BME2301 Biomedical Signals and Systems (I) (EN)

Intro

Systems are continuous-time systems or discrete-time systems, this is depended on the type of the inpust signals the system gets. Since most digital signals are discrete now, the systems today are likely to be discrete-time systems This course is mainly focused of linear time-invariant systems

Basic signals

transformation of independent variable

• time scaling x(t) -> x(t/a) (discrete signals often are not "squeezable" though extendable)
• time shift
• time reversal(actually a special form of scaling)

transformation of discrete signals in the y-axis

(for continous ones just refer to calculus) Fisrt difference: y[n] = x[n] - x[n-1] DT Unit Step Signal -> first difference/Pointwise subtraction-> DT Unit Impulse Signal Running sum of a DT signal: $\sum{}x[k]$ If you first take finite dif, and then the running sum, we don‘t get the same signal(+C)

C(continous)T(time) sinusoidal signals

A sinusoidal signal that is derived from a standard cosine function through time scaling/shifting.etc $x(t) = A \cos(\omega t + \Phi)$

properties

1. periodicity: x(t) equals itself after a time shift of T₀
2. a time shift is equivalent to a phase change $\forall t_0, \exist \Phi_0, \cos(\omega(t - t_0) + \Phi) = \cos(\omega t + \Phi + \Phi _0), vice\space versa$
3. symmetry of sinusoidal signals

CT complex exponential signal

Algebraic definition: $x(t) = C e^{\alpha t}, C, \alpha \isin \Complex$ let $C = |C| e^{j \Phi}$, $\alpha = \gamma + j \omega_0$ and we get

The rectangular/Cartesian form of complex exponential signal

Signal decomposition

almost all real signals can be represented by linear combinations of sinusoidal signals. almost all complex signals can be represented by linear combinations of complex exponential signals.

D(discrete)T(time) complex sinusoidal signal

(lost) properties

1. phase change != time shift(reverse holds)
2. not neccessarily periodic(only when $2 \pi / \Omega_0$)
3. different frequencies may not imply different signals

DT real exponential

Algebraic definition: $x[N] = C \alpha^n, C, \alpha \isin \real, n \isin \N$ can only be seen as the direct sampling of CT real exponential ($Ce^{\beta t}$) when $\alpha > 0$

DT complex exponential

Algebraic definition: $x[n] = C \alpha^n, C, \alpha \isin \Complex$ Rectangular form: $x[n] = |C| |\alpha|^n \cos(\Omega_0 n + \varphi) + j |C| |\alpha|^n \sin(\Omega_0 n + \varphi)$ it is a direct sampling of CT complex exponential signals

Energy and power of signals

energy:

power

where | | means absolute value for real and magnitude for complex

Unit step & unit impulse

DT

step: u[n] = 1 if n>=0 else 0 $u[n]\space or \space u(t)$ impulse: u[n] = 1 if n==0 else 0 $\delta[n]\space or\space \delta(t)$ unit impulse is the first dif of the unit step, unit step is the running sum of unit impulse

CT

it will be ill-defined if we just copy the discrete definitions into the continous field.

Further Introduction to Systems

Equations for descriptions of systems

systems are not normal functions...or to be more precise, they take a function as a variable, and output another...

• Time-shifting sys: $y(t)=x(t-T)$
• Time-scaling sys: $y(t)=x(at)$
• Squarer: $y(t)=(x(t))^2$
• Differentiator: $y(t)= \frac{dx(t)}{dt}$
• Integrator: $y(t)=\int_{-\inf}^{t}x(\tau)d\tau$

System interconnections

• Series interconnection/cascaded interconnection: order (generally) matters
• Parallel interconnection: order does not matter
• Feedback interconnectoin: order matters

Systems properties

System properties are partial knowledge about the input/output relationship: Memoryless, Causality, Invertibility, Stability, Linearity, Time-invariance(The first two are time-dependent and the rest are not)

• Time-dependent properties: it only considers how systems behave along the time dimension
• Time-independent properties: it only considers how systems behave along the signal dimension(about matching one input into another)

Properties

• Memoryless property:
  • it only depends on the current time point
• Causality property:
  • it only depends on previous knowledge
  • second order DT differentiator(y[n] = x[n-1] + x[n+1] - 2x[n]) is not causality
• Invertibility:
  • there is only one input signal for every output signal
  • Running integral have invertibility
• Stability:
  • if a small input generates a small output(bounded input generates bounded output, or BIBO)
  • if x(t) is bounded , there is a positve m that |x(t)| < m for all t
  • squarers are stable
  • A running integral/sum is not stable(note that the bound doesn't apply to t, but to x(t))

Convolution

Convolution only exists on linear and time-invariant systems, it mathematically describes how to predict the output for linear and time-invariant systems, it can be implemented by both hand calculation and computers.

