Signals and Systems formulary

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Unit 1Signals and Systems

An infinitely thin, infinitely tall spike of area 1.

The sifting property: multiplying by $\delta(t-t_0)$ and integrating just picks out $x(t_0)$ .

Any signal splits into an even part (mirror-symmetric) and an odd part (point-symmetric).

A signal is odd only if $x(-t) = -x(t)$ ; a signal starting flat then rising is not.

Seen in Practice exam 3489.

Use $|x|^2$ , not $x^2$ : for a complex value $3 + 4j$ it is $|3+4j|^2 = 25$ .

Seen in Practice exam 3489.

Energy signals have finite energy; power signals (like a never-ending sine) have finite power.

$A$ is amplitude, $\omega$ angular frequency, $\varphi$ phase.

A switch that turns on at $t = 0$ and stays on. Also written $u(t)$ .

Unit 2Time-Domain Analysis (LTI Systems)

Bounded input gives bounded output exactly when the impulse response is absolutely integrable.

The output of any LTI system is its input convolved with the impulse response $h$ .

Flip $h$ , slide it across $x$ , multiply, and add up the overlap at each shift. See the convolution note.

Poke the system with an impulse and $h(t)$ is what comes out.

Once you know $h$ , you know the response to everything (via convolution).

Zero-input response (from initial conditions) plus zero-state response (from the input).

The weighting function (impulse response) is the derivative of the step response $h_\sigma$ .

Because the impulse is the derivative of the step, their responses are too.

Seen in Practice exam 3489.

Unit 3The Laplace Transform

A single forward block with a branch tapped before it is reproduced by a parallel $1/G$ on the branch.

Seen in Practice exam 3489.

Inverse transforms are done by matching a table after partial fractions.

The whole point of transforms: messy convolution in time is just a product in $s$ .

Each derivative brings down a factor of $s$ ; this is how ODEs become polynomials.

Infinite open-loop gain forces the input difference to zero; infinite input impedance forces the input currents to zero.

The *closed-loop* gain is set by the feedback, and is certainly not zero.

Seen in Practice exam 3489.

Turns differential equations into algebra by trading $t$ for the complex frequency $s = \sigma + j\omega$ .

The Laplace transform of the impulse response; output over input in the $s$ -domain.

Unit 4Fourier Series and Transform

$N$ samples give $N$ spectrum points. Zero-padding to more samples gives a finer spectrum.

Seen in Practice exam 3489.

For a decaying geometric $a^k\sigma[k]$ it sums to $\dfrac{1}{1 - a\,e^{-j\Omega}}$ (a geometric series).

Seen in Practice exam 3489.

A periodic signal is a sum of harmonics; $c_k$ is the amount of the $k$ -th one.

Seen in Practice exam 3489.

The Fourier series taken to the limit as the period $T_0 \to \infty$ ; the spectrum becomes continuous.

Seen in Practice exam 3489.

Ideal sampling multiplies the signal by this comb of impulses.

In frequency this makes the spectrum repeat periodically; overlap is aliasing.

Seen in Practice exam 3489.

Rebuild the time signal by adding its frequency components back up.

Energy is the same whether you measure it in time or in frequency.

The time between samples is the reciprocal of the sampling frequency.

Seen in Practice exam 3489.

A pure tone is two impulses in frequency, at $\pm\omega_0$ . A sine is the same but imaginary and antisymmetric.

Seen in Practice exam 3489.

A box in time is a sinc in frequency. At $\omega = 0$ take the limit: $X(0) = A\tau$ (the area).

Seen in Practice exam 3489.

Tap a card to unfold the notes.

Unit 1Signals and Systems

Dirac impulse∫−∞∞x(t) δ(t−t0) dt=x(t0)\int_{-\infty}^{\infty} x(t)\,\delta(t - t_0)\, dt = x(t_0)∫−∞∞​x(t)δ(t−t0​)dt=x(t0​)

Even and odd parts(shows up on exams)xe(t)=12(x(t)+x(−t)),xo(t)=12(x(t)−x(−t))x_e(t) = \tfrac{1}{2}\bigl(x(t) + x(-t)\bigr), \qquad x_o(t) = \tfrac{1}{2}\bigl(x(t) - x(-t)\bigr)xe​(t)=21​(x(t)+x(−t)),xo​(t)=21​(x(t)−x(−t))

Signal energy(shows up on exams)Ex=∫−∞∞∣x(t)∣2 dtE_x = \int_{-\infty}^{\infty} |x(t)|^2 \, dtEx​=∫−∞∞​∣x(t)∣2dt

Signal powerPx=lim⁡T→∞1T∫−T/2T/2∣x(t)∣2 dtP_x = \lim_{T \to \infty} \frac{1}{T}\int_{-T/2}^{T/2} |x(t)|^2 \, dtPx​=T→∞lim​T1​∫−T/2T/2​∣x(t)∣2dt

Sinusoidal signalx(t)=Acos⁡(ωt+φ),ω=2πfx(t) = A\cos(\omega t + \varphi), \qquad \omega = 2\pi fx(t)=Acos(ωt+φ),ω=2πf

Unit stepσ(t)={1t≥00t<0\sigma(t) = \begin{cases} 1 & t \ge 0 \\ 0 & t < 0 \end{cases}σ(t)={10​t≥0t<0​

Unit 2Time-Domain Analysis (LTI Systems)

BIBO stability∫−∞∞∣h(t)∣ dt<∞\int_{-\infty}^{\infty} |h(t)|\, dt < \infty∫−∞∞​∣h(t)∣dt<∞

