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第一次作业链接: 第一次作业 : https://zhuoqing.blog.csdn.net/article/details/123113465
§01 参考答案
1.1 绘制信号波形
(1)必做题
Ⅰ.第一小题
求解:
▲ 带有直流分量正弦振荡信号
from headm import *
t = linspace(-5,5,1000)
ft = 1+cos(2*pi*t)
plt.plot(t,ft)
plt.xlabel("t")
plt.ylabel("f(t)")
plt.grid(True)
plt.tight_layout()
plt.show()
Ⅱ.第二小题
求解:
▲ 幅度调制正弦振荡信号
ft = (1+cos(pi*t)/3)*sin(8*pi*t)
Ⅲ.第三小题
求解:
▲ 双边指数震荡衰减小囊括
t = linspace(-5,5,1000)
ft = 2*sin(20*t)*exp(-abs(t))
Ⅳ.第四小题
求解:
▲ 复合信号
from headm import *
t = linspace(-5,5,1000)
ft = heaviside(sin(2*pi*t), 0)
plt.plot(t,ft)
plt.xlabel("t")
plt.ylabel("f(t)")
plt.grid(True)
plt.tight_layout()
plt.show()
Ⅴ.第五小题
求解:
▲ 有限长指数信号
t = linspace(-5,5,1000)
ft = exp(-0.5*t)*(heaviside(t+1, 0)-heaviside(t-1,0))
Ⅵ.第六小题
求解:
▲ 指数序列
from headm import *
n = arange(0, 10)
fn = (0.5)**n
plt.stem(n, fn)
plt.xlabel("n")
plt.ylabel("f[n]")
plt.grid(True)
plt.tight_layout()
plt.show()
Ⅶ.第七小题
求解:
▲ 左边指数序列信号
n = arange(-10, 0)
fn = exp(n/5)
(2)选做题
Ⅰ.第一小题
求解:
▲ sinc函数连续乘积信号
t = linspace(-15, 15, 1000)
ft = sinc(t/pi)*sinc(t/3/pi)*sinc(t/5/pi)
注:在Python中,sinc(x)定义为 sin c ( x ) = sin ( π x ) / ( π ⋅ x ) \sin c\left( x \right) = \sin \left( {\pi x} \right)/\left( {\pi \cdot x} \right) sinc(x)=sin(πx)/(π⋅x) ,所以在应用时使用 sin c ( x / π ) \sin c\left( {x/\pi } \right) sinc(x/π) 。
Ⅱ.第二小题
求解:
▲ 震荡序列
n=arange(-10, 10)
fn=sin(pi*n*n/3)
plt.stem(n, fn)
Ⅲ.第三小题
求解:
▲ 非周期振荡信号
t = linspace(0.00001, 10, 10000)
ft = sin(2*pi/t)
Ⅳ.第四小题
求解:
▲ 周期波形发散信号
def f0(t):
out = 1/t*(heaviside(t,0)-heaviside(t-1,0))
return out
t = linspace(-3,3,12349)
ft = f0(t-3)+f0(t-2)+f0(t-1)+f0(t)+f0(t+1)+f0(t+2)+f0(t+3)
1.2 写出信号函数表达式
(1)必做题
Ⅰ.第一小题
求解:
-
第一种表达式:
f ( t ) = { 0 , x < − 1 2 x + 2 , − 1 ≤ x < 0 − 2 x + 2 , 0 ≤ x < 1 0 , x > = 1 f\left( t \right) = \left\{ \begin{matrix} {0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x < - 1}\\{2x + 2,\,\,\,\,\,\,\,\,\,\,\,\, - 1 \le x < 0}\\{ - 2x + 2,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \le x < 1}\\{0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x > = 1}\\\end{matrix} \right. f(t)=⎩⎪⎪⎨⎪⎪⎧0,x<−12x+2,−1≤x<0−2x+2,0≤x<10,x>=1 -
第二种表达式:
f ( t ) = ( 2 − 2 ∣ t ∣ ) ⋅ [ u ( t + 1 ) − u ( t − 1 ) ] f\left( t \right) = \left( {2 - 2\left| t \right|} \right) \cdot \left[ {u\left( {t + 1} \right) - u\left( {t - 1} \right)} \right] f(t)=(2−2∣t∣)⋅[u(t+1)−u(t−1)]