• Linearity: defination
  • x₁ -> y₁, x₂ -> y₂, then x₁ + x₂ = y₁ + y₂
  • ax₁(t) -> ay₁(t) for any complex number a
  • (a combined version) a₁x₁ + a₂x₂ -> a₁y₁ + a₂y₂

Systems with the form $y(t) = L\{x(t)\} + y_0(t)$, where $L\{\cdot\}$ represents a linear system, and $y_0(t)$ represents a signal independent of $x(t)$, are called \textit{incrementally linear systems} (to indicate they are linear systems except for addition of another fixed signal), where y₀ is known as the zero-input response

• Time-invariance: aka "shift-invariance"
  • x(t - t₀) -> y(t - t₀) (similar with DT ones)
• $x[n] = \sum_{k=-\infty}^{\infty} x[k]\delta[n-k]$ (a sum of weighted, shifted unit impulses)

Convolution sum&integral

DT LTI systems Let $h[n]$ represent the unit impulse response: $\delta[n] \to h[n]$. for any LTI:

Apply this property to

we have:

however, this wont work with CT systems, for CT it is:

$x(t) \to \int_{-\infty}^{+\infty} x(\tau) h(t - \tau) d\tau = x(t) * h(t)$

we see the change is that it is not just the response times the pulse(since the pulse is infinity, we times an additional d$\tau$, which I suppose can be understood as... the duration?

for DT it is conv sum, and for CT it is intergral, but most of the time we just call it convolution in general.

Some properties

$x(t) * \delta(t-T) = x(t-T)$

$x(t)\delta(t-T) = x(T)\delta(t-T)$

$\int_{-\infty}^{\infty} x(t)\delta(t-T)dt = x(T)$

Mathematical Properties:

• x[n] * h[n] = h[n] * x[n] (commutative property)
• x * {h₁ * h₂} = {x * h₁} * h₂ (associative property) Above makes two cascading LTI systems switchable
• if x(t) -> y(t) then dx(t)/dt -> dy(t)/t (differentiation property) It allow syou to predict the output of a signal(e.a if you know the response of a step signal you know what the response of the impulse signal is like)

Implementation of convolution

RSMS/RSMI procedure

• Reverse $h[k]$, generating $h[-k]$
• Shift the signal $h[-k]$ by $n_0$, generating $h[-(k-n_0)] = h[n_0 - k]$
• Multiply $x[k]$ with $h[n_0 - k]$, generating $x[k]h[n_0 - k]$
• Sum from $k = -\infty$ to $k = +\infty$, generating $y[n_0] = \sum_{k=-\infty}^{\infty} x[k]h[n_0 - k]$ for CT signals it's basically the same, except the last step is integral instead of sum

Use of convolution

Convolution can be used to prove some properties of a system in a more straightforward way

Memoryless?

• Given a system is LTI, the system is also memoryless iff

  $y(t_0) = \int_{-\infty}^{\infty} x(\tau) h(t_0 - \tau) d\tau \text{ depends only on } x(t_0)$

  which is equivalent to say $h(t_0 - \tau) \neq 0$ only if $\tau = t_0$, that is, $h(t) \neq 0$ only if $t = 0$.

• Without loss of generality, $h(t) \neq 0$ only if $t = 0$ means:

  $h(t) = k\delta(t) \text{ for some } k \neq 0$

  Since $x(t) * k\delta(t) = kx(t)$, the only memoryless LTI system is the rescaling system.

Causality?

• General definition: For any $x(t)$, its output $y(t)$, and any time index $t_0$, $y(t_0)$ is dependent only on $x(t)$ for $t \le t_0$

• Given a system is LTI, the above condition means

  $y(t_0) = \int_{-\infty}^{\infty} x(\tau)h(t_0 - \tau)d\tau$ depends only on $x(t)$ for $t \le t_0$

• That is, $h(t_0 - \tau) = 0$ for $t > t_0$. That is, $h(t) = 0$ for $t < 0$

• In other words, an LTI system is causal if

  $h(t) = 0 \text{ for } t < 0$

  Note this is much easier to evaluate than the original definition of causality!

"Initial Zero"

An initial-rest system generates nonzero output values only after the input signal becomes nonzero.

a causal or a memoryless system doesn't necessarily have the initial rest property. But a LTI system have causality if it has initial rest, the reverse also applies.

Stability?

• Recall the definition of stability: bounded input generates bounded output (BIBO)

• Assuming LTI, BIBO holds if and only if the impulse response is absolutely summable (integrable) (integration/sum not infinity)

Invertibility?