Convolutiony(t)=(x∗h)(t)=∫−∞∞x(τ) h(t−τ) dτy(t) = (x * h)(t) = \int_{-\infty}^{\infty} x(\tau)\, h(t - \tau)\, d\tauy(t)=(x∗h)(t)=∫−∞∞​x(τ)h(t−τ)dτ

Impulse responseδ(t)⟶h(t)\delta(t) \longrightarrow h(t)δ(t)⟶h(t)

Total system responsey(t)=yzir(t)+yzsr(t)y(t) = y_{\text{zir}}(t) + y_{\text{zsr}}(t)y(t)=yzir​(t)+yzsr​(t)

Weighting function from step response(shows up on exams)g(t)=ddt hσ(t)g(t) = \frac{d}{dt}\, h_\sigma(t)g(t)=dtd​hσ​(t)

Unit 3The Laplace Transform

Block diagram algebra(shows up on exams)series: H1H2,parallel: H1+H2,feedback: G1+GH\text{series: } H_1 H_2, \quad \text{parallel: } H_1 + H_2, \quad \text{feedback: } \frac{G}{1 + G H}series: H1​H2​,parallel: H1​+H2​,feedback: 1+GHG​

Common Laplace pairsδ(t)→1,σ(t)→1s,e−atσ(t)→1s+a\delta(t) \to 1, \quad \sigma(t) \to \frac{1}{s}, \quad e^{-at}\sigma(t) \to \frac{1}{s + a}δ(t)→1,σ(t)→s1​,e−atσ(t)→s+a1​

Convolution becomes multiplicationy(t)=x(t)∗h(t) ⟺ Y(s)=X(s) H(s)y(t) = x(t) * h(t) \ \Longleftrightarrow\ Y(s) = X(s)\,H(s)y(t)=x(t)∗h(t) ⟺ Y(s)=X(s)H(s)

Differentiation ruleL{x′(t)}=sX(s)−x(0)\mathcal{L}\{x'(t)\} = s X(s) - x(0)L{x′(t)}=sX(s)−x(0)

Ideal op-amp assumptions(shows up on exams)V+−V−=0,i+=i−=0V_+ - V_- = 0, \qquad i_+ = i_- = 0V+​−V−​=0,i+​=i−​=0

Laplace transformX(s)=∫0∞x(t) e−st dtX(s) = \int_{0}^{\infty} x(t)\, e^{-st}\, dtX(s)=∫0∞​x(t)e−stdt

Transfer functionH(s)=Y(s)X(s)H(s) = \frac{Y(s)}{X(s)}H(s)=X(s)Y(s)​

Unit 4Fourier Series and Transform

Discrete Fourier transform(shows up on exams)X[n]=∑k=0N−1x[k] e−j2πnk/N,n=0,1,…,N−1X[n] = \sum_{k=0}^{N-1} x[k]\, e^{-j 2\pi n k / N}, \quad n = 0, 1, \dots, N-1X[n]=k=0∑N−1​x[k]e−j2πnk/N,n=0,1,…,N−1

Discrete-time Fourier transform(shows up on exams)Xd(jΩ)=∑k=−∞∞x[k] e−jΩkX_d(j\Omega) = \sum_{k=-\infty}^{\infty} x[k]\, e^{-j\Omega k}Xd​(jΩ)=k=−∞∑∞​x[k]e−jΩk

Fourier transform(shows up on exams)X(jω)=∫−∞∞x(t) e−jωt dtX(j\omega) = \int_{-\infty}^{\infty} x(t)\, e^{-j\omega t}\, dtX(jω)=∫−∞∞​x(t)e−jωtdt

Ideal sampling (impulse train)(shows up on exams)a(t)=∑k=−∞∞δ(t−k Δt)a(t) = \sum_{k=-\infty}^{\infty} \delta(t - k\,\Delta t)a(t)=k=−∞∑∞​δ(t−kΔt)

Inverse Fourier transformx(t)=12π∫−∞∞X(jω) ejωt dωx(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} X(j\omega)\, e^{j\omega t}\, d\omegax(t)=2π1​∫−∞∞​X(jω)ejωtdω

Parseval / energy theoremEx=∫−∞∞∣x(t)∣2 dt=12π∫−∞∞∣X(jω)∣2 dωE_x = \int_{-\infty}^{\infty} |x(t)|^2 \, dt = \frac{1}{2\pi}\int_{-\infty}^{\infty} |X(j\omega)|^2 \, d\omegaEx​=∫−∞∞​∣x(t)∣2dt=2π1​∫−∞∞​∣X(jω)∣2dω

Sampling interval(shows up on exams)Δt=1fa\Delta t = \frac{1}{f_a}Δt=fa​1​

Transform of a cosine(shows up on exams)Acos⁡(ω0t) ⟷ Aπ (δ(ω−ω0)+δ(ω+ω0))A\cos(\omega_0 t) \ \longleftrightarrow\ A\pi\,\bigl(\delta(\omega - \omega_0) + \delta(\omega + \omega_0)\bigr)Acos(ω0​t) ⟷ Aπ(δ(ω−ω0​)+δ(ω+ω0​))

Transform of a rectangle(shows up on exams)x(t)=A, ∣t∣≤τ2 ⟷ X(jω)=2Asin⁡(ωτ/2)ωx(t) = A,\ |t| \le \tfrac{\tau}{2} \ \longleftrightarrow\ X(j\omega) = \frac{2A\sin(\omega \tau / 2)}{\omega}x(t)=A, ∣t∣≤2τ​ ⟷ X(jω)=ω2Asin(ωτ/2)​