Advanced electrical engineering: 50 upper-level concepts

All 50 cards

1. 1
   Laplace transform (unilateral)
   $F(s) = \displaystyle\int_{0^-}^{\infty} f(t)\,e^{-st}\,dt$

   $s = \sigma + j\omega$ is the complex frequency. Differentiation becomes multiplication by $s$: $\mathcal{L}\{f'\} = sF(s) - f(0^-)$.
2. 2
   Final value theorem
   $\displaystyle\lim_{t \to \infty} f(t) = \lim_{s \to 0} sF(s)$

   Valid only if every pole of $sF(s)$ lies in the open left half-plane, so that $f(t)$ actually settles.
3. 3
   Z-transform
   $X(z) = \displaystyle\sum_{n=-\infty}^{\infty} x[n]\,z^{-n}$

   A delay of one sample multiplies the transform by $z^{-1}$.
4. 4
   Transfer function
   $H(s) = \dfrac{Y(s)}{X(s)}$, the ratio of the output and input transforms with all initial conditions zero

   It is also the Laplace transform of the impulse response.
5. 5
   Poles and zeros of a transfer function
   Zeros are the roots of the numerator of $H(s)$; poles are the roots of the denominator

   On a pole-zero plot, poles are drawn as crosses and zeros as circles.
6. 6
   Stability condition for a continuous-time LTI system
   Every pole of the transfer function lies in the open left half of the s-plane

   A pole in the right half-plane makes the response grow without bound; simple poles on the imaginary axis give sustained oscillation.
7. 7
   Stability condition for a discrete-time LTI system
   Every pole of $H(z)$ lies strictly inside the unit circle, $|z| < 1$

   The mapping $z = e^{sT}$ sends the left half of the s-plane inside the unit circle.
8. 8
   Bode phase of a real left-half-plane pole
   It falls from $0^\circ$ to $-90^\circ$, passing $-45^\circ$ at the corner frequency

   Most of the change happens within a decade either side of the corner, from $0.1\,\omega_c$ to $10\,\omega_c$.
9. 9
   Right-half-plane zero on a Bode plot
   Its magnitude rises at $20\ \mathrm{dB}$ per decade like a normal zero, but its phase falls by $90^\circ$ like a pole

   Systems with one are called non-minimum phase; the step response often starts in the wrong direction.
10. 10
    Standard second-order transfer function
    $H(s) = \dfrac{\omega_n^2}{s^2 + 2\zeta\omega_n s + \omega_n^2}$

    $\omega_n$ is the natural frequency and $\zeta$ the damping ratio; $0 < \zeta < 1$ is underdamped.
11. 11
    Percent overshoot of an underdamped second-order step response
    $\%\mathrm{OS} = 100\,e^{-\zeta\pi/\sqrt{1 - \zeta^2}}$

    It depends only on $\zeta$: about 16% at $\zeta = 0.5$ and 4.3% at $\zeta = 0.707$.
12. 12
    Routh–Hurwitz criterion
    The number of sign changes in the first column of the Routh array equals the number of roots in the right half-plane

    It tests the stability of a characteristic polynomial without solving for its roots.
13. 13
    Nyquist stability criterion
    $Z = N + P$: closed-loop right-half-plane poles equal the clockwise encirclements of $-1$ by $L(s)$ plus the open-loop right-half-plane poles

    Stability needs $Z = 0$, so the plot must encircle $-1$ counterclockwise once for each unstable open-loop pole.
14. 14
    Gain margin
    How much the loop gain can rise before instability: $\mathrm{GM} = \dfrac{1}{|L(j\omega_{pc})|}$, where the phase of $L$ is $-180^\circ$

    $\omega_{pc}$ is the phase-crossover frequency. In decibels, $\mathrm{GM} = -20\log_{10}|L(j\omega_{pc})|$.
15. 15
    Phase margin
    $\mathrm{PM} = 180^\circ + \angle L(j\omega_{gc})$, where $|L(j\omega_{gc})| = 1$

    $\omega_{gc}$ is the gain-crossover frequency. A common design target is $45^\circ$ to $60^\circ$.
16. 16
    Nyquist–Shannon sampling theorem
    A signal band-limited to $B$ hertz can be reconstructed exactly from samples taken at a rate $f_s > 2B$

    Half the sampling rate, $f_s/2$, is called the Nyquist frequency.
17. 17
    Aliasing
    After sampling, frequencies above $f_s/2$ fold back and appear as lower frequencies

