Fourier Transform

Posted on Edited on In

discrete-time frequency: \(\hat{\omega}=\omega T_s\), units are radians per sample

Below diagram show the windowing effect and sampling

NinDFT.drawio

For general window function, we know \(W(e^{j\hat{\omega}})=\frac{1}{T_s}W(j\omega)\),

and \[ \frac{W(j\omega|\omega=0)}{T_s} = \frac{T_sW(e^{j\hat{\omega}}|\hat{\omega}=0)}{T_s} =W(e^{j\hat{\omega}}|\hat{\omega}=0)= \sum_{n=-N_w}^{+N_w}w[n] \]

e.g. \(\frac{W(j\omega|\omega=0)}{T_s} = N\) for Rectangular Window, shown in above figure

warmup

Continuous-time signals \(x_c(t)\) Discrete-time signals \(x[n]\)
Aperiodic signals Continuous Fourier transform Discrete-time Fourier transform
Periodic signals Fourier series Discrete Fourier transform

Continuous Time Fourier Series (CTFS)

\[\begin{align} a_k &= \frac{1}{T}\int_T x(t)e^{-jk(2\pi/T)) t}dt \\ x(t) &= \sum_{k=-\infty}^{+\infty}a_ke^{jk(2\pi/T) t} \end{align}\]

Continuous-Time Fourier transform (CTFT)

\[\begin{align} X(j\omega) &=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}dt \\ x(t)&= \frac{1}{2\pi}\int_{-\infty}^{+\infty}X(j\omega)e^{j\omega t}d\omega \end{align}\]

[https://www.rose-hulman.edu/class/ee/yoder/ece380/Handouts/Fourier%20Transform%20Tables%20w.pdf]

image-20240831104459715

Discrete-Time Fourier Transform (DTFT)

\[\begin{align} X(e^{j\hat{\omega}}) &=\sum_{n=-\infty}^{+\infty}x[n]e^{-j\hat{\omega} n} \\ x[n] &= \frac{1}{2\pi}\int_{2\pi}X(e^{j\hat{\omega}})e^{j\hat{\omega} n}d\hat{\omega} \end{align}\]

DTFT is defined for infinitely long signals as well as finite-length signal

DTFT is continuous in the frequency domain

We could verify that is the correct inverse DTFT relation by substituting the definition of the DTFT and rearranging terms

image-20240831152155093

Discrete-Time Fourier Series (DTFS)

TODO 📅

Discrete Fourier Series (DFS)

TODO 📅

Discrete Fourier Transform (DFT)

Two steps are needed to change the DTFT sum into a computable form:

  1. the continuous frequency variable \(\hat{\omega}\) must be sampled
  2. the limits on the DTFT sum must be finite

\[\begin{align} X[k] &= \sum_{n=0}^{N-1}x[n]e^{-j(2\pi/N)kn}\space\space\space k=0,1,...,N-1 \\ x[n] &= \frac{1}{N}\sum_{k=0}^{N-1}X[k]e^{j(2\pi/N)kn} \space\space\space n=0,1,...,N-1 \end{align}\]

Part of the proof is given by the following step:

image-20240830222204470

DFT \(X[k]\) is a sampled version of the DTFT \(X(e^{j\hat{\omega}})\), where \(\hat{\omega_k} = \frac{2\pi k}{N}\)

impulse train

CTFT:

image-20240830224755336

image-20240911221811991

using time-sampling property

impulse_train.drawio

DTFT:

Given \(x[n]=\sum_{k=-\infty}^{\infty}\delta(n-k)\)

\[\begin{align} X(e^{j\hat{\omega}}) &= X_s(j\frac{\hat{\omega}}{T}) \\ &= \frac{2\pi}{T}\sum_{k=-\infty}^{\infty}\delta(\frac{\hat{\omega}}{T}-\frac{2\pi k}{T}) \\ &= \frac{2\pi}{T}\sum_{k=-\infty}^{\infty}T\delta(\hat{\omega}-2\pi k) \\ &= 2\pi\sum_{k=-\infty}^{\infty}\delta(\hat{\omega}-2\pi k) \end{align}\]

[http://courses.ece.ubc.ca/359/notes/notes_part1_set4.pdf]

impulse scaling

\[ \delta(\alpha t)= \frac{1}{\alpha}\delta( t) \]

where \(\alpha\) is scaling ratio

impulse invariance

\[ h[n] = Th_c(nT) \]