Ⅱ.第二小题
-
第一种表达式:
f ( t ) = [ u ( t ) − u ( t − 1 ) ] + 1.5 [ u ( t − 1 ) − u ( t − 3 ) ] f\left( t \right) = \left[ {u\left( t \right) - u\left( {t - 1} \right)} \right] + 1.5\left[ {u\left( {t - 1} \right) - u\left( {t - 3} \right)} \right] f(t)=[u(t)−u(t−1)]+1.5[u(t−1)−u(t−3)] + 3 [ u ( t − 3 ) − u ( t − 4.5 ) ] + 0.5 u ( t − 4.5 ) + 3\left[ {u\left( {t - 3} \right) - u\left( {t - 4.5} \right)} \right] + 0.5u\left( {t - 4.5} \right) +3[u(t−3)−u(t−4.5)]+0.5u(t−4.5) -
第二种表达式:
f ( t ) = u ( t ) + 0.5 u ( t − 1 ) + 1.5 u ( t − 3 ) − 2.5 u ( t − 4.5 ) f\left( t \right) = u\left( t \right) + 0.5u\left( {t - 1} \right) + 1.5u\left( {t - 3} \right) - 2.5u\left( {t - 4.5} \right) f(t)=u(t)+0.5u(t−1)+1.5u(t−3)−2.5u(t−4.5)
Ⅲ.第三小题
求解:
f ( t ) = E 2 cos ( π T t ) ⋅ [ u ( t + T ) − u ( t − T ) ] + E 2 f\left( t \right) = {E \over 2}\cos \left( { {\pi \over T}t} \right) \cdot \left[ {u\left( {t + T} \right) - u\left( {t - T} \right)} \right] + {E \over 2} f(t)=2Ecos(Tπt)⋅[u(t+T)−u(t−T)]+2E
Ⅳ.第四小题
求解:
f ( t ) = − E sin ( t ) ⋅ u ( t ) f\left( t \right) = - E\sin \left( t \right) \cdot u\left( t \right) f(t)=−Esin(t)⋅u(t)
Ⅴ.第五小题
求解:
f ( t ) = e − 0.1 t ⋅ sin ( t ) ⋅ u ( t ) f\left( t \right) = e^{ - 0.1t} \cdot \sin \left( t \right) \cdot u\left( t \right) f(t)=e−0.1t⋅sin(t)⋅u(t)
(2)选做题
Ⅰ.第一小题
求解:
-
第一种表达式:
f ( t ) = 2 ∑ n = − ∞ + ∞ ( − 1 ) n [ u ( t + 1 − 4 n ) − u ( t − 1 − 4 n ) ] f\left( t \right) = 2\sum\limits_{n = - \infty }^{ + \infty } {\left( { - 1} \right)^n \left[ {u\left( {t + 1 - 4n} \right) - u\left( {t - 1 - 4n} \right)} \right]} f(t)=2n=−∞∑+∞(−1)n[u(t+1−4n)−u(t−1−4n)]
-
第二种表达式:
f ( t ) = 2 s g n [ cos ( π 2 t ) ] f\left( t \right) = 2{\mathop{\rm sgn}} \left[ {\cos \left( { {\pi \over 2}t} \right)} \right] f(t)=2sgn[cos(2πt)] 其中 s g n ( t ) = { 1 , t > 0 0 , t = 0 − 1 , t < 0 {\mathop{\rm sgn}} \left( t \right) = \left\{ \begin{matrix} {1,\,\,\,\,\,\,t > 0}\\{0,\,\,\,\,\,t = 0}\\{ - 1,\,\,\,\,\,t < 0}\\\end{matrix} \right. sgn(t)=⎩⎨⎧1,t>00,t=0−1,t<0
Ⅱ.第二小题