• Recall the definition of invertibility: a system A is invertible if and only if different inputs always generate different outputs.

• When A is invertible, there exists another system A₁, such that the cascaded system (A- A₁) is an identity system.

LCCDE Systems

Linear Constant-Coefficient Differential Equations Systems.

From LCCDE to LCCDE systems

$\sum_{k=0}^{N} a_k \frac{d^k y(t)}{dt^k} = \sum_{k=0}^{M} b_k \frac{d^k x(t)}{dt^k}$

The order of LCCDE refers to the highest derivative of the output y(t) (that is, N).

solving it: homogeneous solution + particular solution

LCCDE have infinite solutions, but LCCDE systems don't, because LCCDE systems have initial state

We can manully select a time, and make it the starting point, this starting point might not have a zero response, and the response the moment before the starting point are seen as the initial state.

LTI and LCCDE systems

Zero-state response: response of the system when there is no initial state = response of the system to the input signal

Zero-input response: response of the system when there is no input = response of the system to the initial state

Solving LCCDE Equations

Solving the solution directly

Solve a 1st order LCCDE: $\frac{dy(t)}{dt} + 2y(t) = e^{3t}u(t)$

Particular solution of this LCCDE strictly satisfies the equation.

• A general form of solution to $\frac{dy_p(t)}{dt} + \alpha y_p(t) = e^{\beta t}u(t)$ is $y_p(t) = K(e^{\beta t} - e^{-\alpha t})u(t)$
• Plugging in the values, we get $\frac{d}{dt} [K(e^{3t} - e^{-2t})u(t)] + 2K(e^{3t} - e^{-2t})u(t) = e^{3t}u(t)$.
• Solving for K, we find $K = \frac{1}{5}$.
• Thus, the particular solution is: $y_p(t) = \frac{1}{5}(e^{3t} - e^{-2t})u(t)$.

Homogeneous solution is the function $y_h(t)$ satisfying $\frac{dy(t)}{dt} + 2y(t) = 0$.

• Homogeneous solutions are of the form $y_h(t) = Ae^{at}$.

• Substituting into the homogeneous equation gives $Aae^{at} + 2Ae^{at} = (a+2)Ae^{at} = 0$. Thus, $a = -2$ and $y_h(t) = Ae^{-2t}$ is the homogeneous solution.

• Note that $A$ can be any number, and the homogeneous equation will still be satisfied.

• There is an infinite number of homogeneous solutions, each in the form $y_h(A, t) = Ae^{-2t}$ for some number $A$.

• The complete solution to the LCCDE is the sum of the particular and homogeneous solutions: $y(t) = y_p(t) + y_h(A, t)$.

for this example, that is: $y(t) = \frac{1}{5}(e^{3t} - e^{-2t})u(t) + Ae^{-2t}$.

• Every LCCDE has infinitely many solutions, each expressible by a sum of the particular solution and one homogeneous solution.

Classical Approach

Wiped out the u(t) and assume it is only in the positive time domain(which is often the case in real life) and solve it in the (I suppose is more like the) normal and intentional way

Take the previous example:

• Modify the equation to $\frac{dy(t)}{dt} + 2y(t) = e^{3t}$, $t > 0$. [$u(t)$ dropped & $t>0$ added]

• The particular solution for the modified input must be of the form $Ke^{3t}$. That leads to $y_p(t) = \frac{1}{5}e^{3t}$ (Instead of $\frac{1}{5}(e^{3t} - e^{-2t})u(t)$ for the original equation)

• The homogeneous solution is still $Ae^{-2t}$. $A$ is to be determined by the initial state.

• The complete solution for the modified input is thus $\frac{1}{5}e^{3t} + Ae^{-2t}$, $t > 0$[Initial state is NOT directly applicable].

• Then, we need to use the continuity assumption about the system output, that is, the output signal is a continuous function at any time (true for most realistic systems).

• By the continuity assumption, $y(0^+) = y(0^-) = \gamma$. Then $(\frac{1}{5}e^{3t} + Ae^{-2t})|_{t=0^+} = \gamma \implies \frac{1}{5} + A = \gamma \implies A = \gamma - \frac{1}{5}$.

