01 · Start hereA second book beside the first
Read a chapter in the original book, then come here when an algebraic step, a limit, or a piece of notation feels too fast. You can also read Sections 2–7 first as a short mathematical preparation.
For every formula, ask three things: What is varying? Where is the signal nonzero? Under what conditions does the expression exist? These questions prevent many errors before any calculation starts.
Examples here are additional explanations. The original offline book and its ZIP are preserved separately. The final reading map links every chapter to the notes most useful alongside it.
02 · Mathematical languageNotation and units
| Symbol | Meaning | Keep in mind |
|---|---|---|
| \(x(t),x[n]\) | Continuous-time signal; discrete-time sequence | \(t\) is real; \(n\) is an integer. Discrete time does not require quantized amplitudes. |
| \(j,\;z^*,\;|z|\) | \(j^2=-1\); complex conjugate; magnitude | \(|z|^2=zz^*\), which is real and nonnegative. |
| \(f,\;\omega\) | Frequency in Hz; angular frequency in rad/s | \(\omega=2\pi f\). |
| \(\Omega\) | Discrete-time angular frequency in rad/sample | For samples spaced \(T_s\) seconds apart, \(\Omega=\omega T_s\), modulo \(2\pi\). |
| \(X(j\omega),X(e^{j\Omega})\) | CT Fourier transform; DT Fourier transform | These notes use \(\omega\) for CT and \(\Omega\) for DT, even if another text uses the same letter for both. |
| \(X(s),X(z)\) | Laplace transform; Z-transform | The region of convergence is part of the answer. |
| \(*\) | Convolution between signals | A superscript star means conjugation instead. Context matters. |
The CT Fourier convention used here is the book’s angular-frequency convention:
That \(1/(2\pi)\) belongs to the inverse because the frequency variable is angular frequency. In the Hz convention the exponent is \(-j2\pi ft\), and the inverse has no such factor.
03 · Before Chapters 3, 10–19, and 23–26Complex numbers make oscillations manageable
Write a complex number in rectangular or polar form:
Use a quadrant-aware angle such as \(\operatorname{atan2}(b,a)\); \(\tan^{-1}(b/a)\) alone can select the wrong quadrant. Multiplication multiplies magnitudes and adds angles. Conjugation reverses the angle.
A sinusoid becomes a pair of exponentials. Differentiation, shifting, and LTI filtering then become operations on scalar factors.
For \(s=-2+j3\),
The magnitude decays as \(e^{-2t}\), while the angle grows at \(3\) rad/s. At \(t=1\), the magnitude is \(e^{-2}\), not \(e^{\sqrt{13}}\). The real part of \(s\) controls growth; its imaginary part controls rotation.
Why exponentials are special to an LTI system
Insert \(x(t)=e^{st}\) into the convolution formula, assuming the defining integral converges:
The output has the same exponential shape and a new multiplier. This is the eigenfunction property. Fourier analysis chooses \(s=j\omega\); Laplace analysis allows a nonzero real part too.
04 · A small calculus toolkitIntegrals, sums, and boundary terms
Three elementary identities do much of the work:
The finite sum works for complex \(r\) too. Multiply the sum by \(r\), subtract, and everything except the first and last terms cancels. The infinite sum adds a convergence condition:
For real \(a>0\),
The term at infinity vanishes because the magnitude is \(e^{-at}\). The oscillation alone does not force the endpoint to vanish. Replacing \(a+j\omega\) by \(s+a\) gives the condition \(\operatorname{Re}s>-a\).
Integration by parts explains the derivative property
When the boundary term vanishes and the transforms are defined, differentiation corresponds to multiplication by \(j\omega\). The property comes with hypotheses; it is not a reason to drop every boundary term in every problem.
For a substitution such as \(v=2t-3\), write \(dt=dv/2\) and change the limits. That small factor is exactly what later produces Fourier scaling factors and impulse scaling.
05 · Before Chapters 4 and 7Transform the support before drawing the waveform
The support is the region where a signal can be nonzero. Suppose \(x(v)\) is supported on \(0\le v\le2\). For \(y(t)=x(2t-4)\), solve:
The pulse is compressed to half its original width and begins at \(t=2\). Factoring \(2t-4=2(t-2)\) reveals the correct shift.
For \(y(t)=x(1-2t)\),
The value originally at \(v=0\) appears at \(t=1/2\); the value originally at \(v=2\) appears at \(t=-1/2\). Tracking two landmarks verifies both the reversal and the width.