f ( t ) = ∑ n = 0 + ∞ 2 − n [ u ( t − 1 + 2 − n ) − u ( t − 1 + 2 − n − 1 ) ] f\left( t \right) = \sum\limits_{n = 0}^{ + \infty } {2^{ - n} \left[ {u\left( {t - 1 + 2^{ - n} } \right) - u\left( {t - 1 + 2^{ - n - 1} } \right)} \right]} f(t)=n=0∑+∞2−n[u(t−1+2−n)−u(t−1+2−n−1)]
▲ 图1.2.2 使用程序绘制公式波形进行验证
from headm import *
t = linspace(-0.25,1.25,10000)
ft = t*0
for n in range(100):
ft = ft + 0.5**n*(heaviside(t-1+0.5**n,0)-heaviside(t-1+0.5**(n+1),0))
plt.plot(t,ft)
plt.xlabel("t")
plt.ylabel("f(t)")
plt.grid(True)
plt.tight_layout()
plt.show()
1.3 判断信号的周期性
(1)必做题
Ⅰ.第一小题
求解: 非周期
f ( t ) = sin ( 10 t ) − cos ( 30 π t ) f\left( t \right) = \sin \left( {10t} \right) - \cos \left( {30\pi t} \right) f(t)=sin(10t)−cos(30πt)
▲ 图1.3.1 f(t)的波形
Ⅱ.第二小题
求解: 周期信号
Ⅲ.第三小题
求解: 周期信号
Ⅳ.第四小题
求解: 周期信号
Ⅴ.第五小题
求解: 非周期信号
▲ 图1.3.2 信号的波形
Ⅵ.第六小题
求解: 周期信号
▲ 序列的波形
n = arange(-30, 30)
fn = cos(pi*(n*n*n)/5)
plt.stem(n,fn)
plt.xlabel("n")
plt.ylabel("f[n]")
plt.grid(True)
plt.tight_layout()
plt.show()
(2)选做题
Ⅰ.第一小题
求解: 非周期信号
▲ 序列的波形
Ⅱ.第二小题
求解
(1) 如果 y [ n ] y\left[ n \right] y[n] 是周期信号, x [ n ] x\left[ n \right] x[n] 不一定是周期信号;
(2) x [ n ] x\left[ n \right] x[n] 是周期信号, y [ n ] y\left[ n \right] y[n] 一定是周期信号;
1.4 冲激信号的抽样特性
Ⅰ.第一小题
∫ − ∞ ∞ f ( τ ) δ ( τ 2 ) d τ = ∫ − ∞ + ∞ f ( τ ) ⋅ 2 δ ( τ ) d τ = 2 f ( 0 ) \int_{ - \infty }^\infty {f\left( \tau \right)\delta \left( { {\tau \over 2}} \right)d\tau } = \int_{ - \infty }^{ + \infty } {f\left( \tau \right) \cdot 2\delta \left( \tau \right)d\tau } = 2f\left( 0 \right) ∫−∞∞f(τ)δ(2τ)dτ=∫−∞+∞f(τ)⋅2δ(τ)dτ=2f(0)
Ⅱ.第二小题
∫ − ∞ + ∞ cos ( 2 τ ) ⋅ δ ( τ − π 3 ) d τ = cos ( 2 ⋅ π 3 ) = − 1 2 \int_{ - \infty }^{ + \infty } {\cos \left( {2\tau } \right) \cdot \delta \left( {\tau - {\pi \over 3}} \right)d\tau } = \cos \left( {2 \cdot {\pi \over 3}} \right) = - {1 \over 2} ∫−∞+∞cos(2τ)⋅δ(τ−3π)dτ=cos(2⋅3π)=−21
Ⅲ.第三小题
∫ − ∞ + ∞ δ ( t − t 0 ) ⋅ u ( t − t 0 3 ) d t = u ( t 0 − t 0 3 ) = u ( 2 t 0 3 ) = { 1 , t 0 > 0 0.5 , t 0 = 0 − 1 , t 0 < 0 \int_{ - \infty }^{ + \infty } {\delta \left( {t - t_0 } \right) \cdot u\left( {t - { {t_0 } \over 3}} \right)dt} = u\left( {t_0 - { {t_0 } \over 3}} \right) = u\left( { { {2t_0 } \over 3}} \right) = \left\{ \begin{matrix} {1,\,\,\,\,t_0 > 0}\\{0.5,\,\,\,t_0 = 0}\\{ - 1,\,\,\,\,t_0 < 0}\\\end{matrix} \right. ∫−∞+∞δ(t−t0)⋅u(t−3t0)dt=u(t0−3t0)=u(32t0)=⎩⎨⎧1,t0>00.5,t0=0−1,t0<0
Ⅳ.第四小题