• Thus, the homogeneous solution for the classical approach is $(\gamma - \frac{1}{5})e^{-2t}$ (Instead of $\gamma e^{-2t}$ for the original equation)

• The total output is then $\frac{1}{5}e^{3t} + (\gamma - \frac{1}{5})e^{-2t} = \frac{1}{5}(e^{3t} - e^{-2t}) + \gamma e^{-2t}$ for $t > 0$. (the total output is equal to that derived from the original equation)

Fourier Series Representation of Periodic Signals

Before we are using impulses as the building block to represent signals (especially in convolution), but the properties of the impulse make it quite complicated to represent some signals, for example, for $x(t) = \cos(\omega_0 t)$ with impulse it can be represented as $x(t) = \int_{-\infty}^{\infty} \cos(\omega_0 \tau) \delta(t - \tau) d\tau$ with complex exponentials it is $x(t) = \frac{1}{2} (e^{j\omega_0 t} + e^{-j\omega_0 t})$ which obviously looks better than the impulse solution

HRCE(Harmonically Related Complex Exponentials)

A set of signals, $\{ \phi_k(t) | k \in \mathbb{Z} \}$, is called a set of Harmonically Related Complex Exponentials if and only if

Where $\omega_0 \neq 0$ is called the fundamental frequency for the set of harmonically related complex exponentials.

• Harmonically Related Complex Exponentials (HRCE) is a set of complex exponential signals.
• These exponential signals are "harmonically related", in that the frequency of $\phi_k(t)$ is $k\omega_0$ ($k$-fold multiple of the fundamental frequency).
• A set of HRCE is known iff its fundamental frequency is known.
• $k$ can be both positive or negative integers.
• they share a common period(T₀)

The Fourier Series Representation

• $a_k$ is called the Fourier series coefficient or the spectral coefficient of $x_{T_0}(t)$.
• $a[k] = \{a_k\}$ is called the Fourier series spectrum of $x_{T_0}(t)$.
• Equation (1) is called the synthesis equation: almost every periodic signal can be synthesized/represented by a linear combination of harmonics in the HRCE.
• Equation (2) is called the analysis equation (spectral analysis): it tells you how to generate the coefficients for the FS representation.

We also have the "Partial Sum":

It's easy to recognize that $\lim_{N \to +\infty} x_N(t) = x(t)$

Displaying The FS Spectrum

• When we have the Fourier series spectrum, we often want to plot the spectrum, because it reflects the frequency components contained in $x_{T_0}(t)$.
• However, notice the Fourier series coefficients are complex numbers, so we need at least two plots to show the spectrum (magnitude & phase).

For example: let's display the FS Spectrum of the following signal

Collecting terms, we obtain

Thus, the Fourier series coefficients for this example are

to polar form

and we will get two plots: both parts are critical for describing a signal, if a signal has shifted in the time domain, the magnitude spectrum will very likely to remain unchanged while the phase spectrum will have changes(and it's not simply a shift in the whole pattern)

Spectrum Analysis for LTI Systems

Eigenfunction

Definition: for a given LTI system, an input signal that generates an output signal that is simply the input multiplied with a constant is called an eigenfunction of the system.

If $x(t) * h(t) = \lambda x(t)$, then $x(t)$ is an eigenfunction of the system $h(t)$, and $\lambda$ is the eigenvalue associated with the eigenfunction. (quite similar for definition for the eigen-things in linear algebra)

complex exponentials are always eigenfunctions of an LTI system.

A New Way of Analyzing LTI Systems

Let $x(t) = e^{j\omega t}$, then the output is $y(t) = \int_{-\infty}^{\infty} x(t-\tau)h(\tau)d\tau = \int_{-\infty}^{\infty} e^{j\omega(t-\tau)}h(\tau)d\tau = e^{j\omega t} \int_{-\infty}^{\infty} e^{-j\omega\tau}h(\tau)d\tau$

Since $\int_{-\infty}^{\infty} e^{-j\omega\tau}h(\tau)d\tau$ is not dependent on $t$, we can write the output as $y(t) = T\{e^{j\omega t}\} = H(j\omega)e^{j\omega t}$ Where $H(j\omega) \triangleq \int_{-\infty}^{\infty} e^{-j\omega\tau}h(\tau)d\tau$ The eigenvalue associated with $e^{j\omega t}$. It is dependent on the frequency $\omega$, the impulse response, but not on $t$.

Fourier series representation decomposes a signal into a linear combination of harmonically related eigenfunction signals. $x_{T_0}(t) = \sum_{k=-\infty}^{+\infty} a_k e^{jk\omega_0 t}$

Then the response of an LTI system to an arbitrary periodic signal would be $y(t) = \sum_{k=-\infty}^{\infty} a_k \mathcal{T}\{e^{jk\omega_0 t}\} = \sum_{k=-\infty}^{\infty} a_k [H(jk\omega_0)e^{jk\omega_0 t}]$ $= \sum_{k=-\infty}^{\infty} [a_k H(jk\omega_0)]e^{jk\omega_0 t} = \sum_{k=-\infty}^{\infty} b_k e^{jk\omega_0 t}$ Output is a periodic signal with the same period & a spectrum of $b_k = a_k H(jk\omega_0)$

Generally the idea is, any input signal can be broken down to a series of eigenfunctions of any LTI system, so the input can be assembled from a series of eigenfunctions, rescaled according to the corresponding eigenvalue.

