编程代考 Afunction x (t)isshownonthegraphwithθ =π.

Afunction x (t)isshownonthegraphwithθ =π.
a) Write x1 (t) polar format
b) Write x1 (t) rectangular format
A function x2 (t ) is defined as x2(t)=−3j⋅e 4 

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c) Plotx2(t)
d) Writeinrectangularformatof x (t)⋅x (t)
− jπ −π  
M = 3.6 ⋅ cos  π  = 4.4498 where M is the magnitude of the length 
 4.4498⋅e  5 
5 jπ+π 
jπ+π  
4.4498⋅e  5  = −3.6000− j2.6155 − jπ −π 
x2(t)=−3j⋅e 4  =2.1213+j2.1213
(−3.6000 − j2.6155)(2.1213 + j2.1213) = −2.0883 − j13.1850

A function x (t ) is given as a) Plot x(−2t+4)
b) Find F{x(−2t+4)}
x (−2t + 4) = x  t − 2 
 −0.5  

F{x(−2t+4)}=∞ x(−2t+4)⋅e−j2π ftdt t=−∞
define k = −2t + 4
dk = −2dt → dt = − 1 dk
−∞ − j2π f    
k−4 1 x(k)⋅e  −2  −2 dk
=2 k=−∞x(k)⋅e −2 ⋅e−j2πf(2)dk
k=∞  1 −j2π f k
=1X f ⋅e−j2πf(2) Rotate,shift,&scale

A function x(t)is shown on the right figure.
Write x (t ) (shown amplitude is 1.) with sin and cos function. The amplitude of the
function is the peak value (not the RMS). Magnitude and phase response of H ( f ) is
shown. Write the H ( f ) Writeh(t)=F−1{H(f)}
Let’s define y(t)= x(t)⋅h(t)find y(t) Plot Y(f)&Y(f)
x(t)=−1⋅sin2π4t−π =−1⋅cos(2π4t) 
b jπ j3π jπ −jπ
) H(f)=3δ(f−f0)⋅e2 +3δ(f+f0)⋅e 2 =3δ(f−f0)⋅e2 +3δ(f+f0)⋅e 2 c) h(t)=6⋅cos(2π f0t)
y(t)=6⋅cos(2π f0t)⋅cos(2π4t) e)
Y(f)=6δ(f−4)+δ(f+4)∗δ(f−f )+δ(f+f ) 4oo
=6δ(f−4−f )+δ(f+4−f )+δ(f−4+f )+δ(f+4+f ) 4oooo
=6δ(f−(4+f))+δ(f+(4−f ))+δ(f−(4−f ))+δ(f+(4+f )) 4oooo