    An anti-aliasing low-pass filter before the sampler prevents it. A $7\ \mathrm{kHz}$ tone sampled at $10\ \mathrm{kHz}$ appears at $3\ \mathrm{kHz}$.
18. 18
    FIR vs IIR filters
    FIR filters have no feedback, are always stable and can have exactly linear phase; IIR filters use feedback and need fewer coefficients for a sharp response, but can be unstable

    An FIR filter with symmetric coefficients has linear phase, so it delays all frequencies equally.
19. 19
    Discrete Fourier transform (DFT)
    $X[k] = \displaystyle\sum_{n=0}^{N-1} x[n]\,e^{-j2\pi kn/N}$
20. 20
    Cost of the FFT vs a direct DFT
    $O(N\log N)$ operations instead of $O(N^2)$

    The common radix-2 Cooley–Tukey algorithm needs $N$ to be a power of 2.
21. 21
    Frequency resolution of an N-point DFT
    $\Delta f = \dfrac{f_s}{N}$

    Bin $k$ sits at $k f_s/N$. Zero-padding interpolates the spectrum but doesn’t add resolution.
22. 22
    Characteristic impedance of a lossless transmission line
    $Z_0 = \sqrt{\dfrac{L'}{C'}}$

    $L'$ and $C'$ are the inductance and capacitance per unit length. Coaxial cables are usually $50$ or $75\ \Omega$.
23. 23
    Reflection coefficient at a load
    $\Gamma = \dfrac{Z_L - Z_0}{Z_L + Z_0}$

    A matched load gives $\Gamma = 0$, a short circuit $-1$ and an open circuit $+1$.
24. 24
    Voltage standing wave ratio (VSWR)
    $\mathrm{VSWR} = \dfrac{1 + |\Gamma|}{1 - |\Gamma|}$

    It runs from 1 for a perfect match to infinity for total reflection.
25. 25
    Quarter-wave transformer
    A $\lambda/4$ line of impedance $Z_0 = \sqrt{Z_{\text{in}} Z_L}$ matches a real load $Z_L$ to $Z_{\text{in}}$

    A quarter-wave line transforms any load to $Z_{\text{in}} = Z_0^2/Z_L$.
26. 26
    Smith chart
    A plot of the reflection coefficient in the complex plane, overlaid with circles of constant normalized resistance and arcs of constant normalized reactance

    Moving along a lossless line rotates the point around the centre; one full turn is half a wavelength.
27. 27
    Gauss’s law in differential form
    $\nabla \cdot \mathbf{D} = \rho_v$

    The electric flux out of a region equals the free charge inside. In free space $\mathbf{D} = \varepsilon_0\mathbf{E}$.
28. 28
    Gauss’s law for magnetism
    $\nabla \cdot \mathbf{B} = 0$

    There are no magnetic monopoles, so magnetic field lines always close on themselves.
29. 29
    Faraday’s law in differential form
    $\nabla \times \mathbf{E} = -\dfrac{\partial \mathbf{B}}{\partial t}$

    A time-varying magnetic field produces a circulating electric field.
30. 30
    Ampère–Maxwell law
    $\nabla \times \mathbf{H} = \mathbf{J} + \dfrac{\partial \mathbf{D}}{\partial t}$

    Maxwell’s displacement current term, $\partial\mathbf{D}/\partial t$, is what makes electromagnetic waves possible.
31. 31
    Intrinsic impedance of free space
    $\eta_0 = \sqrt{\dfrac{\mu_0}{\varepsilon_0}} \approx 377\ \Omega$

    The ratio of the electric to the magnetic field strength in a plane wave travelling through a vacuum.
32. 32
    Skin depth
    $\delta = \sqrt{\dfrac{2}{\omega\mu\sigma}} = \dfrac{1}{\sqrt{\pi f \mu \sigma}}$

    The depth at which current density falls to $1/e$ of its surface value. In copper it is about $9\ \mathrm{mm}$ at $50\ \mathrm{Hz}$ and $65\ \mathrm{\mu m}$ at $1\ \mathrm{MHz}$.
33. 33
    Output voltage of an ideal buck converter
    $V_{\text{out}} = D\,V_{\text{in}}$

    $D$ is the duty cycle, between 0 and 1, in continuous conduction mode.
34. 34
    Output voltage of an ideal boost converter
    $V_{\text{out}} = \dfrac{V_{\text{in}}}{1 - D}$

    In continuous conduction mode; a duty cycle of 0.5 doubles the input voltage.
35. 35
    Pulse-width modulation (PWM)
    Switching at a fixed frequency and varying the duty cycle $D = t_{\text{on}}/T$ to control the average output