Advantages

• First, computational advantages over convolution. For example: what is the output for $x(t) = \cos(\frac{2}{3}t)$ and $h(t) = \frac{\sin(t)}{\pi t}$?
  • Convolution is difficult
  • Frequency domain:
    • $H(j\omega) = \int_{-\infty}^{\infty} e^{-j\omega\tau} \frac{\sin(\tau)}{\pi \tau} d\tau = \begin{cases} 1, & |\omega| < 1 \\ 0, & \text{otherwise} \end{cases}$
    • $a_k$ is nonzero only for $k = -1 \& 1$.
    • Note that since $\omega_0 = \frac{2}{3}$, $H(jk\omega_0) = 1$ only for $k = -1, 0, 1$
• Second, it provides critical insights to help us understand how the system changes each harmonic:
  • If $|H(jk\omega_0)|$ is greater than 1, the $k^{th}$ harmonic gets strengthened in the output spectrum
  • If $|H(jk\omega_0)|$ is less than 1, the $k^{th}$ harmonic gets attenuated (weakened) in the output spectrum

The second one leads to the important concept of "filtering".

Filters

So the filters is mostly a function for H(j$\omega$)

• Type one filters(Frequency-selective filters, or ideal filters): just 1 or 0, no other values

• Type two filters(requency-shaping filters): most reshape, not completely kill or live (actually this is the filter for differentiater)

Performing Fourier Transform on Aperiodic Functions

Fourier transform

previously Analysis Equation $X(j\omega) = \int_{-\infty}^{\infty} x(t)e^{-j\omega t} dt$

Inverse Fourier transform

previously Synthesis Equation $x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega)e^{j\omega t} d\omega$

Fourier Transform Pair

If $X(j\omega) = \mathcal{F}\{x(t)\}$, we say $X(j\omega)$ and $x(t)$ are a Fourier transform pair.

A well-known Fourier transform pair is the rectangle-sinc pair, that is, the Fourier transform of a rectangle signal is a "sinc" signal.

This pair is very important because we often need to analyze the FT of a rectangular signal or IFT of a rectangular spectrum.

Fourier transform pairs are completely invertible, for example, a "sinc" in time domain will be a "rectangle signal" when transformed to the frequency domain.

Properties of Fourier Transform

Performing Fourier Transform on Periodic Signals

To find the Fourier Transform (F.T.) spectrum of a periodic signal $x_{T_0}(t)$, we are looking for a spectrum $X(j\omega)$ that satisfies:

$\qquad X(j\omega) = \int_{-\infty}^{\infty} x_{T_0}(t) e^{-j\omega t} dt \quad (1)$

$\qquad x_{T_0}(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) e^{j\omega t} d\omega \quad (2)$

Let $a_k$ be the Fourier Series (F.S.) spectrum of $x_{T_0}(t)$. If Equation (2) holds, we must have:

$\qquad \frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) e^{j\omega t} d\omega = \sum_{k=-\infty}^{\infty} a_k e^{jk\omega_0 t}$

The left side is a "sum" over all frequencies, while on the right is sum over a discrete set of frequencies

So X(j$\omega$) is some function times $a_k$, and the function is only non-zero on certain frequencies, which is an impulse train.

The equation $\frac{1}{2\pi} \int_{-\infty}^{\infty} X(j\omega) e^{j\omega t} d\omega = \sum_{k=-\infty}^{\infty} a_k e^{jk\omega_0 t}$ indicates that $X(j\omega)$ is an impulse train located at the harmonic frequencies $k\omega_0$.

That is The Fourier Transform for Periodic Signals

$X(j\omega) \triangleq 2\pi \sum_{k=-\infty}^{\infty} a_k \delta(\omega - k\omega_0)$

Linearity

For sure

Time shift

Fourier transform: $\text{If } x(t) \overset{\mathcal{F}}{\longleftrightarrow} X(j\omega), \text{ then } x(t - t_0) \overset{\mathcal{F}}{\longleftrightarrow} e^{-j\omega t_0} X(j\omega)$

which is generally the same with

Fourier series:

$\text{If } x(t) \overset{\text{F.S.}}{\longrightarrow} a_k, \text{ then } x(t - t_0) \overset{\text{F.S.}}{\longrightarrow} e^{-jk\omega_0 t_0} a_k$

Time reversal

Fourier transform: $x(-t) \overset{\mathcal{F}}{\longleftrightarrow} X(-j\omega)$

which is still generally the same with

Fourier series: $x(-t) \overset{\text{F.S.}}{\longrightarrow} a_{-k}$

Two indications(Also works with fourier series):

• If $x(t)$ is even, then $X(j\omega)$ is even, and vice versa.
• If $x(t)$ is odd, then $X(j\omega)$ is odd, and vice versa.