∫ − ∞ + ∞ [ 2 e − t + cos 2 t ] ⋅ δ ′ ( t ) d t = [ 2 e − t + cos 2 t ] ′ ∣ t = 0 = − 2 e − t − 2 sin 2 t ∣ t = 0 = − 2 \int_{ - \infty }^{ + \infty } {\left[ {2e^{ - t} + \cos 2t} \right] \cdot \delta '\left( t \right)dt} = \left. {\left[ {2e^{ - t} + \cos 2t} \right]^\prime } \right|_{t = 0} = \left. { - 2e^{ - t} - 2\sin 2t} \right|_{t = 0} = - 2 ∫−∞+∞[2e−t+cos2t]⋅δ′(t)dt=[2e−t+cos2t]′∣∣∣t=0=−2e−t−2sin2t∣∣t=0=−2
Ⅴ.第五小题
∫ − ∞ + ∞ e − j ω t [ δ ( t ) + δ ( t − t 0 ) ] d t = e − j ω ⋅ 0 + e − j ω ⋅ t 0 = 1 + e − j ω t 0 \int_{ - \infty }^{ + \infty } {e^{ - j\omega t} \left[ {\delta \left( t \right) + \delta \left( {t - t_0 } \right)} \right]dt} = e^{ - j\omega \cdot 0} + e^{ - j\omega \cdot t_0 } = 1 + e^{ - j\omega t_0 } ∫−∞+∞e−jωt[δ(t)+δ(t−t0)]dt=e−jω⋅0+e−jω⋅t0=1+e−jωt0
Ⅵ.第六小题
∫ − ∞ + ∞ e − τ [ δ ( τ − 1 ) + δ ′ ( τ ) ] d τ = e − 1 + ( e − t ) ′ ∣ t = 0 = e − 1 − 1 \int_{ - \infty }^{ + \infty } {e^{ - \tau } \left[ {\delta \left( {\tau - 1} \right) + \delta '\left( \tau \right)} \right]d\tau } = e^{ - 1} + \left. {\left( {e^{ - t} } \right)^\prime } \right|_{t = 0} = e^{ - 1} - 1 ∫−∞+∞e−τ[δ(τ−1)+δ′(τ)]dτ=e−1+(e−t)′∣∣∣t=0=e−1−1
1.5 信号的直流分量
Ⅰ.第一小题
求解:
f D ( t ) = ω π ∫ 0 π ω sin ( ω t ) d t = ω π ⋅ ( − 1 ) 1 ω cos ( ω t ) ∣ t = 0 t = π ω = 2 π f_D \left( t \right) = {\omega \over \pi }\int_0^{
{\pi \over \omega }} {\sin \left( {\omega t} \right)dt} = {\omega \over \pi } \cdot \left. {\left( { - 1} \right){1 \over \omega }\cos \left( {\omega t} \right)} \right|_{t = 0}^{t = {\pi \over \omega }} = {2 \over \pi } fD(t)=πω∫0ωπsin(ωt)dt=πω⋅(−1)ω1cos(ωt)∣∣∣∣t=0t=ωπ=π2
Ⅱ.第二小题
求解: 因为:
cos ( 2 ω t ) = cos 2 ( ω t ) − sin 2 ( ω t ) = 1 − 2 sin 2 ( ω t ) \cos \left( {2\omega t} \right) = \cos ^2 \left( {\omega t} \right) - \sin ^2 \left( {\omega t} \right) = 1 - 2\sin ^2 \left( {\omega t} \right) cos(2ωt)=cos2(ωt)−sin2(ωt)=1−2sin2(ωt) sin 2 ( ω t ) = 1 2 [ 1 − cos ( 2 ω t ) ] \sin ^2 \left( {\omega t} \right) = {1 \over 2}\left[ {1 - \cos \left( {2\omega t} \right)} \right] sin2(ωt)=21[1−cos(2ωt)]
所以: f D ( t ) = 1 2 f_D \left( t \right) = {1 \over 2} fD(t)=21
Ⅲ.第三小题
求解: f D ( t ) = 0 f_D \left( t \right) = 0 fD(t)=0
Ⅳ.第四小题
求解: f D ( t ) = K f_D \left( t \right) = K fD(t)=K
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