[Signal and System] Filters & Bode Plot

文章目录[隐藏]

Filters
Bode Plot
First-order CT system
Reference

Filters

Ideal low-pass filter: zero phase

$$\begin{aligned}
h(t) &=\frac{1}{2 \pi} \int_{-\infty}^{{\infty} \mathrm{H}(j \omega) e}{j \omega t} d \omega \
&=\frac{1}{2 \pi} \int_{-\omega_{c}}^{{\omega_{c}} 1 \cdot e}{j \omega t} d \omega \
&=\left.\frac{1}{2 \pi} \cdot \frac{1}{j t} e^{j \omega t}\right|{-\omega{c}} ^{\omega_{c}} \
&=\frac{1}{2 \pi} \cdot \frac{1}{j t} \cdot 2 j \sin \left(\omega_{c} t\right) \
&=\frac{\sin \omega_{c} t}{\pi t}=\frac{\omega_{c}}{\pi} \operatorname{sinc}\left(\omega_{c} t\right) \
h(n) &=\frac{\sin \omega_{c} n}{\pi n}=\frac{\omega_{c}}{\pi} \operatorname{sinc}\left(\omega_{c} n\right)
\end{aligned}$$

$s(t)=\int_{-\infty}^{t} h(\tau) d \tau$

$s(n)=\sum_{m=-\infty}^{n} h(m)$

Ideal low-pass filter: linear phase

rotate for $-\alpha \omega$

Nonideal low-pass filter: frequencydomain

Pass band: $0-\omega_{p}$, stop band: $\omega>\omega_{s}$ transition: $\omega_{s}-\omega_{p}$
Pass-band ripple: $\delta_{1}$, stop-band ripple: $\delta_{2}$Linear (nearly) linear phase over the passband is desirable.

Nonideal low-pass filter: timedomain

Rise time: $t_{r}$ overshoot: $\Delta$
Ringing frequency: $\omega_{r}$ settling time: $t_{s}$

e.g. Nonideal low-pass filter

Fifth-order Butterworth filter and a fifth-order elliptic filter
Same cutoff frequency
Same passband and stopband rippleThere's trade-off between time-domain characteristic $t_{s}$ and fregurency domain characteristic $\omega_{s}-\omega_{p}$

Ideal vs. non-ideal filter

Gain.

The ideal filter is fixed at a gain of 1 in the passband, which means that the input signal of the passband is "completely" passed, while the gain of the band is fixed at 0, which means that the input signal of the band is "completely" filtered out.
The gain of a non-ideal filter is a function of frequency and not fixed.

Cut-off frequency.

The ideal filter passband can be switched instantaneously between the stopband and the passband.
The non-ideal filter cannot switch instantaneously between the passband and the stopband, its attenuation (gain) varies continuously with frequency, so there is a transition band, and the gain of the stopband cannot reach 0.

Bode Plot

The bode plot is for first and second-order CT system

First-order CT system

$$\begin{aligned}
\text { Differential equation: } & C \frac{d y(t)}{d t}=\frac{x(t)-y(t)}{R} \
& \tau\left(\frac{d y(t)}{d t}\right)+y(t)=x(t), \tau=R C
\end{aligned}$$

$$\begin{aligned}
\text { Frequency response: } &\tau j \omega Y(j \omega)+Y(j \omega)=X(j \omega)\
&H(j \omega)=\frac{Y(j \omega)}{X(j \omega)}=\frac{1}{\underbrace{(j \omega \tau)+1}}_{\text {First order }}
\end{aligned}$$

Basic deduction

$\tau:$ time constant
$t=\tau, h(t)=1 /(\tau e)$
$s(t)=1-1 / e$
$\tau \downarrow, h(t)$ decays more sharply
$s(t)$ rises more sharply

The reason why for the is they be approximated by the domain by the $[\frac1{10\tau},\frac{10}\tau]$ to $-\frac4\pi[log_{10}(\omega\tau)+1]$

Second-order CT system: differential equation

physical meaning

Frequency response $\frac{d^{2} y(t)}{d t}+2 \zeta \omega_{n} \frac{d y(t)}{d t}+\omega_{n}^{2} y(t)=\omega_{n}^{2} x(t)$

\[
\begin{array}{l}j \omega)^{2} Y(j \omega)+2 \zeta \omega_{n}(j \omega) Y(j \omega)+\omega_{n}^{2} Y(j \omega)=\omega_{n}^{2} X(j \omega) \\ H(j \omega)=\frac{\omega_{n}^{2}}{(j \omega)^{2}+2 \zeta \omega_{n}(j \omega)+\omega_{n}^{2}}\notag\end{array}
\]