Conjugacy & Symmetry

• Fourier transform: $\qquad \text{If } x(t) \overset{\mathcal{F}}{\longleftrightarrow} X(j\omega), \text{ then } x^*(t) \overset{\mathcal{F}}{\longleftrightarrow} X^*(-j\omega)$

which is also the same with

• Fourier series: $\qquad \text{If } x(t) \overset{\text{F.S.}}{\longrightarrow} a_k, \text{ then } x^*(t) \overset{\text{F.S.}}{\longrightarrow} a_{-k}^*$

• subcases

  • If $x(t)$ is real, $X(j\omega) = X^*(-j\omega)$ "conjugate symmetric". => Real part is even, and imaginary part is odd.

  • if $x(t)$ is real and even, $X(j\omega)$ is real and even

  • if $x(t)$ is real and odd, $X(j\omega)$ is imaginary and odd.

  • A new indication not mentioned before:

    • Since a real signal $x(t) = x_e(t) + x_o(t)$, $X(j\omega) = \mathcal{F}\{x_e(t)\} + \mathcal{F}\{x_o(t)\} =$ real spectrum + imaginary spectrum.
    • Meanwhile we also have $X(j\omega) = Re\{X(j\omega)\} + j \cdot Im\{X(j\omega)\}$. This means:

    $\qquad Re\{X(j\omega)\} = \mathcal{F}\{x_e(t)\}; \qquad jIm\{X(j\omega)\} = \mathcal{F}\{x_o(t)\};$

below here the properties that start to change

Differentiation and Integration

• Differentiation property of Fourier transform:

  $\text{If } x(t) \overset{\mathcal{F}}{\longleftrightarrow} X(\omega), \text{ then } \frac{dx(t)}{dt} \overset{\mathcal{F}}{\longleftrightarrow} j\omega X(\omega)$

  • The same as for the Fourier series: If $x(t) \overset{\text{F.S.}}{\longrightarrow} a_k$, then

    $\qquad \frac{dx(t)}{dt} \overset{\text{F.S.}}{\longrightarrow} jk\omega_0 a_k$

  • Indication: differentiator is a highpass filter (non-ideal)

• Integration (Different!)

  • If $x(t) \overset{\mathcal{F}}{\longleftrightarrow} X(j\omega)$, then $\int_{-\infty}^{t} x(\tau)d\tau \overset{\mathcal{F}}{\longleftrightarrow} \frac{1}{j\omega}X(j\omega) + \pi X(j0)\delta(\omega)$

  • Fourier series: Suppose $x(t) \overset{\text{F.S.}}{\longrightarrow} a_k$, and $a_0 = 0$, then $\int_{-\infty}^{t} x(\tau)dt \overset{\text{F.S.}}{\longrightarrow} \frac{a_k}{jk\omega_0}$

  • Why this happens? so this is actually the same with the fourier series, but in series we can explicitly say $a_0$ = 0, but this is often not the case for fourier transform, so we have to add this term, which is actually the area under the curve.

Parseval's Theorem

• Fourier transform

  • If $x(t) \leftrightarrow X(j\omega)$ then
    $\int_{-\infty}^{\infty} |x(t)|^2 dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |X(j\omega)|^2 d\omega$

• Fourier series

  • Suppose $x(t) \overset{\text{F.S.}}{\longrightarrow} a_k$
    then $\frac{1}{T_0} \int_{T_0} |x(t)|^2 dt = \sum_{k=-\infty}^{\infty} |a_k|^2$

Time scaling

• If $x(t) \overset{\mathcal{F}}{\longleftrightarrow} X(j\omega)$, then $x(at) \overset{\mathcal{F}}{\longleftrightarrow} \frac{1}{|a|}X(\frac{j\omega}{a})$ for any $a \in \mathbb{R}$ and $a \neq 0$.

  • If $|a| > 1$, $x(t)$ is squeezed to generate $x(at)$, $X(j\omega)$ is stretched and downward scaled to generate $\frac{1}{|a|}X(\frac{j\omega}{a})$.
  • If $|a| < 1$, $x(t)$ is stretched to generate $x(at)$, $X(j\omega)$ is squeezed and upward scaled to generate $\frac{1}{|a|}X(\frac{j\omega}{a})$.