In general, an original landmark at \(v=v_0\) moves to \(t=(v_0-b)/a\) in \(x(at+b)\), provided \(a\ne0\).
06 · Before Chapters 3, 7, and 20An impulse is understood through an integral
The continuous-time impulse is a distribution, not an ordinary function with a very large value at one point. Its defining action on a smooth test function \(\varphi\) is:
Engineering calculations use this sifting property with signals sufficiently well behaved at the impulse location. An impulse at an integration endpoint or multiplied by a discontinuous function requires care about conventions.
Therefore
Locating the impulse at \(t=2\) is only half the calculation. Changing the time scale also changes its area.
A unit step has distributional derivative \(du/dt=\delta(t)\). A jump of size \(A\) at \(t_0\) contributes \(A\delta(t-t_0)\) to the derivative, in addition to any ordinary derivatives between jumps.
07 · Before Chapters 2, 10, and 12Energy, power, and projection
The squared magnitude \(|x|^2=xx^*\) works for real and complex signals. For the usual symmetric time-average convention:
Use sums and \(2N+1\) in discrete time. A nonzero finite-energy signal has zero average power under this convention. A nonzero periodic signal with finite energy per period has infinite total energy and finite positive average power. Some signals have neither finite energy nor finite average power.
For \(x(t)=A\cos(\omega_0t+\phi)\), real \(A\) and \(\omega_0\ne0\), averaging \(\cos^2\theta=(1+\cos2\theta)/2\) over a period gives \(P_x=A^2/2\).
A signal can be treated like a vector
Define an inner product on a chosen interval by \(\langle x,g\rangle=\int x(t)g^*(t)dt\). It measures alignment. The coefficient of the best least-squares approximation \(cg\) is:
Fourier coefficients are this projection formula applied to orthogonal complex exponentials. The conjugate explains the minus sign in the analysis exponent.
08 · Alongside Chapters 6–8Convolution: turn overlap into inequalities
Convolution combines the response to every shifted input contribution:
In the integral, \(t\) is fixed while \(\tau\) varies. Write the conditions that make both factors nonzero, intersect those intervals, and only then integrate.
Let \(x(t)=1\) on \([0,2]\), and \(h(t)=1\) on \([0,1]\), with both zero elsewhere. Then:
The integrand is 1 on this overlap. Its integral is the overlap length:
It rises while the overlap grows, stays flat while the shorter pulse fits inside the longer one, and falls as they separate.
A causal exponential example
For \(x(t)=e^{-at}u(t)\), \(h(t)=e^{-bt}u(t)\), real \(a,b>0\), the overlap is \(0\le\tau\le t\) when \(t\ge0\):
When \(a=b\), integrate the constant integrand after factoring out \(e^{-at}\): \(y(t)=te^{-at}u(t)\). Do not divide by zero in the first expression.
09 · Alongside Chapters 10–16Fourier analysis is projection onto oscillations
Over a period \(T_0\), the exponentials \(e^{jk\omega_0t}\), with \(\omega_0=2\pi/T_0\), are orthogonal:
For \(k\ne m\), the antiderivative has equal values at the endpoints because the phase changes by an integer multiple of \(2\pi\). Multiply the Fourier series by \(e^{-jm\omega_0t}\) and integrate; all terms except \(m\) disappear:
For \(x(t)=3\cos(\omega_0t+\pi/4)\), Euler’s formula gives \(a_1=(3/2)e^{j\pi/4}\), \(a_{-1}=(3/2)e^{-j\pi/4}\), and all other coefficients zero. Parseval then gives \(P_x=|a_1|^2+|a_{-1}|^2=9/2\), matching the time-domain result.
Convergence tells you what equality means
For a piecewise smooth periodic waveform, its Fourier series converges to the waveform at continuity points and to the midpoint of the one-sided limits at a jump. Gibbs overshoot near a jump becomes narrower as terms are added; its limiting relative peak does not disappear. Mean-square convergence concerns integrated squared error and does not require pointwise agreement everywhere.
Read a spectrum in magnitude and phase
For \(x(t)=e^{-at}u(t)\), \(a>0\):
A delay \(x(t-t_0)\) multiplies this spectrum by \(e^{-j\omega t_0}\). The magnitude stays the same while the phase gains the slope \(-t_0\). A phase change can therefore represent a time shift even when the magnitude plot is unchanged.