When \(h[n]\) and \(h_c(t)\) are related through the above equation, i.e., the impulse response of the discrete-time system is a scaled, sampled version of \(h_c(t)\), the discrete-time system is said to be an impulse-invariant version of the continuous-time system

we have \[ H(e^{j\hat{\omega}}) = H_c\left(j\frac{\hat{\omega}}{T}\right) \]

Multiplication

aka Modulation or Windowing Theorem

CTFT: \[ x_1(t)x_2(t)\overset{FT}{\longrightarrow}\frac{1}{2\pi}X_1(\omega)*X_2(\omega) \]

DTFT:

image-20240909215833750

Duality

image-20240921181908992

image-20240921182105935

Conjugate Symmetry

image-20240921181015717

image-20240921181258063

Parseval's Relation

CTFT:

image-20240830230835764

DTFT:

image-20230516022936168

DFT:

image-20240907224043641

Eigenfunctions & frequency response

Complex exponentials are eigenfunctions of LTI systems, that is,

continuous time: \(e^{j\omega t}\to H(j\omega)e^{j\omega t}\)

discrete time: \(e^{j\hat{\omega}n} \to H(e^{j\hat{\omega}})e^{j\hat{\omega}n}\)

where \(H(j\omega)\), \(H(e^{j\hat{\omega}})\) is frequency response of continuous-time systems and discrete-time systems, which is the function of \(\omega\) and \(\hat{\omega}\) \[\begin{align} H(j\omega) &= \int_{-\infty}^{+\infty}h(t)e^{-j\omega t}dt \\ \\ H(e^{j\hat{\omega}}) &= \sum_{n=-\infty}^{+\infty}h[n]e^{-j\hat{\omega} n} \end{align}\]

The frequency response of discrete-time LTI systems is always a periodic function of the frequency variable \(\hat{\omega}\) with period \(2\pi\)

Sampling Theorem

time-sampling theorem: applies to bandlimited signals

spectral sampling theorem: applies to timelimited signals

Aliasing

The frequencies \(f_{\text{sig}}\) and \(Nf_s \pm f_{\text{sig}}\) (\(N\) integer), are indistinguishable in the discrete time domain.

image-20220626000016184

Given below sequence \[ X[n] =A e^{j\omega T_s n} \]

  1. \(kf_s + \Delta f\)

​ \[\begin{align} x[n] &= Ae^{j\left( kf_s+\Delta f \right)2\pi T_sn} + Ae^{j\left( -kf_s-\Delta f \right)2\pi T_sn} \\ &= Ae^{j\Delta f\cdot 2\pi T_sn} + Ae^{-j\Delta f\cdot 2\pi T_sn} \end{align}\]

  1. \(kf_s - \Delta f\)

​ \[\begin{align} x[n] &= Ae^{j\left( kf_s-\Delta f \right)2\pi T_sn} + Ae^{j\left( -kf_s+\Delta f \right)2\pi T_sn} \\ &= Ae^{-j\Delta f\cdot 2\pi T_sn} + Ae^{j\Delta f\cdot 2\pi T_sn} \end{align}\]

complex signal

\[\begin{align} A e^{j(\omega_s + \Delta \omega) T_s n} &= A e^{j(k\omega_s + \Delta \omega) T_s n} \\ A e^{j(\omega_s - \Delta \omega) T_s n} &= A e^{j(k\omega_s - \Delta \omega) T_s n} \end{align}\]

sampling_aliasing.drawio

CTFS & CTFT

Fourier transform of a periodic signal with Fourier series coefficients \(\{a_k\}\) can be interpreted as a train of impulses occurring at the harmonically related frequencies and for which the area of the impulse at the \(k\)th harmonic frequency \(k\omega_0\) is \(2\pi\) times the \(k\)th Fourier series coefficient \(a_k\)

image-20240830225453601

spectral sampling

image-20240831185532202

spectral sampling by \(\omega_0\), and \(\frac{2\pi}{\omega_0} \gt \tau\) \[ X_{n\omega_0}(\omega) = \sum_{n=-\infty}^{\infty}X(n\omega_0)\delta(\omega - n\omega_0) \] Periodic repetition of \(x(t)\) is \[ x_{n\omega_0}(t) = \frac{1}{\omega_0}\sum_{n=-\infty}^{\infty}x(t -n\frac{2\pi}{\omega_0})=\frac{T_0}{2\pi}\sum_{n=-\infty}^{\infty}x(t -nT_0) \]