• The scaling of the signal is opposite to the scaling of the spectrum.

  • Squeezing of signal causes faster variation, thus more high frequency components.
  • Stretching of signal causes slower variation, thus more low frequency components.

• E.g. $\text{rect}(t) \overset{\mathcal{F}}{\longleftrightarrow} 2\text{sinc}(\frac{\omega}{\pi}) \implies \text{rect}(\frac{t}{T_1}) \overset{\mathcal{F}}{\longleftrightarrow} 2|T_1|\text{sinc}(\frac{T_1\omega}{\pi})$, where $\text{rect}(t) = \begin{cases} 1; & |t| < 1 \\ 0; & \text{otherwise} \end{cases}$

Duality

If $x(t) \overset{\mathcal{F}}{\longleftrightarrow} X(j\omega)$, then $X(t) \overset{\mathcal{F}}{\longleftrightarrow} 2\pi x(-j\omega)$, or $\frac{1}{2\pi}X(-t) \overset{\mathcal{F}}{\longleftrightarrow} x(j\omega)$

Convolution

recall this

now we find that this so-called frequency response is the Fourier Transform of the impulse responce, from we can derive the convolution property:

$\text{If } x(t) \stackrel{\mathcal{F}}{\longleftrightarrow} X(j\omega), y(t) \stackrel{\mathcal{F}}{\longleftrightarrow} Y(j\omega), \text{ then}$

$x(t) * y(t) \stackrel{\mathcal{F}}{\longleftrightarrow} X(j\omega)Y(j\omega)$

proof:

Multipication

$r(t) = s(t) \cdot p(t) \stackrel{\mathcal{F}}{\longleftrightarrow} R(\omega) = \frac{1}{2\pi} [S(\omega) * P(\omega)]$

which is actually the duality of the convolution propertyy

Predicting Outputs of a More Specific Systems

(This part is mainly based on the convolution property)

Interconnected Systems

for two systems that has the frequency response of $H_1(j \omega)$ and $H_2(j \omega)$

Cascaded Systems

very easy, $H(j \omega) = H_1(j \omega)H_2(j \omega)$

Parallel Systems

still easy, $H(j \omega) = H_1(j \omega) + H_2(j \omega)$

Feedback Systems

where things get compilcated, take below as an example $Y(j \omega) = Z(j \omega) H_1(j \omega), \space Z(j \omega) = X(j \omega) - H_2(j \omega) Y(j \omega)$

$\text{plug the second into the first: } Y(j \omega) = (X(j \omega) - H_2(j \omega)Y(j \omega))H_1(j \omega)$

$=> H(j \omega) = \frac{Y(j \omega)}{X(j \omega)} = \frac{H_1(j \omega)}{1+ H_1(j \omega)H_2(j \omega)}$

LCCDE Systems

For the LTI subsystem of a stable LCCDE system $H(\omega) = \frac{Y(\omega)}{X(\omega)} = \frac{\sum_{k=0}^{M} b_k (j\omega)^k}{\sum_{k=0}^{N} a_k (j\omega)^k}$

Example: Find impulse response of

Assuming the system is stable

Time-Frequency Analysis of LTI/LCCDE Systems

The Impact of Spectrums and Phases

The spectrum, or |H(j $\omega$)|, officially known as the gain of the system, changes the input spectrum by multiplication. $|Y(j\omega)| = |X(j\omega)||H(j\omega)|$

$\angle H(j\omega)$, formally called phase shift of the system, affects the phase of the input spectrum by addition. $\angle Y(j\omega) = \angle X(j\omega) + \angle H(j\omega)$

So we know that, if the phase is linear, it will delay the signal, which is something don't want.

And there is something we don't want more: if the phase is non-linear, then the delay for different frequencies will be different, causing distortions and errors.

The Bode Plot

While the sepctrums/phases plots we drawn before seems good and straight forward, it is not used a lot in practice, where the filters might encounter frequencies at a very wide range, and the spectrum might also change in a very wide range, and the linear X, Y axis is not enough anymore, so we introduce the bode plot. It's still plots with frequencies as X-axis and spectrum/phase as the Y-axis, but instead of linear, it's log.

Bode Plot

The X-axis of a bode plot is in a log scale, that is the first tick is a, and the next tick is 10a, next 100a, 1000a and so on, an increase of 10 fold is called a decade. The Y-axis is in a 20log10 scale, and one unit in this scale is called a db, so the slope is described as xxdb/decade.

A very good thing of the spectrum bode plot is that it transforms the multiplication to addition, now, we just need to add the signal bode plot with the filter bode plot to get the result bode plot, much easier the multiplication.

