東京大学 情報理工学系研究科 電子情報学専攻 2015年8月実施 専門 第5問
Author
Josuke
Description
Let us denote the Fourier transform of the signal f(t) as F(ω), where t and ω represent a time variable and an angle frequency, respectively.
(1) Show the definition of the Fourier transform F(ω) of the signal f(t).
Also, explain the difference between the Fourier transform and the Fourier series expansion.
(2) Explain why ∣F(ω)∣2 represents the power spectrum, i.e., the power at a certain angle frequency of ω.
(3) Derive the following Parseval's theorem in the Fourier transform and determine k.
∫−∞∞∣f(t)∣2dt=k∫−∞∞∣F(ω)∣2dω,k is a real constant.
You may use ∣f(t)∣2=f(t)f(t), (f(t) is the complex conjugate of f(t)). You may also use the Fourier transform of the following convolution integrals
∫−∞∞f(t−τ)g(τ)dr=k′∫−∞∞F(ω)G(ω)ejωtdω,
j is the imaginary unit and k′ is a real constant.
You may include k′ when answering k.
(4) Explain the physical meaning of the Parseval's theorem in the Fourier transform.
题目描述
记信号 f(t) 的傅里叶变换为 F(ω),其中 t 为时间变量,ω 为角频率。
(1) 写出 f(t) 的傅里叶变换 F(ω) 的定义,并说明傅里叶变换与傅里叶级数展开的区别。
(2) 解释为何 ∣F(ω)∣2 表示功率谱,即角频率 ω 处的功率。
(3) 推导傅里叶变换形式的帕塞瓦尔定理并确定实常数 k:
∫−∞∞∣f(t)∣2dt=k∫−∞∞∣F(ω)∣2dω.
可以使用 ∣f(t)∣2=f(t)f(t),其中 f(t) 是 f(t) 的复共轭;也可以使用卷积积分的傅里叶表示
∫−∞∞f(t−τ)g(τ)dτ=k′∫−∞∞F(ω)G(ω)ejωtdω,
其中 j 为虚数单位,k′ 为实常数。求 k 时可以用 k′ 表示答案。
(4) 说明傅里叶变换中帕塞瓦尔定理的物理意义。
- 傅里叶变换与功率谱:要求从变换定义和复频谱幅值说明 ∣F(ω)∣2 的含义,并区分变换与级数。
- 帕塞瓦尔恒等式:要求结合复共轭和卷积关系推导时域、频域能量等价及归一化常数。
Kai
(1)
F(ω)=∫−∞∞f(t)e−jωtdt
Fourier series can be applied to periodic signal. Fourier transform can be applied to non-periodic signal.
(2)
∣F(ω)∣2=F(ω)F(ω)=(Real{F(ω)})2+(Image{F(ω)})2
Thus ∣F(ω)∣2 represents the power of certain angle frequency ω.
(3)
f(t)∗g(t)f(0)∗g(0)=∫−∞+∞f(t−τ)g(τ)dτ=∫−∞+∞g(t−τ)f(τ)dτ=∫−∞+∞f(−τ)g(τ)dτ=∫−∞+∞g(−τ)f(τ)dτ
if g(τ)=f(−τ)
G(ω)G(ω)=∫−∞+∞g(t)e−jωtdt=∫−∞+∞f(−t)e−jωtdt=∫−∞+∞f(−t)e−jωt=∫−∞+∞f(t)e−jωtdt=F(ω)G(ω)=F(ω)
So
∫−∞∞∣f(t)∣2=k′∫−∞∞∣F(ω)∣2dω
(4)
The energy in time domain equals to k′ times energy in frequency domain.