    Because the switch is either fully on or fully off, it dissipates little power.
36. 36
    Inductor volt-second balance
    In periodic steady state, the average voltage across an inductor over one switching period is zero

    Applying it to the on and off intervals gives the conversion ratios of the buck and boost converters.
37. 37
    Base impedance in a per-unit system
    $Z_{\text{base}} = \dfrac{V_{\text{base}}^2}{S_{\text{base}}}$

    In three-phase systems, use the line-to-line base voltage and the three-phase base power. Then $Z_{\text{pu}} = Z/Z_{\text{base}}$.
38. 38
    Converting a per-unit impedance to a new base
    $Z_{\text{pu,new}} = Z_{\text{pu,old}} \left(\dfrac{V_{\text{base,old}}}{V_{\text{base,new}}}\right)^2 \dfrac{S_{\text{base,new}}}{S_{\text{base,old}}}$

    Needed because nameplates give impedances on each machine’s own rating.
39. 39
    Synchronous speed
    $n_s = \dfrac{120 f}{p}$ revolutions per minute

    $p$ is the number of poles: a four-pole machine on $50\ \mathrm{Hz}$ turns at $1500\ \mathrm{rpm}$.
40. 40
    Power–angle equation of a synchronous generator
    $P = \dfrac{E\,V}{X_s}\sin\delta$

    Per phase, for a round-rotor machine with resistance neglected. $E$ is the internal EMF, $X_s$ the synchronous reactance and $\delta$ the power angle.
41. 41
    Symmetrical components
    Any set of three unbalanced phasors can be written as the sum of balanced positive-sequence, negative-sequence and zero-sequence sets

    They use the operator $a = 1\angle 120^\circ$, and turn an unbalanced fault into three decoupled sequence networks.
42. 42
    Fault current for a single line-to-ground fault
    $I_f = \dfrac{3V_f}{Z_1 + Z_2 + Z_0 + 3Z_f}$

    $Z_1$, $Z_2$ and $Z_0$ are the positive-, negative- and zero-sequence impedances at the fault, $Z_f$ the fault impedance and $V_f$ the pre-fault voltage.
43. 43
    Shannon–Hartley channel capacity
    $C = B\log_2\!\left(1 + \dfrac{S}{N}\right)$ bits per second

    $B$ is the bandwidth in hertz and $S/N$ the signal-to-noise power ratio as a plain ratio, not in decibels.
44. 44
    Amplitude-modulated (AM) signal
    $s(t) = A_c\,[1 + m\cos(\omega_m t)]\cos(\omega_c t)$

    $m$ is the modulation index, kept at or below 1 to avoid overmodulation. The bandwidth is twice the highest message frequency.
45. 45
    Total power of an AM signal
    $P_t = P_c\left(1 + \dfrac{m^2}{2}\right)$

    $P_c$ is the carrier power. Even at $m = 1$, only a third of the power is in the sidebands that carry the message.
46. 46
    Carson’s rule for FM bandwidth
    $B \approx 2(\Delta f + f_m)$

    $\Delta f$ is the peak frequency deviation and $f_m$ the highest modulating frequency; their ratio $\beta = \Delta f/f_m$ is the modulation index.
47. 47
    Thermal (Johnson–Nyquist) noise voltage
    $v_{n,\text{rms}} = \sqrt{4kTRB}$

    $k$ is Boltzmann’s constant, $T$ the temperature in kelvin and $B$ the bandwidth. A $1\ \mathrm{k\Omega}$ resistor at room temperature gives about $4\ \mathrm{nV/\sqrt{Hz}}$.
48. 48
    Thermal noise floor at room temperature
    About $-174\ \mathrm{dBm/Hz}$

    The available noise power is $kTB$: at $290\ \mathrm{K}$ that is $4 \times 10^{-21}\ \mathrm{W}$ per hertz of bandwidth.
49. 49
    Noise figure
    $\mathrm{NF} = 10\log_{10}\dfrac{\mathrm{SNR}_{\text{in}}}{\mathrm{SNR}_{\text{out}}}$, in decibels

    How much a stage degrades the signal-to-noise ratio, defined with the source at $290\ \mathrm{K}$. The linear ratio is the noise factor $F$.
50. 50
    Friis formula for cascaded noise factor
    $F = F_1 + \dfrac{F_2 - 1}{G_1} + \dfrac{F_3 - 1}{G_1 G_2} + \cdots$

    $F_i$ and $G_i$ are each stage’s noise factor and gain as linear ratios. The first stage dominates, which is why receivers start with a low-noise amplifier.