And for the bode plot of phase, while the Y-axis keeps it's original phase, the X is in a log scale, like the spectrum bode plot.

Asymptotic Bode Plots

Unfortunately the exact bode plot, like the ripples on the turning points, are not easy to plot. So we introduce the asymptotic bode plots, it focuses more on the trend and not the details, for example:

$H(j\omega) = \frac{1}{1+j \omega \tau}$

• Spectrum: $20 \text{ lg}|H(j\omega)| = -10 \text{ lg}(1 + (\omega\tau)^2) \quad$
  • When $\omega \ll 1/\tau$, $(*)\approx -10 \text{ lg } 1 = 0$
  • When $\omega \gg 1/\tau$, $(*)\approx -20 \text{ lg}(\omega\tau) = -20 \text{ lg } \omega - 20 \text{ lg } \tau$, which is a straight line with a slope of -20dB/dec and a value of $0@\omega = 1/\tau$
• Phase: $\angle H(j \omega) = -\text{atan}(\omega \tau)$
  • When $\omega \approx 0.1/\tau$, $(**)\approx -\text{atan } 0.1 \approx 0$
  • When $\omega \approx 1/\tau$, $(**)\approx -\text{atan } 1 = -\frac{\pi}{4}$
  • When $\omega \approx 10/\tau$, $(**)\approx -\text{atan } 10 \approx -\frac{\pi}{2}$

so it looks like

How to Draw Asymptotic Bode Polts

(唉懒得装模做样说洋文了这段我就用中文说吧)

反正就是要拆成常数项和$(1+j \frac{\omega}{\omega_0})^{-1或1}$, 然后分子上的每一个$\omega_0$会从$\omega_0$开始给振幅图提供+20db/decade的斜率，给相位图从$0.1\omega_0 - 10\omega_0$提供$+\frac{\pi}{4}$/decade的斜率（如果有平方项就算两次以此类推）

分母上就都是负的

然后常数项若为K，在振幅上决定纵轴截距为$20\lg{|K|}$，在相位上就是正数的话无影响负数的话纵轴截距为$\pm \pi$

Sample Theory

Signal to Sample

$p(t) = \sum_{n=-\infty}^{\infty} \delta(t-nT)$ Where T is the sampling period, then

$x_p(t) = x(t)p(t) = \sum_{-\infty}^{\infty} x(nT)\delta(t-nT)$

where p(t) is called the sampling function, and 2$\pi$/T is called the sampling rate.

Sampled Signals Analysis

first we can discuss about the specturm of the sampled signal, from the multiplication property we know that the spectrum of the signal: $X_{P} (j \omega) = \frac{1}{2 \pi} [X(j \omega) * P(j \omega)], P(j \omega) = \frac{2 \pi}{T} \delta(\omega - n \omega_s)$ $X_p(j\omega) = \ \frac{1}{T}\left[\sum_{n=-\infty}^{\infty} X(j\omega) * \delta(\omega - n\omega_s)\right] = \frac{1}{T}\sum_{n=-\infty}^{\infty} X(j(\omega - n \omega_s))$

so, the sampled signal is a series of original shifted spectrum added up together.

Now, for simplicity, we assume that the original signal is banded-limited (from $-\omega_M$ to $\omega_M$), the sample spectrum are the original spectrum, and move it to $\omega_s$ left or right, and add it to the spectrum and so on.

If we are lucky, the gap is wide enough so there is no overlap, we will have a "spectrum" train, and we just apply a low pass filter to get one, and ift it to get the very original signal back.

But there are times we are not so lucky, where the gap is not that wide and there is an overlap, then spectrums start distorting each other, and now there is no way to recover it perfectly.

So we quantify the "gap" here and we get the Nyquist-Shannon Sampling Theorem:

Let $x(t)$ be any band-limited signal with $X(j\omega) = 0$ for $|\omega| > \omega_M$. Let $\{x(nT)|n \in \mathbb{Z}\}$ be a series of its periodic samples with a sampling frequency of $\omega_s = 2\pi/T$. If and only if $\omega_s > 2\omega_M$, $x(t)$ can be uniquely determined by its samples through the following process:

• Generate the impulse-train sampled signal: $x_p(t) \triangleq \sum_{n=-\infty}^{\infty} x(nT)\delta(t-nT) = x(t)\sum_{n=-\infty}^{\infty} \delta(t-nT)$
• Apply an ideal lowpass filter with a gain of $T$ and a cutoff frequency $\omega_M < \omega_c < \omega_s - \omega_M$. The output of the lowpass filter is $x(t)$