2021年春季学期-信号与系统-第七次作业参考答案-第四小题
本文是 2021年春季学期-信号与系统-第七次作业参考答案 的小题的参考答案。
▌第四小题 ▌
4. Consider the signal
x
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t
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x\left( t \right)
x(t) in the following figure:
(a)Find the fourier transform
X
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X\left( {j\omega } \right)
X(jω) of
x
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x\left( t \right)
x(t)。
(b) Sketch the signal:
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=
x
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∗
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k
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−
∞
∞
δ
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\tilde x\left( t \right) = x\left( t \right) * \sum\limits_{k = - \infty }^\infty {\delta \left( {t - 4k} \right)}
x~(t)=x(t)∗k=−∞∑∞δ(t−4k)(c) Find another signal
g
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t
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g\left( t \right)
g(t) such that
g
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g\left( t \right)
g(t) is not the same as
x
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x\left( t \right)
x(t) and
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g
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K
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\tilde x\left( t \right) = g\left( t \right)K*\sum\limits_{k = - \infty }^\infty {\delta \left( {t - 4k} \right)}
x~(t)=g(t)K∗k=−∞∑∞δ(t−4k)(d) Argue that, alghough
G
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j
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G\left( {j\omega } \right)
G(jω) is different from
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X\left( {j\omega } \right)
X(jω)
G
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=
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G\left( {j{{\pi k} \over 2}} \right) = X\left( {j{{\pi k} \over 2}} \right)
G(j2πk)=X(j2πk)for all intergers
k
k
k, you should not explicitly avaluate
G
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j
ω
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G\left( {j\omega } \right)
G(jω) to answer this question.
还是忍不住多说两句:
(a)参考第4小题中的内容;
(b)这分明就是将x(t)进行周期为4的周期化;
(c)忍住了,就是不提示;
(d)对照X(jω),G(jω)离散化后的表达式,
它们相等,就会得到结论了。
▓ 本题为思考题
▓ 求解:
(a) x(t)是一个偶对称等腰三角形波形,它的 底宽为2,高为1,对应的傅里叶变换为:
X
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ω
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=
S
a
2
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2
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X\left( \omega \right) = Sa^2 \left( {{\omega \over 2}} \right)
X(ω)=Sa2(2ω)
(b) 是对x(t)进行周期为4的周期延拓,形 成如下图所示的周期波形:
(c) 对于任何一个函数 h ( t ) h\left( t \right) h(t) ,它与它自身延迟4相减得到:
h 4 ( t ) = h ( t ) − h ( t − 4 ) h_4 \left( t \right) = h\left( t \right) - h\left( {t - 4} \right) h4(t)=h(t)−h(t−4)
再进行周期延拓,结果为零:
∑ k = − ∞ + ∞ h 4 ( t − 4 k ) − h 4 ( t − 4 − 4 k ) \sum\limits_{k = - \infty }^{ + \infty } {h_4 \left( {t - 4k} \right) - h_4 \left( {t - 4 - 4k} \right)} k=−∞∑+∞h4(t−4k)−h4(t−4−4k) = ∑ k = − ∞ ∞ h 4 ( t − 4 k ) − ∑ k = − ∞ ∞ h 4 ( t − 4 − 4 k ) = 0 = \sum\limits_{k = - \infty }^\infty {h_4 \left( {t - 4k} \right)} - \sum\limits_{k = - \infty }^\infty {h_4 \left( {t - 4 - 4k} \right)} = 0 =k=−∞∑∞h4(t−4k)−k=−∞∑∞h4(t−4−4k)=0
因此任意取h(t)不为零,按照下面方式形成g(t):
g ( t ) = x ( t ) + h ( t ) − h ( t − 4 ) g\left( t \right) = x\left( t \right) + h\left( t \right) - h\left( {t - 4} \right) g(t)=x(t)+h(t)−h(t−4)
∑ k = − ∞ ∞ g ( t − 4 k ) = ∑ k = − ∞ ∞ x ( t − 4 k ) \sum\limits_{k = - \infty }^\infty {g\left( {t - 4k} \right)} = \sum\limits_{k = - \infty }^\infty {x\left( {t - 4k} \right)} k=−∞∑∞g(t−4k)=k=−∞∑∞x(t−4k)
x ~ ( t ) = g ( t ) ∗ ∑ k = − ∞ ∞ δ ( t − 4 k ) \tilde x\left( t \right) = g\left( t \right) * \sum\limits_{k = - \infty }^\infty {\delta \left( {t - 4k} \right)} x~(t)=g(t)∗k=−∞∑∞δ(t−4k)