h (t ) = −3Π  t − 1  2
a) Do the convolution of
And you don’t need to carry out the full integration. Just fill the low and upper limits of integrals in the given spaces.
y(t)= x(t)∗h(t)= ∞ τ=−∞
x(t)h(t −τ )dτ
In this case, you need be careful with what function to flip over based on the above equation.
Where [τ = , τ = ]means
[τ = Low limit , τ = Upper limit ] of integrals
−1 ≤ t < 1 1≤t<2 2≤t<4 y(t)= 2⋅(−3)dτ [τ =−1 τ= y(t)= 2⋅(−3)dτ [τ =t−2 τ= y(t)= 2⋅(−3)dτ [τ =t−2 τ= , τ =t , τ =t b) What is the frequency response of y(t), Y ( f ) b) Y ( f ) = X ( f )⋅ H ( f ) = 2⋅2⋅(1.5)⋅sinc(2⋅(1.5)⋅ f )⋅e− j2π f (0.5)  2⋅(−3)⋅1⋅sinc(2⋅1⋅ f )⋅e− j2π f (1)  = 6⋅sinc(3⋅ f )⋅e− j2π f (0.5)  −6⋅sinc(2⋅ f )⋅e− j2π f (1)   1. [] A continuous signal, x(t), is going to be sampled with the a sampling rate of fs =120 x(t)= sin(260πt)2 −2sin(120πt)  a) FindX(Ω) b) Plot the frequency response (magnitude and phase) of X (Ω) [−2π ≤ Ω ≤ 2π ] x(t)= sin(2π130t)2 −2sin(120πt)  = 11−cos(2⋅(2π130t))−2sin(2π60t) 2 = 11−cos(2π260t)−2sin(2π60t) 2 = 11−cos(2π20t)−2sin(2π60t) 2 F{x(t)}= X (f ) =1δ(f)−1⋅1δ(f −20)+δ(f +20)−1δ(f −60)−δ(f +60) 2 22  j   1δ(f)−1⋅1δ(f −20)+δ(f +20)−1δ(f −60)−δ(f +60) 2 22 j  The terms with red colored disappears by cancelling each other clc; clear; n = 0:3*fs-1; x1 = sin(260*pi*n/fs); x2 = 2*sin(120*pi*n/fs); x = x1.^2 - 2*x2; OM = -pi:0.0001:pi; X = exp(-j*OM'*n)*x'; fq = OM*fs/(2*pi); subplot(2,1,1),plot(fq,abs(X)) subplot(2,1,2),plot(fq,phase(X)) 2. [] One period of the function, x(t),is given as 5Δ t +1 as shown on the right 2 A triangle function can be made of two rect function with two equal width. One of the rect functionsisgivenas 5Δt+1=2Π(t)∗? 2  a) Find the another rect function shown as [?]in the above equation b) Write X k using Fourier Transform method c) FindXkwherek=0,1,2&3 d) Plot the magnitude response of part (c) [?]= 2.5⋅Π(t +1)= 2.5⋅Π(t)∗δ (t +1)  1  k2 k  k2 k Xk =45⋅sinc4 ⋅e−j2π4(−1) =1.25⋅sinc4 ⋅ej2π4        1.2500 + 0.0000i   0.0000 + 1.0132i  X=  k  −0.5066 + 0.0000i  − 0.0000 − 0.1126i   5Δ t +1 = 2Π(t)∗2.5⋅Π(t +1) 2 = 5Π(t)*Π(t)*δ (t +1) Define x(t)= Π(t) X (f )=(2⋅1⋅0.5)sinc(2⋅0.5⋅ f ) = sinc ( f ) F 5Π(t)*Π(t)*δ(t+1) =5⋅sinc(f)2⋅e−j2πf(−1) 1  k2 k  k2 k Xk =45⋅sinc4 ⋅e−j2π4(−1) =1.25⋅sinc4 ⋅ej2π4        1.2500 + 0.0000i   0.0000 + 1.0132i  X=  k  −0.5066 + 0.0000i  − 0.0000 − 0.1126i   k = [0 1 2 3]; X_k = 1.25*((sinc(k/4)).^2).*exp(j*2*pi*k/4) X_k.' subplot(2,1,1),stem(k,abs(X_k),'^','r'); title('|Xk|') subplot(2,1,2),stem(k,phase(X_k),'r'); title('phase') a) Find and plot X ( f ) with all the proper scales in the frequency domain b) Find and plot Y ( f )with all the proper scales in the frequency domain c) Plot Z ( f ) with all the proper scales in the frequency domain 36Λ f  8 36Λ f −5+36Λ f +5 8 8  F 12sinc(4t) {} 2⋅f0 =4→f0 =2; 2⋅A⋅f0 =12→A=3 Multiplication in the time domain means convolution in the frequency domain. F x(t) = F 12sinc(4t)2 { } = 3Π f ∗3Π f   4  4       =36Λ f  8 F{y(t)}=36Λ f *δ(f −5)+δ(f +5) =36Λ f −5+36Λ f +5 8 8  A signal x(t)is given on the right. Another signal h(t)is defined as h(t)= 2Π(t −3) Do the convolution of y(t)= x(t)∗h(t). It means you flip over the h(t). You don’t need to carry out all the integration. Just fill out the low and upper limits and the contents of integrals, just like the quiz that you took. 1.5≤t<2.5 2.5≤t <3.5 3.5≤t<4.5 4.5≤t <7.5 7.5≤t <8.5 t−2.5x1(τ)h(t−τ)dτ τ=−1 t−2.5 x1(τ)h(t−τ)dτ τ =t −3.5 1 x1(τ)h(t−τ)dτ τ=t−3.5 τ=1 +t−2.5 x2 (τ)h(t−τ)dτ dτ t−2.5 x2(τ)h(t−τ)dτ τ=t−3.5 5 x2(τ)h(t−τ)dτ τ =t −3.5 程序代写 CS代考 加微信: powcoder QQ: 1823890830 Email: powcoder@163.com