THEORY EXAMINATION (SEM–VI) 2016-17 DIGITAL SIGNAL PROCESSING

DIGITAL SIGNAL PROCESSING – EEC602

B.Tech (SEM VI) | Section-wise Solved Answers

SECTION – A

(10 × 2 = 20 Marks)

(a) Discrete-time system

A discrete-time system processes input signals defined only at discrete time instants. The system output is calculated using difference equations rather than differential equations.

(b) Periodicity of

x[n]=cos⁡(nπ5)+sin⁡(nπ6)x[n]=\cos\left(\frac{n\pi}{5}\right)+\sin\left(\frac{n\pi}{6}\right)x[n]=cos(5nπ​)+sin(6nπ​)

The individual periods are:
For cosine: N1=10N_1 = 10N1​=10
For sine: N2=12N_2 = 12N2​=12
Overall period = LCM(10,12) = 60
Hence, the signal is periodic with period 60.

(c) Nyquist sampling theorem & reconstruction

Nyquist theorem states that a continuous-time signal band-limited to BBB Hz can be perfectly reconstructed if sampled at a rate ≥ 2B2B2B.
Reconstruction is done using an ideal low-pass filter.

(d) Discrete-time processing of CT signal & vice-versa

CT → DT: Sampling followed by digital processing.
DT → CT: Digital-to-analog conversion followed by smoothing filter.

(e) All-pass system

An all-pass system allows all frequencies to pass with unity magnitude but changes phase response.
Its poles are mirror images of zeros inside the unit circle.

(f) Multirate signal processing

It involves changing sampling rates using decimation (down-sampling) and interpolation (up-sampling) to reduce complexity and improve efficiency.

(g) Sampling & reconstruction of DT signal

Sampling converts CT to DT using periodic impulses. Reconstruction uses interpolation filters to obtain smooth analog signal.

(h) Twiddle factor

The twiddle factor is

WN=e−j2πNW_N = e^{-j\frac{2\pi}{N}}WN​=e−jN2π​

It represents complex roots of unity used in FFT algorithms.

(i) Relationship of DFT with Z-transform

DFT samples the Z-transform on the unit circle at equally spaced points.

(j) 8-point Radix-2 DIT FFT

The sequence is divided into even and odd parts recursively, reducing complexity from N2N^2N2 to Nlog⁡2NN\log_2NNlog2​N.

SECTION – B

(Attempt any five – 5 × 10 = 50 Marks)

(a) 8-point DFT using Radix-2 DIF

Given sequence:

x(n)={1/2,1/2,1/2,1/2,0,0,0,0}x(n)=\{1/2,1/2,1/2,1/2,0,0,0,0\}x(n)={1/2,1/2,1/2,1/2,0,0,0,0}

The DIF algorithm splits the sequence into stages, computes butterflies, and finally produces frequency-domain output with reduced computations.

(b) Z-transform of Hanning window & finite register length

Hanning window is defined as:

w(n)=12[1−cos⁡(2πnN−1)]w(n)=\frac12\left[1-\cos\left(\frac{2\pi n}{N-1}\right)\right]w(n)=21​[1−cos(N−12πn​)]

Its Z-transform is obtained by applying linearity and cosine expansion.
Finite register length causes round-off noise and limit cycles.

(c) Lattice-ladder realization

Given IIR system function, the lattice coefficients are obtained using step-down recursion, and ladder coefficients represent feed-forward paths.

(d) Transposed Direct Form-II

Difference equation:

y(n)=0.5y(n−1)−0.25y(n−2)+x(n)−2x(n−1)+x(n−2)y(n)=0.5y(n-1)-0.25y(n-2)+x(n)-2x(n-1)+x(n-2)y(n)=0.5y(n−1)−0.25y(n−2)+x(n)−2x(n−1)+x(n−2)

In transposed form-II, delays are moved to input side, minimizing memory and improving numerical stability.

(e) FIR LPF using Hamming window (M=7)

Ideal response is truncated and multiplied with Hamming window.
Filter coefficients are computed using

h(n)=hd(n)⋅w(n)h(n)=h_d(n)\cdot w(n)h(n)=hd​(n)⋅w(n) 

(f) Bilinear transformation

Given

H(s)=s+0.1(s+0.1)2+9H(s)=\frac{s+0.1}{(s+0.1)^2+9}H(s)=(s+0.1)2+9s+0.1​

Using

s=2T1−z−11+z−1s=\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}s=T2​1+z−11−z−1​

Analog frequency is warped to digital frequency ωr=π/4\omega_r=\pi/4ωr​=π/4.

(g) Bilinear transformation frequency relationship

The relation is:

ω=2tan⁡−1(ΩT2)\omega = 2\tan^{-1}\left(\frac{\Omega T}{2}\right)ω=2tan−1(2ΩT​)

It avoids aliasing but introduces frequency warping.

(h) IIR filter design methods

Approximation of derivatives: Simple but inaccurate

Impulse invariance: Preserves impulse response but causes aliasing

Bilinear transformation: No aliasing, most widely used

SECTION – C

(Attempt any two – 2 × 15 = 30 Marks)

(3) Cascade & Parallel realization

Given system function is factored into first-order and second-order sections.

Cascade form: Product of sub-systems

Parallel form: Partial-fraction expansion

(4) Cascade & Parallel realization (given transfer function)

Denominator is factored to obtain sections.
Parallel form improves speed, cascade form improves numerical stability.

(5) FIR LPF using rectangular window

Given:

Hd(ejw)=e−j2w, ∣w∣≤π/4H_d(e^{jw}) = e^{-j2w}, \ |w| \le \pi/4Hd​(ejw)=e−j2w, ∣w∣≤π/4

Impulse response hd[n]h_d[n]hd​[n] is calculated using inverse DTFT and truncated using rectangular window.