举出一个例子:
(d)由于:
x
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=
x
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∗
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∞
δ
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\tilde x\left( t \right) = x\left( t \right)*\sum\limits_{k = - \infty }^\infty {\delta \left( {t - 4k} \right)}
x~(t)=x(t)∗k=−∞∑∞δ(t−4k)
x ~ ( t ) = g ( t ) ∗ ∑ k = − ∞ ∞ δ ( t − 4 k ) \tilde x\left( t \right) = g\left( t \right)*\sum\limits_{k = - \infty }^\infty {\delta \left( {t - 4k} \right)} x~(t)=g(t)∗k=−∞∑∞δ(t−4k)
假设 x ~ ( t ) , x ( t ) , g ( t ) \tilde x\left( t \right),x\left( t \right),g\left( t \right) x~(t),x(t),g(t)各自的傅里叶变换为:
而
X
~
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ω
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\tilde X\left( \omega \right)
X~(ω)是对
G
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ω
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G\left( \omega \right)
G(ω)和
X
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ω
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X\left( \omega \right)
X(ω)分别进行间隔为:
ω
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2
π
4
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π
2
\omega _1 = {{2\pi } \over 4} = {\pi \over 2}
ω1=42π=2π
的离散化,即:
X
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2
⋅
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∞
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⋅
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\tilde X\left( \omega \right) = {\pi \over 2} \cdot \sum\limits_{n = - \infty }^\infty {X\left( {{{\pi n} \over 2}} \right) \cdot \delta \left( {\omega - {{\pi n} \over 2}} \right)}
X~(ω)=2π⋅n=−∞∑∞X(2πn)⋅δ(ω−2πn)
=
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⋅
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G
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⋅
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= {\pi \over 2} \cdot \sum\limits_{n = - \infty }^\infty {G\left( {{{\pi n} \over 2}} \right) \cdot \delta \left( {\omega - {{\pi n} \over 2}} \right)}
=2π⋅n=−∞∑∞G(2πn)⋅δ(ω−2πn)
根据上面方程,左右两边奇异函数系数匹配方法,可知: G ( j π k 2 ) = X ( j π k 2 ) G\left( {j{{\pi k} \over 2}} \right) = X\left( {j{{\pi k} \over 2}} \right) G(j2πk)=X(j2πk)
另外,也可以从g(t)的构造方式进行证明。
如果按照前面构造的过程,取任意函数 h ( t ) h\left( t \right) h(t), g ( t ) = x ( t ) + h ( t + 2 ) − h ( t − 2 ) g\left( t \right) = x\left( t \right) + h\left( {t + 2} \right) - h\left( {t - 2} \right) g(t)=x(t)+h(t+2)−h(t−2)
满足条件(c)的假设,而:
令:
由于:
F T [ h ( t + 2 ) − h ( t − 2 ) ] FT\left[ {h\left( {t + 2} \right) - h\left( {t - 2} \right)} \right] FT[h(t+2)−h(t−2)] = H ( ω ) ( e j 2 ω − e − j 2 ω ) = H\left( \omega \right)\left( {e^{j2\omega } - e^{ - j2\omega } } \right) =H(ω)(ej2ω−e−j2ω) = H ( ω ) [ 2 j ⋅ sin 2 ω ] = H\left( \omega \right)\left[ {2j \cdot \sin 2\omega } \right] =H(ω)[2j⋅sin2ω]
所以:
H 4 ( π k 2 ) = H ( π k 2 ) ⋅ 2 j ⋅ sin ( 2 ⋅ π k 2 ) = 0 H_4 \left( {{{\pi k} \over 2}} \right) = H\left( {{{\pi k} \over 2}} \right) \cdot 2j \cdot \sin \left( {2 \cdot {{\pi k} \over 2}} \right) = 0 H4(2πk)=H(2πk)⋅2j⋅sin(2⋅2πk)=0
因此:
G
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G\left( {{{\pi k} \over 2}} \right) = X\left( {{{\pi k} \over 2}} \right) + H_4 \left( {{{\pi k} \over 2}} \right) = X\left( {{{\pi k} \over 2}} \right)
G(2πk)=X(2πk)+H4(2πk)=X(2πk)
【其它各小题参考答案】
- 2021年春季学期-信号与系统-第七次作业参考答案-第一小题
- 2021年春季学期-信号与系统-第七次作业参考答案-第二小题
- 2021年春季学期-信号与系统-第七次作业参考答案-第三小题
- 2021年春季学期-信号与系统-第七次作业参考答案-第四小题
- 2021年春季学期-信号与系统-第七次作业参考答案-第五小题
- 2021年春季学期-信号与系统-第七次作业参考答案-第六小题
- 2021年春季学期-信号与系统-第七次作业参考答案-第七小题
- 2021年春季学期-信号与系统-第七次作业参考答案-第八小题
- 2021年春季学期-信号与系统-第七次作业参考答案-第九小题
- 2021年春季学期-信号与系统-第七次作业参考答案-第十小题
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