10 · Alongside Chapters 17–19 and 23–25A transform expression needs its region of convergence
Laplace and Z-transforms introduce weighting that can make a divergent Fourier integral or sum converge:
Writing \(s=\sigma+j\omega\) reveals a real exponential weight \(e^{-\sigma t}\). Writing \(z=re^{j\Omega}\) reveals a weight \(r^{-n}\). The ROC records which weights are sufficient.
For real \(a\), the same bilateral Laplace expression represents two different signals:
The first integral runs toward positive infinity; the second runs toward negative infinity. Convergence demands opposite inequalities. The algebraic fraction alone cannot tell you which signal was intended.
For real nonzero \(a\), the discrete analogue follows from geometric sums:
Connect the ROC to Fourier analysis and stability
The ordinary frequency response comes from the imaginary axis \(s=j\omega\), or the unit circle \(z=e^{j\Omega}\), when that set lies in the ROC. For an ordinary impulse response, an LTI system is BIBO stable exactly when that response is absolutely integrable in CT or absolutely summable in DT. A finite CT direct-feedthrough term \(D\delta(t)\) is also allowed; it contributes \(Dx(t)\) directly.
For a causal proper rational CT transfer function, after cancellations, BIBO stability requires all poles strictly in the open left half-plane. For a causal rational DT transfer function, its poles must be strictly inside the unit circle. Causality is essential to these shortcuts: without it, the ROC must be supplied explicitly. CT properness matters too: an ideal differentiator \(H(s)=s\) has no finite poles but is not BIBO stable.
11 · Alongside Chapters 9, 19, 25, and 27Initial conditions are another source of response
A transfer function describes the zero-state response: the part caused by the input when the system begins at rest. A stored initial state can produce output even with zero input.
Consider \(y'(t)+ay(t)=x(t)\) for \(t\ge0\), real \(a>0\), with \(y(0^+)=y_0\) and no impulsive input at the origin. Multiplying by the integrating factor \(e^{at}\) gives:
Integrate from 0 to \(t\), then divide by \(e^{at}\):
For \(x(t)=1\) on \(t\ge0\),
At \(t=0^+\), the result is \(y_0\). As \(t\to\infty\), it tends to \(1/a\). Checking both endpoints is often faster than repeating the derivation.
Using the unilateral Laplace convention with the appropriate initial-value term gives \(\mathcal L_+\{y'\}=sY-y_0\) under these assumptions. Hence \((s+a)Y=X+y_0\). Setting \(y_0=0\) yields \(H(s)=1/(s+a)\).
12 · Alongside Chapters 20–22 and 26Sampling identifies frequencies modulo the sample rate
Sampling \(e^{j2\pi ft}\) at \(t=n/f_s\) gives \(e^{j2\pi fn/f_s}\). Replacing \(f\) by \(f+kf_s\), integer \(k\), multiplies each sample by \(e^{j2\pi kn}=1\). Those frequencies are indistinguishable from the samples alone.
At \(f_s=1000\) samples/s, a 700 Hz cosine produces the same samples as a 300 Hz cosine:
Subtracting the sample rate gives \(-300\) Hz; the evenness of cosine makes its samples match positive 300 Hz. A complex exponential at \(-300\) Hz would retain its rotation direction.
If a continuous signal is strictly bandlimited to \(|f|\le B\), choosing \(f_s>2B\) permits ideal reconstruction. With \(\operatorname{sinc}(v)=\sin(\pi v)/(\pi v)\), extended to 1 at zero:
This ideal formula assumes a bandlimited signal and infinitely many exact samples. Real acquisition uses an analog anti-alias filter before sampling. Filtering afterward cannot uniquely undo already-overlapped spectral copies. Using a strict inequality avoids endpoint ambiguities at the Nyquist boundary.
The DFT is a finite sum with a periodic interpretation
With samples at \(f_s\), adjacent bins are spaced \(\Delta f=f_s/N\). Bins above \(N/2\) correspond to negative frequencies after wrapping. The DFT is the DTFT of the finite record evaluated on this grid; its inverse also describes a period-\(N\) extension.
13 · Alongside Chapter 27State space is a coupled version of a first-order equation
A state vector \(\mathbf q(t)\) stores enough information about the past to predict future behavior from future input. A linear time-invariant model is:
For \(m\) states, \(p\) inputs, and \(r\) outputs, the sizes are \(A:m\times m\), \(B:m\times p\), \(C:r\times m\), and \(D:r\times p\). Checking dimensions catches many mistakes.
