.4161\ \text{W} \]

The filter therefore passes about 41.6% of the input power. More interesting is the purity of what survives: the fundamental alone accounts for \(0.4053/0.4161 = 97.4\%\) of the output power, against 81.1% at the input. A single \(RC\) section has turned a square wave into something close to a sinusoid — which is exactly how a crude sine-wave source is built.

4 An ideal lowpass with linear phase

Problem. The same square wave (fundamental 1 kHz) is passed through an ideal lowpass with unity gain, cutoff 4 kHz and phase \(\angle H(j\omega) = -\omega t_d\) with \(t_d = 0.1\) ms. Write the output.

Solution. The input harmonics sit at 1, 3, 5, 7 kHz and so on. Only those below 4 kHz survive, so the fundamental and third harmonic pass with unity gain and everything from the fifth upward is annihilated. The phase term contributes \(-\omega_0 t_d = -2\pi(1000)(10^{-4}) = -0.2\pi = -36^\circ\) at the fundamental and three times that, \(-108^\circ\), at the third harmonic:

Working
\[ y(t) = 1.273\sin\!\big(\omega_0 t - 36^\circ\big) + 0.424\sin\!\big(3\omega_0 t - 108^\circ\big) \]

Because the phase is proportional to frequency, this is not merely a pair of shifted sinusoids: it is the two-term partial sum of the square wave, delayed bodily by 0.1 ms. Writing it as \(y(t) = 1.273\sin\omega_0(t - t_d) + 0.424\sin3\omega_0(t-t_d)\) makes that plain. Truncating the series distorts the shape — Gibbs ears and all — but the linear phase adds no distortion of its own.

Had the cutoff been placed at 2 kHz instead, only the fundamental would have survived and the output would have been a pure sinusoid. The cutoff must be compared against the harmonic frequencies, never against the fundamental alone.

5 Testing two channels for distortion

Problem. The input \(x(t) = \cos(1000t) + 0.5\cos(3000t)\) is applied to two systems. System A has \(|H| = 5\) and \(\angle H = -0.002\,\omega\); system B has \(|H| = 5\) and \(\angle H = -\pi/2\), both across the whole band. Which transmits without distortion?

Solution (A). The phase is linear with slope \(-0.002\), so \(t_d = 2\) ms and the phase shifts are \(-0.002(1000) = -2\) rad and \(-0.002(3000) = -6\) rad:

Working — system A
\[ y_A(t) = 5\cos(1000t - 2) + 2.5\cos(3000t - 6) = 5\cos\!\big(1000(t-0.002)\big) + 2.5\cos\!\big(3000(t-0.002)\big) = 5\,x(t-0.002) \]

Both components are delayed by the same 2 ms, so the output is a scaled, delayed copy: distortionless.

Solution (B). The same \(-\pi/2\) is applied to both components, but the same angle means different delays: \(t_p = (\pi/2)/1000 = 1.571\) ms for the first and \((\pi/2)/3000 = 0.524\) ms for the second.

Working — system B
\[ y_B(t) = 5\cos(1000t - \tfrac{\pi}{2}) + 2.5\cos(3000t - \tfrac{\pi}{2}) = 5\sin(1000t) + 2.5\sin(3000t) \]

There is no \(t_d\) that makes this equal \(5x(t-t_d)\), because that would require \(1000t_d = 3000t_d = \pi/2\). The two harmonics have been pulled out of alignment and the waveform's shape has changed, even though every amplitude ratio is identical to system A's. Flat magnitude is necessary; it is nowhere near sufficient.

6 Distortion from a nonlinearity

Problem. An amplifier is modelled as \(y(t) = x(t) + 0.1\,x^3(t)\) and driven by \(x(t) = \cos\omega_0 t\). Find the output spectrum and the total harmonic distortion. Could any LTI system produce this output?

Solution. Use \(\cos^3\theta = \tfrac34\cos\theta + \tfrac14\cos3\theta\), which follows from \(\cos^3\theta = \cos\theta\cdot\tfrac{1+\cos2\theta}{2}\) and the product-to-sum identity:

Working
\[ y(t) = \cos\omega_0 t + 0.1\Big(\tfrac34\cos\omega_0 t + \tfrac14\cos3\omega_0 t\Big) = 1.075\cos\omega_0 t + 0.025\cos3\omega_0 t \]

The output contains a third harmonic that the input never had, and the fundamental has been slightly boosted as well. With \(A_1 = 1.075\) and \(A_3 = 0.025\) as the only non-zero amplitudes,

Working — THD
\[ \text{THD} = \frac{0.025}{1.075} = 0.0233 = 2.33\% \]

No LTI system can produce this output. The rule \(b_k = H(jk\omega_0)a_k\) forces \(b_3 = H(j3\omega_0)\cdot 0 = 0\) whenever the input has no third harmonic, whatever \(H\) may be. The appearance of \(3\omega_0\) is therefore proof of nonlinearity, and driving the amplifier harder makes it worse: doubling \(A\) multiplies the third-harmonic amplitude \(\varepsilon A^3/4\) by eight while the fundamental barely doubles.