Then, if \(x_{T_0} (t)\), a periodic signal formed by repeating \(x(t)\) every \(T_0\) seconds (\(T_0 \gt \tau\)​), its CTFT is \[ X_{T_0}(\omega) = \frac{2\pi}{T_0} \cdot X_{n\omega_0}(\omega) = \frac{2\pi}{T_0}\sum_{n=-\infty}^{\infty}X(n\omega_0)\delta(\omega - n\omega_0) \] Then \(x_{T_0} (t)\) can be expressed with inverse CTFT as \[\begin{align} x_{T_0} (t) &= \frac{1}{2\pi}\int_{-\infty}^{\infty}X_{T_0}(\omega)e^{j\omega t}d\omega \\ &= \frac{1}{T_0}\sum_{n=-\infty}^{\infty}X(n\omega_0)e^{jn\omega_0 t} =\sum_{n=-\infty}^{\infty}\frac{1}{T_0}X(n\omega_0)e^{jn\omega_0 t} \end{align}\]

i.e. the coefficients of the Fourier series for \(x_{T_0} (t)\) is \(D_n =\frac{1}{T_0}X(n\omega_0)\)

image-20240831190258683

alternative method by direct Fourier series

image-20240831193912709

Why DFT ?

We can use DFT to compute DTFT samples and CTFT samples

image-20240831201335531

\[ \overline{x}(t) = \sum_{n=0}^{N_0-1}x(nT)\delta(t-nT) \] applying the Fourier transform yieds \[ \overline{X}(\omega) = \sum_{n=0}^{N_0-1}x[n]e^{-jn\omega T} \] But \(\overline{X}(\omega)\), the Fourier transform of \(\overline{x}(t)\) is \(X(\omega)/T\), assuming negligible aliasing. Hence, \[ X(\omega) = T\overline{X}(\omega) = T\sum_{n=0}^{N_0-1}x[n]e^{-jn\omega T} \] and \[ X(k\omega_0) = T\sum_{n=0}^{N_0-1}x[n]e^{-jn k\omega_0 T} \] with \(\hat{\omega}_0 = \omega_0 T\) \[ X(k\omega_0) = T\sum_{n=0}^{N_0-1}x[n]e^{-jn k\hat{\omega}_0} \] i.e. the relationship between CTFT and DFT is \(X(k\omega_0) = T\cdot X[k]\), DFT is a tool for computing the samples of CTFT

C/D

Sampling with a periodic impulse train, followed by conversion to a discrete-time sequence

image-20240901155629500

image-20240830231619897

The periodic impulse train is \[ s(t) = \sum_{n=-\infty}^{\infty}\delta(t-nT) \] \(x_s(t)\) can be expressed as \[ x_s(t) = \sum_{n=-\infty}^{\infty}x_c(nT)\delta(t-nT) \] i.e., the size (area) of the impulse at sample time \(nT\) is equal to the value of the continuous-time signal at that time.

\(x_s(t)\)​ is, in a sense, a continuous-time signal (specifically, an impulse train)

samples of \(x_c(t)\) are represented by finite numbers in \(x[n]\) rather than as the areas of impulses, as with \(x_s(t)\)

Frequency-Domain Representation of Sampling

The relationship between the Fourier transforms of the input and the output of the impulse train modulator \[ X_s(j\omega) = \frac{1}{T}\sum_{k=-\infty}^{\infty}X_c(j(\omega -k\omega_s)) \] where \(\omega_s\) is the sampling frequency in radians/s

\(X(e^{j\hat{\omega}})\), the discrete-time Fourier transform (DTFT) of the sequence \(x[n]\), in terms of \(X_s(j\omega)\) and \(X_c(j\omega)\)

continuous-time Fourier transform discrete-time Fourier transform
\(x_s(t) = \sum_{n=-\infty}^{\infty}x_c(nT)\delta(t-nT)\) \(x[n]=x_c(nT)\)
\(X_s(j\omega)=\sum_{n=-\infty}^{\infty}x_c(nT)e^{-j\omega Tn}\) \(X(e^{j\hat{\omega}})=\sum_{n=-\infty}^{\infty}x_c(nT)e^{-j\hat{\omega} n}\)

\[ X(e^{j\omega T}) = \frac{1}{T}\sum_{k=-\infty}^{\infty}X_c(j(\omega-k\omega_s)) \] or equivalently, \[ X(e^{j\hat{\omega}}) = \frac{1}{T}\sum_{k=-\infty}^{\infty}X_c(j(\frac{\hat{\omega}}{T}-\frac{2\pi k}{T})) \]

