Unit - 3
Discrete Fourier Transform
Q1) Draw block diagram for the function using parallel form H(z)=
A1)
H(z)=
Writing above transfer function in standard form for parallel realisation we get
H(z)=-20+
The structure is shown below
Fig 1 Parallel Realisation of H(z)=
Q2) For the following LTI system H(z)= . Realise the cascade form IIR filter.
A2) H(z)=
The above function can be simplified as
H(z)=
Fig 2 Cascade IIR Form
Hence, using the above structure and placing the values of …. And similarly,
Q3) For the system given y(n) -y(n-1) + y(n-2) = x(n) + x(n-1) realise using cascade form?
A3) The system transfer function is given as
H(z) = Y(z)/X(z)
Taking z transform of y(n) -y(n-1) + y(n-2) = x(n) + x(n-1)
Y(z) - z-1Y(z) + z-2 Y(z) = X(z) + z-1 X(z)
H(z)=
Again, simplifying the above function to get into standard cascade form we ca write
H(z) =
= H1(z)+H2(z)
H1(z)=
H2(z)=
The final structure is shown below
Fig 3 Cascade Form of H(z) =
Q4) For the following LTI system H(z)= . Realise the cascade form?
A4) H(z)=
Writing the above in standard form for cascade realisation
H1(z)=
H2(z)=
The cascade structure is shown below
Fig 4 Cascade Form of H(z)=
Q5) Realize the system transfer function using parallel structure H(z)=
A5) H(z)=
Taking Z common and then dividing the above function to convert it into standard form for parallel realisation we get
H(z)=Z [ ++]
The parallel structure is shown below
Fig 5 Parallel Realisation of H(z)=
Q6) Realize the system transfer function using parallel structure H(z)=
A6) Converting the above function to standard form using partial fraction technique
H(z)= +
Solving for A and B we get
A= 10/3
B= -7/3
H(z) = +
H1(z) =
H2(z) =
The parallel form realisation is shown below
Fig 6 Parallel Realisation of H(z)=
Q7) For the transfer function H(z) = . Realise using cascade form?
A7) H(z) =
Writing in standard form
H(z) =
H1(z) =
H2(z) =
The cascade structure is shown below
Fig 7 Cascade Form of H(z) =
Q8) Compute the N-point DFT of x(n)=3δ(n).
A8)
Q9) Compute the N-point DFT of x(n)=7(n−n0)
A9) We know that,
Substituting the value of x(n),
Q10) Perform circular convolution of the two sequences, x1(n)= {2,1,2,-1} and x2(n)= {1,2,3,4}
A10)
Fig: Circular Convolution of x1(n)= {2,1,2,-1} and x2(n)= {1,2,3,4}
(2)
When n=0;
The sum of samples of v0(m) gives x3(0)
⸫ x3(0)=2+4+6-2=10
When n=1;
The sum of samples of v1(m) gives x2(1)
⸫ x3(1)=4 + 1 +8-3=10
(3)When n=2;
The sum of samples of v2(m) gives x3(2)
⸫ x3(2)=6+2+2-4=6
(4) When n=3;
The sum of samples of v3(m) gives x3(3)
⸫ x3(3)=8+ 3+ 4-1= 14
x3(n)={10,10,6,14}
= x1(0) x x2,0(0) + x1(1) x2,0(1) + x1(2) x2,0(2) + x1(3) x2,0(3)
= 2 x 1 + 1 x 4 + 2 x 3 + (-1) x 2 = 2 +4 +6 -2 =10