The scalar exponential becomes the matrix exponential, defined by a convergent power series:
If \(A=\operatorname{diag}(-1,-3)\), then \(e^{At}=\operatorname{diag}(e^{-t},e^{-3t})\). With zero input and initial state \((2,-1)^T\), the state is \((2e^{-t},-e^{-3t})^T\). For \(C=[1\;1]\), the output is \(2e^{-t}-e^{-3t}\).
If \(A=V\Lambda V^{-1}\) is diagonalizable, \(e^{At}=Ve^{\Lambda t}V^{-1}\). Not every matrix is diagonalizable; repeated eigenvalues can produce terms such as \(te^{\lambda t}\).
14 · Alongside Chapters 28–29Expectation averages over possible outcomes
A random variable assigns a number to each outcome. A random process assigns a waveform. Fixing time gives a random variable; fixing an outcome gives one sample waveform.
Use the sum for a discrete distribution and the integral when a density exists. Expectation is linear even for dependent variables. Factoring \(\mathbb E[XY]\) into \(\mathbb E[X]\mathbb E[Y]\) needs an additional condition, such as independence.
Let \(X(t)=A\cos(\omega_0t+\Phi)\), where \(A\) is a fixed real number, \(\omega_0\ne0\), and \(\Phi\) is uniform on \([0,2\pi)\). Averaging over phase gives \(\mathbb E[X(t)]=0\), while \(\mathbb E[X(t)^2]=A^2/2\).
Using \(\cos a\cos b=\tfrac12[\cos(a-b)+\cos(a+b)]\), the phase-dependent second term averages to zero:
The mean is constant and the autocorrelation depends only on lag, so this process is wide-sense stationary.
For a possibly complex WSS process, these notes use \(R_X(\tau)=\mathbb E[X(t+\tau)X^*(t)]\). Its power spectral density is the Fourier transform of this autocorrelation, interpreted in the appropriate ordinary or distributional sense:
For the random-phase cosine above, \(S_X(\omega)=(\pi A^2/2)[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)]\). Integrating it with the \(1/(2\pi)\) factor recovers \(A^2/2\).
For stable LTI filtering of a WSS input with the needed second moments, \(S_Y(\omega)=|H(j\omega)|^2S_X(\omega)\). The magnitude is squared because power involves a signal multiplied by its conjugate.
15 · Try before opening the answersEight short checks
Write one or two lines for each answer. Include the domain or convergence condition whenever it matters.
1. Complex magnitude: find |(1 + j) / (1 − j)| and its angle.
Multiply numerator and denominator by \(1+j\): the ratio is \((1+j)^2/2=j\). Its magnitude is 1 and its angle is \(\pi/2\) modulo \(2\pi\).
2. Support: x(t) is nonzero only on [−1, 2]. Where can x(3 − 2t) be nonzero?
Track the old endpoints as a second check: \(-1\) moves to 2, and 2 moves to \(1/2\).
3. Impulse: evaluate the integral of t² δ(3t − 6) over the real line.
4. Discrete energy: find the energy of x[n] = (1/2)ⁿ u[n].
Average power is zero under the symmetric time-average convention.
5. Convolution: x = [1, 2], h = [1, −1], both starting at n = 0.
Linear convolution is \([1,1,-2]\), at indices 0, 1, 2. The coefficient sums satisfy \((1+2)(1-1)=0=1+1-2\).
6. ROC: a causal system has H(z) = 1 / (1 − 0.8 z⁻¹). Is it BIBO stable?
The causal impulse response is \(0.8^nu[n]\), with ROC \(|z|>0.8\). The unit circle is included, and \(\sum|h[n]|=1/(1-0.8)=5\), so it is BIBO stable.
7. Sampling: a 900 Hz cosine is sampled at 1200 samples/s. Which lower positive frequency has the same samples?
Subtract 1200 Hz to get \(-300\) Hz, then use cosine symmetry: 300 Hz. This equivalence is for the sampled cosine; the continuous waveforms differ between samples.
8. Variance: X takes −2 or +2 with equal probability. Find its mean, mean square, and variance.
\(\mathbb E[X]=0\), \(\mathbb E[X^2]=4\), and \(\operatorname{Var}(X)=4\). Zero mean does not mean zero signal strength.
16 · Use beside the original bookChapter reading map
Every chapter link opens the preserved offline book when the two folders are kept together. The combined ZIP contains both the book and this companion, including all equation-rendering files.