\(X(e^{j\hat{\omega}})\) is a frequency-scaled version of \(X_s(j\omega)\) with the frequency scaling specified by \(\hat{\omega} =\omega T\)

Ref. 9.5 DTFT connection with the CTFT

image-20240831154638540

Here, \(\Omega = \omega T\)

The factor \(\frac{1}{T}\) in \(X(e^{j\hat{\omega}})\) is misleading, actually \(x[n]\) is not scaled by \(\frac{1}{T}\) when taking \(\hat{\omega}\) variable of integration into account \[\begin{align} x_r[n] &= \frac{1}{2\pi} \int_{2\pi}X(e^{j\hat{\omega}})e^{j\hat{\omega} n}d\hat{\omega} \\ &= \frac{1}{2\pi}\int_{2\pi}\frac{1}{T}\sum_{k=-\infty}^{+\infty}X_c \left[ j\left(\frac{\hat{\omega}}{T} - \frac{2\pi k}{T}\right)\right] e^{j\hat{\omega} n}d\hat{\omega} \\ &\approx \frac{1}{2\pi}\frac{1}{T}\int_{2\pi}X_c (\frac{\hat{\omega}}{T} ) e^{j\hat{\omega} n} d\hat{\omega} \\ &=\frac{1}{2\pi} \frac{1}{T}\int_{2\pi} \left[ \int_{\infty}X_c(\Phi)\delta (\Phi - \frac{\hat{\omega}}{T} )d\Phi \right] e^{j\hat{\omega} n} d\hat{\omega} \\ &=\frac{1}{2\pi} \frac{1}{T} \int_{\infty}X_c(\Phi)d\Phi \int_{2\pi}\delta (\Phi - \frac{\hat{\omega}}{T} )e^{j\hat{\omega} n} d\hat{\omega} \\ &=\frac{1}{2\pi} \frac{1}{T} \int_{\infty}X_c(\Phi)d\Phi \int_{2\pi}T\cdot \delta (\Phi T - \hat{\omega} )e^{j\hat{\omega} n} d\hat{\omega} \\ &=\frac{1}{2\pi} \int_{\infty}X_c(\Phi) e^{j\Phi T n}d\Phi \end{align}\]

That is \[\begin{align} x_r[n] &= \frac{1}{2\pi}\int_{2\pi} \frac{1}{T}X_c (\frac{\hat{\omega}}{T} ) e^{j\hat{\omega} n} d\hat{\omega} \\ &= \frac{1}{2\pi} \int_{\infty}X_c(\omega) e^{j\omega T n}d\omega \tag{31} \end{align}\]

assuming Nyquist–Shannon sampling theorem is met

\[\begin{align} x_r[n] &= \frac{1}{2\pi} \int_{\infty}X_c(\omega) e^{j\omega T n}d\omega \\ &= \frac{1}{2\pi} \int_{\infty}X_c(\omega) e^{j\omega t_n}d\omega \\ &= x_c(t_n) \end{align}\]

where \(t_n = T n\), then \(x_r[n] = x_c(nT)\)

Assuming \(x_c(t) = \cos(\omega_0 t)\), \(x_s(t)= \sum_{n=-\infty}^{\infty}x_c(nT)\delta(t-nT)\) and \(x[n]=x_c(nT)\), that is \[\begin{align} x_c(t) & = \cos(\omega_0 t) \\ x_s(t) &= \sum_{n=-\infty}^{\infty}\cos(\omega_0 nT)\delta(t-nT) \\ x[n] &= \cos(\omega_0 nT) \end{align}\]

  • \(X_c(j\omega)\), the Fourier Transform of \(x_c(t)\) \[ X_c(j\omega) = \pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)] \]

  • \(X(e^{j\hat{\omega}})\), the the discrete-time Fourier transform (DTFT) of the sequence \(x[n]\) \[ X(e^{j\hat{\omega}}) =\sum_{k=-\infty}^{+\infty}\pi[\delta(\hat{\omega} - \hat{\omega}_0-2\pi k) + \delta(\hat{\omega} + \hat{\omega}_0-2\pi k)] \]