Review

Chapter Summary

Line spectra

Magnitude and phase plots of \(a_k\) describe a periodic signal completely. For real signals \(|a_k|\) is even and \(\angle a_k\) is odd, and the physical harmonic amplitude is \(2|a_k|\).

Power spectrum

\(P = \sum_k|a_k|^2\) with no cross terms, so power can be attributed line by line. This is what makes essential bandwidth definable.

Eigenfunctions

\(e^{j\omega t}\) passes through any LTI system multiplied by \(H(j\omega) = \int h(\tau)e^{-j\omega\tau}d\tau\), which exists whenever the system is stable.

The central rule

\(b_k = H(jk\omega_0)a_k\): magnitudes multiply, phases add, the period is unchanged, and no new frequency is ever created.

Filtering

Choosing the shape of \(|H(j\omega)|\) selects which harmonics survive. The \(RC\) lowpass has \(|H| = 0.707\) and \(-45^\circ\) at \(\omega_c = 1/RC\).

Distortionless transmission

Requires \(H(j\omega) = Ke^{-j\omega t_d}\) — flat magnitude and phase linear in \(\omega\), so that phase delay and group delay both equal \(t_d\).

Practice

Problems

Problems 1 to 3 exercise the reading of spectra; 4 to 6 push periodic signals through systems; 7 and 8 ask you to prove the results the chapter has been using. Sketch every spectrum before computing anything with it.

  1. Sketch the magnitude and phase spectra of \(x(t) = 3 - 2\cos(400\pi t) + \sin(800\pi t + \pi/4)\). State the fundamental frequency and verify that your magnitude plot is even and your phase plot odd.
  2. A periodic signal has \(a_0 = 1\), \(a_{\pm1} = 0.5e^{\mp j\pi/4}\), \(a_{\pm2} = 0.25\) and no other non-zero coefficients. Write \(x(t)\) in one-sided amplitude–phase form and compute its average power two ways.
  3. For the 25% duty-cycle pulse train of Chapter 10, compute the cumulative fraction of power in the first five harmonics and state the essential bandwidth at the 95% criterion.
  4. A square wave of fundamental 500 Hz and amplitude \(\pm2\) V drives an \(RC\) lowpass with \(R = 1\ \text{k}\Omega\) and \(C = 0.1\ \mu\text{F}\). Find the amplitude and phase of the output fundamental and third harmonic, and the ratio of output to input average power.
  5. The same square wave drives an ideal bandpass filter with unity gain over \(1200 \lt f \lt 1800\) Hz and zero elsewhere. Write the output and explain why it is a pure sinusoid. Then explain why the output would be identically zero if the passband were moved to \(1900 \lt f \lt 2100\) Hz.
  6. A system has \(H(j\omega) = 2e^{-j0.5\omega}\) for \(|\omega| \lt 20\) rad/s and \(H(j\omega) = 0\) beyond. Determine the output for \(x(t) = 1 + \cos(5t) + \cos(30t)\) and say which of the three terms suffers distortion and why.
  7. Starting from the convolution sum, prove that \(x[n] = e^{j\Omega n}\) produces \(y[n] = H(e^{j\Omega})e^{j\Omega n}\), and show that \(H(e^{j\Omega})\) must be periodic in \(\Omega\) with period \(2\pi\).
  8. Prove that if \(y(t) = Kx(t-t_d)\) for every input \(x\), then \(H(j\omega) = Ke^{-j\omega t_d}\). Then show that a system with \(|H| = K\) but \(\angle H = -\pi/2\) at all frequencies cannot be written in that form for any \(t_d\).
Tip: for any "periodic signal into an LTI system" question the procedure never changes — write the input's coefficients, evaluate \(H\) at each harmonic frequency, multiply the magnitudes, add the phases, and reassemble. If you find yourself setting up a convolution integral, you have forgotten which chapter you are in.