Signal and System
How to understand signal \(x(t-t_0)\)?
If \(t_0\) is positive, it is a delayed version of \(x(t)\); If \(t_0\) is negative, it is a advanced version of \(x(t)\). You can use the impulse function as an example.
How to systematically sketch \(x(\alpha t + \beta)\)?
First sketch the waveform of \(x(t + \beta)\), you can use the impulse function to know if it is delay or advance. Then reflect/compress/expand the waveform around origin.
Linear System
The response to \(x_1(t) + x_2(t)\) is \(y_1(t) + y_2(t)\)
Time Invariant System
The response to \(x(t-t_0)\) is \(y(t-t_0)\).
convolution sum
\[x[n] = \sum_{k=-\infty}^{\infty} x[k] \delta[n-k]\] \[x(t) = \int_{-\infty}^{\infty} x(\tau) \delta(t-\tau) d\tau\]note that now \(\delta[n-k]\) and \(\delta(t-\tau)\) is signal, where \(x[k]\) and \(x(\tau)\) are just number.
convolution is commutative
\[x[n]*h[n] = h[n]*x[n]\] \[x(t)*h(t) = h(t)*x(t)\]linear system response
\[y[n] = \sum_{k=-\infty}^{\infty} x[k]h_k[n]\]where \(h_k[n]\) is the response to \(\delta[n-k]\).
\[y(t) = \int_{-\infty}^{\infty} x(\tau) h_{\tau}(t) d\tau\]where \(h_{\tau}(t)\) is the respones to \(\delta(t-\tau)\)
linear time invariant response
\[y[n] = x[n] * h[n] = \sum_{k=-\infty}^{\infty} x[k]h[n-k]\]where \(h[n]\) is the response to \(\delta[n]\).
\[y(t) = x(t)*y(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau\]where \(h(t)\) is the respones to \(\delta(t)\)
initial rest (without proof), what is initial rest?
For differential equation with initial rest, the system is LTI and causal.
\[y(t) = \int_{-\infty}^{t} x(\tau) h(t-\tau) d\tau\]if \(x(t) = x(t)u(t)\)
\[y(t) = \int_{0}^{t} x(\tau) h(t-\tau) d\tau\]LTI system eigen function
The response to \(e^{st}\) (not \(e^{st}u(t)\)) is \(H(s)e^{st}\), \(H(s)= \int_{-\infty}^{\infty} h(\tau)e^{-s\tau} d\tau\).
The response to \(z^n\) (not \(z^n u[n]\)) is \(H(z)z^n\) is \(H(z) = \sum_{k=-\infty}^{\infty} h[k]z^{-k}\).
Fourier Series
Fourier Series
\[x(t) = \sum_{k=-\infty}^{\infty} a_k e^{jk\omega_0 t}\] \[a_k = \dfrac{1}{T}\int_{T} x(t) e^{-jk\omega_0 t} dt\] \[x[n] = \sum_{k=0}^{N-1} a_k e^{jk\omega_0 n}\] \[a_k = \dfrac{1}{N} \sum_{n=0}^{N-1} x[n] e^{-jk\omega_0 n}\]although it is not obvious if or not Fourier series is unique, as an engineer we assume it is unique, meaning there is only one set of \(a_k\) such that
\[x(t) = \sum_{k=-\infty}^{\infty} a_k e^{jk\omega_0 t}\]Fourier Series: Time Shifting
\[x(t-t_0) \overset{FS}{\longleftrightarrow} e^{-jk\omega_0 t_0} a_k\] \[x[n-n_0] \overset{FS}{\longleftrightarrow} e^{-jk\omega_0 n_0} a_k\]Fourier Series: Time Reversal
\[x(-t) \overset{FS}{\longleftrightarrow} a_{-k}\] \[x[-n] \overset{FS}{\longleftrightarrow} a_{-k}\]Fourier Series: even and odd
If \(x(t)\) is even, then \(a_{k}\) is even; If \(x(t)\) is odd, then \(a_{k}\) is odd.
If \(x[n]\) is even, then \(a_{k}\) is even; If \(x[n]\) is odd, then \(a_{k}\) is odd.
