傅里叶级数与变换
引入| English | 释义 | 拼音 |
|---|---|---|
| Fourier transform | 傅里叶变换 | fù lǐ yè biàn huàn |
| frequency response | 频率响应 | pín lǜ xiǎng yìng |
| time–bandwidth trade-off | 时间带宽权衡 | shí jiān dài kuān quán héng |
Why does a click hit every band?
- A sharp click lights up every slider on a graphic equaliser at once, from bass to treble. A held note lights up one.
- The Fourier transform explains why: short events spread across many frequencies. This lesson builds the transform and its central trade-off.
From series to transform
- A periodic signal is a sum of harmonics. Let the period grow without limit and the harmonics merge into a continuous spectrum: the Fourier transform 傅里叶变换.
- $X(\omega) = \displaystyle\int_{-\infty}^{\infty} x(t)\,e^{-j\omega t}\,dt$ and back again, $x(t) = \dfrac{1}{2\pi}\displaystyle\int_{-\infty}^{\infty} X(\omega)\,e^{j\omega t}\,d\omega$.
X(0), the Fourier transform at zero frequency, equals the total area under x(t).
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Put $\omega = 0$: $e^{0} = 1$, so $X(0) = \int x(t)\,dt$. For a pulse that is $A\tau$.
Convolution becomes multiplication
- The property the whole subject rests on: if $y = x * h$, then $Y(\omega) = X(\omega)H(\omega)$.
- $H(\omega)$ is the system's frequency response 频率响应. A filter is designed by shaping it, then multiplying, with no convolution integral.
An input with spectrum X(ω) passes through a system with frequency response H(ω). The output spectrum is
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Convolution in time is multiplication in frequency. That single property is why filters are designed in the frequency domain.
The pulse and its sinc
- A pulse of height $A$ and width $\tau$, centred at 0, transforms to $X(\omega) = A\tau\,\dfrac{\sin(\omega\tau/2)}{\omega\tau/2}$, a sinc.
- At $\omega = 0$ it equals the area $A\tau$. The first null is at $\dfrac{\omega\tau}{2} = \pi$, which is $f = \dfrac{1}{\tau}$.
A 1 ms rectangular pulse has its first spectral null at
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The first null is at $f = \dfrac{1}{\tau} = \dfrac{1}{10^{-3}} = 1$ kHz.
A pulse of height 5 V and width 2 ms. What is X(0), in mV·s?
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$X(0) = A\tau = 5 \times 2 \times 10^{-3} = 0.01$ V·s, which is 10 mV·s.
Time–bandwidth trade-off
- Shorter in time means wider in frequency. A 1 ms click has its first null at 1 kHz; a 0.1 ms click pushes it to 10 kHz.
- This time–bandwidth trade-off 时间带宽权衡 is unavoidable. That is why a click reaches every equaliser band.
A pulse is shortened from 1 ms to 50 µs. Where is its first null now, in kHz?
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$\dfrac{1}{50\,\mu\text{s}} = 20$ kHz: twenty times shorter, twenty times wider.
Worked example: the ideal filter's price
- A brick-wall low-pass passing 0 to 4 kHz has a rectangular $H$. Its impulse response is a sinc in time, $h(t) = \dfrac{\sin(\omega_{c}t)}{\pi t}$, stretching to $\pm\infty$.
- It starts before $t = 0$, so it is non-causal and cannot be built. Cut it to $|t| \le 5$ ms and the edge softens to about $\dfrac{1}{10\,\text{ms}} = 100$ Hz wide, with Gibbs ripple.
Why can an ideal brick-wall low-pass filter not be built?
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The inverse transform of a rectangle is a sinc in time, extending to $-\infty$. A causal system cannot respond before the impulse arrives.
Three common slips
- The wrong sinc argument. It is $\dfrac{\omega\tau}{2}$, dimensionless, not $\omega$.
- Dropping $\dfrac{1}{2\pi}$ from the inverse transform.
- Reading $|X|$ as the whole story. A spectrum needs its phase too.
Match each time-domain feature to its frequency-domain partner.
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Every pair is the time–bandwidth trade-off seen from one side or the other.
Recap
- $X(\omega) = \int x(t)e^{-j\omega t}\,dt$; convolution in time is multiplication in frequency.
- Pulse ↔ sinc, with its first null at $f = \dfrac{1}{\tau}$.
- Narrow in one domain means wide in the other.