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If X(k) is the N-point DFT of a sequence x(n), then what is the DFT of x*(n)?
According to the complex conjugate property of DFT, we have if X(k) is the N-point DFT of a sequence x(n), then what is the DFT of x*(n) is X*(N-k).
If X(k) is the N-point DFT of a sequence x(n), then circular time shift property is that N-point DFT of x((n-l))N is X(k)e-j2πkl/N.
Explanation: According to the circular time shift property of a sequence, If X(k) is the N-point DFT of a sequence x(n), then the N-pint DFT of x((n-l))N is X(k)e-j2πkl/N.
What is the circular convolution of the sequences x1(n)={2,1,2,1} and x2(n)={1,2,3,4}, find using the DFT and IDFT concepts?
Explanation: Given x1(n)={2,1,2,1}=>X1(k)=[6,0,2,0] Given x2(n)={1,2,3,4}=>X2(k)=[10,-2+j2,-2,-2-j2] when we multiply both DFTs we obtain the product
X(k)=X1(k).X2(k)=[60,0,-4,0] By applying the IDFT to the above sequence, we get
x(n)={14,16,14,16}.
What is the circular convolution of the sequences x1(n)={2,1,2,1} and x2(n)={1,2,3,4}?
Explanation: We know that the circular convolution of two sequences is given by the expression
For m=0,x2((-n))4={1,4,3,2}
For m=1,x2((1-n))4={2,1,4,3}
For m=2,x2((2-n))4={3,2,1,4}
For m=3,x2((3-n))4={4,3,2,1}
Now we get x(m)={14,16,14,16}.
If x1(n),x2(n) and x3(m) are three sequences each of length N whose DFTs are given as X1(k),X2(k) and X3(k) respectively and X3(k)=X1(k).X2(k), then what is the expression for x3(m)?
Explanation: If x1(n),x2(n) and x3(m) are three sequences each of length N whose DFTs are given as X1(k),X2(k) and X3(k) respectively and X3(k)=X1(k).X2(k), then according to the multiplication property of DFT we have x3(m) is the circular convolution of x1(n) and x2(n).
. If x(n) is real and odd, then what is the IDFT of the given sequence?
d) None of the mentioned
Explanation: If x(n) is real and odd, that is x(n)=-x(N-n), then XR(k)=0. Hence X(k) is purely imaginary and odd. Since XR(k) reduces to zero, the IDFT reduces to
If x(n) is real and even, then what is the DFT of x(n)?
d) None of the mentioned
Explanation: Given x(n) is real and even, that is x(n)=x(N-n)
We know that XI(k)=0. Hence the DFT reduces to
If x(n) is a real sequence and X(k) is its N-point DFT, then which of the following is true?
Explanation: We know that
Therefore,
X(N-k)=X*(k)=X(-k)
If x(n) is a complex valued sequence given by x(n)=xR(n)+jxI(n), then what is the DFT of xR(n)?
Explanation: Given x(n)=xR(n)+jxI(n)=> xR(n)=1/2(x(n)+x*(n))
Substitute the above equation in the DFT expression
Thus we get,
If X1(k) and X2(k) are the N-point DFTs of x1(n) and x2(n) respectively, then what is the N-point DFT of x(n)=ax1(n)+bx2(n)?
Explanation: We know that, the DFT of a signal x(n) is given by the expression
=>X(k)= aX1(k)+bX2(k).
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