Fourier Series: Multiplication
\[x(t)y(t) \overset{FS}{\longleftrightarrow} h_k = \sum_{l=-\infty}^{\infty} a_l b_{k-l}\] \[x[n]y[n] \overset{FS}{\longleftrightarrow} h_k = \sum_{l=0}^{N-1} a_l b_{k-l}\]Fourier Series: Conjugation
\[x^{*}(t) \overset{FS}{\longleftrightarrow} a_{-k}^{*}\] \[x^{*}[n] \overset{FS}{\longleftrightarrow} a_{-k}^{*}\]Fourier Series: real \(x(t)\), real even \(x(t)\) and real odd \(x(t)\)
for real \(x(t)\), \(a_{k} = a_{-k}^{*}\)
for real even \(x(t)\), \(a_{k}\) is pure real and even.
for real odd \(x(t)\), \(a_{k}\) is pure imaginary and odd.
for real \(x[t]\), \(a_{k} = a_{-k}^{*}\)
for real even \(x[t]\), \(a_{k}\) is pure real and even.
for real odd \(x[t]\), \(a_{k}\) is pure imaginary and odd.
Paeseval’s theorem
\[\dfrac{1}{T}\int_T \vert x(t) \vert^2 dt = \sum_{-\infty}^{\infty} \vert a_k \vert^2\] \[\dfrac{1}{N} \sum_{n=0}^{N-1} \vert x[n] \vert^2 = \sum_{k=0}^{N-1} \vert a_k \vert^2\]filtering
\[y(t) = \sum_{k=-\infty}^{\infty} a_k H(jk\omega_0) e^{jk\omega_0 t}\] \[y[n] = \sum_{n=0}^{N-1} a_k H(e^{jk\omega_0}) e^{jk\omega_0 n}\]obtain frequency respones from ode
\[RC \dfrac{d y(t)}{dt} + y(t) = x(t)\] \[RC \dfrac{d}{dt}[H(j\omega)e^{j\omega t}] + H(j\omega) e^{j\omega t} = e^{j\omega t}\] \[H(j\omega) = \dfrac{1}{1+RCj\omega}\]convergence of fourier series
book_signals_and_systems, section 3.4
Fourier Transform
Fourier Transform
\[x(t) = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega\] \[X(\omega) = \int_{-\infty}^{\infty} x(t) e^{-j\omega t} dt\] \[X(\omega) = \sum_{k=-\infty}^{\infty} 2\pi a_k \delta(\omega - k\omega_0)\]We assume Fourier transform is unique, e.g., there is at most one \(X(\omega)\) such that
\[x(t) = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} X(\omega) e^{j\omega t} d\omega\]Fourier Transform: Differentiation and Integration
\[\dfrac{d x(t)}{dt} \overset{FT}{\longleftrightarrow} j\omega X(\omega)\] \[\int_{-\infty}^{t} x(\tau) d\tau \overset{FT}{\longleftrightarrow} \dfrac{1}{j\omega} X(\omega) + \pi X(0)\delta(\omega)\]Fourier Transform: Duality
\[g(t) \overset{FT}{\longleftrightarrow} f(\omega)\] \[f(t) \overset{FT}{\longleftrightarrow} 2\pi g(-\omega)\]Fourier Transform: Parseval’s Relation
\[\int_{-\infty}^{\infty} \vert x(t) \vert^2 dt = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} \vert X(\omega) \vert^2 d\omega\]Fourier Transform: conlvlution property
\[h(t)*x(t) \overset{FT}{\longleftrightarrow} H(\omega)X(\omega)\]Fourier Transform: modulation property
\[s(t)p(t) \overset{FT}{\longleftrightarrow} \dfrac{1}{2\pi} [S(\omega)*P(\omega)]\]DTFT
\[x[n] = \dfrac{1}{2\pi}\int_{0}^{2\pi} X(\Omega) e^{j\Omega n} d\Omega\] \[X(\Omega) = \sum_{n=-\infty}^{\infty} x[n] e^{-j\Omega n}\] \[X(\Omega) = \sum_{k=-\infty}^{\infty} 2\pi a_k \delta(\Omega - k\omega_0)\]