  • \(X_s(j\omega)\), the Fourier Transform of \(x_s(t)\) \[ X_s(j\omega)= \frac{1}{T}\sum_{k=-\infty}^{+\infty}\pi[\delta(\omega - \omega_0-k\omega_s) + \delta(\omega + \omega_0-k\omega_s)] \]

Express \(X(e^{j\hat{\omega}})\) in terms of \(X_s(j\omega)\) and \(X_c(j\omega)\) \[ X(e^{j\hat{\omega}}) = \frac{1}{T}\sum_{k=-\infty}^{+\infty}\pi[\delta(\frac{\hat{\omega}}{T} - \omega_0-k\omega_s) + \delta(\frac{\hat{\omega}}{T} + \omega_0-k\omega_s)] \] Inverse \(X(e^{j\hat{\omega}})\) \[\begin{align} x_r[n] &= \frac{1}{2\pi} \int_{2\pi}X(e^{j\hat{\omega}}) e^{j\hat{\omega} n} d\hat{\omega} \\ &= \frac{1}{2\pi}\int_{2\pi} \pi[\delta(\frac{\hat{\omega}}{T} - \omega_0) + \delta(\frac{\hat{\omega}}{T} + \omega_0)]e^{j\hat{\omega} n} d\frac{\hat{\omega}}{T} \\ &= \frac{1}{2\pi}\int_{2\pi} \pi[\delta(\frac{\hat{\omega}}{T} - \omega_0)e^{j\hat{\omega}_0 n} + \delta(\frac{\hat{\omega}}{T} + \omega_0)e^{-j\hat{\omega}_0 n}] d\frac{\hat{\omega}}{T} \\ &= \frac{1}{2}[ e^{j\hat{\omega}_0 n}\int_{2\pi} [\delta(\frac{\hat{\omega}}{T} - \omega_0)d\frac{\hat{\omega}}{T} + e^{-j\hat{\omega}_0 n}\int_{2\pi} [\delta(\frac{\hat{\omega}}{T} + \omega_0)d\frac{\hat{\omega}}{T}] \\ &= \frac{1}{2}[ e^{j\hat{\omega}_0 n} + e^{-j\hat{\omega}_0 n} ] \\ &= \cos(\hat{\omega}_0 n) \end{align}\]

or follow EQ.(31)

\[\begin{align} x_r[n] &= \frac{1}{2\pi} \int_{\infty}X_c(\omega) e^{j\omega T n}d\omega \\ &= \frac{1}{2\pi} \int_{\infty} \pi[\delta(\omega - \omega_0) + \delta(\omega + \omega_0)]e^{j\omega T n}d\omega \\ &= \frac{1}{2}(e^{j\omega_0 T n}+e^{-j\omega_0 T n}) \\ &= \cos(\hat{\omega}_0 n) \end{align}\]

where \(\hat{\omega}_0 = \omega_0 T\)

D/C

image-20240831161852787

image-20240831162625943

image-20240831162559492

zero padding

This option increases \(N_0\), the number of samples of \(x(t)\), by adding dummy samples of 0 value. This addition of dummy samples is known as zero padding.

We should keep in mind that even if the fence were transparent, we would see a reality distorted by aliasing.

Zero padding only allows us to look at more samples of that imperfect reality

Transfer function

sampled impulse response

The below equation demonstrates how to obtain continuous Fourier Transform from DTFT . \[ X_c(\omega) = T \cdot X(\omega) \]

\(T\) is sample period, follow previous equation

useful functions

  • using fft

    The outputs of the DFT are samples of the DTFT

  • using freqz

    modeling as FIR filter, and the impulse response sequence of an FIR filter is the same as the sequence of filter coefficients, we can express the frequency response in terms of either the filter coefficients or the impulse response

    fft is used in freqz internally

Example

Question:

How to obtain continuous system transfer function from sampled impulse

Answer:

using above mentioned functions

First order lowpass filter with 3-dB frequency 1Hz

image-20220501020004068

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clear all;
clc;

%% continuous system
s = tf('s');
h = 2*pi/(2*pi+s);
[mag, phs, wout] = bode(h);
fct = wout(:)/2/pi;
Hct_dB = 20*log10(mag(:));


fstep = 0.01;           % freq resolution
fnyqst = 100;
Ts = 1/(2*fnyqst);
Fs = 1/Ts;              % sampling freq
Ns = ceil(Fs/fstep);    % samping points
fstep = Fs/Ns;          % update fstep
t = (0:Ns-1)*Ts;        % sampling time points

y = impulse(h, t);      % impulse resp

%% modelling as discrete system
Y = fft(y);                 % dft
Hfft = Y * Ts;              % !!! multiply Ts
Hfft_dB = 20*log10(abs(Hfft(1:Ns/2+1)));
ffft = (1:Ns/2+1)*fstep - fstep;


[Hfir, ffir] = freqz(y, 1, [], 1/Ts);   % modelling as FIR
Hfir = Hfir * Ts;           % !!! multiply Ts
Hfir_dB = 20*log10(abs(Hfir));

%% plot
semilogx(fct, Hct_dB, 'k', ffft, Hfft_dB, 'r.-', ffir, Hfir_dB, 'b--');
legend('bode(s)', 'fft', 'FIR model')
xlabel('Freq(Hz)');
ylabel('dB');
xlim([1e-2 1e2]);
grid on;
title('frequency response of different methods');

Gotcha

A remarkable fact of linear systems is that the complex exponentials are eigenfunctions of a linear system, as the system output to these inputs equals the input multiplied by a constant factor.

  • Both amplitude and phase may change
  • but the frequency does not change

For an input \(x(t)\), we can determine the output through the use of the convolution integral, so that with \(x(t) = e^{st}\) \[\begin{align} y(t) &= \int_{-\infty}^{+\infty}h(\tau)x(t-\tau)d\tau \\ &= \int_{-\infty}^{+\infty} h(\tau) e^{s(t-\tau)}d\tau \\ &= e^{st}\int_{-\infty}^{+\infty} h(\tau) e^{-s\tau}d\tau \\ &= e^{st}H(s) \end{align}\]

Take the input signal to be a complex exponential of the form \(x(t)=Ae^{j\phi}e^{j\omega t}\)

\[\begin{align} y(t) &= h(t)*x(t) \\ &= H(j\omega)Ae^{j\phi}e^{j\omega t} \end{align}\]

The frequency response at \(-\omega\) is the complex conjugate of the frequency response at \(+\omega\), given \(h(t)\) is real

\[\begin{align} H^*(t) &= \left(\int_{-\infty}^{+\infty}h(t)e^{-j\omega t}dt\right)^* \\ &= \int_{-\infty}^{+\infty}h^*(t)e^{+j\omega t}dt \\ &= \int_{-\infty}^{+\infty}h(t)e^{-j(-\omega t)}dt \\ &= H(-j\omega) \end{align}\]

The real cosine signal is actually composed of two complex exponential signals: one with positive frequency and the other with negative \[ cos(\omega t + \phi) = \frac{e^{j(\omega t + \phi)} + e^{-j(\omega t + \phi)}}{2} \]

The sinusoidal response is the sum of the complex-exponential response at the positive frequency \(\omega\) and the response at the corresponding negative frequency \(-\omega\) because of LTI systems's superposition property

  • input: \[\begin{align} x(t) &= A cos(\omega t + \phi) \\ &= \frac{1}{2}Ae^{\phi}e^{\omega t} + \frac{1}{2}Ae^{-\phi}e^{-\omega t} \end{align}\]

  • output with \(H(j\omega)=Ge^{j\theta}\): \[\begin{align} y(t) &= H(j\omega)\frac{1}{2}Ae^{\phi}e^{\omega t} + H(-j\omega)\frac{1}{2}Ae^{-\phi}e^{-\omega t} \\ &= Ge^{j\theta}\frac{1}{2}Ae^{\phi}e^{\omega t} + Ge^{-j\theta}\frac{1}{2}Ae^{-\phi}e^{-\omega t} \\ &= GAcos(\omega t + \phi + \theta) \end{align}\]

Its phase shift is \(\theta\) and gain is \(G\), which is same with \(H(j\omega)\).

reference

Alan V Oppenheim, Ronald W. Schafer. Discrete-Time Signal Processing, 3rd edition

B.P. Lathi, Roger Green. Linear Systems and Signals (The Oxford Series in Electrical and Computer Engineering) 3rd Edition

Alan V. Oppenheim, Alan S. Willsky, and S. Hamid Nawab. 1996. Signals & systems (2nd ed.)

James H. McClellan, Ronald Schafer, and Mark Yoder. 2015. DSP First (2nd. ed.). Prentice Hall Press, USA